Electrochemistry

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Electrochemical Cells for Every Occasion Perhaps near you right now are a laptop, an MP3 player, and
a cell phone—just a few of the devices whose energy comes from an electrochemical cell. In this chapter,
you’ll look at the two faces of electrochemistry—reactions in cells, such as those in batteries, that do electrical
work and reactions in cells that require electrical work to occur. Both are indispensable to our modern way of life.
Electrochemistry: Chemical Change

and Electrical Work
21.1 Redox Reactions and 21.3 Cell Potential: Output of a Voltaic Cell 21.6 Corrosion: A Case of Environmental
Electrochemical Cells Standard Cell Potentials Electrochemistry
Review of Oxidation-Reduction Strengths of Oxidizing and Reducing Corrosion of Iron
Concepts Agents Protecting Against Corrosion
Half-Reaction Method for Balancing 21.4 Free Energy and Electrical Work 21.7 Electrolytic Cells: Using Electrical
Redox Reactions Standard Cell Potential and K Energy to Drive Nonspontaneous
Electrochemical Cells Effect of Concentration on Ecell Reactions
21.2 Voltaic Cells: Using Spontaneous Changes in Ecell During Cell Operation Construction and Operation
Reactions to Generate Electrical Concentration Cells Predicting Electrolysis Products
Energy 21.5 Electrochemical Processes Stoichiometry of Electrolysis
Construction and Operation in Batteries
Cell Notation Primary (Nonrechargeable) Batteries
Why Does the Cell Work? Secondary (Rechargeable) Batteries
Fuel Cells
hermodynamics has many applications other than expanding gases inside steam
T engines. In fact, some applications are probably within your reach right now, in
the form of battery-operated devices—laptop computer, palm organizer, DVD
remote, and, of course, wristwatch and calculator—or in the form of metal-plated
Concepts & Skills to Review
before you study this chapter
• redox terminology (Section 4.5
and Interchapter Topic 5)
jewelry or silverware. The operation and creation of these objects, and the many
• balancing redox reactions
similar ones you use daily, involve the principles of electrochemistry, one of the
(Section 4.5)
most important areas of applied thermodynamics. • activity series of the metals
Electrochemistry is the study of the relationship between chemical change (Section 4.6)
and electrical work. It is typically investigated through the use of electrochemi- • free energy, work, and equilibrium
cal cells, systems that incorporate a redox reaction to produce or utilize electri- (Sections 20.3 and 20.4)
cal energy. The common objects just mentioned highlight the essential difference • Q vs. K (Section 17.4) and G vs.
between the two types of electrochemical cells: G (Section 20.4)
• One type of cell does work by releasing free energy from a spontaneous reac-
tion to produce electricity. A battery houses this type of cell.
• The other type of cell does work by absorbing free energy from a source of
electricity to drive a nonspontaneous reaction. Such cells are used to plate a
thin layer of metal on objects and, in major industrial processes, produce some
compounds and nonmetals and recover many metals from their ores.
IN THIS CHAPTER . . . We review redox concepts and highlight a method for
balancing redox equations (mentioned briefly in Chapter 4) that is particularly
useful for electrochemical cells. An overview of the two cell types follows. The
first type releases free energy to do electrical work, and we see how and why
it operates. The cell’s electrical output and its relation to the relative strengths
of the redox species concern us next. We then examine the free energy change
and equilibrium nature of the cell reaction and how they relate to the cell out-
put. The concentration cells and batteries we consider are useful examples of The Electrochemical Future Is Here
this type of cell, and corrosion is a destructive electrochemical process that As the combustion products of coal and
operates by the same principle. Next, we focus on cells that absorb free energy gasoline continue to threaten our atmo-
to do electrical work and see how they are used to isolate elements from their sphere, a new generation of electrochemi-
compounds and how to determine the identity and amount of product formed. cal devices is being developed. Battery-
Finally, we examine the redox system that generates energy in living cells. gasoline hybrid cars are already common,
achieving much higher mileage than tra-
ditional gasoline-powered cars. Soon,
electric cars, powered by banks of ad-
21.1 REDOX REACTIONS AND ELECTROCHEMICAL CELLS vanced fuel cells, may reduce the need for
Whether an electrochemical process releases or absorbs free energy, it always gasoline to a minimum. Every major car
involves the movement of electrons from one chemical species to another in an company has a fuel-cell prototype in op-
oxidation-reduction (redox) reaction. In this section, we review the redox process eration. And these devices, once used
principally on space missions, are now
and describe the half-reaction method of balancing redox reactions that was men-
being produced for everyday residential
tioned in Section 4.5. Then we see how such reactions are used in the two types and industrial applications as well.
of electrochemical cells.
A Quick Review of Oxidation-Reduction Concepts

In electrochemical reactions, as in any redox process, oxidation is the loss of elec-
trons, and reduction is the gain of electrons. An oxidizing agent is the species that
does the oxidizing, taking electrons from the substance being oxidized. A reduc-
ing agent is the species that does the reducing, giving electrons to the substance
being reduced. After the reaction, the oxidized substance has a higher (more pos-
itive or less negative) oxidation number (O.N.), and the reduced substance has a
lower (less positive or more negative) one. Keep in mind three key points:
• Oxidation (electron loss) always accompanies reduction (electron gain).
• The oxidizing agent is reduced, and the reducing agent is oxidized.
• The total number of electrons gained by the atoms/ions of the oxidizing agent
always equals the total number lost by the atoms/ions of the reducing agent.
923
924 Chapter 21 Electrochemistry: Chemical Change and Electrical Work
Figure 21.1 A summary of redox termi- PROCESS Zn(s) + 2H+(aq) Zn2+(aq) + H2 (g)
nology. In the reaction between zinc and
hydrogen ion, Zn is oxidized and H is OXIDATION
• One reactant loses electrons. Zinc loses electrons.
reduced.
• Reducing agent is oxidized. Zinc is the reducing agent and
becomes oxidized.
• Oxidation number increases. The oxidation number of Zn
increases from 0 to +2.
REDUCTION
• Other reactant gains electrons. Hydrogen ion gains electrons.
• Oxidizing agent is reduced. Hydrogen ion is the oxidizing
agent and becomes reduced.
• Oxidation number decreases. The oxidation number of H
decreases from +1 to 0.
Figure 21.1 presents these ideas for the aqueous reaction between zinc metal
and a strong acid. Be sure you can identify the oxidation and reduction parts of
a redox process. If you’re having trouble, you may want to review the full dis-
cussion in Chapter 4 and the summary in Topic 5 of the Interchapter.
Half-Reaction Method for Balancing Redox Reactions

In Chapter 4, two methods for balancing redox reactions were mentioned—the
oxidation number method and the half-reaction method—but only the first was
discussed in detail (see Sample Problem 4.10, p. 163). The essential difference
between the two methods is that the half-reaction method divides the overall
redox reaction into oxidation and reduction half-reactions. Each half-reaction is
balanced for atoms and charge. Then, one or both are multiplied by some integer
to make electrons gained equal electrons lost, and the half-reactions are recom-
bined to give the balanced redox equation. The half-reaction method offers sev-
eral advantages for studying electrochemistry:
• It separates the oxidation and reduction steps, which reflects their actual phys-
ical separation in electrochemical cells.
• It is readily applied to redox reactions that take place in acidic or basic solu-
tion, which is common in these cells.
• It (usually) does not require assigning O.N.s. (In cases where the half-reactions
are not obvious, we assign O.N.s to determine which atoms undergo a change
and write half-reactions with the species that contain those atoms.)
In general, we begin with a “skeleton” ionic reaction, which shows only the
species that are oxidized and reduced. If the oxidized form of a species is on the
left side of the skeleton reaction, the reduced form of that species must be on the
right, and vice versa. Unless H2O, H, and OH are being oxidized or reduced,
they do not appear in the skeleton reaction. The following steps are used in bal-
ancing a redox reaction by the half-reaction method:
Step 1. Divide the skeleton reaction into two half-reactions, each of which con-
tains the oxidized and reduced forms of one of the species. (Which half-reaction
is the oxidation and which the reduction becomes clear in the next step.)
Step 2. Balance the atoms and charges in each half-reaction.
• Atoms are balanced in order: atoms other than O and H, then O, and then H.
• Charge is balanced by adding electrons (e). They are added to the left in the
reduction half-reaction because the reactant gains them; they are added to the
right in the oxidation half-reaction because the reactant loses them.
Step 3. If necessary, multiply one or both half-reactions by an integer to make
the number of e gained in the reduction equal the number lost in the oxidation.
21.1 Redox Reactions and Electrochemical Cells 925
Step 4. Add the balanced half-reactions, and include states of matter.

Step 5. Check that the atoms and charges are balanced.
We’ll balance a redox reaction that occurs in acidic solution first and then go
through Sample Problem 21.1 to balance one in basic solution.
Balancing Redox Reactions in Acidic Solution When a redox reaction occurs in
acidic solution, H2O molecules and H ions are available for balancing. Even
though we’ve usually used H3O to indicate the proton in water, we use H in
this chapter because it makes the balanced equations less complex. However, after
using H in this example, we’ll balance the same equation with H3O just to
show you that the only difference is in the number of water molecules needed.
Let’s balance the redox reaction between dichromate ion and iodide ion to
form chromium(III) ion and solid iodine, which occurs in acidic solution (Figure
21.2). The skeleton ionic reaction shows only the oxidized and reduced species:
Cr2O72(aq) I(aq) ±£ Cr3(aq) I2 (s) [acidic solution]
Step 1. Divide the reaction into half-reactions, each of which contains the Figure 21.2 The redox reaction be-
oxidized and reduced forms of one species. The two chromium species make up tween dichromate ion and iodide ion.
one half-reaction, and the two iodine species make up the other: When Cr2O72 (left) and I (center) are
mixed in acid solution, they react to form
Cr2O72 ±£ Cr3 Cr3 and I2 (right).
I ±£ I2
Step 2. Balance atoms and charges in each half-reaction. We use H2O to bal-
ance O atoms, H to balance H atoms, and e to balance positive charges.
• For the Cr2O72/Cr3 half-reaction:
a. Balance atoms other than O and H. We balance the two Cr on the left with
a coefficient 2 on the right:
Cr2O72 ±£ 2Cr3
b. Balance O atoms by adding H2O molecules. Each H2O has one O atom, so
we add seven H2O on the right to balance the seven O in Cr2O72:
Cr2O72 ±£ 2Cr3 7H2O
c. Balance H atoms by adding H ions. Each H2O contains two H, and we
added seven H2O, so we add 14 H ions on the left:
14H Cr2O72 ±£ 2Cr3 7H2O
d. Balance charge by adding electrons. Each H ion has a 1 charge, and 14
H plus Cr2O72 gives 12 on the left. Two Cr3 give 6 on the right. There
is an excess of 6 on the left, so we add six e on the left:
6e 14H Cr2O72 ±£ 2Cr3 7H2O
This half-reaction is balanced, and we see it is the reduction because electrons
appear on the left, as reactants: the reactant Cr2O72 gains electrons (is
reduced), so Cr2O72 is the oxidizing agent. (Note that the O.N. of Cr decreases
from 6 on the left to 3 on the right.)
• For the I/I2 half-reaction:
a. Balance atoms other than O and H. Two I atoms on the right require a coef-
ficient 2 on the left:
2I ±£ I2
b. Balance O atoms with H2O. Not needed; there are no O atoms.
c. Balance H atoms with H. Not needed; there are no H atoms.
d. Balance charge with e. To balance the 2 on the left, we add two e on
the right:
2I ±£ I2 2e
This half-reaction is balanced, and it is the oxidation because electrons appear
on the right, as products: the reactant I loses electrons (is oxidized), so I is
the reducing agent. (Note that the O.N. of I increases from 1 to 0.)
Step 3. Multiply each half-reaction, if necessary, by an integer so that the num-

ber of e lost in the oxidation equals the number of e gained in the reduction.
Two e are lost in the oxidation and six e are gained in the reduction, so we
multiply the oxidation by 3:
3(2I ±£ I2 2e)
6I ±£ 3I2 6e
Step 4. Add the half-reactions together, canceling substances that appear on
both sides, and include states of matter. In this example, only the electrons cancel:
6e 14H Cr2O72 ±£ 2Cr3 7H2O
6I ±£ 3I2 6e

6I (aq) 14H (aq) Cr2O7 (aq) ±£ 3I2 (s) 7H2O(l) 2Cr3(aq)
2
Step 5. Check that atoms and charges balance:

Reactants (6I, 14H, 2Cr, 7O; 6) ±£ products (6I, 14H, 2Cr, 7O; 6)
Balancing a Redox Reaction Using H3O Now let’s balance the reduction half-
reaction with H3O as the supplier of H atoms. In step 2, balancing Cr atoms and
balancing O atoms with H2O are the same as before, so we start with
Cr2O72 ±£ 2Cr3 7H2O
Now we use H3O instead of H to balance H atoms. Because each H3O is an

H bonded to an H2O, we balance the 14 H on the right with 14 H3O on the

left and immediately balance the H2O that are part of those 14 H3O by adding
14 more H2O on the right:
14H3O Cr2O72 ±£ 2Cr3 7H2O 14H2O
Simplifying the right side by taking the sum of the H2O molecules gives
14H3O Cr2O72 ±£ 2Cr3 21H2O
None of this affects the redox change, so balancing the charge still requires six
e on the left, and we obtain the balanced reduction half-reaction:
6e 14H3O Cr2O72 ±£ 2Cr3 21H2O
Adding the balanced oxidation half-reaction gives the balanced redox equation:
6I(aq) 14H3O(aq) Cr2O72(aq) ±£ 3I2 (s) 21H2O(l) 2Cr3(aq)
Note, as mentioned earlier, that the only difference is the number of H2O mol-
ecules: there are 14 more H2O (21 instead of 7) from the 14 H3O.
To depict the reaction even more accurately, we could show the metal ion in
its hydrated form, which would also affect only the total number of H2O:
6I(aq) 14H3O(aq) Cr2O72(aq) ±£ 3I2 (s) 9H2O(l) 2Cr(H2O) 63(aq)
Although these added steps present the species in solution more accurately, they
change only the number of water molecules and make balancing more difficult;
thus, they detract somewhat from the key chemical event—the redox change.
Therefore, we’ll employ the method that uses H to balance H atoms.
Balancing Redox Reactions in Basic Solution As you just saw, in acidic solution,
H2O molecules and H (or H3O) ions are available for balancing. As Sample
Problem 21.1 shows, in basic solution, H2O molecules and OH ions are avail-
able. Only one additional step is needed to balance a redox equation that takes
place in basic solution. It appears after both half-reactions have first been bal-
anced as if they took place in acidic solution (steps 1 and 2), the e lost have
been made equal to the e gained (step 3), and the half-reactions have been com-
bined (step 4). At this point, we add one OH ion to both sides of the equation
for every H ion present. (We label this step “4 Basic.”) The H ions on one side
are combined with the added OH ions to form H2O, and OH ions appear on
the other side of the equation. Excess H2O molecules are canceled, and states of
matter are identified. Finally, we check that atoms and charges balance (step 5).
siL48593_ch21_922-979 29:11:07 03:41pm Page 927
21.1 Redox Reactions and Electrochemical Cells 927
SAMPLE PROBLEM 21.1 Balancing Redox Reactions by the Half-Reaction

Method
PROBLEM Permanganate ion is a strong oxidizing agent, and its deep purple color makes
it useful as an indicator in redox titrations (see Figure 4.15, p. 165). It reacts in basic solu-
tion with the oxalate ion to form carbonate ion and solid manganese dioxide. Balance the
skeleton ionic equation for the reaction between NaMnO4 and Na2C2O4 in basic solution:
MnO4(aq) C2O42(aq) ±£ MnO2 (s) CO32(aq) [basic solution]
PLAN We proceed through step 4 as if this took place in acidic solution. Then, we add the
appropriate number of OH ions and cancel excess H2O molecules (step 4 Basic).
SOLUTION
1. Divide into half-reactions.
MnO4 ±£ MnO2 C2O42 ±£ CO32
2. Balance.
a. Atoms other than O and H, a. Atoms other than O and H,
Not needed C2O42 ±£ 2CO32
b. O atoms with H2O, b. O atoms with H2O,
MnO4 ±£ MnO2 2H2O 2H2O C2O42 ±£ 2CO32
c. H atoms with H, c. H atoms with H,

4H MnO4 ±£ MnO2 2H2O 2H2O C2O42 ±£ 2CO32 4H

d. Charge with e , d. Charge with e,

3e 4H MnO4 ±£ MnO2 2H2O 2H2O C2O42 ±£ 2CO32 4H 2e
[reduction] [oxidation]
3. Multiply each half-reaction, if necessary, by some integer to make e lost equal e gained.
2(3e 4H MnO4 ±£ MnO2 2H2O ) 3(2H2O C2O42 ±£ 2CO32 4H 2e )
6e 8H 2MnO4 ±£ 2MnO2 4H2O 6H2O 3C2O42 ±£ 6CO32 12H 6e
4. Add half-reactions, and cancel substances appearing on both sides.
The six e cancel, eight H cancel to leave four H on the right, and four H2O cancel
to leave two H2O on the left:
6e 8H 2MnO4 ±£ 2MnO2 4H2O
2 6H2O 3C2O42 ±£ 6CO32 4 12H 6e
2MnO4 2H2O 3C2O42 ±£ 2MnO2 6CO32 4H
4 Basic. Add OH to both sides to neutralize H, and cancel H2O.
Adding four OH to both sides forms four H2O on the right, two of which cancel the
two H2O on the left, leaving two H2O on the right:
2MnO4 2H2O 3C2O42 4OH ±£ 2MnO2 6CO32 [4H 4OH]
2MnO4 2H2O 3C2O42 4OH ±£ 2MnO2 6CO32 2 4 H2O
Including states of matter gives the final balanced equation:
2MnO4(aq) 3C2O42(aq) 4OH(aq) ±£ 2MnO2 (s) 6CO32(aq) 2H2O(l)
5. Check that atoms and charges balance.
(2Mn, 24O, 6C, 4H; 12) ±£ (2Mn, 24O, 6C, 4H; 12)
COMMENT As a final step, we can obtain the balanced molecular equation for this reac-
tion by noting the number of moles of each anion in the balanced ionic equation and
adding the correct number of moles of spectator ions (in this case, Na) to obtain neutral
compounds. Thus, for instance, balancing the charge of 2 mol of MnO4 requires 2 mol
of Na, so we have 2NaMnO4. The balanced molecular equation is
2NaMnO4 (aq) 3Na2C2O4 (aq) 4NaOH(aq) ±£
2MnO2 (s) 6Na2CO3 (aq) 2H2O(l)
FOLLOW-UP PROBLEM 21.1 Write a balanced molecular equation for the reaction
between KMnO4 and KI in basic solution. The skeleton ionic reaction is
MnO4(aq) I(aq) ±£ MnO42(aq) IO3(aq) [basic solution]
The half-reaction method reveals a great deal about redox processes and is
essential to understanding electrochemical cells. The major points are
• Any redox reaction can be treated as the sum of a reduction and an oxidation
half-reaction.
• Atoms and charge are conserved in each half-reaction.
• Electrons lost in one half-reaction are gained in the other.
• Although the half-reactions are treated separately, electron loss and electron
gain occur simultaneously.
An Overview of Electrochemical Cells

We distinguish two types of electrochemical cells based on the general thermo-
dynamic nature of the reaction:
1. A voltaic cell (or galvanic cell) uses a spontaneous reaction (G 0) to gen-
erate electrical energy. In the cell reaction, the difference in chemical potential
energy between higher energy reactants and lower energy products is converted
into electrical energy. This energy is used to operate the load—flashlight bulb,
CD player, car starter motor, or other electrical device. In other words, the sys-
tem does work on the surroundings. All batteries contain voltaic cells.
2. An electrolytic cell uses electrical energy to drive a nonspontaneous reaction
(G 0). In the cell reaction, electrical energy from an external power sup-
ply converts lower energy reactants into higher energy products. Thus, the sur-
roundings do work on the system. Electroplating and recovering metals from
ores involve electrolytic cells.
The two types of cell have certain design features in common (Figure 21.3).
Two electrodes, which conduct the electricity between cell and surroundings, are
dipped into an electrolyte, a mixture of ions (usually in aqueous solution) that
Figure 21.3 General characteristics of VOLTAIC CELL ELECTROLYTIC CELL

voltaic and electrolytic cells. A voltaic Energy is released from Energy is absorbed to drive
cell (A) generates energy from a sponta- spontaneous redox reaction nonspontaneous redox reaction
neous reaction (G 0), whereas an
electrolytic cell (B) requires energy to System does work on Surroundings (power supply)
drive a nonspontaneous reaction surroundings do work on system (cell)
(G 0). In both types of cell, two elec-
e– Surroundings e– e– Power e–
trodes dip into electrolyte solutions, and
Anode (Load) Cathode Anode supply Cathode
an external circuit provides the means
(oxidation) (reduction) (oxidation) (reduction)
for electrons to flow between them. Energy
Most important, notice that oxidation Energy
takes place at the anode and reduction
takes place at the cathode, but the rela-
tive electrode charges are opposite in
the two cells. (–) (+) (+) (–)
Electrolyte
Electrolyte X+ Electrolyte Y+ with A– and B+
Oxidation half-reaction Oxidation half-reaction

X X+ + e– A– A + e–
Reduction half-reaction Reduction half-reaction

e– + Y+ Y e– + B+ B
Overall (cell) reaction Overall (cell) reaction

X + Y+ X+ + Y; ΔG < 0 A– + B+ A + B; ΔG > 0
A B
21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 929
are involved in the reaction or that carry the charge. An electrode is identified as
either anode or cathode depending on the half-reaction that takes place there:
• The oxidation half-reaction occurs at the anode. Electrons are lost by the sub-
stance being oxidized (reducing agent) and leave the cell at the anode.
• The reduction half-reaction occurs at the cathode. Electrons are gained by the
substance being reduced (oxidizing agent) and enter the cell at the cathode.
Which Half-Reaction Occurs at Which
As shown in Figure 21.3, the relative charges of the electrodes are opposite in
Electrode? Here are some memory aids to
the two types of cell. As you’ll see in the following sections, these opposite help you remember which half-reaction
charges result from the different phenomena that cause the electrons to flow. occurs at which electrode:
1. The words anode and oxidation start
Section Summary with vowels; the words cathode and re-
An oxidation-reduction (redox) reaction involves the transfer of electrons from a reduc- duction start with consonants.
ing agent to an oxidizing agent. • The half-reaction method of balancing divides the 2. Alphabetically, the A in anode comes
overall reaction into half-reactions that are balanced separately and then recombined. before the C in cathode, and the O in ox-
• The two types of electrochemical cells are based on redox reactions. In a voltaic idation comes before the R in reduction.
cell, a spontaneous reaction generates electricity and does work on the surroundings; 3. Look at the first syllables and use your
in an electrolytic cell, the surroundings supply electricity that does work to drive a imagination:
nonspontaneous reaction. In both types, two electrodes dip into electrolyte solutions; ANode, OXidation; REDuction, CAThode
oxidation occurs at the anode, and reduction occurs at the cathode. : AN OX and a RED CAT
21.2 VOLTAIC CELLS: USING SPONTANEOUS REACTIONS

TO GENERATE ELECTRICAL ENERGY
If you put a strip of zinc metal in a solution of Cu2 ion, the blue color of the
solution fades as a brown-black crust of Cu metal forms on the Zn strip (Figure
21.4). Judging from what we see, the reaction involves the reduction of Cu2 ion
to Cu metal, which must be accompanied by the oxidation of Zn metal to Zn2
ion. The overall reaction consists of two half-reactions:
Cu2(aq) 2e ±£ Cu(s) [reduction]
Zn(s) ±£ Zn2 (aq) 2e [oxidation]
Zn(s) Cu2(aq) ±£ Zn2(aq) Cu(s) [overall reaction]
Figure 21.4 The spontaneous reaction

between zinc and copper(II) ion. When a
strip of zinc metal is placed in a solution
of Cu2 ion, a redox reaction begins (left),
in which the zinc is oxidized to Zn2 and
the Cu2 is reduced to copper metal. As
the reaction proceeds (right), the deep
blue color of the solution of hydrated
Cu2 ion lightens, and the Cu “plates out”
on the Zn and falls off in chunks. (The Cu
appears black because it is very finely di-
vided.) At the atomic scale, each Zn atom
loses two electrons, which are gained by
a Cu2 ion. The process is summarized
with symbols in the balanced equation.
Zn(s) + Cu 2+(aq) Zn2+(aq) + Cu(s)

In the remainder of this section, we examine this spontaneous reaction as the basis
of a voltaic (galvanic) cell.
Construction and Operation of a Voltaic Cell

Animation: Operation of a Voltaic Cell Electrons are being transferred in the Zn/Cu2 reaction (Figure 21.4), but the sys-
tem does not generate electrical energy because the oxidizing agent (Cu2) and
the reducing agent (Zn) are in the same beaker. If, however, the half-reactions are
physically separated and connected by an external circuit, the electrons are trans-
ferred by traveling through the circuit and an electric current is produced.
This separation of half-reactions is the essential idea behind a voltaic cell
(Figure 21.5A). The components of each half-reaction are placed in a separate
container, or half-cell, which consists of one electrode dipping into an electrolyte
solution. The two half-cells are joined by the circuit, which consists of a wire and
a salt bridge (the inverted U tube in the figure; we will discuss its function
shortly). In order to measure the voltage generated by the cell, a voltmeter is
inserted in the path of the wire connecting the electrodes. A switch (not shown)
closes (completes) or opens (breaks) the circuit. By convention, the oxidation half-
cell (anode compartment) is shown on the left and the reduction half-cell (cathode
compartment) on the right. Here are the key points about the Zn/Cu2 voltaic cell:
1. The oxidation half-cell. In this case, the anode compartment consists of a
zinc bar (the anode) immersed in a Zn2 electrolyte (such as a solution of zinc
sulfate, ZnSO4). The zinc bar is the reactant in the oxidation half-reaction, and it
conducts the released electrons out of its half-cell.
2. The reduction half-cell. In this case, the cathode compartment consists of
a copper bar (the cathode) immersed in a Cu2 electrolyte [such as a solution of
copper(II) sulfate, CuSO4]. Copper metal is the product in the reduction half-
reaction, and the bar conducts electrons into its half-cell.
3. Relative charges on the electrodes. The electrode charges are determined
by the source of electrons and the direction of electron flow through the circuit.
In this cell, zinc atoms are oxidized at the anode to Zn2 ions and electrons. The
Zn2 ions enter the solution, while the electrons enter the bar and then the wire.
The electrons flow left to right through the wire to the cathode, where Cu2 ions
in the solution accept them and are reduced to Cu atoms. As the cell operates,
electrons are continuously generated at the anode and consumed at the cathode.
Therefore, the anode has an excess of electrons and a negative charge relative to
the cathode. In any voltaic cell, the anode is negative and the cathode is positive.
4. The purpose of the salt bridge. The cell cannot operate unless the circuit
is complete. The oxidation half-cell originally contains a neutral solution of
Zn2 and SO42 ions, but as Zn atoms in the bar lose electrons, the solution
would develop a net positive charge from the Zn2 ions entering. Similarly, in
the reduction half-cell, the neutral solution of Cu2 and SO42 ions would
develop a net negative charge as Cu2 ions leave the solution to form Cu atoms.
If the half-cells do not remain neutral, the resulting charge imbalance would stop
cell operation. To avoid this situation and enable the cell to operate, the two half-
cells are joined by a salt bridge, which acts as a “liquid wire,” allowing ions to
flow through both compartments and complete the circuit. The salt bridge shown
in Figure 21.5A is an inverted U tube containing a solution of the nonreacting
ions Na and SO42 in a gel. The solution cannot pour out, but ions can diffuse
through it into or out of the half-cells.
To maintain neutrality in the reduction half-cell (right; cathode compartment)
as Cu2 ions change to Cu atoms, Na ions move from the salt bridge into the
solution (and some SO42 ions move from the solution into the salt bridge). Sim-
ilarly, to maintain neutrality in the oxidation half-cell (left; anode compartment)
siL48593_ch21_922-979 29:11:07 03:41pm Page 931
e–
2e– gained
2e– lost per Cu2 + ion
per Zn atom reduced
oxidized
Cu2+
Zn Cu e–
Zn2+
B
Voltmeter
e– e– Figure 21.5 A voltaic cell based on the
Anode Salt bridge Cathode zinc-copper reaction. A, The anode half-
Na+ cell (oxidation) consists of a Zn electrode
Zn (–) 2–
SO4
(+) Cu
dipping into a Zn2 solution. The two
electrons generated in the oxidation of
each Zn atom move through the Zn bar
and the wire, and into the Cu electrode,
which dips into a Cu2 solution in the
cathode half-cell (reduction). There, the
electrons reduce Cu2 ions. Thus, elec-
Cu
trons flow left to right through electrodes
Zn 2+ Cu 2+ and wire. A salt bridge contains unreac-
tive Na and SO42 ions that maintain
neutral charge in the electrolyte solutions:
Oxidation half-reaction anions in the salt bridge flow to the left,
Zn(s ) Zn2+(aq) + 2e– and cations flow to the right. The volt-
meter registers the electrical output of the
Reduction half-reaction cell. B, After the cell runs for several
Cu2+(aq) + 2e– Cu(s) hours, the Zn anode weighs less because
Zn atoms have been oxidized to aqueous
Overall (cell) reaction Zn2 ions, and the Cu cathode weighs
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) more because aqueous Cu2 ions have
A been reduced to Cu metal.
as Zn atoms change to Zn2 ions, SO42 ions move from the salt bridge into that
solution (and some Zn2 ions move from the solution into the salt bridge). Thus,
as Figure 21.5A shows, the circuit is completed as electrons move left to right
through the wire, while anions move right to left and cations move left to right
through the salt bridge.
5. Active vs. inactive electrodes. The electrodes in the Zn/Cu2 cell are active
because the metal bars themselves are components of the half-reactions. As the
cell operates, the mass of the zinc electrode gradually decreases, and the [Zn2]
in the anode half-cell increases. At the same time, the mass of the copper elec-
trode increases, and the [Cu2] in the cathode half-cell decreases; we say that the
Cu2 “plates out” on the electrode. Look at Figure 21.5B to see how the elec-
trodes look, removed from their half-cells, after several hours of operation.
For many redox reactions, there are no reactants or products capable of serv-
ing as electrodes, so inactive electrodes are used. Most commonly, inactive elec-
trodes are rods of graphite or platinum: they conduct electrons into or out of the
half-cells but cannot take part in the half-reactions. In a voltaic cell based on
the following half-reactions, for instance, the reacting species cannot act as
electrodes:
2I (aq) ±£ I2 (s) 2e [anode; oxidation]
MnO4(aq) 8H (aq) 5e ±£ Mn2(aq) 4H2O(l)

