ch21 Electrochem SILBERBERG
ch21 Electrochem SILBERBERG
ch21 Electrochem SILBERBERG
21-1
Figure 21.1 A summary of redox terminology, as applied to the
reaction of zinc with hydrogen ion.
0 +1 +2 0
Zn(s) + 2H (aq) Zn (aq) + H2(g)
+ 2+
21-2
Half-Reaction Method for
Balancing Redox Reactions
The half-reaction method divides a redox reaction into its
oxidation and reduction half-reactions.
- This reflects their physical separation in electrochemical cells.
21-3
Steps in the Half-Reaction Method
Divide the skeleton reaction into two half-reactions, each
of which contains the oxidized and reduced forms of one
of the species.
Balance the atoms and charges in each half-reaction.
First balance atoms other than O and H, then O, then H.
Charge is balanced by adding electrons (e-) to the left side in
the reduction half-reaction and to the right side in the oxidation
half-reaction.
If necessary, multiply one or both half-reactions by an
integer so that
number of e- gained in reduction = number of e- lost in oxidation
Add the balanced half-reactions, and include states of
matter.
21-4
Balancing Redox Reactions in Acidic Solution
21-5
Balance H atoms by adding H+ ions:
14H+ + Cr2O72- 2Cr3+ + 7H2O
21-6
Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
The reduction half-reaction shows that 6e- are gained; the oxidation
half-reaction shows only 2e- being lost and must be multiplied by 3:
3(2I- I2 + 2e-)
6I- 3I2 + 6e-
21-7
Balancing Redox Reactions in Basic Solution
An acidic solution contains H+ ions and H2O. We use H+
ions to balance H atoms.
A basic solution contains OH- ions and H2O. To balance
H atoms, we proceed as if in acidic solution, and then
add one OH- ion to both sides of the equation.
For every OH- ion and H+ ion that appear on the same
side of the equation we form an H2O molecule.
21-8
Sample Problem 21.1 Balancing a Redox Reaction in Basic
Solution
21-9
Sample Problem 21.1
Step 2: Balance the atoms and charges in each half-reaction.
Balance atoms other than O and H:
MnO4- MnO2 C2O42- 2CO32-
21-10
Sample Problem 21.1
21-11
Sample Problem 21.1
Basic. Add OH- to both sides of the equation to neutralize H+, and
cancel H2O.
21-12
Electrochemical Cells
21-13
Figure 21.2 General characteristics of (A) voltaic and (B) electrolytic
cells.
21-14
Spontaneous Redox Reactions
A strip of zinc metal in a solution of Cu2+ ions will react
spontaneously:
Cu2+(aq) + 2e- Cu(s) [reduction]
Zn(s) Zn2+(aq) + 2e- [oxidation]
Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s)
21-15
Construction of a Voltaic Cell
21-16
Operation of the Voltaic Cell
Oxidation (loss of e-) occurs at the anode, which is
therefore the source of e-.
Zn(s) Zn2+(aq) + 2e-
21-17
Figure 21.4B A voltaic cell based on the zinc-copper reaction.
21-18
Charges of the Electrodes
21-19
The Salt Bridge
The salt bridge completes the electrical circuit and allows
ions to flow through both half-cells.
21-20
Flow of Charge in a Voltaic Cell
Zn2+ SO42-
Cations move through the salt Anions move through the salt
bridge from the anode solution bridge from the cathode solution
to the cathode solution. to the anode solution.
21-21
Active and Inactive Electrodes
An active electrode is an active component in its half-
cell and is a reactant or product in the overall reaction.
21-22
Figure 21.5 A voltaic cell using inactive electrodes.
21-23
Notation for a Voltaic Cell
The components of each half-cell are written in the same
order as in their half-reactions.
The single line shows a phase The double line shows that the half-
boundary between the cells are physically separated.
components of a half-cell.
21-24
Notation for a Voltaic Cell
21-25
Sample Problem 21.2 Describing a Voltaic Cell with Diagram
and Notation
PLAN: From the given contents of the half-cells, we write the half-
reactions. To determine which is the anode compartment
(oxidation) and which is the cathode (reduction), we note the
relative electrode charges. Electrons are released into the
anode during oxidation, so it has a negative charge. Since Cr
is negative, it must be the anode, and Ag is the cathode.