[cathode; reduction]
e– Voltmeter Therefore, each half-cell consists of inactive electrodes immersed in an electrolyte

e–
solution that contains all the reactant species involved in that half-reaction (Fig-
ure 21.6). In the anode half-cell, I ions are oxidized to solid I2. The electrons
Anode Cathode
that are released flow into the graphite anode, through the wire, and into the
C (–) K+ (+) C
NO3– graphite cathode. From there, the electrons are consumed by MnO4 ions, which
are reduced to Mn2 ions. (A KNO3 salt bridge is used.)
As Figures 21.5A and 21.6 show, there are certain consistent features in the
diagram of any voltaic cell. The physical arrangement includes the half-cell con-
tainers, electrodes, wire, and salt bridge, and the following details appear:
I2 • Components of the half-cells: electrode materials, electrolyte ions, and other
substances involved in the reaction
I– MnO4–, H+, Mn 2+
• Electrode name (anode or cathode) and charge. By convention, the anode com-
partment always appears on the left.
Oxidation half-reaction
• Each half-reaction with its half-cell and the overall cell reaction
2I–(aq) I2(s) + 2e–
• Direction of electron flow in the external circuit
Reduction half-reaction • Nature of ions and direction of ion flow in the salt bridge
MnO4–(aq) + 8H+(aq) + 5e–
You’ll see how to specify these details and diagram a cell shortly.
Mn2+(aq) + 4H2O(l )
Overall (cell) reaction

Notation for a Voltaic Cell
2MnO4–(aq) + 16H+(aq) + 10I –(aq) A useful shorthand notation describes the components of a voltaic cell. For exam-
2Mn2+(aq) + 5I 2(s) + 8H2O(l ) ple, the notation for the Zn/Cu2 cell is
Zn(s) 0 Zn2(aq) 0 0 Cu2(aq) 0 Cu(s)
Figure 21.6 A voltaic cell using inactive
electrodes. The reaction between I and Key parts of the notation are
MnO4 in acidic solution does not have
species that can be used as electrodes,
• The components of the anode compartment (oxidation half-cell) are written to
so inactive graphite (C) electrodes are the left of the components of the cathode compartment (reduction half-cell).
used. • A single vertical line represents a phase boundary. For example, Zn(s)
Zn2(aq) indicates that the solid Zn is a different phase from the aqueous Zn2.
A comma separates the half-cell components that are in the same phase.
For example, the notation for the voltaic cell housing the reaction between I
and MnO4 shown in Figure 21.6 is
graphite 0 I(aq) I2 (s) 0 0 H(aq), MnO4(aq), Mn2(aq) 0 graphite
That is, in the cathode compartment, H, MnO4, and Mn2 ions are all in
aqueous solution with solid graphite immersed in it. Often, we specify the con-
centrations of dissolved components; for example, if the concentrations of
Zn2 and Cu2 are 1 M, we write
Zn(s) 0 Zn2(1 M) 0 0 Cu2(1 M) 0 Cu(s)
• Half-cell components usually appear in the same order as in the half-reaction,
and electrodes appear at the far left and far right of the notation.
• A double vertical line indicates the separated half-cells and represents the phase
boundary on either side of the salt bridge (the ions in the salt bridge are omit-
ted because they are not part of the reaction).
SAMPLE PROBLEM 21.2 Describing a Voltaic Cell with a Diagram and

Notation
PROBLEM Draw a diagram, show balanced equations, and write the notation for a voltaic
cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell
with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates
that the Cr electrode is negative relative to the Ag electrode.
PLAN From the given contents of the half-cells, we can write the half-reactions. We must
determine which is the anode compartment (oxidation) and which is the cathode (reduc-
tion). To do so, we must find the direction of the spontaneous redox reaction, which is
given by the relative electrode charges. Electrons are released into the anode during oxi-
e– Voltmeter e–
dation, so it has a negative charge. We are told that Cr is negative, so it must be the anode;
and, therefore, Ag is the cathode.
SOLUTION Writing the balanced half-reactions. The Ag electrode is positive, so the half- Anode Cathode
reaction consumes e: Cr
(–) K+ –
(+)
Ag
NO3
Ag(aq) e ±£ Ag(s) [reduction; cathode]
The Cr electrode is negative, so the half-reaction releases e:
Cr(s) ±£ Cr3(aq) 3e [oxidation; anode]
Writing the balanced overall cell reaction. We triple the reduction half-reaction in order
to balance e, and then we combine the half-reactions to obtain the overall reaction:
Cr(s) 3Ag(aq) ±£ Cr3(aq) 3Ag(s)
Cr 3+ Ag+
Determining direction of electron and ion flow. The released e in the Cr electrode (neg-
ative) flow through the external circuit to the Ag electrode (positive). As Cr3 ions enter
the anode electrolyte, NO3 ions enter from the salt bridge to maintain neutrality. As Ag
Cr(s) Cr3+(aq) + 3e–
ions leave the cathode electrolyte and plate out on the Ag electrode, K ions enter from
the salt bridge to maintain neutrality. The diagram of this cell is shown in the margin.
Reduction half-reaction
Writing the cell notation:
Ag+(aq) + e–
Cr(s) 0 Cr3(aq) 0 0 Ag(aq) 0 Ag(s)
Ag(s)
CHECK Always be sure that the half-reactions and cell reaction are balanced, the half-cells Overall (cell) reaction
contain all components of the half-reactions, and the electron and ion flow are shown. You Cr(s) + 3Ag+(aq) Cr3+(aq) + 3Ag(s)
should be able to write the half-reactions from the cell notation as a check.
COMMENT The key to diagramming a voltaic cell is to use the direction of the sponta-
neous reaction to identify the oxidation (anode; negative) and reduction (cathode; posi-
tive) half-reactions.
FOLLOW-UP PROBLEM 21.2 In one compartment of a voltaic cell, a graphite rod
dips into an acidic solution of K2Cr2O7 and Cr(NO3)3; in the other compartment, a tin bar
dips into a Sn(NO3)2 solution. A KNO3 salt bridge joins them. The tin electrode is nega-
tive relative to the graphite. Draw a diagram of the cell, show the balanced equations, and
write the cell notation.
Why Does a Voltaic Cell Work?

By placing a lightbulb in the circuit or looking at the voltmeter, we can see that
the Zn/Cu2 cell generates electrical energy. But what principle explains how the
reaction takes place, and why do electrons flow in the direction shown?
Let’s examine what happens when the switch is open and no reaction is occur-
ring. In each half-cell, we can consider the metal electrode to be in equilibrium
with the metal ions in the electrolyte and the electrons residing in the metal:
A Zn2(aq) 2e(in Zn metal)
Zn(s) B
A Cu2(aq) 2e(in Cu metal)
Cu(s) B
From the direction of the overall spontaneous reaction, we know that Zn gives up
its electrons more easily than Cu does; thus, Zn is a stronger reducing agent. Electron Flow and Water Flow Con-
Therefore, the equilibrium position of the Zn half-reaction lies farther to the right: sider this analogy between electron “pres-
Zn produces more electrons than Cu does. You might think of the electrons in the sure” and water pressure. A U tube (cell)
Zn electrode as being subject to a greater electron “pressure” than those in the is separated into two arms (two half-cells)
Cu electrode, a greater potential energy (referred to as electrical potential) ready by a stopcock (switch), and the two arms
to “push” them through the circuit. Close the switch, and electrons flow from the contain water at different heights (the
Zn to the Cu electrode to equalize this difference in electrical potential. The flow half-cells contain half-reactions with dif-
disturbs the equilibrium at each electrode. The Zn half-reaction shifts to the right ferent electrical potentials). Open the
to restore the electrons flowing out, and the Cu half-reaction shifts to the left to stopcock (close the switch), and the dif-
remove the electrons flowing in. Thus, the spontaneous reaction occurs as a result ferent heights (potential difference) be-
come equal as water flows (electrons flow
of the different abilities of these metals to give up their electrons and the ability
and current is generated).
of the electrons to flow through the circuit.
siL48593_ch21_922-979 29:11:07 03:41pm Page 934
Section Summary
A voltaic cell consists of oxidation (anode) and reduction (cathode) half-cells, con-
nected by a wire to conduct electrons and a salt bridge to maintain charge neu-
trality as the cell operates. • Electrons move from anode (left) to cathode (right),
while cations move from the salt bridge into the cathode half-cell and anions from
the salt bridge into the anode half-cell. • The cell notation shows the species and
their phases in each half-cell, as well as the direction of current flow. • A voltaic
cell operates because species in the two half-cells differ in their tendency to lose
electrons.
21.3 CELL POTENTIAL: OUTPUT OF A VOLTAIC CELL

The purpose of a voltaic cell is to convert the free energy change of a sponta-
neous reaction into the kinetic energy of electrons moving through an external
circuit (electrical energy). This electrical energy is proportional to the difference
in electrical potential between the two electrodes, which is called the cell
potential (Ecell); it is also called the voltage of the cell or the electromotive force
(emf).
Electrons flow spontaneously from the negative to the positive electrode, that
is, toward the electrode with the more positive electrical potential. Thus, when
the cell operates spontaneously, there is a positive cell potential:
Ecell 7 0 for a spontaneous process (21.1)
The more positive Ecell is, the more work the cell can do, and the farther the reac-
tion proceeds to the right as written. A negative cell potential, on the other hand,
is associated with a nonspontaneous cell reaction. If Ecell 0, the reaction has
reached equilibrium and the cell can do no more work. (There is a clear rela-
tionship between Ecell, K, and G that we’ll discuss in Section 21.4.)
How are the units of cell potential related to those of energy available to do
work? As you’ve seen, work is done when charge moves between electrode com-
partments that differ in electrical potential. The SI unit of electrical potential is
the volt (V), and the SI unit of electrical charge is the coulomb (C). By defini-
tion, for two electrodes that differ by 1 volt of electrical potential, 1 joule of
energy is released (that is, 1 joule of work can be done) for each coulomb of
charge that moves between the electrodes. Thus,
1 V 1 J/C (21.2)
Table 21.1 lists the voltages of some commercial and natural voltaic cells. Next,
we’ll see how to measure cell potential.
Table 21.1 Voltages of Some Voltaic Cells

Voltaic Cell Voltage ( V )
Common alkaline flashlight battery 1.5
Lead-acid car battery (6 cells 12 V) 2.0
Calculator battery (mercury) 1.3
Lithium-ion laptop battery 3.7
Electric eel (⬃5000 cells in 6-ft eel 750 V) 0.15
Nerve of giant squid (across cell membrane) 0.070
siL48593_ch21_922-979 12:11:07 09:22am Page 935
21.3 Cell Potential: Output of a Voltaic Cell 935
Standard Cell Potentials Cu half-cell

The measured potential of a voltaic cell is affected by changes in concentration (cathode; reduction)
as the reaction proceeds and by energy losses due to heating of the cell and the
external circuit. Therefore, in order to compare the output of different cells, we
obtain a standard cell potential (Ecell), the potential measured at a specified tem-
perature (usually 298 K) with no current flowing* and all components in their
standard states: 1 atm for gases, 1 M for solutions, the pure solid for electrodes.
When the zinc-copper cell that we diagrammed in Figure 21.5 begins operating
Zn half-cell
under standard state conditions, that is, when [Zn2] [Cu2] 1 M, the cell
(anode; oxidation)
produces 1.10 V at 298 K (Figure 21.7):
Zn(s) Cu2(aq; 1 M) ±£ Zn2(aq; 1 M) Cu(s) E°cell 1.10 V
Standard Electrode (Half-Cell) Potentials Just as each half-reaction makes up part Figure 21.7 Measurement of a
of the overall reaction, the potential of each half-cell makes up a part of the over- standard cell potential. The zinc-copper
all cell potential. The standard electrode potential (Ehalf-cell) is the potential cell, operating at 298 K under standard-
state conditions, produces a voltage of
associated with a given half-reaction (electrode compartment) when all the com-
1.10 V.
ponents are in their standard states.
By convention, a standard electrode potential always refers to the half-
reaction written as a reduction. For the zinc-copper reaction, for example, the
standard electrode potentials for the zinc half-reaction (Ezinc, anode compartment)
and for the copper half-reaction (Ecopper, cathode compartment) refer to the
processes written as reductions:
Zn2(aq) 2e ±£ Zn(s) E°zinc (E°anode ) [reduction]
Cu2(aq) 2e ±£ Cu(s) E°copper (E°cathode ) [reduction]
The overall cell reaction involves the oxidation of zinc at the anode, not the reduc-
tion of Zn2, so we reverse the zinc half-reaction:
Zn(s) ±£ Zn2(aq) 2e [oxidation]
Cu (aq) 2e ±£ Cu(s)
2
[reduction]
The overall redox reaction is the sum of these half-reactions: Animation: Galvanic Cell
Zn(s) Cu2(aq) ±£ Zn2(aq) Cu(s)
Because electrons flow spontaneously toward the copper electrode (cathode), it
must have a more positive Ehalf-cell than the zinc electrode (anode). Therefore, to
obtain a positive Ecell, we subtract Ezinc from Ecopper:
E°cell E°copper E°zinc
We can generalize this result for any voltaic cell: the standard cell potential is
the difference between the standard electrode potential of the cathode (reduction)
half-cell and the standard electrode potential of the anode (oxidation) half-cell:
E°cell E°cathode (reduction) E°anode (oxidation) (21.3)
For a spontaneous reaction at standard conditions, Ecell 0.
Determining E half-cell: The Standard Hydrogen Electrode What portion of Ecell
for the zinc-copper reaction is contributed by the anode half-cell (oxidation of Zn)
and what portion by the cathode half-cell (reduction of Cu2)? That is, how can
we know half-cell potentials if we can only measure the potential of the complete
cell? Half-cell potentials, such as Ezinc and Ecopper, are not absolute quantities,
but rather are values relative to that of a standard. This standard reference
*The current required to operate modern digital voltmeters makes a negligible difference in the
value of Ecell.
half-cell has its standard electrode potential defined as zero (Ereference 0.00 V).
The standard reference half-cell is a standard hydrogen electrode, which con-
sists of a specially prepared platinum electrode immersed in a 1 M aqueous solu-
tion of a strong acid, H(aq) [or H3O(aq)], through which H2 gas at 1 atm is
bubbled. Thus, the reference half-reaction is
2H(aq; 1 M) 2e B
A H2 (g; 1 atm) E°reference 0.00 V
Now we can construct a voltaic cell consisting of this reference half-cell and
another half-cell whose potential we want to determine. With Ereference defined as
zero, the overall Ecell allows us to find the unknown standard electrode potential,
Eunknown. When H2 is oxidized, the reference half-cell is the anode, and so reduc-
tion occurs at the unknown half-cell:
E°cell E°cathode E°anode E°unknown E°reference E°unknown 0.00 V E°unknown
When H is reduced, the reference half-cell is the cathode, and so oxidation
occurs at the unknown half-cell:
E°cell E°cathode E°anode E°reference E°unknown 0.00 V E°unknown E°unknown
Figure 21.8 shows a voltaic cell that has the Zn/Zn2 half-reaction in one com-
partment and the H/H2 (or H3O/H2) half-reaction in the other. The zinc elec-
trode is negative relative to the hydrogen electrode, so we know that the zinc is
being oxidized and is the anode. The measured Ecell is 0.76 V, and we use this
value to find the unknown standard electrode potential, Ezinc:
2H(aq) 2e ±£ H2 (g) E°reference 0.00 V [cathode; reduction]
Zn(s) ±£ Zn2(aq) 2e E°zinc ? V [anode; oxidation]
Zn(s) 2H(aq) ±£ Zn2(aq) H2 (g) E°cell 0.76 V
E°cell E°cathode E°anode E°reference E°zinc
E°zinc E°reference E°cell 0.00 V 0.76 V 0.76 V
e– 0.76 V e–
Voltmeter
Anode Cathode H2
H2 (g)
Zn (–) Salt bridge
(+)
e– 1 atm
H2 bubble Pt surface
2e– lost
per Zn atom
oxidized Pt wire 2H2O
Zn 2 + Pt
2e– gained e–
per H2 formed
Zn
1 M Zn 2 + 1 M H3O+ 2H3O+
Zn(s) Zn2+(aq ) + 2e–
Reduction half-reaction
2H3O + (aq ) + 2e– H2(g ) + 2H2O(l )

Zn(s) + 2H3O + (aq ) Zn 2+(aq) + H2(g) + 2H2O(l )
Figure 21.8 Determining an unknown E half-cell with the standard which enters the H2 bubble. The Zn/Zn2 half-cell potential is negative
2
reference (hydrogen) electrode. A voltaic cell has the Zn/Zn half- (anode), and the cell potential is 0.76 V. The potential of the standard
reaction in one half-cell and the hydrogen reference half-reaction in the reference electrode is defined as 0.00 V, so the cell potential equals the
other. The magnified view of the hydrogen half-reaction shows two negative of the anode potential; that is,
H3O ions being reduced to two H2O molecules and an H2 molecule, 0.76 V 0.00 V E°zinc so E°zinc 0.76 V
siL48593_ch21_922-979 17:11:07 11:47am Page 937
Now let’s return to the zinc-copper cell and use the measured value of Ecell
(1.10 V) and the value we just found for Ezinc to calculate Ecopper:
E°cell E°cathode E°anode E°copper E°zinc
E°copper E°cell E°zinc 1.10 V (0.76 V) 0.34 V
By continuing this process of constructing cells with one known and one unknown
electrode potential, we can find many other standard electrode potentials. Let’s go
over these ideas once more with a sample problem.
SAMPLE PROBLEM 21.3 Calculating an Unknown Ehalf-cell from Ecell

PROBLEM A voltaic cell houses the reaction between aqueous bromine and zinc metal:
Br2(aq) Zn(s) ±£ Zn2(aq) 2Br(aq) E°cell 1.83 V
Calculate E°bromine, given E°zinc 0.76 V.
PLAN E°cell is positive, so the reaction is spontaneous as written. By dividing the reaction
into half-reactions, we see that Br2 is reduced and Zn is oxidized; thus, the zinc half-cell
contains the anode. We use Equation 21.3 to find E°unknown (E°bromine).
SOLUTION Dividing the reaction into half-reactions:
Br2(aq) 2e ±£ 2Br(aq) E°unknown E°bromine ? V
Zn(s) ±£ Zn2(aq) 2e E°zinc 0.76 V
Calculating E°bromine:
E°cell E°cathode E°anode E°bromine E°zinc
E°bromine E°cell E°zinc 1.83 V (0.76 V) 1.07 V
CHECK A good check is to make sure that calculating E°bromine E°zinc gives E°cell:
1.07 V (0.76 V) 1.83 V.
COMMENT Keep in mind that, whichever is the unknown half-cell, reduction is the cath-
ode half-reaction and oxidation is the anode half-reaction. Always subtract E°anode from
E°cathode to get E°cell.
FOLLOW-UP PROBLEM 21.3 A voltaic cell based on the reaction between aqueous
Br2 and vanadium(III) ions has Ecell 1.39 V:
Br2 (aq) 2V3 (aq) 2H2O(l) ±£ 2VO2 (aq) 4H (aq) 2Br (aq)
What is Evanadium, the standard electrode potential for the reduction of VO2 to V3?
Relative Strengths of Oxidizing and Reducing Agents

One of the things we can learn from measuring potentials of voltaic cells is the
relative strengths of the oxidizing and reducing agents involved. Three oxidizing
agents present in the voltaic cell just discussed are Cu2, H, and Zn2. We can
rank their relative oxidizing strengths by writing each half-reaction as a gain of
electrons (reduction), with its corresponding standard electrode potential:
Cu2(aq) 2e ±£ Cu(s) E° 0.34 V
2H(aq) 2e ±£ H2 (g) E° 0.00 V
Zn2(aq) 2e ±£ Zn(s) E° 0.76 V
The more positive the E value, the more readily the reaction (as written) occurs;
thus, Cu2 gains two electrons more readily than H, which gains them more
readily than Zn2. In terms of strength as an oxidizing agent, therefore, Cu2
H Zn2. Moreover, this listing also ranks the strengths of the reducing agents:
Zn H2 Cu. Notice that this list of half-reactions in order of decreasing half-
cell potential shows, from top to bottom, the oxidizing agents (reactants) decreas-
ing in strength and the reducing agents (products) increasing in strength; that is,
Cu2 (top left) is the strongest oxidizing agent, and Zn (bottom right) is the
strongest reducing agent.
By combining many pairs of half-cells into voltaic cells, we can create a list
of reduction half-reactions and arrange them in decreasing order of standard elec-
trode potential (from most positive to most negative). Such a list, called an emf
series or a table of standard electrode potentials, appears in Appendix D, with a
few examples in Table 21.2. There are several key points to keep in mind:
• All values are relative to the standard hydrogen (reference) electrode:
2H(aq; 1 M) 2e B
A H2 (g; 1 atm) E°reference 0.00 V
• By convention, the half-reactions are written as reductions, which means that
only reactants are oxidizing agents and only products are reducing agents.
• The more positive the Ehalf-cell, the more readily the half-reaction occurs.
• Half-reactions are shown with an equilibrium arrow because each can occur as
a reduction or an oxidation (that is, take place at the cathode or anode, respec-
tively), depending on the Ehalf-cell of the other half-reaction.
• As Appendix D (and Table 21.2) is arranged, the strength of the oxidizing agent
(reactant) increases going up (bottom to top), and the strength of the reducing
agent (product) increases going down (top to bottom).
Table 21.2 Selected Standard Electrode Potentials (298 K)

Half-Reaction
E half-cell (V)
F2(g) 2e B
A 2F(aq) 2.87
Cl2(g) 2e B

A 2Cl(aq) 1.36
MnO2(s) 4H (aq) 2e B

A Mn2(aq) 2H2O(l) 1.23
NO3 (aq) 4H (aq) 3e B

A NO(g) 2H2O(l) 0.96
Strength of oxidizing agent
Strength of reducing agent

Ag(aq) e B A Ag(s) 0.80
Fe3(aq) e BA Fe2(aq) 0.77
O2(g) 2H2O(l) 4e B
A 4OH(aq) 0.40
Cu (aq) 2e B
2
A Cu(s) 0.34
2H(aq) 2e B A H2(g) 0.00
N2(g) 5H(aq) 4e B A N2H5(aq) 0.23
Fe (aq) 2e B
2
A Fe(s) 0.44
Zn2(aq) 2e B A Zn(s) 0.76
2H2O(l) 2e BA H2(g) 2OH(aq) 0.83
Na (aq) e B

A Na(s) 2.71
Li(aq) e BA Li(s) 3.05
Thus, F2(g) is the strongest oxidizing agent (has the largest positive E), which
means F(aq) is the weakest reducing agent. Similarly, Li(aq) is the weakest
oxidizing agent (has the most negative E), which means Li(s) is the strongest
reducing agent. (You may have noticed an analogy to conjugate acid-base pairs:
a strong acid forms a weak conjugate base, and vice versa, just as a strong oxi-
dizing agent forms a weak reducing agent, and vice versa.)
If you forget the ranking in the table, just rely on your chemical knowledge
of the elements. You know that F2 is very electronegative and typically occurs as
F. It is easily reduced (gains electrons), so it must be a strong oxidizing agent
(high, positive E). Similarly, Li metal has a low ionization energy and typically
occurs as Li. Therefore, it is easily oxidized (loses electrons), so it must be a
strong reducing agent (low, negative E).
Writing Spontaneous Redox Reactions Appendix D can be used to write sponta-
neous redox reactions, which is useful for constructing voltaic cells.
Every redox reaction is the sum of two half-reactions, so there is a reducing
agent and an oxidizing agent on each side. In the zinc-copper reaction, for
instance, Zn and Cu are the reducing agents, and Cu2 and Zn2 are the oxidiz-
ing agents. The stronger oxidizing and reducing agents react spontaneously to
form the weaker oxidizing and reducing agents:
Zn(s) Cu2(aq) ±£ Zn2(aq) Cu(s)
stronger stronger weaker weaker
reducing agent oxidizing agent oxidizing agent reducing agent
Here, too, note the similarity to acid-base chemistry. The stronger acid and base
spontaneously form the weaker base and acid, respectively. The members of a
conjugate acid-base pair differ by a proton: the acid has the proton and the base
does not. The members of a redox pair, or redox couple, such as Zn and Zn2,
differ by one or more electrons: the reduced form (Zn) has the electrons and the
oxidized form (Zn2) does not. In acid-base reactions, we compare acid and base
strength using Ka and Kb values. In redox reactions, we compare oxidizing and
reducing strength using E values.
Based on the order of the E values in Appendix D, the stronger oxidizing
agent (species on the left) has a half-reaction with a larger (more positive or less
negative) E value, and the stronger reducing agent (species on the right) has a
half-reaction with a smaller (less positive or more negative) E value. Therefore,
a spontaneous reaction (Ecell 0) will occur between an oxidizing agent and any
reducing agent that lies below it in the list. For instance, Zn (right) lies below
Cu2 (left), and Cu2and Zn react spontaneously. In other words, for a sponta-
neous reaction to occur, the half-reaction higher in the list proceeds at the cath-
ode as written, and the half-reaction lower in the list proceeds at the anode in
reverse. This pairing ensures that the stronger oxidizing agent (higher on the left)
and stronger reducing agent (lower on the right) will be the reactants.
However, if we know the electrode potentials, we can write a spontaneous
redox reaction even if Appendix D is not available. Let’s choose a pair of half-
reactions from the appendix and, without referring to their relative positions in
the list, use them to write a spontaneous redox reaction:
Ag(aq) e ±£ Ag(s) E°silver 0.80 V
Sn2(aq) 2e ±£ Sn(s) E°tin 0.14 V
There are two steps involved:
1. Reverse one of the half-reactions into an oxidation step such that the differ-
ence of the electrode potentials (cathode minus anode) gives a positive Ecell.
Note that when we reverse the half-reaction, we need not reverse the sign of
Ehalf-cell because the minus sign in Equation 21.3 (Ecell Ecathode Eanode)
will do that.
2. Add the rearranged half-reactions to obtain a balanced overall equation. Be
sure to multiply by coefficients so that e lost equals e gained and to cancel
species common to both sides.
(You may be tempted in this particular case to add the two half-reactions as writ-
ten, because you obtain a positive Ecell, but you would then have two oxidizing
agents forming two reducing agents, which cannot occur.)
We want to pair the stronger oxidizing and reducing agents as reactants. The
larger (more positive) E value for the silver half-reaction means that Ag is a
stronger oxidizing agent (gains electrons more readily) than Sn2, and the smaller
(more negative) E value for the tin half-reaction means that Sn is a stronger
reducing agent (loses electrons more readily) than Ag. Therefore, we reverse the
tin half-reaction (but not the sign of Etin):
Sn(s) ±£ Sn2(aq) 2e E°tin 0.14 V
Subtracting Ehalf-cell of the tin half-reaction (anode, oxidation) from Ehalf-cell
of the silver half-reaction (cathode, reduction) gives a positive Ecell; that is,
0.80 V (0.14 V) 0.94 V.
With the half-reactions written in the correct direction, we must next make
sure that the number of electrons lost in the oxidation equals the number gained
in the reduction. In this case, we double the silver (reduction) half-reaction.
Adding the half-reactions and applying Equation 21.3 gives the balanced equa-
tion and Ecell:
2Ag(aq) 2e ±£ 2Ag(s) E°silver 0.80 V [reduction]
Sn(s) ±£ Sn2(aq) 2e E°tin 0.14 V [oxidation]
Sn(s) 2Ag(aq) ±£ Sn2(aq) 2Ag(s) E°cell E°silver E°tin 0.94 V
With the reaction spontaneous as written, the stronger oxidizing and reducing
agents are reactants, which confirms that Sn is a stronger reducing agent than Ag,
and Ag is a stronger oxidizing agent than Sn2.
A very important point to note is that, when we doubled the coefficients of
the silver half-reaction to balance the number of electrons, we did not double its
E value—it remained 0.80 V. That is, changing the balancing coefficients of a
half-reaction does not change its E value. The reason is that a standard electrode
potential is an intensive property, one that does not depend on the amount of sub-
stance present. The potential is the ratio of energy to charge. When we change
the coefficients, thus changing the amount of substance, the energy and the charge
change proportionately, so their ratio stays the same. (Recall that density, which
is also an intensive property, does not change with the amount of substance
because the mass and the volume change proportionately.)
SAMPLE PROBLEM 21.4 Writing Spontaneous Redox Reactions and

Ranking Oxidizing and Reducing Agents by
Strength
PROBLEM (a) Combine the following three half-reactions into three balanced equations (A,
B, and C) for spontaneous reactions, and calculate Ecell for each. (b) Rank the relative
strengths of the oxidizing and reducing agents.
(1) NO3(aq) 4H(aq) 3e ±£ NO(g) 2H2O(l) E° 0.96 V
(2) N2 (g) 5H(aq) 4e ±£ N2H5(aq) E° 0.23 V
(3) MnO2 (s) 4H(aq) 2e ±£ Mn2(aq) 2H2O(l) E° 1.23 V
PLAN (a) To write the redox equations, we combine the possible pairs of half-reactions:
(1) and (2), (1) and (3), and (2) and (3). They are all written as reductions, so the oxidizing
agents appear as reactants and the reducing agents appear as products. In each pair, we
reverse the reduction half-reaction that has the smaller (less positive or more negative) E
value to an oxidation to obtain a positive Ecell. We make e lost equal e gained, with-
out changing the magnitude of the E value, add the half-reactions together, and then apply
Equation 21.3 to find Ecell. (b) Because each reaction is spontaneous as written, the
stronger oxidizing and reducing agents are the reactants. To obtain the overall ranking, we
first rank the relative strengths within each equation and then compare them.
SOLUTION (a) Combining half-reactions (1) and (2) gives equation (A). The E value for
half-reaction (1) is larger (more positive) than that for (2), so we reverse (2) to obtain a
positive Ecell:
(1) NO3(aq) 4H(aq) 3e ±£ NO(g) 2H2O(l) E° 0.96 V
(rev 2) N2H5(aq) ±£ N2(g) 5H(aq) 4e E° 0.23 V
To make e lost equal e gained, we multiply (1) by four and the reversed (2) by three;
then add half-reactions and cancel appropriate numbers of common species (H and e):
4NO3(aq) 16H(aq) 12e ±£ 4NO(g) 8H2O(l) E° 0.96 V
3N2H5(aq) ±£ 3N2(g) 15H(aq) 12e E° 0.23 V
(A) 3N2H5(aq) 4NO3(aq) H(aq) ±£ 3N2(g) 4NO(g) 8H2O(l)
Ecell 0.96 V (0.23 V) 1.19 V
Combining half-reaction (1) and half-reaction (3) gives equation (B). Half-reaction (1) must
be reversed:
(rev 1) NO(g) 2H2O(l) ±£ NO3(aq) 4H(aq) 3e E° 0.96 V
(3) MnO2(s) 4H(aq) 2e ±£ Mn2(aq) 2H2O(l) E° 1.23 V
We multiply reversed (1) by two and (3) by three, then add and cancel:
2NO(g) 4H2O(l) ±£ 2NO3(aq) 8H(aq) 6e E° 0.96 V
3MnO2(s) 12H(aq) 6e ±£ 3Mn2(aq) 6H2O(l) E° 1.23 V
(B) 3MnO2(s) 4H(aq) 2NO(g) ±£ 3Mn2(aq) 2H2O(l) 2NO3(aq)
Ecell 1.23 V 0.96 V 0.27 V
Combining half-reaction (2) and half-reaction (3) gives equation (C). Half-reaction (2) must
be reversed:
(rev 2) N2H5(aq) ±£ N2(g) 5H(aq) 4e E° 0.23 V
(3) MnO2(s) 4H (aq) 2e ±£ Mn2(aq) 2H2O(l)

E° 1.23 V
We multiply reaction (3) by two, add the half-reactions, and cancel:
N2H5(aq) ±£ N2(g) 5H(aq) 4e E° 0.23 V
2MnO2(s) 8H (aq) 4e

±£ 2Mn2(aq) 4H2O(l) E° 1.23 V
(C) N2H5(aq) 2MnO2(s) 3H(aq) ±£ N2(g) 2Mn2(aq) 4H2O(l)
Ecell 1.23 V (0.23 V) 1.46 V
(b) Ranking the oxidizing and reducing agents within each equation:
Equation (A): Oxidizing agents: NO3 N2 Reducing agents: N2H5 NO
Equation (B): Oxidizing agents: MnO2 NO3 Reducing agents: NO Mn2
Equation (C): Oxidizing agents: MnO2 N2 Reducing agents: N2H5 Mn2
Determining the overall ranking of oxidizing and reducing agents. Comparing the relative
strengths from the three balanced equations gives
Oxidizing agents: MnO2 NO3 N2
Reducing agents: N2H5 NO Mn2
CHECK As always, check that atoms and charge balance on each side of the equation. A
good way to check the ranking and equations is to list the given half-reactions in order
of decreasing E° value:
MnO2(s) 4H(aq) 2e ±£ Mn2(aq) 2H2O(l) E° 1.23 V
NO3(aq) 4H(aq) 3e ±£ NO(g) 2H2O(l) E° 0.96 V
N2(g) 5H(aq) 4e ±£ N2H5(aq) E° 0.23 V
Then the oxidizing agents (reactants) decrease in strength going down the list, so the reduc-
ing agents (products) decrease in strength going up. Moreover, each of the three sponta-
neous reactions (A, B, and C) should combine a reactant with a product that is lower down
on this list.
FOLLOW-UP PROBLEM 21.4 Is the following reaction spontaneous as written?
3Fe (aq) ±£ Fe(s) 2Fe3(aq)
2
If not, write the equation for the spontaneous reaction, calculate Ecell, and rank the three
species of iron in order of decreasing reducing strength.
Relative Reactivities of Metals In Chapter 4, we discussed the activity series of

the metals (see Figure 4.21, p. 171), which ranks metals by their ability to “dis-
place” one another from aqueous solution. Now you’ll see why this displacement
occurs, as well as why many, but not all, metals react with acid to form H2, and
why a few metals form H2 even in water.
1. Metals that can displace H2 from acid. The standard hydrogen half-
reaction represents the reduction of H ions from an acid to H2:
2H(aq) 2e ±£ H2(g) E° 0.00 V

To see which metals reduce H (referred to as “displacing H2”) from acids,
choose a metal, write its half-reaction as an oxidation, combine this half-reaction
with the hydrogen half-reaction, and see if E°cell is positive. What you find is that
the metals Li through Pb, those that lie below the standard hydrogen (reference)
half-reaction in Appendix D, give a positive E°cell when reducing H. Iron, for
example, reduces H from an acid to H2:
Fe(s) ±£ Fe2(aq) 2e E° 0.44 V [anode; oxidation]
2H (aq) 2e

±£ H2(g) E° 0.00 V [cathode; reduction]

Fe(s) 2H (aq) ±£ H2(g) Fe2(aq) E°cell 0.00 V (0.44 V) 0.44 V
The lower the metal in the list, the stronger it is as a reducing agent; therefore,
the more positive its half-cell potential when the half-reaction is reversed, and the
higher the E°cell for its reduction of H to H2. If E°cell for the reduction of H is
more positive for metal A than it is for metal B, metal A is a stronger reducing
agent than metal B and a more active metal.
2. Metals that cannot displace H2 from acid. Metals that are above the
standard hydrogen (reference) half-reaction cannot reduce H from acids. When
we reverse the metal half-reaction, the E°cell is negative, so the reaction does not
occur. For example, the coinage metals—copper, silver, and gold, which are in
Group 1B(11)—are not strong enough reducing agents to reduce H from acids:
Ag(s)±£ Ag(aq) e E° 0.80 V [anode; oxidation]
2H (aq) 2e

±£ H2(g) E° 0.00 V [cathode; reduction]
2Ag(s) 2H (aq) ±£ 2Ag(aq) H2(g)

E°cell 0.00 V 0.80 V 0.80 V
The higher the metal in the list, the more negative is its E°cell for the reduction of
H to H2, the lower is its reducing strength, and the less active it is. Thus, gold
is less active than silver, which is less active than copper.
3. Metals that can displace H2 from water. Metals active enough to reduce
H2O lie below that half-reaction:
2H2O(l) 2e ±£ H2(g) 2OH(aq) E 0.42 V
(The value shown here is the nonstandard electrode potential because, in pure
water, [OH] is 1.0 107 M, not the standard-state value of 1 M.) For example,
consider the reaction of sodium in water (with the Na/Na half-reaction reversed and
doubled):
2Na(s)±£ 2Na(aq) 2e E° 2.71 V [anode; oxidation]
2H2O(l) 2e ±£ H2(g) 2OH(aq) E0 0.42 V [cathode; reduction]
2Na(s) 2H2O(l) ±£ 2Na(aq) H2(g) 2OH(aq)
Ecell 0.42 V (2.71 V) 2.29 V
Oxidation half-reaction The alkali metals [Group 1A(1)] and the larger alkaline earth metals [Group
Ca(s) Ca2+(aq) + 2e– 2A(2)] can reduce water, or displace H2 from H2O (Figure 21.9).
4. Metals that can displace other metals from solution. We can also predict
Reduction half-reaction whether one metal can reduce the aqueous ion of another metal. Any metal that
2H2O(l ) + 2e– H2(g) + 2OH–(aq) is lower in the list in Appendix D can reduce the ion of a metal that is higher up,
and thus displace that metal from solution. For example, zinc can displace iron
Ca(s) + 2H2O(l ) Ca(OH)2(aq) + H2(g)
from solution:
Zn(s)±£ Zn2(aq) 2e E° 0.76 V [anode; oxidation]
Figure 21.9 The reaction of calcium in
Fe (aq) 2e
2
±£ Fe(s) E° 0.44 V [cathode; reduction]
water.Calcium is one of the metals active
enough to displace H2 from H2O. Zn(s) Fe (aq) ±£ Zn2(aq) Fe(s)
2
E°cell 0.44 V (0.76 V) 0.32 V
21.4 Free Energy and Electrical Work 943
This particular reaction has tremendous economic importance in protecting iron

from rusting, as you’ll see shortly. The reducing power of metals has other, more
personal, consequences, as the margin note points out. Al Al3+ + 3e– Foil
Section Summary Saliva

Ag/Sn/Hg
The output of a cell is called the cell potential (Ecell) and is measured in volts
(1 V 1 J/C). • When all substances are in their standard states, the output is the Filling
standard cell potential (Ecell). Ecell 0 for a spontaneous reaction at standard-state O2 + 4H+ + 4e–
conditions. • By convention, a standard electrode potential (Ehalf-cell) refers to the 2H2O
reduction half-reaction. Ecell equals Ehalf-cell of the cathode minus Ehalf-cell of the
anode. • Using a standard hydrogen (reference) electrode (Ereference = 0 V), Ehalf-cell The Pain of a Dental Voltaic Cell
values can be measured and used to rank oxidizing (or reducing) agents (see Appen- Have you ever felt a jolt of pain when bit-
dix D). • Spontaneous redox reactions combine stronger oxidizing and reducing ing down with a filled tooth on a scrap of
agents to form weaker ones. • A metal can reduce another species (H, H2O, or an foil left on a piece of food? Here’s the rea-
ion of another metal) if Ecell for the reaction is positive. son. The aluminum foil acts as an active
anode (E of Al 1.66 V), saliva as
the electrolyte, and the filling (usually a
silver/tin/mercury alloy) as an inactive
21.4 FREE ENERGY AND ELECTRICAL WORK cathode. O2 is reduced to water, and the
In Chapter 20, we discussed the relationship of useful work, free energy, and the short circuit between the foil in contact
with the filling creates a current that is
equilibrium constant. In this section, we examine this relationship in the context
sensed by the nerve of the tooth.
of electrochemical cells and see the effect of concentration on cell potential.
Standard Cell Potential and the Equilibrium Constant

As you know from Section 20.3, a spontaneous reaction has a negative free energy
change (G 0), and you’ve just seen that a spontaneous electrochemical reac-
tion has a positive cell potential (Ecell 0). Note that the signs of G and Ecell
are opposite for a spontaneous reaction. These two indications of spontaneity are
proportional to each other:
G
Ecell
Let’s determine this proportionality constant by focusing on the electrical work
done (w, in joules), which is the product of the potential (Ecell, in volts) and the
amount of charge that flows (in coulombs). The value used for Ecell is measured
with no current flowing and, therefore, no energy lost to heating the cell. Thus,
Ecell is the maximum voltage possible for the cell, and the work is the maximum
work possible (wmax). For work done by the cell on the surroundings, this quan-
tity is negative:*
wmax Ecell charge
Equation 20.9, p. 902, shows that the maximum work done on the surroundings
is G:
wmax Ecell charge ¢G and so ¢G Ecell charge
The charge that flows through the cell equals the number of moles of electrons
(n) transferred times the charge of 1 mol of electrons (symbol F):
charge
Charge moles of e or charge nF
mol e
The charge of 1 mol of electrons is the Faraday constant (F), named in honor
of Michael Faraday, the 19th-century British scientist who pioneered the study of
electrochemistry:
96,485 C
F
mol e
*Recall from Chapter 20 that only a reversible process can do maximum work. For no current
to flow and the process to be reversible, Ecell must be opposed by an equal potential in the
measuring circuit: if the opposing potential is infinitesimally smaller, the cell reaction goes for-
ward; if it is infinitesimally larger, the reaction goes backward.
Because 1 V 1 J/C, we have 1 C 1 J/V, and

J
F 9.65 104 (3 sf) (21.4)
Vmol e
Substituting for charge, the proportionality constant is nF:
G nFEcell (21.5)
When all of the components are in their standard states, we have
G° nFE°cell (21.6)
Using this relationship, we can relate the standard cell potential to the equi-
librium constant of the redox reaction. Recall from Equation 20.12 that
G° RT ln K
Substituting for G° from Equation 21.6 gives
nFE°cell RT ln K
Solving for E°cell gives
RT
E°cell ln K (21.7)
nF
Figure 21.10 summarizes the interconnections among the standard free energy
change, the equilibrium constant, and the standard cell potential. Notice that, even
though these relationships hold under standard-state conditions, they are still
extremely useful. Recall from Chapter 20 that determining K required that we
know G, either from H and S values or from Gf values. Now, for redox
reactions, we have a direct experimental method for determining K and G:
measuring Ecell.
ΔG Reaction Parameters at the Standard State

Reaction at
standard-state
ΔG K
E cell conditions
<0 >1 >0 Spontaneous
ΔG
ll
ce
E
0 1 0 At equilibrium
=
nF
>0 <1 <0 Nonspontaneous

–R
=–
Tl

B
nK
ΔG
Figure 21.10 The interrelation-

ship of G °, E °cell, and K. A, Any
one of these three central thermo-
dynamic parameters can be used
to find the other two. B, The signs

E cell K
of G° and E°cell determine the re-
action direction at standard-state
= RT ln K
E cell
A nF conditions.
It is common practice to simplify Equation 21.7 in calculations by

• Substituting the known value of 8.314 J/(mol rxnK) for the constant R
• Substituting the known value of 9.65 104 J/(Vmol e) for the constant F
• Substituting the standard temperature of 298.15 K for T, but keeping in mind
that the cell can run at other temperatures.
• Multiplying by 2.303 to convert from natural to common (base-10) logarithms.
This conversion shows that a 10-fold change in K makes Ecell change by 1.
siL48593_ch21_922-979 29:11:07 03:41pm Page 945
Thus, when n moles of e are transferred per mole of reaction in the bal-
anced equation, this simplified relation between Ecell and K gives
J
8.314 298.15 K
RT RT mol rxnK
E°cell ln K 2.303 log K 2.303 log K
a9.65104 b
nF nF n mol e J
mol rxn Vmol e
And, we have
0.0592 V nE°cell
E°cell log K or log K (at 298.15 K) (21.8)
n 0.0592 V
SAMPLE PROBLEM 21.5 Calculating K and G from E cell

PROBLEM Lead can displace silver from solution:
Pb(s) 2Ag(aq) ±£ Pb2(aq) 2Ag(s)
As a consequence, silver is a valuable byproduct in the industrial extraction of lead from
its ore. Calculate K and G° at 298.15 K for this reaction.
PLAN We divide the spontaneous redox equation into the half-reactions and use values from
Appendix D to calculate E°cell. Then, we substitute this result into Equation 21.8 to find K
and into Equation 21.6 to find G°.
SOLUTION Writing the half-reactions with their E° values:
(1) Ag(aq) e ±£ Ag(s) E° 0.80 V
(2) Pb2(aq) 2e ±£ Pb(s) E° 0.13 V
Calculating E°cell: We double (1), reverse (2), add the half-reactions, and subtract E°lead from
E°silver:
2Ag(aq) 2e ±£ 2Ag(s) E° 0.80 V
Pb(s) ±£ Pb2(aq) 2e E° 0.13 V
Pb(s) 2Ag(aq) ±£ Pb2(aq) 2Ag(s) E°cell 0.80 V (0.13 V) 0.93 V
Calculating K with Equations 21.7 and 21.8:
RT RT
E°cell
ln K 2.303 log K
nF nF
The adjusted half-reactions show that 2 mol of e are transferred per mole of reaction as
written, so n 2. Then, performing the substitutions for R and F that we just discussed
with the cell running at 25°C (298.15 K), we have
0.0592 V
E°cell log K 0.93 V
2
0.93 V 2
So, log K 31.42 and K 2.61031
0.0592 V
Calculating G° (Equation 21.6):
2 mol e 96.5 kJ
G° nFE°cell 0.93 V 1.8102 kJ/mol rxn
mol rxn Vmol e
CHECK The three variables are consistent with the reaction being spontaneous at standard-
state conditions: E°cell 0, G° 0, and K 1. Be sure to round and check the order
of magnitude: in the G° calculation, for instance, G° 2 100 1 200, so
the overall math seems right. Another check would be to obtain G° directly from its rela-
tion with K:
G° RT ln K 8.314 J/mol rxnK 298.15 K ln (2.61031)
1.8105 J/mol rxn 1.8102 kJ/mol rxn
FOLLOW-UP PROBLEM 21.5 When cadmium metal reduces Cu2 in solution,
Cd forms in addition to copper metal. Given that G 143 kJ/mol rxn, calculate K
2
at 25C. What is Ecell of a voltaic cell that uses this reaction?

siL48593_ch21_922-979 29:11:07 03:41pm Page 946
The Effect of Concentration on Cell Potential

So far, we’ve considered cells with all components in their standard states and
found standard cell potential (Ecell) from standard half-cell potentials (Ehalf-cell).
However, most cells do not start with all components in their standard states, and
even if they did, the concentrations change after a few moments of operation.
Moreover, in all practical voltaic cells, such as batteries, reactant concentrations
are far from standard-state values. Clearly, we must be able to determine Ecell,
the cell potential under nonstandard conditions.
To do so, let’s derive an expression for the relation between cell potential and
concentration based on the relation between free energy and concentration. Recall
from Chapter 20 (Equation 20.13) that G equals G (the free energy change
when the system moves from standard-state concentrations to equilibrium) plus
RT ln Q (the free energy change when the system moves from nonstandard-state
to standard-state concentrations):
¢G ¢G° RT ln Q
G is related to Ecell and G to Ecell (Equations 21.5 and 21.6), so we substi-
tute for them and get
nFEcell nFE°cell RT ln Q
Dividing both sides by nF, we obtain the Nernst equation, developed by the
Walther Hermann Nernst (1864–1941) German chemist Walther Hermann Nernst in 1889:
This great German physical chemist was RT
only 25 years old when he developed the Ecell E°cell ln Q (21.9)
nF
equation for the relationship between cell
voltage and concentration. His career, The Nernst equation says that a cell potential under any conditions depends on
which culminated in the Nobel Prize in the potential at standard-state concentrations and a term for the potential at
chemistry in 1920, included formulating nonstandard-state concentrations. How do changes in Q affect cell potential?
the third law of thermodynamics, estab- From Equation 21.9, we see that
lishing the principle of the solubility
product, and contributing key ideas to • When Q 1 and thus [reactant] [product], ln Q 0, so Ecell Ecell.
photochemistry and the Haber process. • When Q 1 and thus [reactant] [product], ln Q 0, so Ecell Ecell.
He is shown here in later years making a
point to a few of his celebrated col-
• When Q 1 and thus [reactant] [product], ln Q 0, so Ecell Ecell.
leagues: from left to right, Nernst, Albert As before, to obtain a simplified form of the Nernst equation for use in cal-
Einstein, Max Planck, Robert Millikan, culations, let’s substitute known values of R and F, operate the cell at 298.15 K,
and Max von Laue. and convert to common (base-10) logarithms:
RT RT
Ecell E°cell ln Q E°cell 2.303 log Q
nF nF
J
8.314 298.15 K
mol rxnK
E°cell 2.303 log Q
n mol e
a9.65 10 b
4 J
mol rxn Vmol e
And we obtain:
0.0592 V
Ecell E°cell log Q (at 298.15 K) (21.10)
n
Remember that the expression for Q contains only those species with con-
centrations (and/or pressures) that can vary; thus, solids do not appear, even
when they are the electrodes. For example, in the reaction between cadmium and
silver ion, the Cd and Ag electrodes do not appear in the expression for Q:
[Cd2]
Cd(s) 2Ag(aq) ±£ Cd2(aq) 2Ag(s) Q
[Ag]2
SAMPLE PROBLEM 21.6 Using the Nernst Equation to Calculate Ecell

PROBLEM In a test of a new reference electrode, a chemist constructs a voltaic cell con-
sisting of a Zn/Zn2 half-cell and an H2/H half-cell under the following conditions:
[Zn2] 0.010 M [H] 2.5 M PH2 0.30 atm
Calculate Ecell at 298.15 K.
PLAN To apply the Nernst equation and determine Ecell, we must know Ecell and Q. We
write the spontaneous reaction, calculate Ecell from standard electrode potentials (Appen-
dix D), and use the given pressure and concentrations to find Q. (Recall that the ideal gas
law allows us to use P at constant T as another way of writing concentration, n/V.) Then
we substitute into Equation 21.10.
SOLUTION Determining the cell reaction and Ecell:
2H(aq) 2e ±£ H2(g) E 0.00 V
Zn(s) ±£ Zn2(aq) 2e E 0.76 V

2H (aq) Zn(s) ±£ H2(g) Zn (aq) 2
Ecell 0.00 V (0.76 V) 0.76 V
Calculating Q:
PH2 [Zn2] 0.30 0.010
Q 4.8 104
[H]2 2.52
Solving for Ecell at 25°C (298.15 K), with n 2:
RT 0.0592 V
Ecell Ecell 2.303 log Q Ecell log Q
nF n
0.76 V c log (4.8 104 ) d 0.76 V (0.0982 V) 0.86 V

0.0592 V
2
CHECK After you check the arithmetic, reason through the answer: Ecell Ecell
(0.86 0.76) because the log Q term was negative, which is consistent with Q 1.
FOLLOW-UP PROBLEM 21.6 Consider a voltaic cell based on the following reac-
tion: Fe(s) Cu2(aq) ±£ Fe2(aq) Cu(s). If [Cu2] 0.30 M, what must [Fe2]
be to increase Ecell by 0.25 V above Ecell at 25C?
Changes in Potential During Cell Operation

Let’s see how the potential of the zinc-copper cell changes as concentrations
change during cell operation. In this case, the only concentrations that change are
[reactant] [Cu2] and [product] [Zn2]:
[Zn2 ]
Zn(s) Cu2 (aq) ±£ Zn2 (aq) Cu(s) Q
[Cu2 ]
The positive E°cell (1.10 V) means that this reaction proceeds spontaneously from
standard-state conditions, at which [Zn2] [Cu2] 1 M (Q 1), to some
point at which [Zn2] [Cu2] (Q 1). Now, suppose that we start the cell
when [Zn2] [Cu2] (Q 1), for example, when [Zn2] 1.0 104 M
and [Cu2] 2.0 M. Thus, the cell potential is higher than the standard cell
potential:
[Zn2] 1.0 104
a b
0.0592 V 0.0592 V
Ecell E°cell log 1.10 V log
2 [Cu2] 2 2.0
1.10 V c (4.30) d 1.10 V 0.127 V 1.23 V

0.0592 V
2
As the cell operates, [Zn2] increases (as the Zn electrode deteriorates) and
2
[Cu ] decreases (as Cu plates out on the Cu electrode). Although the changes
during this process occur smoothly, if we keep Equation 21.10 in mind, we can
1.30
Changes in Ecell and Concentration
Q⬍
1.20 1 Stage in
cell Relative 0.0592 V
E cell = 1.10 V ––––––– log Q
operation Q [P] and [R] n
Ecell (V)
1.10
1. E > E <1 [P] < [R] <0
1.00
Q⬎ 2. E = E =1 [P] = [R] =0
1
3. E < E >1 [P] > [R] >0
4. E = 0 =K [P] >> [R] =E
0.90
A 10– 5 10–4 10–3 10–2 10–1 1 10 102 103 104 B
[Zn2+]
Q=
[Cu2+]
Figure 21.11 The relation between

Ecell and log Q for the zinc-copper cell. identify four general stages of operation. Figure 21.11A shows the first three. The
A, A plot of Ecell vs. Q (on a logarithmic main point to note is that as the cell operates, its potential decreases:
scale) for the zinc-copper cell shows a
linear decrease. When Q 1 (left), [reac- Stage 1. Ecell E°cell when Q 1: When the cell begins operation, [Cu2]
tant] is relatively high, and the cell can do [Zn2], so the [(0.0592 V/n) log Q] term 0 and Ecell E°cell.
relatively more work. When Q 1, Ecell
E°cell. When Q 1 (right), [product] is rela- As cell operation continues, [Zn2] increases and [Cu2] decreases; thus, Q
tively high, and the cell can do relatively becomes larger, the [(0.0592 V/n) log Q] term becomes less negative (more pos-
less work. B, A summary of the changes itive), and Ecell decreases.
in Ecell as the cell operates, including the
changes in [Zn2], denoted [P] for [product], Stage 2. Ecell E°cell when Q 1: At the point when [Cu2] [Zn2], Q 1,
and [Cu2], denoted [R] for [reactant]. so the [(0.0592 V/n) log Q] term 0 and Ecell E°cell.
Stage 3. Ecell E°cell when Q 1: As the [Zn2]/[Cu2] ratio continues to
increase, the [(0.0592 V/n) log Q] term 0, so Ecell E°cell.
Stage 4. Ecell 0 when Q K: Eventually, the [(0.0592 V/n) log Q] term
becomes so large that it equals E°cell, which means that Ecell is zero. This
occurs when the system reaches equilibrium: no more free energy is
released, so the cell can do no more work. At this point, we say that a
battery is “dead.”
Figure 21.11B summarizes these four key stages in the operation of a voltaic cell.
Let’s find K for the zinc-copper cell. At equilibrium, Equation 21.10 becomes
0 E°cell a b log K, which rearranges to E°cell
0.0592 V 0.0592 V
log K
n n
Note that this result is identical to Equation 21.8, which we obtained from G.
Solving for K of the zinc-copper cell (Ecell 1.10 V),
2 E°cell
log K , so K 10(2 1.10 V)/0.0592 V 1037.16 1.4 1037
0.0592 V
Thus, the zinc-copper cell does work until the [Zn2]/[Cu2] ratio is very high.
To conclude, let’s examine cell potential in terms of the starting Q/K ratio:
• If Q/K 1, Ecell is positive for the reaction as written. The smaller the Q/K
ratio, the greater the value of Ecell, and the more electrical work the cell can do.
• If Q/K 1, Ecell 0. The cell is at equilibrium and can no longer do work.
• If Q/K 1, Ecell is negative for the reaction as written. The reverse reaction
will take place, and the cell will do work until Q/K equals 1 at equilibrium.
Concentration Cells
If you mix a concentrated solution of a salt with a dilute solution of the salt, the
final concentration equals some intermediate value. A concentration cell employs
this phenomenon to generate electrical energy. The two solutions are in separate
half-cells, so they do not mix; rather, their concentrations become equal as the
cell operates.
How a Concentration Cell Works Suppose both compartments of a voltaic cell

house the Cu/Cu2 half-reaction. The cell reaction is the sum of identical half-
reactions, written in opposite directions, so the standard half-cell potentials can-
cel (Ecopper Ecopper) and Ecell is zero. This occurs because standard electrode
potentials are based on concentrations of 1 M. In a concentration cell, however,
the half-reactions are the same but the concentrations are different. As a result,
even though Ecell equals zero, the nonstandard cell potential, Ecell, does not equal
zero because it depends on the ratio of concentrations.
In Figure 21.12A, a concentration cell has 0.10 M Cu2 in the anode half-
cell and 1.0 M Cu2, a 10-fold higher concentration, in the cathode half-cell:
Cu(s) ±£ Cu2(aq; 0.10 M) 2e [anode; oxidation]
Cu (aq; 1.0 M) 2e ±£ Cu(s)
2
The overall cell reaction is the sum of the half-reactions:
Cu2(aq; 1.0 M) ±£ Cu2(aq; 0.10 M) Ecell ?
The cell potential at the initial concentrations of 0.10 M (dilute) and 1.0 M (con-
centrated), with n 2, is obtained from the Nernst equation:
[Cu2]dil
0Va b
0.0592 V 0.0592 V 0.10 M
Ecell E°cell log 2
log
2 [Cu ]conc 2 1.0 M
0V c (1.00) d 0.0296 V

0.0592 V
2
As you can see, because Ecell for a concentration cell equals zero, Ecell for non-
standard conditions depends entirely on the [(0.0592 V/n) log Q] term.
What is actually going on as this cell operates? In the half-cell with dilute
electrolyte (anode), the Cu atoms in the electrode give up electrons and become
Cu2 ions, which enter the solution and make it more concentrated. The electrons
released at the anode flow to the cathode compartment. There, Cu2 ions in the
concentrated solution pick up the electrons and become Cu atoms, which plate
out on the electrode, so that solution becomes less concentrated. As in any voltaic
cell, Ecell decreases until equilibrium is attained, which happens when [Cu2] is
the same in both half-cells (Figure 21.12B). The same final concentration would
result if we mixed the two solutions, but no electrical work would be done.
e– 0.0296 V e– e– 0.00 V e–
Voltmeter Voltmeter
Anode Cathode Anode Cathode
(–) (+) (+) Cu
Cu Salt bridge Cu Cu (–) Salt bridge
2e– 2e–
+
+ Cu2
Cu Cu2 Cu
0.10 M Cu 2+ 1.0 M Cu 2+ 0.55 M Cu 2+ 0.55 M Cu 2+

B
Cu(s) Cu2+(aq; 0.10 M) + 2e– Figure 21.12 A concentration cell based on the
Cu/Cu2 half-reaction. A, The half-reactions are
Reduction half-reaction the same, so Ecell 0. The cell operates because
Cu2+(aq; 1.0 M) + 2e– Cu(s) the half-cell concentrations are different, which
makes Ecell 0 in this case. B, The cell operates
Overall (cell) reaction until the half-cell concentrations are equal. Note
Cu2+(aq; 1.0 M ) Cu2+(aq; 0.10 M ) the change in electrodes (exaggerated here for
A clarity) and the identical color of the solutions.
siL48593_ch21_922-979 14:12:07 03:53pm Page 950 fdfd ve403:MHQY042:siL5ch21:
SAMPLE PROBLEM 21.7 Calculating the Potential of a Concentration Cell

PROBLEM A concentration cell consists of two Ag/Ag half-cells. In half-cell A, electrode
A dips into 0.010 M AgNO3; in half-cell B, electrode B dips into 4.0104 M AgNO3.
What is the cell potential at 298.15 K? Which electrode has a positive charge?
PLAN The standard half-cell reactions are identical, so Ecell is zero, and we calculate Ecell
from the Nernst equation. Because half-cell A has a higher [Ag], Ag ions will be
reduced and plate out on electrode A. In half-cell B, Ag will be oxidized and Ag ions
will enter the solution. As in all voltaic cells, reduction occurs at the cathode, which is
positive.
SOLUTION Writing the spontaneous reaction: The [Ag] decreases in half-cell A and
increases in half-cell B, so the spontaneous reaction is
Ag(aq; 0.010 M) [half-cell A] ±£ Ag(aq; 4.0104 M) [half-cell B]
Calculating Ecell, with n 1:
[Ag]dil 4.0104
a0.0592 b
0.0592 V
Ecell E°cell log 0 V V log
1 [Ag]conc 0.010
0.0828 V
Reduction occurs at the cathode, electrode A: Ag(aq; 0.010 M) e ±£ Ag(s). Thus,
electrode A has a positive charge due to a relative electron deficiency.
FOLLOW-UP PROBLEM 21.7 A concentration cell is built using two Au/Au3 half-
cells. In half-cell A, [Au ] 7.0104 M, and in half-cell B, [Au3] 2.5102 M.
3
What is Ecell, and which electrode is negative?
Concentration Cells in Your Nerve Applications of Concentration Cells Chemists, biologists, and environmental
Cells The nerve cells that mediate every scientists apply the principle of a concentration cell in a host of applications.
thought, movement, and other bodily The most important is the measurement of unknown ion concentrations, particu-
process function by the principle of a con- larly [H]. Suppose we construct a concentration cell based on the H2/H half-
centration cell. The nerve membrane is reaction, in which the cathode compartment houses the standard hydrogen
imbedded with a host of specialized en- electrode and the anode compartment has the same apparatus dipping into an
zyme “gates” that use energy from ATP unknown [H] in solution. The half-reactions and overall reaction are
hydrolysis to separate an interior solution
of low [Na] and high [K] from an exte- H2 (g; 1 atm) ±£ 2H(aq; unknown) 2e [anode; oxidation]
rior solution of high [Na] and low [K]. 2H (aq; 1 M) 2e ±£ H2 (g; 1 atm)

As a result of the differences in [Na] and 2H(aq; 1 M) ±£ 2H(aq; unknown) Ecell ?
[K], the membrane is more positive out- 2
side than inside. One-third of the body’s As for the Cu/Cu concentration cell, Ecell is zero; however, the half-cells dif-
ATP is used to create and maintain these fer in [H], so Ecell is not zero. From the Nernst equation, with n 2, we obtain
concentration differences. (The 1997 No- 0.0592 V [H]2unknown
bel Prize in chemistry was shared by Jens Ecell E°cell log 2
2 [H ]standard
C. Skou for elucidation of this enzyme
mechanism.) When the nerve membrane Substituting 1 M for [H]standard and 0 V for Ecell gives
is stimulated, Na ions spontaneously 0.0592 V [H ]2unknown 0.0592 V
rush in, and in 0.001 s, the inside of the Ecell 0 V log log [H ]2unknown
2 12 2
membrane becomes more positive than
the outside. This event is followed by K Because log x2 2 log x (see Appendix A), we obtain
Ecell c (2 log [H ]unknown ) d 0.0592 V log [H ]unknown

ions spontaneously rushing out; after an- 0.0592 V
other 0.001 s, the membrane is again 2
more positive outside. These large
changes in charge in one region of the
Substituting log [H] pH, we have
membrane stimulate the neighboring re- Ecell 0.0592 V pH
gion and the electrical impulse moves Thus, by measuring Ecell, we can find the pH.
down the length of the cell. In the routine measurement of pH, a concentration cell incorporating two
hydrogen electrodes is too bulky and difficult to maintain. Instead, as was pointed
out in Chapter 18, a pH meter is used. As shown in Figure 21.13A, two separate
electrodes dip into the solution being tested. One of them is a glass electrode,
Pt Figure 21.13 The laboratory mea-

Reference Glass Reference
Glass (calomel) electrode (calomel)
surement of pH. A, The glass electrode
electrode electrode electrode (left) is a self-contained Ag/AgCl half-cell
immersed in an HCl solution of known
concentration and enclosed by a thin glass
membrane. It monitors the external [H] in
Hg
the solution relative to its fixed internal
Paste of [H]. The saturated calomel electrode
AgCl on Hg2Cl2 Test
Ag on Pt (right) acts as a reference. B, Most modern
in Hg solution
Test laboratories use a combination electrode,
KCl pH = ?
1 M HCl solution which houses both the glass and reference
pH = ? solution
B electrodes in one tube.
Thin glass Porous
A membrane ceramic plug
which consists of an Ag/AgCl half-reaction immersed in an HCl solution of fixed

concentration (usually 1.000 M) and enclosed by a thin (0.05 mm) membrane
made of a special glass that is highly sensitive to the presence of H ions. The
other electrode is a reference electrode, typically a saturated calomel electrode.
It consists of a platinum wire immersed in a paste of Hg2Cl2 (calomel), liquid
Hg, and saturated KCl solution. The glass electrode monitors the solution’s [H]
relative to its own fixed internal [H], and the instrument converts the potential
difference between the glass and reference electrodes into a measure of pH. In
modern instruments, a combination electrode is used, which houses both elec-
trodes in one tube (Figure 21.13B).
The pH electrode is one example of an ion-selective (or ion-specific) elec-
trode. Electrodes have been designed with highly specialized membranes in order
to selectively measure the concentrations of many different ions in industrial,
environmental, and biological samples. Recent advances allow measurement in
the picomolar-to-femtomolar (10121015 M) range. Table 21.3 shows a few of
the many ions for which ion-selective electrodes are available. Minimicroanalysis Electronic minia-
turization has greatly broadened the appli-
cations of electrochemistry. Environ-
Table 21.3 Some Ions Measured with Ion-Specific Electrodes mental chemists use pocket-sized pH
meters in the field to study natural waters
Species Detected Typical Sample and soil. Physiologists and biochemists
employ electrodes so small that they can
NH3/NH4 Industrial wastewater, seawater
CO2/HCO3 Blood, groundwater be surgically implanted in a single biolog-
F Drinking water, urine, soil, industrial stack gases ical cell to study ion channels and recep-
Br Grain, plant tissue tors. In the photo, a microelectrode is
I Milk, pharmaceuticals recording the electrical impulses of a single
NO3 Soil, fertilizer, drinking water neuron in a monkey’s visual cortex.
K Blood serum, soil, wine
H Laboratory solutions, soil, natural waters
Section Summary
A spontaneous process is indicated by a negative G or a positive Ecell, which are
related: G nFEcell. The G of the cell reaction represents the maximum amount
of electrical work the cell can do. • Because the standard free energy change, G,
is related to Ecell and to K, we can use Ecell to determine K. • For nonstandard con-
ditions, the Nernst equation shows that Ecell depends on Ecell and a correction term
based on Q. Ecell is high when Q is small (high [reactant]), and it decreases as the
cell operates. At equilibrium, G and Ecell are zero, which means that Q K. • Con-
centration cells have identical half-reactions, but solutions of differing concentration;
thus, they generate electrical energy as the concentrations become equal. • Ion-specific
electrodes, such as the pH electrode, measure the concentration of one species.
21.5 ELECTROCHEMICAL PROCESSES IN BATTERIES

Because of their compactness and mobility, batteries have a major influence on
our way of life. In industrialized countries, each person uses an average of 10
batteries per year! A battery, strictly speaking, is a self-contained group of voltaic
cells arranged in series (plus-to-minus-to-plus, and so on), so that their individ-
ual voltages are added together. In everyday speech, however, the term may also
be applied to a single voltaic cell. Batteries are ingeniously engineered devices
that house rather unusual half-reactions and half-cells, but they operate through
the same electrochemical principles we’ve been discussing. In this section, we
examine the three categories of batteries—primary, secondary, and fuel cells—
and note important examples, including some newer designs, of each.
Primary (Nonrechargeable) Batteries

A primary battery cannot be recharged, so it is discarded when the components
have reached their equilibrium concentrations, that is, when the cell is “dead.”
We’ll discuss the alkaline battery, with a brief historical reference to the dry cell,
the mercury and silver “button” batteries, and the primary lithium battery.
Alkaline Battery The precursor of today’s ubiquitous alkaline battery, the com-
mon dry cell, or Leclanché cell, was invented in the 1860s and became a famil-
iar household item during the first three-quarters of the 20th century. The anode
of the dry cell was a zinc can that housed a mixture of MnO2 and a weakly acidic
electrolyte paste, consisting of NH4Cl, ZnCl2, H2O, and starch. Powdered graphite
was used to improve conductivity, and the cathode was an inactive graphite rod.
Even today, the cathode half-reaction is not completely understood, but it involved
the reduction of MnO2(s) to Mn2O3(s) and an acid-base reaction between NH4
and OH. At a high current drain, this reaction generated ammonia gas, which
could build up and cause a serious voltage drop. Moreover, because the zinc anode
reacted with the acidic NH4 ions, dry cells had a short shelf life.
Although more expensive than the dry cell, the alkaline battery avoids these
drawbacks. The same electrode materials, zinc and manganese dioxide, are used,
but the electrolyte is a paste of KOH and water. The half-reactions are essentially
the same, but use of the KOH paste eliminates the NH3 gas buildup and main-
tains the Zn electrode (Figure 21.14):
Anode (oxidation): Zn(s) 2OH(aq) ±£ ZnO(s) H2O(l) 2e
Cathode (reduction): MnO2 (s) 2H2O(l) 2e ±£ Mn(OH) 2 (s) 2OH(aq)
Overall (cell) reaction:
Zn(s) MnO2 (s) H2O(l) ±£ ZnO(s) Mn(OH) 2 (s) Ecell 1.5 V
Like the dry cell, the alkaline battery powers portable radios, toys, flashlights, and
so on, is safe, and comes in many sizes. It has no voltage drop, longer shelf life,
and better performance in terms of power capability and stored energy.
Figure 21.14 Alkaline battery. Positive button

Steel case
MnO2 in KOH paste
Zn (anode)
Graphite rod (cathode)
Absorbent/separator
Negative end cap

21.5 Electrochemical Processes in Batteries 953
Mercury and Silver (Button) Batteries Mercury and silver batteries are quite sim-
ilar. Both use a zinc container as the anode (reducing agent) in a basic medium.
The mercury battery employs HgO as the oxidizing agent, the silver uses Ag2O,
and both use a steel can around the cathode. The solid reactants are compacted
with KOH and separated with moist paper. The half-reactions are
Anode (oxidation): Zn(s) 2OH(aq) ±£ ZnO(s) H2O(l) 2e
Cathode (reduction) (mercury): HgO(s) H2O(l) 2e ±£ Hg(l) 2OH(aq)
Cathode (reduction) (silver): Ag2O(s) H2O(l) 2e ±£ 2Ag(s) 2OH(aq)
Overall (cell) reaction (mercury):
Zn(s) HgO(s) ±£ ZnO(s) Hg(l) Ecell 1.3 V
Overall (cell) reaction (silver):
Zn(s) Ag2O(s) ±£ ZnO(s) 2Ag(s) Ecell 1.6 V
Both cells are manufactured as small button-sized batteries. The mercury cell is
used in calculators (Figure 21.15). Because of its very steady output, the silver
cell is used in watches, cameras, heart pacemakers, and hearing aids. Their major
disadvantages are toxicity of discarded mercury and high cost of silver cells.
Anode cap
Figure 21.15 Silver button battery.
Cathode can
Zn in KOH gel
(anode) (–)
Gasket
Separator
Pellet of Ag2O
in graphite
(cathode) (+)
Primary Lithium Batteries The primary lithium battery is widely used in watches
and implanted medical devices. It offers an extremely high energy/mass ratio, pro-
ducing 1 mol of e (1 F) from less than 7 g of metal ( of Li 6.941 g/mol).
The anode is lithium metal foil, which requires a nonaqueous electrolyte. The Heart
cathode is one of several metal oxides in which lithium ions lie between oxide pacemaker
layers. Some implantable medical devices have a silver vanadium oxide (SVO;
AgV2O5.5) cathode and can provide power for several years, but at a low rate
because energy storage is limited (Figure 21.16). The half-reactions are
Anode (oxidation): 3.5Li(s) ±£ 3.5Li 3.5e
Cathode (reduction): AgV2O5.5 3.5Li 3.5e

±£ Li3.5AgV2O5.5 Stainless Feedthrough pin (+)
steel Insulative seal
Overall (cell) reaction: AgV2O5.5 3.5Li(s) ±£ Li3.5AgV2O5.5 case (–) Multiplate
Electrolyte
cell stack
fill hole
Secondary (Rechargeable) Batteries
In contrast to primary batteries, a secondary, or rechargeable, battery is recharged
when it runs down by supplying electrical energy to reverse the cell reaction and
re-form reactant. In other words, in this type of battery, the voltaic cells are peri- Cathode
odically converted to electrolytic cells to restore nonequilibrium concentrations of lead bridge
the cell components. By far the most widely used secondary battery is the com-
mon car battery. Two newer types are the nickel–metal hydride battery and the
lithium-ion battery, a secondary version of a lithium battery.
Lead-Acid Battery A typical lead-acid car battery has six cells connected in series, Polypropylene
each of which delivers about 2.1 V for a total of about 12 V. Each cell contains separator
two lead grids loaded with the electrode materials: high-surface-area (spongy) Pb in
Lithium (anode) SVO (cathode)
the anode and high-surface-area PbO2 in the cathode. The grids are immersed in an
electrolyte solution of 4.5 M H2SO4. Fiberglass sheets between the grids prevent Figure 21.16 Lithium battery.
– shorting due to physical contact (Figure 21.17). When the cell discharges, it gener-
ates electrical energy as a voltaic cell:
Anode (oxidation): Pb(s) HSO4(aq) ±£ PbSO4 (s) H 2e
+ Cathode (positive): Cathode (reduction):
lead grids filled
with PbO2 PbO2 (s) 3H (aq) HSO4(aq) 2e ±£ PbSO4 (s) 2H2O(l)
Notice that both half-reactions produce Pb2 ions, one through the oxidation of
Pb, the other through the reduction of PbO2. The Pb2 forms PbSO4(s) at both
electrodes by reacting with HSO4.
Overall (cell) reaction (discharge):
PbO2 (s) Pb(s) 2H2SO4 (aq) ±£ 2PbSO4 (s) 2H2O(l) Ecell 2.1 V
When the cell is recharged, it uses electrical energy as an electrolytic cell, and
Anode (negative): the half-cell and overall reactions are reversed.
similar grids filled
with spongy lead Overall (cell) reaction (recharge):
2PbSO4 (s) 2H2O(l) ±£ PbO2 (s) Pb(s) 2H2SO4 (aq)
H2SO4 electrolyte
For more than a century, car and truck owners have relied on the lead-acid
Figure 21.17 Lead-acid battery. battery to provide the large burst of current to the starter motor needed to start
the engine—and to do so for years in both hot and cold weather. Nevertheless,
there are problems with the lead-acid battery, mainly loss of capacity and safety
concerns. Loss of capacity arises from several factors, including corrosion of the
positive (Pb) grid, detachment of active material as a result of normal mechani-
cal bumping, and the formation of large crystals of PbSO4, which make recharg-
ing more difficult. Most of the safety concerns have been remedied in modern
batteries. Older batteries had a cap on each cell for monitoring electrolyte den-
sity and replacing water lost on overcharging. During recharging, some water
could be electrolyzed to H2 and O2, which could explode if sparked, and splatter
H2SO4. Modern batteries are sealed, so they don’t require addition of water during
normal operation, and they use flame attenuators to reduce the explosion hazard.
Nickel–Metal Hydride (Ni-MH) Battery Concerns about the toxicity of the nickel-
cadmium (nicad) battery are leading to its replacement by the nickel–metal
hydride battery. The anode half-reaction oxidizes the hydrogen absorbed within a
metal alloy (designated M; e.g., LaNi5) in a basic (KOH) electrolyte, while
nickel(III) in the form of NiO(OH) is reduced at the cathode (Figure 21.18):
Anode (oxidation): MH(s) OH(aq) ±£ M(s) H2O(l) e
Cathode (reduction): NiO(OH)(s) H2O(l) e ±£ Ni(OH)2(s) OH(aq)
Overall (cell) reaction: MH(s) NiO(OH)(s) ±£ M(s) Ni(OH)2(s)
Ecell 1.4 V
The cell reaction is reversed during recharging. The Ni-MH battery is common in
cordless razors, camera flash units, and power tools. It is lightweight, has high
power, and is nontoxic, but it may discharge excessively during long-term storage.
Figure 21.18 Nickel–metal hydride NiO(OH) (anode)

battery. Separator
MH (cathode)
(–)
Insulator
Gasket
(+)
Heat shrink tube

siL48593_ch21_922-979 17:11:07 11:48am Page 955
21.5 Electrochemical Processes in Batteries 955
Lithium-Ion Battery The secondary lithium-ion battery has an anode of Li

atoms that lie between sheets of graphite (designated LixC6). The cathode is a
lithium metal oxide, such as LiMn2O4 or LiCoO2, and a typical electrolyte is
1 M LiPF6 in an organic solvent, such as dimethyl carbonate (often mixed with
methylethyl carbonate). Electrons flow through the circuit, while solvated Li
ions flow from anode to cathode within the cell, as shown in Figure 21.19. The
cell reactions are
Anode (oxidation): LixC6 ±£ xLi xe C6(s)
Cathode (reduction): Li1xMn2O4(s) xLi xe ±£ LiMn2O4(s) (–) (+)
Overall (cell) reaction: LixC6 Li1xMn2O4(s) ±£ LiMn2O4(s) C6(s) Anode Cathode
Ecell 3.7 V Li+
Li in LiMn2O4
The cell reaction is reversed during recharging. The lithium-ion battery powers graphite Electrolyte
countless laptop computers, cell phones, and camcorders. Its key drawbacks are LiPF6 in
(CH3)2CO3
expense and the flammability of the organic solvent.
Fuel Cells
In contrast to primary and secondary batteries, a fuel cell, sometimes called a flow Figure 21.19 Lithium-ion battery.
battery, is not self-contained. The reactants (usually a combustible fuel and oxy-
gen) enter the cell, and the products leave the cell, generating electricity through
the controlled oxidation of the fuel. In other words, fuel cells use combustion
to produce electricity. The fuel does not burn because, as in other batteries, the
half-reactions are separated, and the electrons are transferred through an exter-
nal circuit.
The most common fuel cell being developed for use in cars is the proton
exchange membrane (PEM) cell, which uses H2 as the fuel and has an operating
temperature of around 80C (Figure 21.20). The cell reactions are
Anode (oxidation): 2H2(g) ±£ 4H(aq) 4e
Cathode (reduction): O2(g) 4H(aq) 4e ±£ 2H2O(g)
Overall (cell) reaction: 2H2(g) O2(g) ±£ 2H2O(g) Ecell 1.2 V
The reactions in fuel cells have much lower rates than those in
other batteries, so they require an electrocatalyst to decrease the
activation energy (Section 16.8). The PEM cell electrodes are com- H (g) in O2(g ) in
2 e– e–
posites consisting of nanoparticles of a Pt-based catalyst deposited Anode Cathode
– –
on graphite. These are embedded in a polymer electrolyte mem- (–) e e (+)
brane having a perfluoroethylene backbone (—[F2C—CF2]—) n with
attached sulfonic acid groups (RSO3) that play a key role in fer- ELECTROLYTE
rying protons from anode to cathode. At the anode, two H2 mol- (polymer membrane:
sulfonated
ecules adsorb onto the catalyst and are split and oxidized. From perfluoroethylene)

each H2, two e travel through the wire to the cathode, while two
H become hydrated and migrate through the electrolyte as H3O. H3O+
According to a proposed multistep mechanism, at the cathode, an H3O+

O2 molecule adsorbs onto the catalyst, which provides an e to H3O+
form O2. One H3O donates its H to the O2, forming HO2 (that
H3O+
is, HO—O). The O—O bond stretches and breaks as another H3O
gives its H and the catalyst provides another e: the first H2O has
formed. In similar fashion, a third H and e attach to the freed
O atom to form OH, and a fourth H and e are transferred to form Pt-based catalyst
deposited on graphite
the second H2O. Both water molecules desorb and leave the cell. H2O(l ) out
Hydrogen fuel cells have been used for years to provide electricity and pure Figure 21.20 Hydrogen fuel cell.
water during space flights. In the very near future, similar ones will supply elec-
tric power for transportation, residential, and commercial needs. Already, every
major car manufacturer has a fuel-cell prototype. By themselves, these cells pro-
duce no pollutants, and they convert about 75% of the fuel’s bond energy into
useable power, in contrast to 40% for a coal-fired power plant and 25% for a
gasoline-powered car engine. Of course, their overall environmental impact will
depend on how the H2 is obtained; for example, electrolyzing water with solar
power will have a negligible impact, whereas electrolyzing it with electricity from
a coal-fired plant will have a sizeable one. Despite steady progress, current fuel-
cell research remains focused on lowering costs by improving membrane con-
ductivity and developing more efficient electrocatalysts.
Section Summary
Batteries contain several voltaic cells in series and are classified as primary (e.g., alka-
line, mercury, silver, and lithium), secondary (e.g., lead-acid, nickel–metal hydride, and
lithium-ion), or fuel cell. • Supplying electricity to a rechargeable (secondary) battery
reverses the redox reaction, forming more reactant for further use. • Fuel cells gen-
erate a current through the controlled oxidation of a fuel such as H2.
21.6 CORROSION: A CASE OF ENVIRONMENTAL

ELECTROCHEMISTRY
By now, you may be thinking that spontaneous electrochemical processes are
always beneficial, but consider the problem of corrosion, the natural redox
process that oxidizes metals to their oxides and sulfides. In chemical terms, cor-
rosion is the reverse of isolating a metal from its oxide or sulfide ore; in electro-
chemical terms, the process shares many similarities with the operation of a
voltaic cell. Damage from corrosion to cars, ships, buildings, and bridges runs
into tens of billions of dollars annually, so it is a major problem in much of the
world. Although we focus here on the corrosion of iron, many other metals, such
as copper and silver, also corrode.
The Corrosion of Iron

The most common and economically destructive form of corrosion is the rusting
of iron. About 25% of the steel produced in the United States is made just to
replace steel already in use that has corroded. Contrary to the simplified equation
presented earlier, rust is not a direct product of the reaction between iron and oxy-
gen but arises through a complex electrochemical process. Let’s look at the facts
of iron corrosion and then use the features of a voltaic cell to explain them:
1. Iron does not rust in dry air: moisture must be present.
2. Iron does not rust in air-free water: oxygen must be present.
3. The loss of iron and the depositing of rust often occur at different places on
the same object.
4. Iron rusts more quickly at low pH (high [H]).
5. Iron rusts more quickly in contact with ionic solutions.
6. Iron rusts more quickly in contact with a less active metal (such as Cu) and
more slowly in contact with a more active metal (such as Zn).
Picture the magnified surface of a piece of iron or steel (Figure 21.21).
Strains, ridges, and dents in contact with water are typically the sites of iron loss
(fact 1). These sites are called anodic regions because the following half-reaction
occurs there:
Fe(s) ±£ Fe2(aq) 2e [anodic region; oxidation]
Once the iron atoms lose electrons, the damage to the object has been done, and
a pit forms where the iron is lost.
The freed electrons move through the external circuit—the piece of iron
itself—until they reach a region of relatively high O2 concentration (fact 2), near
21.6 Corrosion: A Case of Environmental Electrochemistry 957
3 The Fe 2+ migrates through

the drop and reacts with 3 Water droplet
O
2 2
O2 and H2O to form rust
n H2O 2 H2O 2 Electrons at the
Fe (inactive)
2Fe2+
Rust cathode reduce
Fe2O3 • nH2O 4H+ O2 O2 to H2O
2Fe 4e–
4 (cathodic
(anodic region) region)
1 Oxidation of Fe forms a pit and yields

IRON BAR electrons, which travel through the metal
A B
Figure 21.21 The corrosion of iron. A, A close-up view of an iron matic depiction of a small area of the surface, showing the steps in the
surface. Corrosion usually occurs at a surface irregularity. B, A sche- corrosion process.
the surface of a surrounding water droplet, for instance. At this cathodic region,
the electrons released from the iron atoms reduce O2 molecules:
O2 (g) 4H(aq) 4e ±£ 2H2O(l) [cathodic region; reduction]
Notice that this overall redox process is complete; thus, the iron loss has occurred
without any rust forming:
2Fe(s) O2 (g) 4H(aq) ±£ 2Fe2(aq) 2H2O(l)
Rust forms through another redox reaction in which the reactants make direct
contact. The Fe2 ions formed originally at the anodic region disperse through
the surrounding water and react with O2, often at some distance from the pit (fact
3). The overall reaction for this step is
2Fe2(aq) 12O2 (g) (2n)H2O(l) ±£ Fe2O3nH2O(s) 4H(aq)
[The inexact coefficient n for H2O in the above equation appears because rust,
Fe2O3nH2O, is a form of iron(III) oxide with a variable number of waters of
hydration.] The rust deposit is really incidental to the damage caused by loss of
iron—a chemical insult added to the original injury.
Adding the previous two equations together shows the overall equation for
the rusting of iron:
2Fe(s) 32O2 (g) nH2O(l) 4H (aq) ±£ Fe2O3nH2O(s) 4H (aq)
The canceled H ions are shown to emphasize that they act as a catalyst; that is,
they speed the process as they are used up in one step of the overall reaction and
created in another. As a result of this action, rusting is faster at low pH (high
[H]) (fact 4). Ionic solutions speed rusting by improving the conductivity of the
aqueous medium near the anodic and cathodic regions (fact 5). The effect of ions
is especially evident on ocean-going vessels (Figure 21.22) and on the under-
bodies and around the wheel wells of cars driven in cold climates, where salts
are used to melt ice on slippery roads.
In many ways, the components of the corrosion process resemble those of a
voltaic cell:
• Anodic and cathodic regions are separated in space.
• The regions are connected via an external circuit through which the electrons
travel.
• In the anodic region, iron behaves like an active electrode, whereas in the Figure 21.22 Enhanced corrosion at
sea. The high ion concentration of seawater
cathodic region, it is inactive. leads to its high conductivity, which en-
• The moisture surrounding the pit functions somewhat like a salt bridge, a hances the corrosion of iron in the hulls
means for ions to ferry back and forth and keep the solution neutral. and anchors of ocean-going vessels.
Figure 21.23 The effect of metal-metal 2 Electrons from Fe

2 Electrons from Zn
contact on the corrosion of iron. A, When reduce O2 to H2O
Water droplet reduce O2 to H2O
iron is in contact with a less active metal, 2H2O
4H+ O2
2Zn2+
such as copper, the iron loses electrons
more readily (is more anodic), so it cor- 2H2O 2Fe2+ 4H+ O2
4e– Cu
rodes faster. B, When iron is in contact 2Zn
(inactive (anode)
with a more active metal, such as zinc, 2Fe cathode) 4e–
the zinc acts as the anode and loses (anode) 1 Zn gives up Fe
electrons. Therefore, the iron is cathodic, 1 Fe gives up electrons electrons to (inactive cathode)
so it does not corrode. The process is to Cu cathode Fe cathode
known as cathodic protection.
A Enhanced corrosion B Cathodic protection
Protecting Against the Corrosion of Iron

A common approach to preventing or limiting corrosion is to eliminate contact
with the corrosive factors. The simple act of washing off road salt removes the
ionic solution from auto bodies. Iron objects are frequently painted to keep out
O2 and moisture, but if the paint layer chips, rusting proceeds. More permanent
coatings include chromium plated on plumbing fixtures. In the “blueing” of gun
barrels, wood stoves, and other steel objects, an adherent coating of Fe3O4 (mag-
netite) is bonded to the surface.
The only fact regarding corrosion that we have not yet addressed concerns
the relative activity of other metals in contact with iron (fact 6), which leads to
the most effective way to prevent corrosion. The essential idea is that iron func-
tions as both anode and cathode in the rusting process, but it is lost only at the
anode. Therefore, anything that makes iron behave more like the anode increases
corrosion. As you can see in Figure 21.23A, when iron is in contact with a less
active metal (weaker reducing agent), such as copper, its anodic function is
enhanced. As a result, when iron plumbing is connected directly to copper plumb-
ing, the iron pipe corrodes rapidly.
On the other hand, anything that makes iron behave more like the cathode
Mg rod prevents corrosion. In cathodic protection. iron makes contact with a more active
metal (stronger reducing agent), such as zinc. The iron becomes cathodic and
remains intact, while the zinc acts as the anode and loses electrons (Figure
Iron pipe 21.23B). Coating steel with a “sacrificial” layer of zinc is the basis of the galva-
nizing process. In addition to blocking physical contact with H2O and O2, the zinc
is “sacrificed” (oxidized) instead of the iron.
Sacrificial anodes are employed to protect iron and steel structures (pipes,
Figure 21.24 The use of sacrificial tanks, oil rigs, and so on) in marine and moist underground environments. The
anodes to prevent iron corrosion. metals that are most frequently used for this purpose are magnesium and alu-
In cathodic protection, an active metal, minum because they are much more active than iron. As a result, they act as
such as magnesium or aluminum, is
connected to underground iron pipes to
the anode while iron acts as the cathode (Figure 21.24). Another advantage of
prevent their corrosion. The active metal these metals is that they form adherent oxide coatings, which slows their own
is sacrificed instead of the iron. corrosion.
Section Summary
Corrosion damages metal structures through a natural electrochemical change. • Iron
corrosion occurs in the presence of oxygen and moisture and is increased by high
[H], high [ion], or contact with a less active metal, such as Cu. Fe is oxidized and
O2 is reduced in one redox reaction, while rust (hydrated form of Fe2O3) is formed in
another reaction that often takes place at a different location. Because Fe functions
as both anode and cathode in the process, an iron or steel object can be protected
by physically covering its surface or joining it to a more active metal (such as Zn, Mg,
or Al), which acts as the anode in place of the Fe.
siL48593_ch21_922-979 14:12:07 03:54pm Page 959 fdfd ve403:MHQY042:siL5ch21:
21.7 Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions 959
21.7 ELECTROLYTIC CELLS: USING ELECTRICAL ENERGY

TO DRIVE NONSPONTANEOUS REACTIONS
Up to now, we’ve been considering voltaic cells, those that generate electrical
energy from a spontaneous redox reaction. The principle of an electrolytic cell is
exactly the opposite: electrical energy from an external source drives a non-
spontaneous reaction.
Construction and Operation of an Electrolytic Cell

Let’s examine the operation of an electrolytic cell by constructing one from a
voltaic cell. Consider the tin-copper voltaic cell in Figure 21.25A. The Sn anode
will gradually become oxidized to Sn2 ions, and the Cu2 ions will gradually
be reduced and plate out on the Cu cathode because the cell reaction is sponta-
neous in that direction:
For the voltaic cell
Sn(s) ±£ Sn2(aq) 2e [anode; oxidation]
Cu2(aq) 2e ±£ Cu(s) [cathode; reduction]
Sn(s) Cu2(aq) ±£ Sn2(aq) Cu(s) E°cell 0.48 V and ¢G° 93 kJ
Therefore, the reverse cell reaction is nonspontaneous and never happens of its
own accord, as the negative Ecell and positive G indicate:
Cu(s) Sn2(aq) ±£ Cu2(aq) Sn(s) E°cell 0.48 V and ¢G° 93 kJ
However, we can make this process happen by supplying from an external source
an electric potential greater than Ecell. In effect, we have converted the voltaic
cell into an electrolytic cell and changed the nature of the electrodes—anode is
now cathode, and cathode is now anode (Figure 21.25B):
For the electrolytic cell
Cu(s) ±£ Cu2(aq) 2e [anode; oxidation]
Sn2(aq) 2e ±£ Sn(s) [cathode; reduction]
Cu(s) Sn2(aq) ±£ Cu2(aq) Sn(s) [overall (cell) reaction]
e– 0.48 V e– e– External source e– Figure 21.25 The tin-copper reaction

Voltmeter greater than 0.48 V as the basis of a voltaic and an elec-
Anode Cathode Cathode Anode
trolytic cell. A, At standard conditions,
(–) (+) (–) (+) the spontaneous reaction between Sn
Sn Salt bridge Cu Sn Salt bridge Cu and Cu2 generates 0.48 V in a voltaic
cell. B, If more than 0.48 V is supplied,
the same apparatus and components
2e– 2e– 2e– become an electrolytic cell, and the
2e–
nonspontaneous reaction between Cu
+ and Sn2 occurs. Note the changes in
+ Cu2
+ Sn2 +
Sn Sn2 Cu Sn Cu2 Cu electrode charges and direction of elec-
tron flow.
1 M Sn2+ 1 M Cu 2+ 1 M Sn2+ 1 M Cu 2+
Oxidation half-reaction Oxidation half-reaction

Sn(s) Sn2+(aq ) + 2e– Cu(s) Cu2+(aq) + 2e–
Reduction half-reaction Reduction half-reaction

Cu2+(aq ) + 2e– Cu(s) Sn2+(aq) + 2e– Sn(s )
Overall (cell) reaction Overall (cell) reaction

Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s) Cu(s) + Sn2+(aq) Cu2+(aq) + Sn(s)
A Voltaic cell B Electrolytic cell

Note that in an electrolytic cell, as in a voltaic cell, oxidation takes place at the
anode and reduction takes place at the cathode, but the direction of electron flow
and the signs of the electrodes are reversed.
To understand these changes, keep in mind the cause of the electron flow:
• In a voltaic cell, electrons are generated at the anode, so it is negative, and
electrons are consumed at the cathode, so it is positive.
• In an electrolytic cell, the electrons come from the external power source,
which supplies them to the cathode, so it is negative, and removes them from
the anode, so it is positive.
A rechargeable battery functions as a voltaic cell when it is discharging and
as an electrolytic cell when it is recharging, so it provides a good way to com-
pare these two cell types and the changes in the processes and charges of the elec-
trodes. Figure 21.26 shows these changes in the lead-acid battery. In the discharge
mode (voltaic cell), oxidation occurs at electrode I, thus making the negative elec-
trode the anode. In the recharge mode (electrolytic cell), oxidation occurs at elec-
trode II, thus making the positive electrode the anode. Similarly, the cathode is
positive during discharge (electrode II) and negative during recharge (electrode I).
To reiterate, regardless of whether the cell is discharging or recharging, oxidation
occurs at the anode and reduction occurs at the cathode.
Figure 21.26 The processes occurring dur-

ing the discharge and recharge of a lead-acid
battery. When the lead-acid battery is discharging
(top), it behaves like a voltaic cell: the anode is
negative (electrode I), and the cathode is positive
(electrode II). When it is recharging (bottom), it
behaves like an electrolytic cell: the anode is pos-
itive (electrode II), and the cathode is negative
(electrode I).
VOLTAIC (Discharge)
Starter
e– motor e–
Oxidation half-reaction at I Reduction half-reaction at II

e– Pb(s) + HSO –(aq) PbO2(s) + 3H+(aq) + HSO4–(aq) + 2e– e
–
4
PbSO4(s) + H+ + 2e– PbSO4(s) + 2H2O(l )
(–) (+)
Anode Cathode
Switch I LEAD-ACID CELL II
Cathode Anode
(–) (+)
e– Reduction half-reaction at I Oxidation half-reaction at II

PbSO4(s) + H+ + 2e– PbSO4(s) + 2H2O(l ) e–
Pb(s) + HSO4–(aq) PbO2(s) + 3H+(aq) + HSO4–(aq) + 2e–
e– Power e–
supply
ELECTROLYTIC (Recharge)
siL48593_ch21_922-979 12:11:07 11:21am Page 961
Table 21.4 Comparison of Voltaic and Electrolytic Cells

Electrode
Cell Type ⌬G Ecell Name Process Sign
Voltaic 0 0 Anode Oxidation

Voltaic 0 0 Cathode Reduction
Electrolytic 0 0 Anode Oxidation
Electrolytic 0 0 Cathode Reduction
Table 21.4 summarizes the processes and signs in the two types of electro-
chemical cells.
Predicting the Products of Electrolysis

Electrolysis, the splitting (lysing) of a substance by the input of electrical energy,
is often used to decompose a compound into its elements. Electrolytic cells are
involved in key industrial production steps for some of the most commercially
important elements, including chlorine, aluminum, and copper, as you’ll see in
Chapter 22. The first laboratory electrolysis of H2O to H2 and O2 was performed
in 1800, and the process is still used to produce these gases in ultrahigh purity.
The electrolyte in an electrolytic cell can be the pure compound (such as H2O or
a molten salt), a mixture of molten salts, or an aqueous solution of a salt. The
products obtained depend on atomic properties and several other factors, so let’s
examine some actual cases.
Electrolysis of Pure Molten Salts Many electrolytic applications involve isolating
a metal or nonmetal from a molten salt. Predicting the product at each electrode
is simple if the salt is pure because the cation will be reduced and the anion oxi-
dized. The electrolyte is the molten salt itself, and the ions move through the cell
because they are attracted by the oppositely charged electrodes.
Consider the electrolysis of molten (fused) calcium chloride. The two species
present are Ca2 and Cl, so Ca2 ion is reduced and Cl ion is oxidized:
2Cl(l) ±£ Cl2 (g) 2e [anode; oxidation]
Ca (l) 2e ±£ Ca(s)
2
Ca2(l) 2Cl(l) ±£ Ca(s) Cl2 (g) [overall]
Metallic calcium is prepared industrially this way, as are several other active met-
als, such as Na and Mg, and the halogens Cl2 and Br2. We examine the details
of these and several other electrolytic processes in Chapter 22.
Electrolysis of Mixed Molten Salts More typically, the electrolyte is a mixture of
molten salts, which is then electrolyzed to obtain a particular metal. When we
have a choice of product, how can we tell which species will react at which elec-
trode? The general rule for all electrolytic cells is that the more easily oxidized
species (stronger reducing agent) reacts at the anode, and the more easily reduced
species (stronger oxidizing agent) reacts at the cathode.
It’s important to realize that for electrolysis of mixtures of molten salts, we
cannot use tabulated E values to tell the relative strength of the oxidizing and
reducing agents. Those values refer to the change from aqueous ion to free ele-
ment, Mn(aq) ne ±£ M(s), under standard-state conditions, but there are
no aqueous ions in a molten salt. Instead, we rely on our knowledge of periodic
atomic trends to predict which of the ions present gains or loses electrons more
easily (Sections 8.4 and 9.5).
siL48593_ch21_922-979 17:11:07 11:48am Page 962
SAMPLE PROBLEM 21.8 Predicting the Electrolysis Products of a Molten

Salt Mixture
PROBLEM A chemical engineer melts a naturally occurring mixture of NaBr and MgCl2 and
decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and
write balanced half-reactions and the overall cell reaction.
PLAN We have to determine which metal and nonmetal will form more easily at the elec-
trodes. We first list the ions as oxidizing or reducing agents. If one metal holds its elec-
trons more tightly than another, it has a higher ionization energy (IE). Therefore, as a
cation, it gains electrons more easily; it is the stronger oxidizing agent and is reduced at
the cathode. Similarly, if one nonmetal holds its electrons less tightly than another, it has
a lower electronegativity (EN). Therefore, as an anion, it loses electrons more easily; it is
the stronger reducing agent and is oxidized at the anode.
SOLUTION Listing the ions as oxidizing or reducing agents:
The possible oxidizing agents are Na and Mg2.
The possible reducing agents are Br and Cl.
Determining the cathode product (more easily reduced cation): Mg is to the right of Na
in Period 3. IE increases from left to right, so Mg has a higher IE. It takes more energy
to remove an e from Mg than from Na, so it follows that Mg2 has a greater attraction
for e and thus is more easily reduced (stronger oxidizing agent):
Mg2(l) 2e ±£ Mg(l) [cathode; reduction]
Determining the anode product (more easily oxidized anion): Br is below Cl in Group
7A(17). EN decreases down the group, so Br has a lower EN than Cl. Therefore, it fol-
lows that Br holds its e less tightly than Cl, so Br is more easily oxidized (stronger
reducing agent):
2Br(l) ±£ Br2(g) 2e [anode; oxidation]
Writing the overall cell reaction:
Mg2(l) 2Br(l) ±£ Mg(l) Br2(g) [overall]
H2 O2
COMMENT The cell temperature must be high enough to keep the salt mixture molten. In
this case, the temperature is greater than the melting point of Mg, so it appears as a liq-
uid in the equation, and greater than the boiling point of Br2, so it appears as a gas.
FOLLOW-UP PROBLEM 21.8 A sample of AlBr3 contaminated with KF is melted
and electrolyzed. Determine the electrode products and write the overall cell reaction.
Electrolysis of Water and Nonstandard Half-Cell Potentials Before we can ana-

lyze the electrolysis products of aqueous salt solutions, we must examine the elec-
trolysis of water itself. Extremely pure water is difficult to electrolyze because
very few ions are present to conduct a current. If we add a small amount of a salt
that cannot be electrolyzed in water (such as Na2SO4), however, electrolysis pro-
ceeds rapidly. A glass electrolytic cell with separated gas compartments is used
2H2O(l ) O2(g) + 4H+(aq) + 4e–
to keep the H2 and O2 gases from mixing (Figure 21.27). At the anode, water is
oxidized as the O.N. of O changes from 2 to 0:
Reduction half-reaction 2H2O(l) ±£ O2 (g) 4H(aq) 4e E 0.82 V [anode; oxidation]
2H2O(l ) + 2e– H2(g) + 2OH–(aq)
At the cathode, water is reduced as the O.N. of H changes from 1 to 0:
Overall (cell) reaction 2H2O(l) 2e ±£ H2 (g) 2OH(aq) E 0.42 V [cathode; reduction]
2H2O(l ) 2H2(g) + O2(g)
Doubling the cathode half-reaction to equate e loss and gain, adding the half-
Figure 21.27 The electrolysis of water. reactions (which involves combining the H and OH into H2O and canceling
A certain volume of oxygen forms e and excess H2O), and calculating Ecell gives the overall reaction:
through the oxidation of H2O at the an- 2H2O(l) ±£ 2H2 (g) O2 (g) Ecell 0.42 V 0.82 V 1.24 V [overall]
ode (right), and twice that volume of
hydrogen forms through the reduction of Notice that these electrode potentials are not written with a standard-state
H2O at the cathode (left). symbol (degree sign) because they are not standard electrode potentials. The [H]
and [OH] are 1.0 107 M rather than the standard-state value of 1 M. These E
values are obtained by applying the Nernst equation. For example, the calcula-
tion for the anode potential (with n 4) is
0.0592 V
Ecell E°cell log (PO2 [H]4 )
4
The standard potential for the oxidation of water is 1.23 V (from Appendix D)
and PO2 1 atm in the half-cell, so we have
Ecell 1.23 V e [log 1 4 log (1.0 107 )] f 0.82 V

0.0592 V
4
In aqueous ionic solutions, [H] and [OH] are approximately 107 M also, so
we use these nonstandard Ecell values to predict electrode products.
Electrolysis of Aqueous Salt Solutions and the Phenomenon of Overvoltage
Aqueous salt solutions are mixtures of ions and water, so we have to compare
the various electrode potentials to predict the electrode products. When two half-
reactions are possible at an electrode,
• The reduction with the less negative (more positive) electrode potential occurs.
• The oxidation with the less positive (more negative) electrode potential occurs.
What happens, for instance, when a solution of potassium iodide is elec-
trolyzed? The possible oxidizing agents are K and H2O, and their reduction half-
reactions are
K(aq) e ±£ K(s) E° 2.93 V
2H2O(l) 2e ±£ H2 (g) 2OH(aq) E 0.42 V [reduction]
The less negative electrode potential for water means that it is much easier to
reduce than K, so H2 forms at the cathode. The possible reducing agents are I
and H2O, and their oxidation half-reactions are
2I(aq) ±£ I2 (s) 2e E° 0.53 V [oxidation]
2H2O(l) ±£ O2 (g) 4H(aq) 4e E 0.82 V
The less positive electrode potential for I means that a lower potential is needed
to oxidize it than to oxidize H2O, so I2 forms at the anode.
However, the products predicted from this type of comparison of electrode
potentials are not always the actual products. For gases such as H2(g) and O2(g)
to be produced at metal electrodes, an additional voltage is required. This incre-
ment above the expected voltage is called the overvoltage, and it is 0.4 to 0.6 V
for these gases. The overvoltage results from kinetic factors, such as the large
activation energy (Section 16.5) required for gases to form at the electrode.
Overvoltage has major practical significance. A multibillion-dollar example is
the industrial production of chlorine from concentrated NaCl solution. Water is
easier to reduce than Na, so H2 forms at the cathode, even with an overvoltage
of 0.6 V:
Na(aq) e ±£ Na(s) E° 2.71 V
2H2O(l) 2e ±£ H2 (g) 2OH(aq) E 0.42 V ( 1 V with overvoltage)
[reduction]
But Cl2 does form at the anode, even though the electrode potentials themselves
would lead us to predict that O2 should form:
2H2O(l) ±£ O2 (g) 4H(aq) 4e E 0.82 V (1.4 V with overvoltage)
2Cl(aq) ±£ Cl2 (g) 2e E° 1.36 V [oxidation]
An overvoltage of 0.6 V makes the potential needed to form O2 slightly above
that for Cl2. Keeping the [Cl] high also favors Cl2 formation. Thus, chlorine,
which is one of the 10 most heavily produced industrial chemicals, can be formed
from plentiful natural sources of aqueous sodium chloride (see Chapter 22).
From these and other examples, we can determine which elements can be pre-
pared electrolytically from aqueous solutions of their salts:
1. Cations of less active metals are reduced to the metal, including gold, silver,
copper, chromium, platinum, and cadmium.
2. Cations of more active metals are not reduced, including those in Groups 1A(1)
and 2A(2), and Al from 3A(13). Water is reduced to H2 and OH instead.
3. Anions that are oxidized, because of overvoltage from O2 formation, include
the halides ([Cl] must be high), except for F.
4. Anions that are not oxidized include F and common oxoanions, such as
SO42, CO32, NO3, and PO43, because the central nonmetal in these
oxoanions is already in its highest oxidation state. Water is oxidized to O2 and
H instead.
SAMPLE PROBLEM 21.9 Predicting the Electrolysis Products of Aqueous

Salt Solutions
PROBLEM What products form during electrolysis of aqueous solutions of the following
salts: (a) KBr; (b) AgNO3; (c) MgSO4?
PLAN We identify the reacting ions and compare their electrode potentials with those of
water, taking the 0.4 to 0.6 V overvoltage into consideration. The reduction half-reaction
with the less negative electrode potential occurs at the cathode, and the oxidation half-
reaction with the less positive electrode potential occurs at the anode.
SOLUTION
(a) K(aq) e ±£ K(s) E° 2.93 V
2H2O(l) 2e ±£ H2 (g) 2OH(aq) E 0.42 V
Despite the overvoltage, which makes E for the reduction of water between 0.8 and
1.0 V, H2O is still easier to reduce than K, so H2(g) forms at the cathode.
2Br(aq) ±£ Br2 (l) 2e E° 1.07 V

2H2O(l) ±£ O2 (g) 4H (aq) 4e E 0.82 V
Because of the overvoltage, which makes E for the oxidation of water between 1.2 and
1.4 V, Br is easier to oxidize than water, so Br2(l) forms at the anode (see photo).
(b) Ag(aq) e ±£ Ag(s) E° 0.80 V
2H2O(l) 2e ±£ H2 (g) 2OH (aq) E 0.42 V
Electrolysis of aqueous KBr.
As the cation of an inactive metal, Ag is a better oxidizing agent than H2O, so
Ag(s) forms at the cathode. NO3 cannot be oxidized, because N is already in its highest
(5) oxidation state. Thus, O2(g) forms at the anode:
2H2O(l) ±£ O2 (g) 4H(aq) 4e
(c) Mg (aq) 2e ±£ Mg(s)
2
E° 2.37 V

Like K in part (a), Mg2 cannot be reduced in the presence of water, so H2(g) forms at
the cathode. The SO42 ion cannot be oxidized because S is in its highest (6) oxida-
tion state. Thus, H2O is oxidized, and O2(g) forms at the anode:
2H2O(l) ±£ O2 (g) 4H (aq) 4e
FOLLOW-UP PROBLEM 21.9 Write half-reactions showing the products you pre-
dict will form in the electrolysis of aqueous AuBr3.
The Stoichiometry of Electrolysis: The Relation Between

Amounts of Charge and Product
As you’ve seen, the charge flowing through an electrolytic cell yields products at
the electrodes. In the electrolysis of molten NaCl, for example, the power source
supplies electrons to the cathode, where Na ions migrate to pick them up and
become Na metal. At the same time, the power source pulls from the anode the
electrons that Cl ions release as they become Cl2 gas. It follows that the more
electrons picked up by Na ions and released by Cl ions, the greater the amounts
of Na and Cl2 that form. This relationship was first determined experimentally by
Michael Faraday and is referred to as Faraday’s law of electrolysis: the amount
of substance produced at each electrode is directly proportional to the quantity
of charge flowing through the cell.
Each balanced half-reaction shows the amounts (mol) of reactant, electrons,
and product involved in the change, so it contains the information we need to
answer such questions as “How much material will form as a result of a given
quantity of charge?” or, conversely, “How much charge is needed to produce a
given amount of material?” To apply Faraday’s law,
1. Balance the half-reaction to find the number of moles of electrons needed per The Father of Electrochemistry and
mole of product. Much More The investigations by Michael
Faraday (1791–1867) into the mass of an
2. Use the Faraday constant (F 9.65 104 C/mol e) to find the corresponding element that is equivalent to a given
charge. amount of charge established electro-
3. Use the molar mass to find the charge needed for a given mass of product. chemistry as a quantitative science, but
In practice, to supply the correct amount of electricity, we need some means his breakthroughs in physics are even
more celebrated. In 1821, Faraday devel-
of finding the charge flowing through the cell. We cannot measure charge directly,
oped the precursor of the electric motor,
but we can measure current, the charge flowing per unit time. The SI unit of cur- and his later studies of currents induced
rent is the ampere (A), which is defined as a charge of 1 coulomb flowing through by electric and magnetic fields eventually
a conductor in 1 second: led to the electric generator and the trans-
1 ampere 1 coulomb/second or 1 A 1 C/s (21.11) former. After his death, this self-educated
blacksmith’s son was rated by Albert Ein-
Thus, the current multiplied by the time gives the charge: stein as the peer of Newton, Galileo, and
C Maxwell. He is shown here delivering
Current time charge or A s sC
s one of his famous lectures, “Chemical
History of a Candle.”
Therefore, we find the charge by measuring the current and the time during which
the current flows. This, in turn, relates to the amount of product formed. Figure
21.28 summarizes these relationships.
MASS (g) AMOUNT (mol) AMOUNT (mol)

of substance of substance of electrons CHARGE CURRENT
oxidized or (g/mol) oxidized or balanced transferred Faraday (C) time (s) (A)
reduced half-reaction constant
reduced
(C/mol e–)
Figure 21.28 A summary diagram for

the stoichiometry of electrolysis.
Problems based on Faraday’s law often ask you to calculate current, mass of
material, or time. The electrode half-reaction provides the key to solving these
problems because it is related to the mass for a certain quantity of charge.
As an example, let’s consider a typical problem in practical electrolysis: how
long does it take to produce 3.0 g of Cl2(g) by electrolysis of aqueous NaCl using
a power supply with a current of 12 A? The problem asks for the time needed to
produce a certain mass, so let’s first relate mass to number of moles of electrons
to find the charge needed. Then, we’ll relate the charge to the current to find the
time.
We know the mass of Cl2 produced, so we can find the amount (mol) of Cl2.
The half-reaction tells us that the loss of 2 mol of electrons produces 1 mol of
chlorine gas:
2Cl(aq) ±£ Cl2 (g) 2e
We use this relationship as a conversion factor, and multiplying by the Faraday

constant gives us the total charge:
1 mol Cl2 2 mol e 9.65 104 C
Charge (C) 3.0 g Cl2 8.2 103 C
70.90 g Cl2 1 mol Cl2 1 mol e
Now we use the relationship between charge and current to find the time needed:
charge(C) 1s
Time (s) 8.2 103 C 6.8 102 s (11 min)
current (A, or C/s) 12 C
Note that the entire calculation follows Figure 21.28 until the last step:
grams of Cl2 ⇒ moles of Cl2 ⇒ moles of e⇒ coulombs ⇒ seconds
Sample Problem 21.10 demonstrates the steps as they appear in Figure 21.28.
SAMPLE PROBLEM 21.10 Applying the Relationship Among Current,

Time, and Amount of Substance
PROBLEM A technician is plating a faucet with 0.86 g of chromium from an electrolytic
bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is
needed?
PLAN To find the current, we divide the charge by the time; therefore, we need to find the
Mass (g) of Cr needed charge. First, we write the half-reaction for Cr3 reduction. From it, we know the number
of moles of e required per mole of Cr. As the roadmap shows, to find the charge, we
divide by ᏹ (g/mol)
convert the mass of Cr needed (0.86 g) to amount (mol) of Cr. The balanced half-
reaction gives the amount (mol) of e transferred. Then, we use the Faraday constant
(9.65 104 C/mol e) to find the charge and divide by the time (12.5 min, converted to
Amount (mol) of Cr needed seconds) to obtain the current.
SOLUTION Writing the balanced half-reaction:
3 mol eⴚ 1 mol Cr
Cr3(aq) 3e ±£ Cr(s)
Combining steps to find amount (mol) of e transferred for mass of Cr needed:
Amount (mol) of eⴚ transferred 1 mol Cr 3 mol e
Moles of e transferred 0.86 g Cr 0.050 mol e
52.00 g Cr 1 mol Cr
1 mol eⴚ 9.65104 C
Calculating the charge:
9.65 104 C
Charge (C) 0.050 mol e 4.8 103 C
Charge (C) 1 mol e
divide by time (convert min to s)
Calculating the current:
charge (C) 4.8 103 C 1 min
Current (A) 6.4 C/s 6.4 A
time (s) 12.5 min 60 s
Current (A)
CHECK Rounding gives
(0.9 g)(1 mol Cr/50 g)(3 mol e/1 mol Cr) 5 102 mol e
then (5 102 mol e)(1 105 C/mol e) 5 103 C
and (5 103 C/12 min)(1 min/60 s) 7 A
COMMENT For the sake of introducing Faraday’s law, the details of the electroplating
process have been simplified here. Actually, electroplating chromium is only 30% to 40%
efficient and must be run at a particular temperature range for the plate to appear bright.
Nearly 10,000 metric tons (2 108 mol) of chromium is used annually for electroplating.
FOLLOW-UP PROBLEM 21.10 Using a current of 4.75 A, how many minutes does
it take to plate onto a sculpture 1.50 g of Cu from a CuSO4 solution?
The following Chemical Connections essay links several themes of this chap-
ter in the setting of a living cell.
siL48593_ch21_922-979 20:11:07 14:31pm Page 967
Chemical Connections to Biological Energetics

Cellular Electrochemistry and the Production of ATP
iological cells apply the principles of electrochemical cells to logical oxidizing agent called NAD (nicotinamide adenine
B generate energy. The complex multistep process can be di-

vided into two parts:
1. Bond energy in food is used to generate an electrochemical

dinucleotide) acquires these protons and electrons in the process
of oxidizing the molecules in food. To show this process, we use
the following half-reaction (without canceling an H on both
sides):
potential.
NAD(aq) 2H(aq) 2e ±£ NADH(aq) H(aq)
2. The potential is used to create the bond energy of the high-
energy molecule adenosine triphosphate (ATP; see Chemical At the mitochondrial inner membrane, the NADH and H transfer
Connections, pp. 908–909). the two e to the first redox couple of the ETC and release the two
H. The electrons are transported down the chain of redox cou-
The redox species that accomplish these steps are part of the ples, where they finally reduce O2 to H2O. The overall process,
electron-transport chain (ETC), which lies on the inner mem- with standard electrode potentials,* is
branes of mitochondria, the subcellular particles that produce the NADH(aq) H(aq) ±£ NAD(aq) 2H(aq) 2e
cell’s energy (Figure B21.1).
E°¿ 0.315 V
The ETC is a series of large molecules (mostly proteins),
each of which contains a redox couple (the oxidized and reduced
1
2 O2 (aq) 2e 2H (aq) ±£ H2O(l) E°¿ 0.815 V
forms of a species), such as Fe3/Fe2, that passes electrons down NADH(aq) H (aq) 1
±£ NAD (aq) H2O(l)
2 O2 (aq)
the chain. At three points along the chain, large potential differ-
ences supply enough free energy to convert adenosine diphos- overall 0.815 V (0.315 V) 1.130 V
E°¿
phate (ADP) into ATP. Thus, for each mole of NADH that enters the ETC, the free-
energy equivalent of 1.14 V is available:
Bond Energy to Electrochemical Potential ¢G°¿ nFE°¿
Cells utilize the energy in food by releasing it in controlled steps (2 mol e/mol NADH)(96.5 kJ/Vmol e) (1.130 V)
rather than all at once. The reaction that ultimately powers the 218 kJ/mol NADH
ETC is the oxidation of hydrogen to form water: (continued)
H2 12O2 ±£ H2O
However, instead of H2 gas, which does not occur in organisms, *In biological systems, standard potentials are designated E and the
the hydrogen takes the form of two H ions and two e. A bio- standard states include a pH of 7.0 ([H] 1107 M).
Outside mitochondrion
Cell Mitochondrion Intermembrane space
Inner membrane
Outer membrane
Intermembrane space
Outer membrane
Matrix
Inner membrane
Matrix
A ETC components
Figure B21.1 The mitochondrion. A, Mitochondria are subcellular in an electron micrograph. C, The components of the electron-transport
particles outside the cell nucleus. B, They have a smooth outer mem- chain are attached to the inner membrane.
brane and a highly folded inner membrane, shown schematically and
967
Chemical Connections continued

Note that this aspect of the process functions like a voltaic cell: a
spontaneous reaction, the reduction of O2 to H2O, is used to gen- NADH + H+
erate a potential. In contrast to a laboratory voltaic cell, in which
the overall process occurs in one step, this process occurs in many NAD+
COMPLEX
small steps. Figure B21.2 is a greatly simplified diagram of the E = 0.36 V
Free energy relative to O2

three key steps in the ETC that generate the high potential to pro- ox
ΔG = –69 kJ/mol
duce ATP. Q
In the cell, each of these steps is part of a complex consisting red COMPLEX
E = 0.19 V Fe3+
of several components, most of which are proteins. Electrons are ΔG = –37 kJ/mol Cyt
passed from one redox couple to the next down the chain, such Fe2+
that the reduced form of the first couple reduces the oxidized form
COMPLEX
of the second, and so forth. Most of the ETC components are iron-
containing proteins, and the redox change consists of the oxida- E = 0.58 V
ΔG = –112 kJ/mol
tion of Fe2 to Fe3 in one component accompanied by the
1O
reduction of Fe3 to Fe2 in another: 2 2
Fe2 (in A) Fe3 (in B) ±£ Fe3 (in A) Fe2 (in B) H2O

In other words, metal ions within the ETC proteins are the actual
Electron flow
species undergoing the redox reactions.
Figure B21.2 The main energy-yielding steps in the electron-
Electrochemical Potential to Bond Energy transport chain (ETC). The electron carriers in the ETC undergo oxida-
tion and reduction as they pass electrons to one another along the
At the three points shown in Figure B21.2, the large potential dif- chain. At the three points shown, the difference in potential E (or free
ference is used to form ATP: energy, G) is large enough to be used for ATP production. (A com-
ADP3(aq) HPO42(aq) H(aq) ±£ ATP4(aq) H2O(l) plex consists of many components, mostly proteins; Q is a large or-
ganic molecule; and Cyt is the abbreviation for a cytochrome, a protein
¢G°¿ 30.5 kJ/mol
that contains a metal-ion redox couple, such as Fe3/Fe2, which is
Note that the free energy that is released at each of the three the actual electron carrier.)
ATP-producing points exceeds 30.5 kJ, the free energy that must
be absorbed to form ATP. Thus, just as in an electrolytic cell, an
electrochemical potential is supplied to drive a nonspontaneous
reaction.
So far, we’ve followed the flow of electrons through the Outside mitochondrion
members of the ETC, but where have the released protons gone?
The answer is the key to how electrochemical potential is con- Outer membrane
verted to bond energy in ATP. As the electrons flow and the redox Intermembrane space Inner
couples change oxidation states, free energy released at the three high [H+] membrane
key steps is used to force H ions into the intermembrane space,
so that the [H] of the intermembrane space soon becomes higher +++
+++++
++ H+
+ ++ ++
than that of the matrix (Figure B21.3). In other words, this aspect +++ + ++ Cyt +
++++++++ Q
of the process acts as an electrolytic cell by using the free energy –– ––––––
–– ––
supplied by the three steps to create what amounts to an H con- 2e– ––
– – ––
–––
–– – – – – –– – – H+ H+ H+ 1O H O
centration cell across the membrane. 2 2 2
When the [H] difference across the membrane reaches ADP + Pi ATP
about 2.5-fold, it triggers the membrane to let H ions flow back NADH NAD+ Matrix
H+ low [H+]
through spontaneously (in effect, closing the switch and allowing
the concentration cell to operate). The free energy released in this
spontaneous process drives the nonspontaneous ATP formation Figure B21.3 Coupling electron transport to proton transport
via a mechanism that is catalyzed by the enzyme ATP synthase. to ATP synthesis. The purpose of the ETC is to convert the free
energy released from food molecules into the stored free energy of
(Paul D. Boyer and John E. Walker shared half of the 1997 Nobel
ATP. It accomplishes this by transporting electrons along the chain
Prize in chemistry for elucidating this mechanism.) (curved yellow line), while protons are pumped out of the mitochon-
Thus, the mitochondrion uses the “electron-motive force” of drial matrix. This pumping creates an [H] difference and generates a
redox couples on the membrane to generate a “proton-motive potential across the inner membrane (in effect, a concentration cell).
force” across the membrane, which converts a potential difference When this potential reaches a “trigger” value, H flows back into the
to bond energy. inner space, and the free energy released drives the formation of ATP.
968
Chapter Review Guide 969
Section Summary
An electrolytic cell uses electrical energy to drive a nonspontaneous reaction. • Oxi-
dation occurs at the anode and reduction at the cathode, but the direction of elec-
tron flow and the charges of the electrodes are opposite those in voltaic cells. • When
two products can form at each electrode, the more easily oxidized substance reacts
at the anode and the more easily reduced at the cathode. • The reduction or oxi-
dation of water takes place at nonstandard conditions. • Overvoltage causes the
actual voltage required to be unexpectedly high and can affect the electrode product that
forms. • The amount of product that forms depends on the quantity of charge flow-
ing through the cell, which is related to the magnitude of the current and the time it
flows. • Biological redox systems combine aspects of voltaic, concentration, and
electrolytic cells to convert bond energy in food into electrochemical potential and
then into the bond energy of ATP.
Chapter Perspective
The field of electrochemistry is one of the many areas in which the principles of
thermodynamics lead to practical benefits. As you’ve seen, electrochemical cells can
use a reaction to generate energy or use energy to drive a reaction. Such processes
are central not only to our mobile way of life, but also to our biological existence. In
Chapter 22, we examine the electrochemical (and other) methods used by industry to
convert raw natural resources into some of the materials modern society finds
indispensable.
CHAPTER REVIEW GUIDE The following sections provide many aids to help you study this chapter.
(Numbers in parentheses refer to pages, unless noted otherwise.)
Learning Objectives These are concepts and skills you should know after studying this chapter.
Relevant section and/or sample problem (SP) numbers 12. How the relative reactivity of a metal is determined by its re-
appear in parentheses. ducing power and is related to the negative of its Ehalf-cell (Section
21.3)
Understand These Concepts 13. How Ecell (the nonstandard cell potential) is related to G
1. The meanings of oxidation and reduction; why an oxidizing (maximum work) and the charge (moles of electrons times the
agent is reduced and a reducing agent is oxidized (Section 21.1; Faraday constant) flowing through the cell (Section 21.4)
also Section 4.5) 14. The interrelationship of G, Ecell, and K (Section 21.4)
2. How the half-reaction method is used to balance redox reac- 15. How Ecell changes as the cell operates (Q changes) (Section
tions in acidic or basic solution (Section 21.1) 21.4)
3. The distinction between voltaic and electrolytic cells in terms 16. Why a voltaic cell can do work until Q K (Section 21.4)
of the sign of G (Section 21.1) 17. How a concentration cell does work until the half-cell con-
4. How voltaic cells use a spontaneous reaction to release electri- centrations are equal (Section 21.4)
cal energy (Section 21.2) 18. The distinction between primary (nonrechargeable) and sec-
5. The physical makeup of a voltaic cell: arrangement and com- ondary (rechargeable) batteries and the nature of fuel cells
position of half-cells, relative charges of electrodes, and purpose (Section 21.5)
of a salt bridge (Section 21.2) 19. How corrosion occurs and is prevented; the similarities be-
6. How the difference in reducing strength of the electrodes de- tween a corroding metal and a voltaic cell (Section 21.6)
termines the direction of electron flow (Section 21.2) 20. How electrolytic cells use nonspontaneous redox reactions
7. The correspondence between a positive Ecell and a spontaneous driven by an external source of electricity (Section 21.7)
cell reaction (Section 21.3) 21. How atomic properties (ionization energy and electronegativ-
8. The usefulness and significance of standard electrode poten- ity) determine the products of the electrolysis of molten salt mix-
tials (Ehalf-cell) (Section 21.3) tures (Section 21.7)
9. How Ehalf-cell values are combined to give Ecell (Section 21.3) 22. How the electrolysis of water influences the products of
10. How the standard reference electrode is used to find an un- aqueous electrolysis; the importance of overvoltage (Section 21.7)
known Ehalf-cell (Section 21.3) 23. The relationship between the quantity of charge flowing
11. How an emf series (e.g., Table 21.2 or Appendix D) is used to through the cell and the amount of product formed (Section 21.7)
write spontaneous redox reactions (Section 21.3)
Learning Objectives (continued)

Master These Skills 7. Predicting whether a metal can displace hydrogen or another
metal from solution (Section 21.3)
1. Balancing redox reactions by the half-reaction method (Sec-
8. Using the interrelationship of G, Ecell, and K to calculate one
tion 21.1 and SP 21.1)
of the three given the other two (Section 21.4 and SP 21.5)
2. Diagramming and notating a voltaic cell (Section 21.2 and SP
9. Using the Nernst equation to calculate the nonstandard cell po-
21.2)
tential (Ecell) (SP 21.6)
3. Combining Ehalf-cell values to obtain Ecell (Section 21.3)
10. Calculating Ecell of a concentration cell (SP 21.7)
4. Using Ecell and a known Ehalf-cell to find an unknown Ehalf-cell
11. Predicting the products of the electrolysis of a mixture of
(SP 21.3)
molten salts (SP 21.8)
5. Manipulating half-reactions to write a spontaneous redox reac-
12. Predicting the products of the electrolysis of aqueous salt so-
tion and calculate its Ecell (SP 21.4)
lutions (SP 21.9)
6. Ranking the relative strengths of oxidizing and reducing agents
13. Calculating the current (or time) needed to produce a given
in a redox reaction (SP 21.4)
amount of product by electrolysis (SP 21.10)
Key Terms These important terms appear in boldface in the chapter and are defined again in the Glossary.
electrochemistry (923) Section 21.2 standard cell potential (Ecell) Section 21.5
electrochemical cell (923) half-cell (930) (935) battery (952)
Section 21.1 salt bridge (930) standard electrode (half-cell) fuel cell (955)
half-reaction method (924) Section 21.3 potential (Ehalf-cell) (935) Section 21.6
voltaic (galvanic) cell (928) cell potential (Ecell) (934) standard reference half-cell corrosion (956)
electrolytic cell (928) voltage (934) (standard hydrogen
Section 21.7
electrode (928) electromotive force (emf) electrode) (936)
electrolysis (961)
electrolyte (928) (934) Section 21.4 overvoltage (963)
anode (929) volt (V) (934) Faraday constant (F) (943)
ampere (A) (965)
cathode (929) coulomb (C) (934) Nernst equation (946)
concentration cell (948)
Key Equations and Relationships Numbered and screened concepts are listed for you to refer to or memorize.
21.1 Relating a spontaneous process to the sign of the cell poten- 21.7 Finding the equilibrium constant from the standard cell po-
tial (934): tential (944):
Ecell 0 for a spontaneous process RT
E°cell ln K
21.2 Relating electric potential to energy and charge in SI units nF
(934): 21.8 Substituting known values of R, F, and T into Equation 21.7
Potential energy/charge or 1 V 1 J/C and converting to common logarithms (945):
21.3 Relating standard cell potential to standard electrode poten- 0.0592 V nE°cell
tials in a voltaic cell (935): E°cell log K or log K (at 298.15 K)
n 0.0592 V
Ecell Ecathode (reduction) E anode (oxidation)
21.9 Calculating the nonstandard cell potential (Nernst equation)
21.4 Defining the Faraday constant (944): (946):
J RT
F 9.65 104 (3 sf) Ecell E°cell ln Q
Vmol e nF
21.5 Relating the free energy change to the cell potential (944): 21.10 Substituting known values of R, F, and T into the Nernst
G nFEcell equation and converting to common logarithms (946):
21.6 Finding the standard free energy change from the standard 0.0592 V
cell potential (944): Ecell E°cell log Q (at 298.15 K)
n
G nFEcell
21.11 Relating current to charge and time (965):
Current charge/time or 1 A 1 C/s
Chapter Review Guide 971
Highlighted Figures and Tables These figures (F) and tables (T) provide a visual review of key ideas.
F21.1 Summary of redox terminology (924) F21.21 The corrosion of iron (957)
F21.3 Voltaic and electrolytic cells (928) F21.25 Tin-copper reaction in voltaic and electrolytic cells (959)
F21.5 A voltaic cell based on the zinc-copper reaction (931) T21.4 Comparison of voltaic and electrolytic cells (961)
F21.10 The interrelationship of G, Ecell, and K (944) F21.28 Summary of the stoichiometry of electrolysis (965)
F21.11 Ecell and log Q for the zinc-copper cell (948)
Brief Solutions to FOLLOW-UP PROBLEMS Compare your solutions to these calculation steps and answers.
21.1 6KMnO4(aq) 6KOH(aq) KI(aq) ±£ 21.6 Fe(s) ±£ Fe2(aq) 2e E 0.44 V
6K2MnO4(aq) KIO3(aq) 3H2O(l) Cu (aq) 2e ±£ Cu(s)
2
E 0.34 V
21.2 Sn(s) ±£ Sn2(aq) 2e Fe(s) Cu2 (aq) ±£ Fe2(aq) Cu(s) Ecell 0.78 V
[anode; oxidation] So Ecell 0.78 V 0.25 V 1.03 V
6e 14H(aq) Cr2O72(aq) ±£ 2Cr3(aq) 7H2O(l) 0.0592 V [Fe2]
[cathode; reduction] 1.03 V 0.78 V log

2 [Cu2]
3Sn(s) Cr2O7 (aq) 14H (aq) ±£
2 2
[Fe ]
3Sn2(aq) 2Cr3(aq) 7H2O(l) [overall] 3.6 109
[Cu2]
Cell notation:
[Fe2] 3.6 109 0.30 M 1.1 109 M
Sn(s) Sn2(aq) H(aq), Cr2O72(aq), Cr3(aq) graphite
21.7 Au3(aq; 2.5 102 M) [B] ±£
e– Voltmeter e– Au3(aq; 7.0 104 M) [A]
7.0 104
Ecell 0 V a b 0.0306 V
0.0592 V
Anode Cathode log
3 2.5 102
Sn (–) NO3– K+ (+) C
The electrode in A is negative, so it is the anode.
21.8 Oxidizing agents: K and Al3.
Reducing agents: F and Br.
Al is above and to the right of K in the periodic table, so it has a
higher IE:
Al3(l) 3e ±£ Al(s) [cathode; reduction]
Sn 2+ Cr 3+, H+, Cr2O72–
Br is below F in Group 7A(17), so it has a lower EN:
21.3 Br2(aq) 2e ±£ 2Br(aq) Ebromine 1.07 V 2Br(l) ±£ Br2(g) 2e [anode; oxidation]
[cathode] 2Al3(l) 6Br(l) ±£ 2Al(s) 3Br2(g) [overall]
2V3(aq) 2H2O(l) ±£ 2VO2(aq) 4H(aq) 2e 21.9 The reduction with the more positive electrode potential is
Evanadium ? [anode] Au3(aq) 3e ±£ Au(s); E 1.50 V
Evanadium Ebromine Ecell 1.07 V 1.39 V 0.32 V [cathode; reduction]
21.4 Fe2(aq) 2e ±£ Fe(s) E 0.44 V Because of overvoltage, O2 will not form at the anode, so Br2 will
2[Fe2(aq) ±£ Fe3(aq) e] E 0.77 V form:
2Br(aq) ±£ Br2(l) 2e; E 1.07 V
3Fe2(aq) ±£ 2Fe3(aq) Fe(s)
[cathode; oxidation]
Ecell 0.44 V 0.77 V 1.21 V
21.10 Cu2(aq) 2e ±£ Cu(s); therefore,
The reaction is nonspontaneous. The spontaneous reaction is 2 mol e/1 mol Cu 2 mol e/63.55 g Cu
2Fe3(aq) Fe(s) ±£ 3Fe2(aq) Ecell 1.21 V 2 mol e
Fe Fe2 Fe3 Time (min) 1.50 g Cu
63.55 g Cu
21.5 Cd(s) Cu2(aq) ±£ Cd2(aq) Cu(s) 9.65 104 C 1s 1 min
G RT ln K 8.314 J/mol rxnK 298 K ln K
1 mol e 4.75 C 60 s
143 kJ/mol rxn; K 1.2 1025 16.0 min
0.0592 V
E°cell log (1.2 1025 ) 0.742 V
2
PROBLEMS
Problems with colored numbers are answered in Appendix E (d) Which species is the reducing agent?
and worked in detail in the Student Solutions Manual. Prob- (e) From which species to which does electron transfer occur?
lem sections match those in the text and provide the numbers (f) Write the balanced molecular equation, with Na as the spec-
of relevant sample problems. Most offer Concept Review tator ion.
Questions, Skill-Building Exercises (grouped in pairs covering 21.12 Balance the following skeleton reactions and identify the ox-
the same concept), and Problems in Context. The Compre- idizing and reducing agents:
hensive Problems are based on material from any section or (a) ClO3(aq) I(aq) ±£ I2(s) Cl(aq) [acidic]
previous chapter.
(b) MnO4(aq) SO32(aq) ±£
Note: Unless stated otherwise, all problems refer to systems MnO2(s) SO42(aq) [basic]
at 298 K (25C).
(c) MnO4 (aq) H2O2(aq) ±£ Mn2(aq) O2(g) [acidic]
21.13 Balance the following skeleton reactions and identify the ox-
Redox Reactions and Electrochemical Cells idizing and reducing agents:
(Sample Problem 21.1) (a) O2(g) NO(g) ±£ NO3(aq) [acidic]
(b) CrO42(aq) Cu(s) ±£
Concept Review Questions Cr(OH)3(s) Cu(OH)2(s) [basic]
21.1 Define oxidation and reduction in terms of electron transfer
(c) AsO43(aq) NO2(aq) ±£
and change in oxidation number.
AsO2(aq) NO3(aq) [basic]
21.2 Why must an electrochemical process involve a redox reaction?
21.3 Can one half-reaction in a redox process take place indepen- 21.14 Balance the following skeleton reactions and identify the ox-
dently of the other? Explain. idizing and reducing agents:
21.4 Water is used to balance O atoms in the half-reaction method. (a) Cr2O72(aq) Zn(s) ±£Zn2(aq) Cr3(aq) [acidic]
Why can’t O2 ions be used instead? (b) Fe(OH)2(s) MnO4(aq) ±£
21.5 During the redox balancing process, what step is taken to en- MnO2(s) Fe(OH)3(s) [basic]
sure that e loss equals e gain? (c) Zn(s) NO3(aq) ±£ Zn2(aq) N2(g) [acidic]
21.6 How are protons removed when balancing a redox reaction 21.15 Balance the following skeleton reactions and identify the ox-
in basic solution? idizing and reducing agents:
21.7 Are spectator ions used to balance the half-reactions of a re- (a) BH4(aq) ClO3(aq) ±£
dox reaction? At what stage might spectator ions enter the bal-
H2BO3(aq) Cl(aq) [basic]
ancing process?
(b) CrO4 (aq) N2O(g) ±£ Cr3(aq) NO(g) [acidic]
2
21.8 Which type of electrochemical cell has Gsys 0? Which
type shows an increase in free energy? (c) Br2(l) ±£ BrO3(aq) Br(aq) [basic]
21.9 Which statements are true? Correct any that are false. 21.16 Balance the following skeleton reactions and identify the ox-
(a) In a voltaic cell, the anode is negative relative to the cathode. idizing and reducing agents:
(b) Oxidation occurs at the anode of a voltaic or electrolytic cell. (a) Sb(s) NO3(aq) ±£ Sb4O6(s) NO(g) [acidic]
(c) Electrons flow into the cathode of an electrolytic cell. (b) Mn2(aq) BiO3(aq) ±£
(d) In a voltaic cell, the surroundings do work on the system. MnO4(aq) Bi3(aq) [acidic]
(e) A metal that plates out of an electrolytic cell appears on the
(c) Fe(OH)2(s) Pb(OH)3 (aq) ±£
cathode.
Fe(OH)3(s) Pb(s) [basic]
(f) The cell electrolyte provides a solution of mobile electrons.
Skill-Building Exercises (grouped in similar pairs) idizing and reducing agents:
21.10 Consider the following balanced redox reaction: (a) NO2(g) ±£ NO3(aq) NO2(aq) [basic]
16H(aq) 2MnO4(aq) 10Cl(aq) ±£ (b) Zn(s) NO3(aq) ±£ Zn(OH)42(aq) NH3(g) [basic]
2Mn2(aq) 5Cl2(g) 8H2O(l) (c) H2S(g) NO3(aq) ±£ S8(s) NO(g) [acidic]
(a) Which species is being oxidized?
(b) Which species is being reduced?
idizing and reducing agents:
(c) Which species is the oxidizing agent?
(d) Which species is the reducing agent? (a) As4O6(s) MnO4(aq) ±£
(e) From which species to which does electron transfer occur? AsO43(aq) Mn2(aq) [acidic]
(f) Write the balanced molecular equation, with K and SO42 (b) P4(s) ±£ HPO3 (aq) PH3(g) [acidic]
2
as the spectator ions. (c) MnO4(aq) CN(aq) ±£

21.11 Consider the following balanced redox reaction: MnO2(s) CNO(aq) [basic]
2CrO2(aq) 2H2O(l) 6ClO(aq) ±£ 21.19 Balance the following skeleton reactions and identify the ox-
2CrO42(aq) 3Cl2(g) 4OH(aq) idizing and reducing agents:
(a) Which species is being oxidized? (a) SO32(aq) Cl2(g) ±£ SO42(aq) Cl(aq) [basic]
(b) Which species is being reduced? (b) Fe(CN)63(aq) Re(s) ±£
(c) Which species is the oxidizing agent? Fe(CN)64(aq) ReO4(aq) [basic]
Problems 973
(c) MnO4(aq) HCOOH(aq) ±£ 21.28 A voltaic cell is constructed with an Ag/Ag half-cell and a
Mn2(aq) CO2(g) [acidic] Pb/Pb2 half-cell. The silver electrode is positive.
Problems in Context (a) Write balanced half-reactions and the overall reaction.
21.20 In many residential water systems, the aqueous Fe3con- (b) Diagram the cell, labeling electrodes with their charges and
centration is high enough to stain sinks and turn drinking water showing the directions of electron flow in the circuit and of
light brown. The iron content is analyzed by first reducing the cation and anion flow in the salt bridge.
Fe3 to Fe2 and then titrating with MnO4 in acidic solution. 21.29 Consider the following voltaic cell:
Balance the skeleton reaction of the titration step:
Fe2(aq) MnO4(aq) ±£ Mn2(aq) Fe3(aq) Voltmeter
21.21 Aqua regia, a mixture of concentrated HNO3 and HCl, was
developed by alchemists as a means to “dissolve” gold. The pro- Fe Ni
Salt bridge
cess is a redox reaction with the simplified skeleton reaction:
Au(s) NO3(aq) Cl(aq) ±£ AuCl4(aq) NO2(g)
(a) Balance the reaction by the half-reaction method.
(b) What are the oxidizing and reducing agents?
(c) What is the function of HCl in aqua regia? 1 M Fe2+
Voltaic Cells: Using Spontaneous Reactions

to Generate Electrical Energy (a) In which direction do electrons flow in the external circuit?
(Sample Problem 21.2) (b) In which half-cell does oxidation occur?
(c) In which half-cell do electrons enter the cell?
Concept Review Questions (d) At which electrode are electrons consumed?
21.22 Consider the following general voltaic cell: (e) Which electrode is negatively charged?
(f) Which electrode decreases in mass during cell operation?
Voltmeter (g) Suggest a solution for the cathode electrolyte.
(h) Suggest a pair of ions for the salt bridge.
A E
C (i) For which electrode could you use an inactive material?
(j) In which direction do anions within the salt bridge move to
maintain charge neutrality?
(k) Write balanced half-reactions and an overall cell reaction.
21.30 Consider the following voltaic cell:
B D
Voltmeter
Identify the (a) anode, (b) cathode, (c) salt bridge, (d) electrode Zn Co
at which e leave the cell, (e) electrode with a positive charge, Salt bridge
and (f) electrode that gains mass as the cell operates (assuming
that a metal plates out).
21.23 Why does a voltaic cell not operate unless the two compart-
ments are connected through an external circuit?
1 M Co 2+
21.24 What purpose does the salt bridge serve in a voltaic cell, and
how does it accomplish this purpose?
21.25 What is the difference between an active and an inactive (a) In which direction do electrons flow in the external circuit?
electrode? Why are inactive electrodes used? Name two sub- (b) In which half-cell does reduction occur?
stances commonly used for inactive electrodes. (c) In which half-cell do electrons leave the cell?
21.26 When a piece of metal A is placed in a solution containing (d) At which electrode are electrons generated?
ions of metal B, metal B plates out on the piece of A. (e) Which electrode is positively charged?
(a) Which metal is being oxidized? (f) Which electrode increases in mass during cell operation?
(b) Which metal is being displaced? (g) Suggest a solution for the anode electrolyte.
(c) Which metal would you use as the anode in a voltaic cell in- (h) Suggest a pair of ions for the salt bridge.
corporating these two metals? (i) For which electrode could you use an inactive material?
(d) If bubbles of H2 form when B is placed in acid, will they (j) In which direction do cations within the salt bridge move to
form if A is placed in acid? Explain. maintain charge neutrality?
(k) Write balanced half-reactions and an overall cell reaction.
Skill-Building Exercises (grouped in similar pairs)
21.27 A voltaic cell is constructed with an Sn/Sn2 half-cell and a 21.31 A voltaic cell is constructed with an Fe/Fe2 half-cell and an
2
Zn/Zn half-cell. The zinc electrode is negative. Mn/Mn2 half-cell. The iron electrode is positive.
(a) Write balanced half-reactions and the overall reaction. (a) Write balanced half-reactions and the overall reaction.
(b) Diagram the cell, labeling electrodes with their charges and (b) Diagram the cell, labeling electrodes with their charges and
showing the directions of electron flow in the circuit and of showing the directions of electron flow in the circuit and of
cation and anion flow in the salt bridge. cation and anion flow in the salt bridge.
21.32 A voltaic cell is constructed with a Cu/Cu2 half-cell and an 21.44 Balance each skeleton reaction, calculate E°cell, and state
Ni/Ni2 half-cell. The nickel electrode is negative. whether the reaction is spontaneous:
(a) Write balanced half-reactions and the overall reaction. (a) Ag(s) Cu2(aq) ±£ Ag(aq) Cu(s)
(b) Diagram the cell, labeling electrodes with their charges and (b) Cd(s) Cr2O72(aq) ±£ Cd2(aq) Cr3(aq)
showing the directions of electron flow in the circuit and of (c) Ni2(aq) Pb(s) ±£ Ni(s) Pb2(aq)
cation and anion flow in the salt bridge.
21.45 Balance each skeleton reaction, calculate E°cell, and state
21.33 Write the cell notation for the voltaic cell that incorporates whether the reaction is spontaneous:
each of the following redox reactions: (a) Cu(aq) PbO2(s) SO42(aq) ±£
(a) Al(s) Cr3(aq) ±£ Al3(aq) Cr(s) PbSO4(s) Cu2(aq) [acidic]
(b) Cu2(aq) SO2(g) 2H2O(l) ±£ (b) H2O2(aq) Ni (aq) ±£ O2(g) Ni(s) [acidic]
2
Cu(s) SO42(aq) 4H(aq)
(c) MnO2(s) Ag(aq) ±£ MnO4(aq) Ag(s) [basic]
21.34 Write a balanced equation from each cell notation:
(a) Mn(s) Mn2(aq) Cd2(aq) Cd(s) 21.46 Use the following half-reactions to write three spontaneous
(b) Fe(s) Fe2(aq) NO3(aq) NO(g) Pt(s) reactions, calculate E°cell for each reaction, and rank the strengths
of the oxidizing and reducing agents:
Cell Potential: Output of a Voltaic Cell (1) Al3(aq) 3e ±£ Al(s) E° 1.66 V
(Sample Problems 21.3 and 21.4) (2) N2O4(g) 2e ±£ 2NO2(aq) E° 0.867 V
Concept Review Questions (3) SO42(aq) H2O(l) 2e ±£ SO32(aq) 2OH(aq)
21.35 How is a standard reference electrode used to determine un- E° 0.93 V
known E°half-cell values? 21.47 Use the following half-reactions to write three spontaneous
21.36 What does a negative E°cell indicate about a redox reaction? reactions, calculate E°cell for each reaction, and rank the strengths
What does it indicate about the reverse reaction? of the oxidizing and reducing agents:
21.37 The standard cell potential is a thermodynamic state func-
(1) Au(aq) e ±£ Au(s) E° 1.69 V
tion. How are E° values treated similarly to H°, G°, and S°
values? How are they treated differently? (2) N2O(g) 2H(aq) 2e ±£ N2(g) H2O(l)
E° 1.77 V
Skill-Building Exercises (grouped in similar pairs)
(3) Cr3(aq) 3e ±£ Cr(s) E° 0.74 V
21.38 In basic solution, Se2 and SO32 ions react spontaneously:
2Se2(aq) 2SO32(aq) 3H2O(l) ±£ 21.48 Use the following half-reactions to write three spontaneous
2Se(s) 6OH(aq) S2O32(aq) E°cell 0.35 V reactions, calculate E°cell for each reaction, and rank the strengths
(a) Write balanced half-reactions for the process.
(b) If E°sulfite is 0.57 V, calculate E°selenium. (1) 2HClO(aq) 2H(aq) 2e ±£ Cl2(g) 2H2O(l)
21.39 In acidic solution, O3 and Mn2 ion react spontaneously: E° 1.63 V
O3(g) Mn2(aq) H2O(l) ±£ (2) Pt2(aq) 2e ±£ Pt(s) E° 1.20 V
O2(g) MnO2(s) 2H(aq) E°cell 0.84 V (3) PbSO4(s) 2e ±£ Pb(s) SO42(aq) E° 0.31 V
(a) Write the balanced half-reactions. 21.49 Use the following half-reactions to write three spontaneous
(b) Using Appendix D to find E°ozone, calculate E°manganese. reactions, calculate E°cell for each reaction, and rank the strengths
21.40 Use the emf series (Appendix D) to arrange the species. (1) I2(s) 2e ±£ 2I(aq) E° 0.53 V
(a) In order of decreasing strength as oxidizing agents: Fe3, (2) S2O82(aq) 2e ±£ 2SO42(aq) E° 2.01 V
Br2, Cu2
(3) Cr2O72(aq) 14H(aq) 6e ±£
(b) In order of increasing strength as oxidizing agents: Ca2,
Cr2O72, Ag 2Cr3(aq) 7H2O(l) E° 1.33 V
21.41 Use the emf series (Appendix D) to arrange the species. Problems in Context
(a) In order of decreasing strength as reducing agents: SO2, 21.50 When metal A is placed in a solution of a salt of metal B, the
PbSO4, MnO2 surface of metal A changes color. When metal B is placed in acid
(b) In order of increasing strength as reducing agents: Hg, Fe, Sn solution, gas bubbles form on the surface of the metal. When
metal A is placed in a solution of a salt of metal C, no change is
observed in the solution or on the metal A surface. Will metal C
whether the reaction is spontaneous:
cause formation of H2 when placed in acid solution? Rank met-
(a) Co(s) H(aq) ±£ Co2(aq) H2(g)
als A, B, and C in order of decreasing reducing strength.
(b) Mn2(aq) Br2(l) ±£ MnO4(aq) Br(aq) [acidic]
21.51 When a clean iron nail is placed in an aqueous solution of
(c) Hg22(aq) ±£ Hg2(aq) Hg(l)
copper(II) sulfate, the nail becomes coated with a brownish
black material.
whether the reaction is spontaneous:
(a) What is the material coating the iron?
(a) Cl2(g) Fe2(aq) ±£ Cl(aq) Fe3(aq)
(b) What are the oxidizing and reducing agents?
(b) Mn2(aq) Co3(aq) ±£ MnO2(s) Co2(aq) [acidic]
(c) Can this reaction be made into a voltaic cell?
(c) AgCl(s) NO(g) ±£
(d) Write the balanced equation for the reaction.
Ag(s) Cl(aq) NO3(aq) [acidic]
(e) Calculate E°cell for the process.
Problems 975
Free Energy and Electrical Work (a) What is the initial Ecell?
(Sample Problems 21.5 to 21.7) (b) What is [Ni2] when Ecell reaches 0.03 V?
(c) What are the equilibrium concentrations of the ions?
Concept Review Questions 21.71 A voltaic cell with Mn/Mn2 and Cd/Cd2 half-cells has the
21.52 (a) How do the relative magnitudes of Q and K relate to the following initial concentrations: [Mn2] 0.090 M; [Cd2]
signs of G and Ecell? Explain.
0.060 M.
(b) Can a cell do work when Q/K 1 or Q/K 1? Explain.
(a) What is the initial Ecell?
21.53 A voltaic cell consists of a metal A/A electrode and a metal (b) What is Ecell when [Cd2] reaches 0.050 M?
B/B electrode, with the A/A electrode negative. The initial
(c) What is [Mn2] when Ecell reaches 0.055 V?
[A]/[B] is such that Ecell E°cell.
(d) What are the equilibrium concentrations of the ions?
(a) How do [A] and [B] change as the cell operates?
(b) How does Ecell change as the cell operates? 21.72 A concentration cell consists of two H2/H half-cells. Half-
(c) What is [A]/[B] when Ecell E°cell? Explain. cell A has H2 at 0.95 atm bubbling into 0.10 M HCl. Half-cell B
(d) Is it possible for Ecell to be less than E°cell? Explain. has H2 at 0.60 atm bubbling into 2.0 M HCl. Which half-cell
21.54 Explain whether Ecell of a voltaic cell will increase or de- houses the anode? What is the voltage of the cell?
crease with each of the following changes: 21.73 A concentration cell consists of two Sn/Sn2 half-cells. The
(a) Decrease in cell temperature electrolyte in compartment A is 0.13 M Sn(NO3)2. The elec-
(b) Increase in [active ion] in the anode compartment trolyte in B is 0.87 M Sn(NO3)2. Which half-cell houses the
(c) Increase in [active ion] in the cathode compartment cathode? What is the voltage of the cell?
(d) Increase in pressure of a gaseous reactant in the cathode com-
partment Electrochemical Processes in Batteries
21.55 In a concentration cell, is the more concentrated electrolyte Concept Review Questions
in the cathode or the anode compartment? Explain. 21.74 What is the direction of electron flow with respect to the an-
Skill-Building Exercises (grouped in similar pairs) ode and the cathode in a battery? Explain.
21.56 What is the value of the equilibrium constant for the reaction 21.75 In the everyday batteries used for flashlights, toys, etc., no
between each pair at 25°C? salt bridge is evident. What is used in these cells to separate the
(a) Ni(s) and Ag(aq) (b) Fe(s) and Cr3(aq) anode and cathode compartments?
21.57 What is the value of the equilibrium constant for the reaction 21.76 Both a D-sized and an AAA-sized alkaline battery have an
between each pair at 25°C? output of 1.5 V. What property of the cell potential allows this to
(a) Al(s) and Cd2(aq) (b) I2(s) and Br(aq) occur? What is different about these two batteries?
21.58 What is the value of the equilibrium constant for the reaction Problems in Context
between each pair at 25°C?
21.77 Many common electrical devices require the use of more
than one battery.
(a) Ag(s) and Mn2(aq) (b) Cl2(g) and Br(aq)
(a) How many alkaline batteries must be placed in series to light
21.59 What is the value of the equilibrium constant for the reaction
a flashlight with a 6.0-V bulb?
between each pair at 25°C?
(b) What is the voltage requirement of a camera that uses six sil-
(a) Cr(s) and Cu2(aq) (b) Sn(s) and Pb2(aq)
ver batteries?
21.60 Calculate G° for each of the reactions in Problem 21.56. (c) How many volts can a car battery deliver if two of its
21.61 Calculate G° for each of the reactions in Problem 21.57. anode/cathode cells are shorted?
21.62 Calculate G° for each of the reactions in Problem 21.58. Corrosion: A Case of Environmental Electrochemistry
21.63 Calculate G° for each of the reactions in Problem 21.59.
Concept Review Questions
21.64 What are E°cell and G° of a redox reaction at 25°C for 21.78 During reconstruction of the Statue of Liberty, Teflon spac-
which n 1 and K 5.0 104? ers were placed between the iron skeleton and the copper plates
21.65 What are E°cell and G° of a redox reaction at 25°C for which that cover the statue. What purpose do these spacers serve?
n 1 and K 5.0 106? 21.79 Why do steel bridge-supports rust at the waterline but not
21.66 What are E°cell and G° of a redox reaction at 25°C for above or below it?
which n 2 and K 65? 21.80 After the 1930s, chromium replaced nickel for corrosion re-
21.67 What are E°cell and G° of a redox reaction at 25°C for which sistance and appearance on car bumpers and trim. How does
n 2 and K 0.065? chromium protect steel from corrosion?
21.81 Which of the following metals are suitable for use as sacri-
21.68 A voltaic cell consists of a standard hydrogen electrode in ficial anodes to protect against corrosion of underground iron
one half-cell and a Cu/Cu2 half-cell. Calculate [Cu2] when pipes? If any are not suitable, explain why:
Ecell is 0.22 V. (a) Aluminum
21.69 A voltaic cell consists of an Mn/Mn2 half-cell and a (b) Magnesium
Pb/Pb2 half-cell. Calculate [Pb2] when [Mn2] is 1.4 M and (c) Sodium
Ecell is 0.44 V. (d) Lead
21.70 A voltaic cell with Ni/Ni2 and Co/Co2 half-cells has the (e) Nickel
following initial concentrations: [Ni2] 0.80 M; [Co2] (f) Zinc
0.20 M. (g) Chromium
Electrolytic Cells: Using Electrical Energy to Drive 21.96 Identify those elements that can be prepared by electrolysis
Nonspontaneous Reactions of their aqueous salts: fluorine, manganese, iron, cadmium.
(Sample Problems 21.8 to 21.10) 21.97 What product forms at each electrode in the aqueous elec-
Concept Review Questions trolysis of the following salts: (a) LiF; (b) SnSO4?
Note: Unless stated otherwise, assume that the electrolytic 21.98 What product forms at each electrode in the aqueous elec-
cells in the following problems operate at 100% efficiency. trolysis of the following salts: (a) ZnBr2; (b) Cu(HCO3)2?
21.82 Consider the following general electrolytic cell: 21.99 What product forms at each electrode in the aqueous elec-
trolysis of the following salts: (a) Cr(NO3)3; (b) MnCl2?
Power 21.100 What product forms at each electrode in the aqueous elec-
supply trolysis of the following salts: (a) FeI2; (b) K3PO4?
(+) (–)
21.101 Electrolysis of molten MgCl2 is the final production step in
the isolation of magnesium from seawater by the Dow process
(Section 22.4). Assuming that 45.6 g of Mg metal forms,
Molten (a) How many moles of electrons are required?
MX (b) How many coulombs are required?
(c) How many amps will produce this amount in 3.50 h?
21.102 Electrolysis of molten NaCl in a Downs cell is the major
(a) At which electrode does oxidation occur? isolation step in the production of sodium metal (Section 22.4).
(b) At which electrode does elemental M form? Assuming that 215 g of Na metal forms,
(c) At which electrode are electrons being released by ions? (a) How many moles of electrons are required?
(d) At which electrode are electrons entering the cell? (b) How many coulombs are required?
21.83 A voltaic cell consists of Cr/Cr3 and Cd/Cd2 half-cells (c) How many amps will produce this amount in 9.50 h?
with all components in their standard states. After 10 minutes of 21.103 How many grams of radium can form by passing 235 C
operation, a thin coating of cadmium metal has plated out on the through an electrolytic cell containing a molten radium salt?
cathode. Describe what will happen if you attach the negative 21.104 How many grams of aluminum can form by passing 305 C
terminal of a dry cell (1.5 V) to the cell cathode and the positive through an electrolytic cell containing a molten aluminum salt?
terminal to the cell anode.
21.84 Why are Ehalf-cell values for the oxidation and reduction of 21.105 How many seconds does it take to deposit 65.5 g of Zn on
water different from E°half-cell values for the same processes? a steel gate when 21.0 A is passed through a ZnSO4 solution?
21.85 In an aqueous electrolytic cell, nitrate ions never react at the 21.106 How many seconds does it take to deposit 1.63 g of Ni on
anode, but nitrite ions do. Explain. a decorative drawer handle when 13.7 A is passed through a
21.86 How does overvoltage influence the products in the elec- Ni(NO3)2 solution?
trolysis of aqueous salts? Problems in Context
Skill-Building Exercises (grouped in similar pairs) 21.107 A professor adds Na2SO4 to water to facilitate its electroly-
21.87 In the electrolysis of molten NaBr, sis in a lecture demonstration. (a) What is the purpose of the
(a) What product forms at the anode? Na2SO4? (b) Why is the water electrolyzed instead of the salt?
(b) What product forms at the cathode? 21.108 Subterranean brines in parts of the United States are rich in
21.88 In the electrolysis of molten BaI2, iodides and bromides and serve as an industrial source of these
(a) What product forms at the negative electrode? elements. In one recovery method, the brines are evaporated to
(b) What product forms at the positive electrode? dryness and then melted and electrolyzed. Which halogen is
more likely to form from this treatment? Why?
21.89 In the electrolysis of a molten mixture of KI and MgF2, 21.109 Zinc plating (galvanizing) is an important means of corro-
identify the product that forms at the anode and at the cathode. sion protection. Although the process is done customarily by
21.90 In the electrolysis of a molten mixture of CsBr and SrCl2, dipping the object into molten zinc, the metal can also be elec-
identify the product that forms at the negative electrode and at troplated from aqueous solutions. How many grams of zinc can
the positive electrode. be deposited on a steel tank from a ZnSO4 solution when a
21.91 In the electrolysis of a molten mixture of NaCl and CaBr2, 0.855-A current flows for 2.50 days?
identify the product that forms at the anode and at the cathode. Comprehensive Problems
21.92 In the electrolysis of a molten mixture of RbF and CaCl2,
identify the product that forms at the negative electrode and at 21.110 The MnO2 used in alkaline batteries can be produced by an
the positive electrode. electrochemical process of which one half-reaction is
Mn2(aq) 2H2O(l) ±£ MnO2(s) 4H(aq) 2e
21.93 Identify those elements that can be prepared by electrolysis If a current of 25.0 A is used, how many hours are needed to pro-
of their aqueous salts: copper, barium, aluminum, bromine.
duce 1.00 kg of MnO2? At which electrode is the MnO2 formed?
21.94 Identify those elements that can be prepared by electrolysis 21.111 Car manufacturers are developing engines that use H2 as
of their aqueous salts: strontium, gold, tin, chlorine.
fuel. In Iceland, Sweden, and other parts of Scandinavia, where
21.95 Identify those elements that can be prepared by electrolysis hydroelectric plants produce inexpensive electric power, the H2
of their aqueous salts: lithium, iodine, zinc, silver. can be made industrially by the electrolysis of water.
siL48593_ch21_922-979 12:11:07 09:22am Page 977
Problems 977
(a) How many coulombs are needed to produce 3.5106 L of H2 operates for 10 h with an average current of 5.8 A. If 53.4 g of
gas at 12.0 atm and 25°C? (Assume the ideal gas law applies.) copper is deposited, at what efficiency is the cell operating?
(b) If the coulombs are supplied at 1.44 V, how many joules are 21.121 Commercial electrolysis is performed on both molten NaCl
produced? and aqueous NaCl solutions. Identify the anode product, cathode
(c) If the combustion of oil yields 4.0104 kJ/kg, what mass of product, species reduced, and species oxidized for the (a) molten
oil must be burned to yield the number of joules in part (b)? electrolysis and (b) aqueous electrolysis.
21.112 The overall cell reaction occurring in an alkaline battery is 21.122 To examine the effect of ion removal on cell voltage, a
Zn(s) MnO2(s) H2O(l) ±£ ZnO(s) Mn(OH)2(s) chemist constructs two voltaic cells, each with a standard hydro-
(a) How many moles of electrons flow per mole of reaction? gen electrode in one compartment. One cell also contains a
(b) If 4.50 g of zinc is oxidized, how many grams of manganese Pb/Pb2 half-cell; the other contains a Cu/Cu2 half-cell.
dioxide and of water are consumed? (a) What is E° of each cell at 298 K?
(c) What is the total mass of reactants consumed in part (b)? (b) Which electrode in each cell is negative?
(d) How many coulombs are produced in part (b)? (c) When Na2S solution is added to the Pb2 electrolyte, solid
(e) In practice, voltaic cells of a given capacity (coulombs) are PbS forms. What happens to the cell voltage?
heavier than the calculation in part (c) indicates. Explain. (d) When sufficient Na2S is added to the Cu2 electrolyte, CuS
forms and [Cu2] drops to 11016 M. Find the cell voltage.
21.113 An inexpensive and accurate method of measuring the
21.123 Electrodes used in electrocardiography are disposable, and
quantity of electricity passing through a circuit is to pass it
many incorporate silver. The metal is deposited in a thin layer on
through a solution of a metal ion and weigh the metal deposited.
a small plastic “button,” and then some is converted to AgCl:
A silver electrode immersed in an Ag solution weighs 1.7854 g
before the current passes and weighs 1.8016 g after the current Ag(s) Cl(aq) B A AgCl(s) e
has passed. How many coulombs have passed? (a) If the surface area of the button is 2.0 cm2 and the thickness
21.114 Brass, an alloy of copper and zinc, can be produced by si- of the silver layer is 7.5106 m, calculate the volume (in cm3)
multaneously electroplating the two metals from a solution con- of Ag used in one electrode.
taining their 2 ions. If 65.0% of the total current is used to (b) The density of silver metal is 10.5 g/cm3. How many grams
plate copper, while 35.0% goes to plating zinc, what is the mass of silver are used per electrode?
percent of copper in the brass? (c) If Ag is plated on the button from an Ag solution with a cur-
21.115 Compare and contrast a voltaic cell and an electrolytic cell rent of 12.0 mA, how many minutes does the plating take?
with respect to each of the following: (d) If bulk silver costs $13.00 per troy ounce (31.10 g), what is
(a) Sign of the free energy change the cost (in cents) of the silver in one disposable electrode?
(b) Nature of the half-reaction at the anode 21.124 Commercial aluminum production is done by the electrol-
(c) Nature of the half-reaction at the cathode ysis of a bath containing Al2O3 dissolved in molten Na3AlF6.
(d) Charge on the electrode labeled “anode” Why isn’t it done by electrolysis of an aqueous AlCl3 solution?
(e) Electrode from which electrons leave the cell 21.125 Comparing the standard electrode potentials (E°) of the
21.116 A thin circular-disk earring 4.00 cm in diameter is plated Group 1A(1) metals Li, Na, and K with the negative of their first
with a coating of gold 0.25 mm thick from an Au3 bath. ionization energies reveals a discrepancy:
(a) How many days does it take to deposit the gold on one side Ionization process reversed: M(g) e B A M(g) (IE)
of one earring if the current is 0.013 A (d of gold 19.3 g/cm3)? Electrode reaction: M(aq) e B A M(s) (E°)
(b) How many days does it take to deposit the gold on both sides
of the pair of earrings? Metal IE (kJ/mol) E (V)
(c) If the price of gold is $320 per troy ounce (31.10 g), what is
the total cost of the gold plating? Li 520 3.05
21.117 (a) How many minutes does it take to form 10.0 L of O2 Na 496 2.71
measured at 99.8 kPa and 28°C from water if a current of 1.3 A K 419 2.93
passes through the electrolytic cell? (b) What mass of H2 forms?
21.118 Trains powered by electricity, including subways, use direct Note that the electrode potentials do not decrease smoothly
current. One conductor is the overhead wire (or “third rail” for down the group, as the ionization energies do. You might expect
subways), and the other is the rails upon which the wheels run. that if it is more difficult to remove an electron from an atom to
The rails are on supports in contact with the ground. To mini- form a gaseous ion (larger IE), then it would be less difficult to
mize corrosion, should the overhead wire or the rails be con- add an electron to an aqueous ion to form an atom (smaller E°),
nected to the positive terminal? Explain. yet Li(aq) is more difficult to reduce than Na(aq). Applying
21.119 A silver button battery used in a watch contains 0.75 g of Hess’s law, use an approach similar to that for a Born-Haber cy-
zinc and can run until 80% of the zinc is consumed. (a) How cle to break down the process occurring at the electrode into
many days can the battery run at a current of 0.85 microamps three steps and label the energy involved in each step. How can
(106 amps)? (b) When the battery dies, 95% of the Ag2O has you account for the discrepancy?
been consumed. How many grams of Ag was used to make the 21.126 To improve conductivity in the electroplating of automo-
battery? (c) If Ag costs $13.00 per troy ounce (31.10 g), what is bile bumpers, a thin coating of copper separates the steel from a
the cost of the Ag consumed each day the watch runs? heavy coating of chromium.
21.120 Like any piece of apparatus, an electrolytic cell operates at (a) What mass of Cu is deposited on an automobile trim piece if
less than 100% efficiency. A cell depositing Cu from a Cu2 bath plating continues for 1.25 h at a current of 5.0 A?
(b) If the area of the trim piece is 50.0 cm2, what is the thickness 21.136 You are investigating a particular chemical reaction. State
of the Cu coating (d of Cu 8.95 g/cm3)? all the types of data available in standard tables that enable you
21.127 In Appendix D, standard electrode potentials range from to calculate the equilibrium constant for the reaction at 298 K.
about 3 to 3 V. Thus, it might seem possible to use a half-cell 21.137 In an electric power plant, personnel monitor the O2 content
from each end of this range to construct a cell with a voltage of of boiler feed water to prevent corrosion of the boiler tubes. Why
approximately 6 V. However, most commercial aqueous voltaic does Fe corrode faster in steam and hot water than in cold water?
cells have E° values of 1.5 to 2 V. Why are there no aqueous cells 21.138 A voltaic cell using Cu/Cu2 and Sn/Sn2 half-cells is set
with significantly higher potentials? up at standard conditions, and each compartment has a volume
21.128 Tin is used to coat “tin” cans used for food storage. If the of 345 mL. The cell delivers 0.17 A for 48.0 h. (a) How many
tin is scratched and the iron of the can exposed, will the iron cor- grams of Cu(s) are deposited? (b) What is the [Cu2] remaining?
rode more or less rapidly than if the tin were not present? Inside 21.139 The Ehalf-cell for the reduction of water is very different
the can, the tin itself is coated with a clear varnish. Explain. from the E°half-cell value. Calculate the Ehalf-cell value when H2 is
21.129 Commercial electrolytic cells for producing aluminum op- in its standard state.
erate at 5.0 V and 100,000 A. 21.140 From the skeleton equations below, create a list of balanced
(a) How long does it take to produce exactly 1 metric ton half-reactions in which the strongest oxidizing agent is on top
(1000 kg) of aluminum? and the weakest is on the bottom:
(b) How much electrical power (in kilowatt-hours, kWh) is U3(aq) Cr3(aq) ±£ Cr2(aq) U4(aq)
used [1 W 1 J/s; 1 kWh 3.6 103 kJ]? Fe(s) Sn2(aq) ±£ Sn(s) Fe2(aq)
(c) If electricity costs 0.90¢ per kWh and cell efficiency is Fe(s) U4(aq) ±£ no reaction
90.%, what is the cost of electricity to produce exactly 1 lb of Cr3(aq) Fe(s) ±£ Cr2(aq) Fe2(aq)
aluminum? Cr2(aq) Sn2(aq) ±£ Sn(s) Cr3(aq)
21.130 Magnesium bars are connected electrically to underground 21.141 You are given the following three half-reactions:
iron pipes to serve as sacrificial anodes. (1) Fe3(aq) e B A Fe2(aq)

(a) Do electrons flow from the bar to the pipe or the reverse? (2) Fe (aq) 2e B
2
A Fe(s)
(b) A 12-kg Mg bar is attached to an iron pipe, and it takes 8.5 yr (3) Fe3(aq) 3e B A Fe(s)
for the Mg to be consumed. What is the average current flowing (a) Use E°half-cell values for (1) and (2) to find E°half-cell for (3).
between the Mg and the Fe during this period? (b) Calculate ΔG° for (1) and (2) from their E°half-cell values.
21.131 Bubbles of H2 form when metal D is placed in hot H2O. (c) Calculate ΔG° for (3) from (1) and (2).
No reaction occurs when D is placed in a solution of a salt of metal (d) Calculate E°half-cell for (3) from its ΔG°.
E, but D is discolored and coated immediately when placed in a (e) What is the relationship between the E°half-cell values for (1)
solution of a salt of metal F. What happens if E is placed in a so- and (2) and the E°half-cell value for (3)?
lution of a salt of metal F? Rank metals D, E, and F in order of 21.142 Use the half-reaction method to balance the equation for
increasing reducing strength. the conversion of ethanol to acetic acid in acid solution:
21.132 Calcium is obtained industrially by electrolysis of molten CH3CH2OH Cr2O72 ±£ CH3COOH Cr3
CaCl2 and is used in aluminum alloys. How many coulombs are 21.143 When zinc is refined by electrolysis, the desired half-
needed to produce 10.0 g of Ca metal? If a cell runs at 15 A, how reaction at the cathode is
many minutes will it take to produce 10.0 g of Ca(s)? Zn2(aq) 2e ±£ Zn(s)
21.133 In addition to reacting with gold (see Problem 21.21), aqua
A competing reaction, which lowers the yield, is the formation
regia is used to bring other precious metals into solution. Bal-
of hydrogen gas:
ance the skeleton equation for the reaction with Pt:
2H(aq) 2e ±£ H2(g)
Pt(s) NO3(aq) Cl(aq) ±£ PtCl62(aq) NO(g) If 91.50% of the current flowing results in zinc being deposited,
21.134 The following reactions are used in batteries: while 8.50% produces hydrogen gas, how many liters of H2,
I 2H2(g) O2(g) ±£ 2H2O(l) Ecell 1.23 V measured at STP, form per kilogram of zinc?
II Pb(s) PbO2(s) 2H2SO4(aq) ±£ 21.144 A chemist designs an ion-specific probe for measuring
2PbSO4(s) 2H2O(l) Ecell 2.04 V [Ag] in an NaCl solution saturated with AgCl. One half-cell
III 2Na(l) FeCl2(s) ±£ 2NaCl(s) Fe(s) Ecell 2.35 V has an Ag-wire electrode immersed in the unknown AgCl-
Reaction I is used in fuel cells, II in the automobile lead-acid saturated NaCl solution. It is connected through a salt bridge to
battery, and III in an experimental high-temperature battery for the other half-cell, which has a calomel reference electrode
powering electric vehicles. The aim is to obtain as much work as [a platinum wire immersed in a paste of mercury and calomel
possible from a cell, while keeping its weight to a minimum. (Hg2Cl2)] in a saturated KCl solution. The measured Ecell is
(a) In each cell, find the moles of electrons transferred and G. 0.060 V.
(b) Calculate the ratio, in kJ/g, of wmax to mass of reactants for (a) Given the following standard half-reactions, calculate [Ag].
each of the cells. Which has the highest ratio, which the lowest, Calomel: Hg2Cl2(s) 2e ±£
and why? (Note: For simplicity, ignore the masses of cell com- 2Hg(l) 2Cl(aq) E° 0.24 V

ponents that do not appear in the cell as reactants, including elec- Silver: Ag (aq) e ±£ Ag(s) E° 0.80 V
trode materials, electrolytes, separators, cell casing, wiring, etc.) (Hint: Assume [Cl] is so high that it is essentially constant.)
21.135 A current is applied to two electrolytic cells in series. In the (b) A mining engineer wants an ore sample analyzed with the
first, silver is deposited; in the second, a zinc electrode is con- Ag-selective probe. After pretreating the ore sample, the
sumed. How much Ag is plated out if 1.2 g of Zn dissolves? chemist measures the cell voltage as 0.53 V. What is [Ag]?
Problems 979
21.145 Use Appendix D to calculate the Ksp of AgCl. (b) Calculate Ecell when an additional 10.0 mL of 0.500 M NH3
21.146 Black-and-white photographic film is coated with silver is added.
halides. Because silver is expensive, the manufacturer monitors (c) In the second concentration cell, 10.0 mL of 0.500 M NaOH
the Ag content of the waste stream, [Ag]waste, from the plant is added to one half-cell; the precipitate Cu(OH)2 forms (Ksp
with an Ag-selective electrode at 25°C. A stream of known 2.2 1020). Calculate E°cell.
Ag concentration, [Ag]standard, is passed over the electrode in (d) What would the molarity of NaOH have to be for the addition
turn with the waste stream and the data recorded by a computer. of 10.0 mL to result in an E°cell of 0.340 V?
(a) Write the equations relating the nonstandard cell potential to 21.154 Two voltaic cells are to be joined so that one will run the
the standard cell potential and [Ag] for each solution. other as an electrolytic cell. In the first cell, one half-cell has Au
(b) Combine these into a single equation to find [Ag]waste. foil in 1.00 M Au(NO3)3, and the other half-cell has a Cr bar in
(c) Rewrite the equation from part (b) to find [Ag]waste in ng/L. 1.00 M Cr(NO3)3. In the second cell, one half-cell has a Co bar
(d) If Ewaste is 0.003 V higher than Estandard, and the standard so- in 1.00 M Co(NO3)2, and the other half-cell has a Zn bar in
lution contains 1000. ng/L, what is [Ag]waste? 1.00 M Zn(NO3)2. (a) Calculate E°cell for each cell. (b) Calculate
(e) Rewrite the equation in part (b) to find [Ag]waste for a sys- the total potential if the two cells are connected as voltaic cells in
tem in which T changes and Twaste and Tstandard may be different. series. (c) When the electrode wires are switched in one of the
21.147 Calculate the Kf of Ag(NH3)2 from cells, which cell will run as the voltaic cell and which as the elec-
Ag(aq) e B A Ag(s) E° 0.80 V trolytic cell? (d) Which metal ion is being reduced in each cell?
(e) If 2.00 g of metal plates out in the voltaic cell, how much
Ag(NH3)2 (aq) e B A Ag(s) 2NH3(aq) E° 0.37 V
21.148 Even though the toxicity of cadmium has become a con- metal ion plates out in the electrolytic cell?
cern, nickel-cadmium (nicad) batteries are still used commonly 21.155 A voltaic cell has one half-cell with a Cu bar in a 1.00 M
in many devices. The overall cell reaction is Cu2salt, and the other half-cell with a Cd bar in the same vol-
Cd(s) 2NiO(OH)(s) 2H2O(l) ±£ ume of a 1.00 M Cd2salt. (a) Find E°cell, ΔG°, and K. (b) As the
2Ni(OH)(s) Cd(OH)2(s) cell operates, [Cd2] increases; find Ecell and ΔG when [Cd2] is
1.95 M. (c) Find Ecell, ΔG, and [Cu2] at equilibrium.
A certain nicad battery weighs 18.3 g and has a capacity of
300. mAh (that is, the cell can store charge equivalent to a cur- 21.156 Gasoline is a mixture of hydrocarbons, but the heat released
rent of 300. mA flowing for 1 h). when it burns is close to that of octane, C8H18(l) (ΔH°f
(a) What is the capacity of this cell in coulombs? 250.1 kJ/mol). As an alternative to gasoline, research is underway
(b) What mass of reactants is needed to deliver 300. mAh? to use H2 from the electrolysis of water in fuel cells to power cars.
(c) What percentage of the cell mass consists of reactants? (a) Calculate ΔH° when 1.00 gal of gasoline (d 0.7028 g/mL)
21.149 The zinc-air battery is a less expensive alternative for silver burns to produce carbon dioxide gas and water vapor.
batteries in hearing aids. The cell reaction is (b) How many liters of H2 at 25°C and 1.00 atm must burn to
produce this quantity of energy?
2Zn(s) O2(g) ±£ 2ZnO(s) (c) How long would it take to produce this amount of H2 by elec-
1
A new battery weighs 0.275 g. The zinc accounts for exactly 10 trolysis with a current of 1.00 103 A at 6.00 V?
of the mass, and the oxygen does not contribute to the mass be- (d) How much power in kilowatthours (kWh) is required to
cause it is supplied by the air. generate this amount of H2? (1 W 1 J/s, 1 J 1 CV, and
(a) How much electricity (in C) can the battery deliver? 1 kWh 3.6 106 J.)
(b) How much free energy (in J) is released if Ecell is 1.3 V? (e) If the cell is 88.0% efficient and electricity costs 0.950¢ per
21.150 Use Appendix D to create an activity series of Mn, Fe, Ag, kWh, what is the cost of producing the amount of H2 equivalent
Sn, Cr, Cu, Ba, Al, Na, Hg, Ni, Li, Au, Zn, and Pb. Rank these to 1.00 gal of gasoline?
metals in order of decreasing reducing strength, and divide them 21.157 Balance the following redox reactions:
into three groups: those that displace H2 from water, those that (a) In acidic solution, xenon trioxide reacts with iodide ion to
displace H2 from acid, and those that cannot displace H2. produce xenon gas, triiodide ion (I3), and water.
21.151 Both Ti and V are reactive enough to displace H2 from (b) In basic solution, the hydrogen xenate ion (HXeO4) dispro-
water. The difference in their E°half-cell values is 0.43 V. Given portionates to xenon gas, perxenate ion (XeO64), water, and
V(s) Cu2(aq) ±£ V2(aq) Cu(s) ΔG° 298 kJ/mol oxygen gas.
use Appendix D to calculate the E°half-cell values for V and Ti. (c) In basic solution, bismuthate ion (BiO3) reacts with man-
21.152 For the reaction ganese(II) ion to produce bismuth(III) and permanganate ions.
S4O62(aq) 2I(aq) ±£ I2(s) S2O32(aq) (d) In basic solution, oxygen difluoride reacts to produce fluo-
ΔG° 87.8 kJ/mol ride ion, oxygen gas, and water.
(a) Identify the oxidizing and reducing agents. (b) Calculate (e) In acidic solution, manganese(IV) oxide reacts with sulfite
E°cell. (c) For the reduction half-reaction, write a balanced equa- ion to form manganese(II) ion, water, and dithionate ion
tion, give the oxidation number of each element, and calculate (S2O62).
E°half-cell. (f) In one of the few reactions known for astatine, the oxidation
21.153 Two concentration cells are prepared, both with 90.0 mL of astinide ion by chlorine gas in acid solution produces astinate
of 0.0100 M Cu(NO3)2 and a Cu bar in each half-cell. ion (AtO3) and chloride ion.
(a) In the first concentration cell, 10.0 mL of 0.500 M NH3 is 21.158 If the Ecell of the following cell is 0.915 V, what is the pH in
added to one half-cell; the complex ion Cu(NH3)42 forms, and the anode compartment?
Ecell is 0.129 V. Calculate Kf for the formation of the complex ion. Pt(s) H2(1.00 atm) H(aq) Ag(0.100 M) Ag(s)
siL48593_appe_A015-A048 5:12:07 02:03am Page 41
Appendix E Answers to Selected Problems A-41
Maximum slope (equivalence point) is at Vavg = 40.00 mL. 93.1 JK; yes, the positive sign of
S is expected because there is
[BH][OH] a net increase in the number of gas molecules. 20.37
S
19.128 Kb
[B] 311 JK; yes, the negative entropy change matches the de-
Rearranging to isolate [OH]: crease in moles of gas. 20.39 75.6 JK 20.41 242 JK
[B] 20.44 97.2 JK 20.46 A spontaneous process has
Suniv 0.
[OH] Kb Since the absolute temperature is always positive,
Gsys must be
[BH]
Taking the negative log: negative (
Gsys 0) for a spontaneous process. 20.48
Hrxn is
[B] positive and
Ssys is positive. Melting is an example. 20.49 The
log [OH] log Kb log entropy changes little within a phase. As long as the substance
[BH]
[BH] does not change phase, the value of
S is relatively unaffected
Therefore, pOH pKb log by temperature. 20.50(a) 1138.0 kJ (b) 1379.4 kJ
[B]
19.129 4.05 19.136 H2CO3HCO3 and H2PO4HPO42; (c) 224 kJ 20.52(a) 1138 kJ (b) 1379 kJ (c) 226 kJ
[HPO42][H2PO4] 5.8 19.139 3.8 19.141(a) 58.2 mL 20.54(a) Entropy decreases (
S is negative) because the number
(b) 7.4 mL (c) 6.30 19.143 170 mL 19.146(a) 65 mol of moles of gas decreases. The combustion of CO releases en-
(b) 6.28 (c) 4.0103 g 19.148 5.68 19.150 3.9109 g ergy (
H is negative). (b) 257.2 kJ or 257.3 kJ, depending
Pb2100 mL blood 19.152 No NaCl will precipitate. on the method 20.56(a) 0.409 kJmol K (b) 197 kJmol
19.153(a) A and D (b) pHA 4.35; pHB 8.67; pHC 2.67; 20.58(a)
Hrxn 90.7 kJ;
Srxn 221 JK (b) at 28C,
G
pHD 4.57 (c) C, A, D, B (d) B 24.3 kJ; at 128C,
G 2.2 kJ; at 228C,
G 19.9 kJ
(c) For the substances in their standard states, the reaction is non-
spontaneous at 28C, near equilibrium at 128C, and spontaneous
Chapter 20
at 228C. 20.60
H 30910 J,
S 93.15 JK, T 331.8 K
20.2 A spontaneous process occurs by itself, whereas a nonspon- 20.62(a)
Hrxn 241.826 kJ,
Srxn 44.4 JK,
Grxn
taneous process requires a continuous input of energy to make it 228.60 kJ (b) Yes. The reaction will become nonsponta-
happen. It is possible to cause a nonspontaneous process to neous at higher temperatures. (c) The reaction is spontaneous
occur, but the process stops once the energy source is removed. below 5.45103 K. 20.64(a)
G is a relatively large positive
A reaction that is nonspontaneous under one set of conditions value. (b) K 1. Q depends on initial conditions, not equi-
may be spontaneous under a different set of conditions. librium conditions. 20.67 The standard free energy change,
20.5 The transition from liquid to gas involves a greater increase
G, applies when all components of the system are in their stan-
in dispersal of energy and freedom of motion than does the tran- dard states;
G
G when all concentrations equal 1 M and all
sition from solid to liquid. 20.6 In an exothermic reaction, partial pressures equal 1 atm. 20.68(a) 1.7106

Ssurr 0. In an endothermic reaction,
Ssurr 0. A chemical (b) 3.891034 (c) 1.261048 20.70(a) 6.5710173
cold pack for injuries is an example of an application using a (b) 4.461015 (c) 3.46104 20.72 4.891051
spontaneous endothermic process. 20.8(a), (b), and (c) 20.74 3.36105 20.76 2.7104 Jmol; no
20.10(a) and (b) 20.12(a) positive (b) negative (c) negative 20.78(a) 2.9104 Jmol (b) The reverse direction, formation
20.14(a) positive (b) positive (c) positive 20.16(a) negative of reactants, is spontaneous, so the reaction proceeds to the left.
(b) negative (c) positive 20.18(a) positive (b) negative (c) 7.0103 Jmol; the reaction proceeds to the left to reach
(c) positive 20.20(a) positive (b) negative (c) positive equilibrium. 20.80(a) no T (b) 163 kJ (c) 1102 kJmol
20.22(a) Butane. The double bond in 2-butene restricts freedom 20.83(a) spontaneous (b) (c) (d) (e) , not spon-
of rotation. (b) Xe(g). It has the greater molar mass. (c) CH4(g). taneous (f) 20.87(a) 2.3102 (b) Administer oxygen-rich
Gases have greater entropy than liquids. 20.24(a) C2H5OH(l). It air to counteract the CO poisoning. 20.90 370. kJ
is a more complex molecule. (b) KClO3(aq). Ions in solution 20.92(a) 2N2O5(g) 6F2(g) ±£ 4NF3(g) 5O2(g)
have their energy more dispersed than those in a solid. (c) K(s). (b)
Grxn 569 kJ (c)
Grxn 5.60102 kJmol
It has a greater molar mass. 20.26(a) Diamond graphite 20.95
Hrxn 137.14 kJ;
Srxn 120.3 JK;
Grxn
charcoal. Freedom of motion is least in the network solid; more 101.25 kJ 20.100(a) 1.2031023 molecules ATPg glucose
freedom between graphite sheets; most freedom in amorphous (b) 3.0731023 molecules ATPg tristearin 20.106(a) 1.67103
solid. (b) Ice liquid water water vapor. Entropy increases Jmol (b) 7.37103 Jmol (c) 4.04103 Jmol (d) 0.19
as a substance changes from solid to liquid to gas. (c) O atoms 20.110(a) 465 K (b) 6.59104 (c) The reaction rate is
O2 O3. Entropy increases with molecular complexity. higher at the higher temperature. The shorter time required
20.28(a) ClO4(aq) ClO3(aq) ClO2(aq); decreasing mo- (kinetics) overshadows the lower yield (thermodynamics).
lecular complexity (b) NO2(g) NO(g) N2(g). N2 has lower
standard molar entropy because it consists of two of the same Chapter 21
atoms; the other species have two different types of atoms. NO2
is more complex than NO. (c) Fe3O4(s) Fe2O3(s) 21.1 Oxidation is the loss of electrons and results in a higher oxi-
Al2O3(s). Fe3O4 is more complex and more massive. Fe2O3 is dation number; reduction is the gain of electrons and results in a
more massive than Al2O3. 20.31 For a system at equilibrium, lower oxidation number. 21.3 No, one half-reaction cannot take

Suniv
Ssys
Ssurr 0. For a system moving to equilib- place independently because there is a transfer of electrons from
rium,
Suniv 0. 20.32 S°Cl2O(g) 2S°HClO(g) S°H2O(g) ¢S°rxn one substance to another. If one substance loses electrons, an-
other substance must gain them. 21.6 To remove H ions from
20.33(a) negative;
S 172.4 JK (b) positive;
S
an equation, add an equal number of OH ions to both sides to
141.6 JK (c) negative;
S 837 JK 20.35
S
neutralize the H ions and produce water. 21.8 Spontaneous
A-42 Appendix E Answers to Selected Problems
reactions, for which

Gsys 0, take place in voltaic cells (also (b) e flow
called galvanic cells). Nonspontaneous reactions, for which

Gsys 0, take place in electrolytic cells. 21.10(a) Cl
(b) MnO4 (c) MnO4 (d) Cl (e) from Cl to MnO4 Zn
()
Sn
()
(f) 8H2SO4(aq) 2KMnO4(aq) 10KCl(aq) ±£
2MnSO4(aq) 5Cl2(g) 8H2O(l) 6K2SO4(aq) Anion Cation
21.12(a) ClO3(aq) 6H(aq) 6I(aq) ±£ flow flow
Cl(aq) 3H2O(l) 3I2(s) 1M 1M

Zn2 Sn2
Oxidizing agent is ClO3 and reducing agent is I.

(b) 2MnO4(aq) H2O(l) 3SO32(aq) ±£

2MnO2(s) 3SO42(aq) 2OH(aq)
Oxidizing agent is MnO4 and reducing agent is SO32. 21.29(a) left to right (b) left (c) right (d) Ni (e) Fe
(c) 2MnO4(aq) 6H(aq) 5H2O2(aq) ±£ (f) Fe (g) 1 M NiSO4 (h) K and NO3 (i) neither (j) from
2Mn2(aq) 8H2O(l) 5O2(g) right to left
Oxidizing agent is MnO4 and reducing agent is H2O2. (k) Oxidation: Fe(s) ±£ Fe2(aq) 2e
21.14(a) Cr2O72(aq) 14H(aq) 3Zn(s) ±£ Reduction: Ni2(aq) 2e ±£ Ni(s)
2Cr3(aq) 7H2O(l) 3Zn2(aq) Overall: Fe(s) Ni2(aq) ±£ Fe2(aq) Ni(s)
Oxidizing agent is Cr2O72 and reducing agent is Zn. 21.31(a) Reduction: Fe2(aq) 2e ±£ Fe(s)
(b) MnO4(aq) 3Fe(OH)2(s) 2H2O(l) ±£ Oxidation: Mn(s) ±£ Mn2(aq) 2e
MnO2(s) 3Fe(OH)3(s) OH(aq) Overall: Fe2(aq) Mn(s) ±£ Fe(s) Mn2(aq)
Oxidizing agent is MnO4 and reducing agent is (b) e flow
Fe(OH)2.
(c) 2NO3(aq) 12H(aq) 5Zn(s) ±£ Mn Fe
N2(g) 6H2O(l) 5Zn2(aq) () ()

Oxidizing agent is NO3 and reducing agent is Zn.
21.16(a) 4NO3(aq) 4H(aq) 4Sb(s) ±£ Anion
flow
Cation
flow
4NO(g) 2H2O(l) Sb4O6(s)
Oxidizing agent is NO3 and reducing agent is Sb. 1M 1M
Mn2 Fe2
(b) 5BiO3(aq) 14H(aq) 2Mn2(aq) ±£
5Bi3(aq) 7H2O(l) 2MnO4(aq)
Oxidizing agent is BiO3 and reducing agent is Mn2.
(c) Pb(OH)3(aq) 2Fe(OH)2(s) ±£ 21.33(a) Al(s) Al3(aq) Cr3(aq) Cr(s)
Pb(s) 2Fe(OH)3(s) OH(aq) (b) Pt(s) SO2(g) SO42(aq), H(aq) Cu2(aq) Cu(s)
Oxidizing agent is Pb(OH)3 and reducing agent is 21.36 A negative Ecell indicates that the redox reaction is not
Fe(OH)2. spontaneous, that is,
G 0. The reverse reaction is sponta-
21.18(a) 5As4O6(s) 8MnO4(aq) 18H2O(l) ±£ neous with Ecell 0. 21.37 Similar to other state functions, E
20AsO43(aq) 8Mn2(aq) 36H(aq) changes sign when a reaction is reversed. Unlike
G,
H, and
Oxidizing agent is MnO4 and reducing agent is As4O6. S, E (the ratio of energy to charge) is an intensive property.
(b) P4(s) 6H2O(l) ±£ When the coefficients in a reaction are multiplied by a factor, the
2HPO32(aq) 2PH3(g) 4H(aq) values of
G,
H, and S are multiplied by that factor. How-
P4 is both the oxidizing agent and the reducing agent. ever, E does not change because both the energy and charge are
(c) 2MnO4(aq) 3CN(aq) H2O(l) ±£ multiplied by the factor and thus their ratio remains unchanged.
2MnO2(s) 3CNO(aq) 2OH(aq) 21.38(a) Oxidation: Se2(aq) ±£ Se(s) 2e
Oxidizing agent is MnO4 and reducing agent is CN. Reduction: 2SO32(aq) 3H2O(l) 4e ±£
21.21(a) Au(s) 3NO3(aq) 4Cl(aq) 6H(aq) ±£ S2O32(aq) 6OH(aq)
AuCl4(aq) 3NO2(g) 3H2O(l) (b) Oxidizing agent is (b) Eanode Ecathode Ecell 0.57 V 0.35 V 0.92 V
NO3 and reducing agent is Au. (c) HCl provides chloride ions 21.40(a) Br2 Fe3 Cu2 (b) Ca2 Ag Cr2O72
that combine with the gold(III) ion to form the stable AuCl4 ion. 21.42(a) Co(s) 2H(aq) ±£ Co2(aq) H2(g)
21.22(a) A (b) E (c) C (d) A (e) E (f) E 21.25 An ac- Ecell 0.28 V; spontaneous
tive electrode is a reactant or product in the cell reaction. An (b) 2Mn2(aq) 5Br2(l) 8H2O(l) ±£
inactive electrode does not take part in the reaction and is present 2MnO4(aq) 10Br(aq) 16H(aq)
only to conduct a current. Platinum and graphite are commonly Ecell 0.44 V; not spontaneous
used as inactive electrodes. 21.26(a) A (b) B (c) A (c) Hg22(aq) ±£ Hg2(aq) Hg(l)
(d) Hydrogen bubbles will form when metal A is placed in acid. Ecell 0.07 V; not spontaneous
Metal A is a better reducing agent than metal B, so if metal B re- 21.44(a) 2Ag(s) Cu2(aq) ±£ 2Ag(aq) Cu(s)
duces H in acid, then metal A will also. Ecell 0.46 V; not spontaneous
21.27(a) Oxidation: Zn(s) ±£ Zn2(aq) 2e (b) Cr2O72(aq) 3Cd(s) 14H(aq) ±£
Reduction: Sn2(aq) 2e ±£ Sn(s) 2Cr3(aq) 3Cd2(aq) 7H2O(l)
Overall: Zn(s) Sn2(aq) ±£ Zn2(aq) Sn(s) Ecell 1.73 V; spontaneous
Appendix E Answers to Selected Problems A-43
(c) Pb(s) Ni2(aq) ±£ Pb2(aq) Ni(s) the anode; calcium metal forms at the cathode. 21.93 copper
Ecell 0.12 V; not spontaneous and bromine 21.95 iodine, zinc, and silver
21.46 3N2O4(g) 2Al(s) ±£ 6NO2(aq) 2Al3(aq) 21.97(a) Anode: 2H2O(l) ±£ O2(g) 4H(aq) 4e
Ecell 0.867 V (1.66 V) 2.53 V Cathode: 2H2O(l) 2e ±£ H2(g) 2OH(aq)
2Al(s) 3SO42(aq) 3H2O(l) ±£ (b) Anode: 2H2O(l) ±£ O2(g) 4H(aq) 4e
2Al3(aq) 3SO32(aq) 6OH(aq) Cathode: Sn2(aq) 2e ±£ Sn(s)
Ecell 2.59 V 21.99(a) Anode: 2H2O(l) ±£ O2(g) 4H(aq) 4e
SO42(aq) 2NO2(aq) H2O(l) ±£ Cathode: NO3(aq) 4H(aq) 3e ±£
SO32(aq) N2O4(g) 2OH(aq) NO(g) 2H2O(l)
Ecell 0.06 V (b) Anode: 2Cl(aq) ±£ Cl2(g) 2e
Oxidizing agents: Al3 N2O4 SO42 Cathode: 2H2O(l) 2e ±£ H2(g) 2OH(aq)
Reducing agents: SO32 NO2 Al 21.101(a) 3.75 mol e (b) 3.62105 C (c) 28.7 A
21.48 2HClO(aq) Pt(s) 2H(aq) ±£ 21.103 0.275 g Ra 21.105 9.20103 s 21.107(a) The sodium
Cl2(g) Pt2(aq) 2H2O(l) and sulfate ions make the water conductive so that current will
Ecell 0.43 V flow through the water, facilitating electrolysis. Pure water,
2HClO(aq) Pb(s) SO42(aq) 2H(aq) ±£ which contains very low (107 M) concentrations of H and
Cl2(g) PbSO4(s) 2H2O(l) OH, conducts electricity very poorly. (b) The reduction of
Ecell 1.94 V H2O has a more positive half-potential than does the reduction of
Pt2(aq) Pb(s) SO42(aq) ±£ Pt(s) PbSO4(s) Na; the oxidation of H2O is the only reaction possible because
Ecell 1.51 V SO42 cannot be oxidized. Thus, it is easier to reduce H2O than
Oxidizing agent: PbSO4 Pt2 HClO Na and easier to oxidize H2O than SO42. 21.109 62.6 g Zn
Reducing agent: Cl2 Pt Pb 21.111(a) 3.31011 C (b) 4.81011 J (c) 1.2104 kg
21.50 Yes; C A B 21.53 A(s) B(aq) ±£ A(aq) 21.114 64.3 mass % Cu 21.116(a) 8 days (b) 32 days
B(s) with Q [A][B]. (a) [A] increases and [B] (c) $250 21.119 (a) 2.4104 days (b) 2.1 g (c) 3.4105
decreases. (b) Ecell decreases. (c) Ecell Ecell (RTnF) dollars 21.122(a) PbPb2: Ecell 0.13 V; CuCu2: Ecell
ln ([A][B]); Ecell Ecell when (RTnF) ln ([A][B]) 0. 0.34 V (b) The anode (negative electrode) is Pb. The anode in
This occurs when ln ([A][B]) 0, that is, [A] equals [B]. the other cell is platinum in the standard hydrogen electrode.
(d) Yes, when [A] [B]. 21.55 In a concentration cell, the (c) The precipitation of PbS decreases [Pb2], which increases
overall reaction decreases the concentration of the more concen- the potential. (d) 0.13 V 21.125 The three steps equivalent
trated electrolyte because it is reduced in the cathode compartment. to the overall reaction M(aq) e ±£ M(s) are
21.56(a) 31035 (b) 41031 21.58(a) 11067 (b) 6109 (1) M(aq) ±£ M(g)
H is
Hhydration
21.60(a) 2.03105 J (b) 1.73105 J 21.62(a) 3.82105 J (2) M(g) e ±£ M(g)
H is IE
(b) 5.6104 J 21.64
G 2.7104 J; E 0.28 V (3) M(g) ±£ M(s)
H is
Hatomization
21.66 Ecell 0.054 V;
G 1.0104 J 21.68 8.8105 M The energy for step 3 is similar for all three elements, so the dif-
21.70(a) 0.05 V (b) 0.50 M (c) [Co2] 0.91 M; [Ni2] ference in energy for the overall reaction depends on the values
0.09 M 21.72 A; 0.083 V 21.74 Electrons flow from the an- for
Hhydration and IE. The Li ion has a much greater hydration
ode, where oxidation occurs, to the cathode, where reduction energy than Na and K because it is smaller, with large charge
occurs. The electrons always flow from the anode to the cathode density that holds the water molecules more tightly. The energy
no matter what type of battery. 21.76 A D-sized alkaline battery required to remove the waters surrounding the Li offsets the
is much larger than an AAA-sized one, so the D-sized battery lower ionization energy, making the overall energy for the reduc-
contains greater amounts of the cell components. The cell poten- tion of lithium larger than expected. 21.127 The very high and
tial is an intensive property and does not depend on the amounts very low standard electrode potentials involve extremely reactive
of the cell components. The total charge, however, depends on substances, such as F2 (a powerful oxidizer) and Li (a powerful
the amount of cell components, so the D-sized battery produces reducer). These substances react directly with water because any
more charge than the AAA-sized battery. 21.78 The Teflon aqueous cell with a voltage of more than 1.23 V has the ability to
spacers keep the two metals separated so that the copper cannot electrolyze water into hydrogen and oxygen.
conduct electrons that would promote the corrosion (rusting) of 21.129(a) 1.073105 s (b) 1.5104 kWh (c) 6.8¢
the iron skeleton. 21.81 Sacrificial anodes are made of metals 21.131 F D E. If metal E and a salt of metal F are mixed,
with E less than that of iron, 0.44 V, so they are more easily the salt is reduced, producing metal F because E has the greatest
oxidized than iron. Only (b), (f), and (g) will work for iron; reducing strength of the three metals.
(a) will form an oxide coating that prevents further oxidation; 21.134(a) Cell I: 4 mol electrons;
G 4.75105 J
(c) would react with groundwater quickly; (d) and (e) are less Cell II: 2 mol electrons;
G 3.94105 J
easily oxidized than iron. 21.83 To reverse the reaction requires Cell III: 2 mol electrons;
G 4.53105 J
0.34 V with the cell in its standard state. A 1.5 V cell supplies (b) Cell I: 13.2 kJg
more than enough potential, so the cadmium metal is oxidized to Cell II: 0.613 kJg
Cd2 and chromium plates out. 21.85 The oxidation number Cell III: 2.62 kJg
of N in NO3 is 5, the maximum O.N. for N. In the nitrite ion, Cell I has the highest ratio (most energy released per
NO2, the O.N. of N is 3, so nitrogen can be further oxidized. gram) because the reactants have very low mass, while
21.87(a) Br2 (b) Na 21.89 I2 gas forms at the anode; magne- Cell II has the lowest ratio because the reactants have
sium (liquid) forms at the cathode. 21.91 Bromine gas forms at large masses.
A-44 Appendix E Answers to Selected Problems
21.138(a) 9.7 g Cu (b) 0.56 M Cu2 H2PO4 ions, which the plant can absorb. Through excretion and
21.140Sn2(aq) 2e ±£ Sn(s) decay, organisms return soluble phosphate compounds to the
Cr3(aq) e ±£ Cr2(aq) cycle. 22.17(a) 1.1103 L (b) 4.2102 m3 22.18(a) The
Fe2(aq) 2e ±£ Fe(s) iron ions form an insoluble salt, Fe3(PO4)2, that decreases the
U4(aq) e ±£ U3(aq) yield of phosphorus. (b) 8.8 t 22.20(a) Roasting involves
21.144(a) 3.6109 M (b) 0.3 M heating the mineral in air at high temperatures to convert the
21.146(a) Nonstandard cell: mineral to the oxide. (b) Smelting is the reduction of the metal
Ewaste Ecell (0.0592 V1) log [Ag]waste oxide to the free metal using heat and a reducing agent such as
Standard cell: coke. (c) Flotation is a separation process in which the ore is
Estandard Ecell (0.0592 V1) log [Ag]standard removed from the gangue by exploiting the difference in density
Estandard Ewaste
(b) [Ag]waste antilog a b[Ag]standard
in the presence of detergent. The gangue sinks to the bottom and
0.0592 the lighter ore-detergent mix is skimmed off the top. (d) Refin-
Estandard Ewaste ing is the final step in the purification process to yield the pure
(c) CAg, waste antilog a bCAg, standard, metal. 22.25(a) Slag is a byproduct of steel-making and con-
0.0592
where C is concentration in ngL tains the impurity SiO2. (b) Pig iron is the impure product of iron
(d) 900 ngL metallurgy (containing 3–4% C and other impurities). (c) Steel
(e) [Ag]waste refers to iron alloyed with other elements to attain desirable
nF properties. (d) The basic-oxygen process is used to purify pig
(Estandard Ewaste ) Tstandard log [Ag]standard iron and obtain carbon steel. 22.27 Iron and nickel are more
antilog £ §
2.303R
easily oxidized and less easily reduced than copper. They are
Twaste separated from copper in the roasting step and converted to slag.
21.148 (a) 1.08103 C (b) 0.629 g Cd, 1.03 g NiO(OH), 0.202 g In the electrorefining process, all three metals are in solution, but
H2O; total mass of reactants 1.86 g (c) 10.1% 21.150 Li only Cu2 ions are reduced at the cathode to form Cu(s).
Ba Na Al Mn Zn Cr Fe Ni Sn Pb Cu 22.30 Le Châtelier’s principle says that the system shifts toward
Ag Hg Au. Metals with potentials lower than that of wa- formation of K as the gaseous metal leaves the cell.
ter (0.83 V) can displace H2 from water: Li, Ba, Na, Al, and 22.31(a) Ehalf-cell 3.05 V, 2.93 V, and 2.71 V for Li,
Mn. Metals with potentials lower than that of hydrogen (0.00 V) K, and Na, respectively. In all of these cases, it is energetically
can displace H2 from acid: Li, Ba, Na, Al, Mn, Zn, Cr, Fe, Ni, more favorable to reduce H2O to H2 than to reduce M to M.
Sn, and Pb. Metals with potentials greater than that of hydrogen (b) 2RbX Ca ±£ CaX2 2Rb, where
H IE1(Ca)
(0.00 V) cannot displace H2: Cu, Ag, Hg, and Au. IE2(Ca) 2IE1(Rb) 929 kJmol. Based on the IEs and posi-
21.153(a) 5.31011 (b) 0.20 V (c) 0.43 V (d) 8.2104 M tive
H for the forward reaction, it seems more reasonable that
NaOH 21.156(a) 1.18105 kJ (b) 1.20104 L Rb metal will reduce Ca2 than the reverse. (c) If the reaction
(c) 2.31106 s (d) 3.84103 kWh (e) 4.15103 cents is carried out at a temperature greater than the boiling point of
21.158 2.94 Rb, the product mixture will contain gaseous Rb, which can be
removed from the reaction vessel; this would cause a shift in
Chapter 22 equilibrium to form more Rb as product. (d) 2CsX Ca ±£
CaX2 2Cs, where
H IE1(Ca) IE2(Ca) 2IE1(Cs)
22.2 Fe from Fe2O3; Ca from CaCO3; Na from NaCl; Zn from
983 kJmol. This reaction is more unfavorable than for Rb,
ZnS 22.3(a) Differentiation refers to the processes involved in
but Cs has a lower boiling point. 22.32(a) 4.5104 L
the formation of Earth into regions (core, mantle, and crust) of
(b) 1.30108 C (c) 1.69106 s 22.35(a) Mg2 is more diffi-
differing composition. Substances separated according to their
cult to reduce than H2O, so H2(g) would be produced instead of
densities, with the more dense material in the core and the less
Mg metal. Cl2(g) forms at the anode due to overvoltage.
dense in the crust. (b) O, Si, Al, and Fe (c) O 22.7 Plants
(b) The
Hf of MgCl2(s) is 641.6 kJmol. High temperature
produced O2, slowly increasing the oxygen concentration in the
favors the reverse (endothermic) reaction, the formation of mag-
atmosphere and creating an oxidative environment for metals.
nesium metal and chlorine gas. 22.37(a) Sulfur dioxide is the
The oxygen-free decay of plant and animal material created large
reducing agent and is oxidized to the 6 state (SO42).
fossil fuel deposits. 22.9 Fixation refers to the process of con-
(b) HSO4(aq) (c) H2SeO3(aq) 2SO2(g) H2O(l) ±£
verting a substance in the atmosphere into a form more readily
Se(s) 2HSO4(aq) 2H(aq) 22.42(a) O.N. for Cu: in
usable by organisms. Carbon and nitrogen; fixation of carbon
Cu2S, 1; in Cu2O, 1; in Cu, 0 (b) Cu2S is the reducing
dioxide gas by plants and fixation of nitrogen gas by nitrogen-
agent, and Cu2O is the oxidizing agent. 22.44(a) 1.3106 C
fixing bacteria. 22.12 Atmospheric nitrogen is fixed by three
(b) 1.2103 A 22.47 2ZnS(s) C(graphite) ±£ 2Zn(s)
pathways: atmospheric, industrial, and biological. Atmospheric
CS2(g);
Grxn 463 kJ. Since
Grxn is positive, this reaction is
fixation requires high-temperature reactions (e.g., initiated by
not spontaneous at standard-state conditions. 2ZnO(s) C(s) ±£
lightning) to convert N2 into NO and other oxidized species.
2Zn(s) CO2(g);
Grxn 242.0 kJ. This reaction is also not
Industrial fixation involves mainly the formation of ammonia,
spontaneous, but is less unfavorable. 22.48 The formation of
NH3, from N2 and H2. Biological fixation occurs in nitrogen-
sulfur trioxide is very slow at ordinary temperatures. Increasing
fixing bacteria that live on the roots of legumes. Human activity
the temperature can speed up the reaction, but because the reac-
is an example of industrial fixation. It contributes about 17% of
tion is exothermic, increasing the temperature decreases the
the nitrogen fixed. 22.14(a) the atmosphere (b) Plants excrete
yield. Adding a catalyst increases the rate of the reaction, so a
acid from their roots to convert PO43 ions into more soluble
lower temperature can be used to enhance the yield.

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siL48593_ch21_922-979 8:11:07 04:52am Page 922 fdfd ve403:MHQY042:siL5ch21:

Electrochemistry: Chemical Change

A Quick Review of Oxidation-Reduction Concepts

924 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Half-Reaction Method for Balancing Redox Reactions

21.1 Redox Reactions and Electrochemical Cells 925

Step 4. Add the balanced half-reactions, and include states of matter.

926 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Step 3. Multiply each half-reaction, if necessary, by an integer so that the num-

Step 5. Check that atoms and charges balance:

H bonded to an H2O, we balance the 14 H on the right with 14 H3O on the

21.1 Redox Reactions and Electrochemical Cells 927

SAMPLE PROBLEM 21.1 Balancing Redox Reactions by the Half-Reaction

928 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

An Overview of Electrochemical Cells

Figure 21.3 General characteristics of VOLTAIC CELL ELECTROLYTIC CELL

Oxidation half-reaction Oxidation half-reaction

Reduction half-reaction Reduction half-reaction

Overall (cell) reaction Overall (cell) reaction

21.2 VOLTAIC CELLS: USING SPONTANEOUS REACTIONS

Figure 21.4 The spontaneous reaction

Zn(s) + Cu 2+(aq) Zn2+(aq) + Cu(s)

930 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Construction and Operation of a Voltaic Cell

932 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

e– Voltmeter Therefore, each half-cell consists of inactive electrodes immersed in an electrolyte

Overall (cell) reaction

SAMPLE PROBLEM 21.2 Describing a Voltaic Cell with a Diagram and

Why Does a Voltaic Cell Work?

934 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

21.3 CELL POTENTIAL: OUTPUT OF A VOLTAIC CELL

Table 21.1 Voltages of Some Voltaic Cells

21.3 Cell Potential: Output of a Voltaic Cell 935

Standard Cell Potentials Cu half-cell

936 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Overall (cell) reaction

21.3 Cell Potential: Output of a Voltaic Cell 937

SAMPLE PROBLEM 21.3 Calculating an Unknown Ehalf-cell from Ecell

Relative Strengths of Oxidizing and Reducing Agents

938 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Table 21.2 Selected Standard Electrode Potentials (298 K)

Strength of reducing agent

21.3 Cell Potential: Output of a Voltaic Cell 939

940 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

SAMPLE PROBLEM 21.4 Writing Spontaneous Redox Reactions and

21.3 Cell Potential: Output of a Voltaic Cell 941

Relative Reactivities of Metals In Chapter 4, we discussed the activity series of

942 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

21.4 Free Energy and Electrical Work 943

This particular reaction has tremendous economic importance in protecting iron

Section Summary Saliva

Standard Cell Potential and the Equilibrium Constant

944 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Because 1 V  1 J/C, we have 1 C  1 J/V, and

ΔG Reaction Parameters at the Standard State

>0 <1 <0 Nonspontaneous

Figure 21.10 The interrelation-

It is common practice to simplify Equation 21.7 in calculations by

21.4 Free Energy and Electrical Work 945

Because 1 V 1 J/C, we have 1 C 1 J/V, and

SAMPLE PROBLEM 21.5 Calculating K and G from E cell

0.76 V c log (4.8 104 ) d 0.76 V (0.0982 V) 0.86 V

1.10 V c (4.30) d 1.10 V 0.127 V 1.23 V

0V c (1.00) d 0.0296 V

Ecell c (2 log [H ]unknown ) d 0.0592 V log [H ]unknown

Voltaic 0 0 Anode Oxidation

Ecell 1.23 V e [log 1 4 log (1.0 107 )] f 0.82 V