21-26
Sample Problem 21.2
SOLUTION:
The half-reactions are:
Ag+(aq) + e- Ag(s) [reduction; cathode]
Cr(s) Cr3+(aq) + 3e- [oxidation; anode]
The balanced overall equation is:
3Ag+ + Cr(s) 3Ag(s) + Cr3+(aq)
21-27
Electrical Potential and the Voltaic Cell
When the switch is closed and no reaction is occurring,
each half-cell is in an equilibrium state:
Zn(s) Zn2+(aq) + 2e- (in Zn metal)
Cu(s) Cu2+(aq) + 2e- (in Cu metal)
Zn is a stronger reducing agent than Cu, so the position of
the Zn equilibrium lies farther to the right.
Zn has a higher electrical potential than Cu. When the
switch is closed, e- flow from Zn to Cu to equalize the
difference in electrical potential
The spontaneous reaction occurs as a result of the different
abilities of these metals to give up their electrons.
21-28
Cell Potential
21-29
Table 21.1 Voltages of Some Voltaic Cells
21-30
Figure 21.6 Measuring the standard cell potential of a zinc-
copper cell.
21-31
Standard Electrode Potentials
The standard electrode potential (Ehalf-cell) is the potential
of a given half-reaction when all components are in their
standard states.
21-32
The Standard Hydrogen Electrode
Half-cell potentials are measured relative to a standard
reference half-cell.
21-33
Figure 21.7 Determining an unknown Ehalf-cell with the standard
reference (hydrogen) electrode.
Oxidation half-reaction
Zn(s) Zn2+(aq) + 2e
Reduction half-reaction
2H3O+(aq) + 2e- H2(g) + 2H2O(l)
SOLUTION:
Br2(aq) + 2e- 2Br-(aq) [reduction; cathode]
Zn(s) Zn2+(aq) + 2e- [oxidation; anode] Ezinc = -0.76 V
21-35
Sample Problem 21.3
Ebromine = 1.07 V
21-36
Comparing Ehalf-cell values
21-37
Table 21.2 Selected Standard Electrode Potentials (298 K)
Half-Reaction E(V)
F2(g) + 2e 2F(aq) +2.87
Cl2(g) + 2e 2Cl(aq) +1.36
MnO2(g) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l) +1.23
NO3-(aq) + 4H+(aq) + 3e NO(g) + 2H2O(l) +0.96
21-38
Writing Spontaneous Redox Reactions
21-39
Using half-reactions to write a spontaneous redox reaction:
Sn2+(aq) + 2e- Sn(s) Etin = -0.14 V
Ag+(aq) + e- Ag(s) Esilver = 0.80 V
21-40
Step 2: Multiply the half-reactions if necessary so that the number
of e- lost is equal to the number or e- gained.
2Ag+(aq) + 2e- 2Ag(s) Esilver = 0.80 V
Note that we multiply the equation but not the value for E.
21-41
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 for spontaneous reactions (A, B,
and C), 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
21-43
Sample Problem 21.4
For (1) and (3), equation (1) has the smaller, less positive E value:
21-44
Sample Problem 21.4
For (2) and (3), equation (2) has the smaller, less positive E value:
(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
21-45
Sample Problem 21.4:
(b) We first rank the oxidizing and reducing agents within each
equation, then we can compare Ecell values.
Equation (A) Oxidizing agents: NO3- > N2
Reducing Agents: N2H5+ > NO
Equation (B) Oxidizing agents: MnO2 > NO3-
Reducing Agents: N2H5+ > NO
Equation (C) Oxidizing agents: MnO2 > N2
Reducing Agents: N2H5+ > Mn2+
21-46
The Activity Series of the Metals
21-47
The Activity Series of the Metals
21-48
The Activity Series of the Metals
21-49
The Activity Series of the Metals
21-50
Figure 21.8 The reaction of calcium in water.
21-51
Free Energy and Electrical Work
n = mol of e- transferred
G = -nFEcell F is the Faraday constant
= 9.65x104 J/Vmol e-
RT ln K or 0.0592 V log K
Ecell = Ecell =
nF n
for T = 298.15 K
21-52
Figure 21.10 The interrelationship of G, Ecell, and K.
Reaction at standard-
G K Ecell state conditions
<0 >1 >0 spontaneous
G
ll
ce
0 1 0 at equilibrium
E
=-
- nF
RT
=
ln
G
Ecell K
RT ln K
Ecell =
nF
21-53
Sample Problem 21.5 Calculating K and G from Ecell
PROBLEM: Lead can displace silver from solution, and silver occurs in
trace amounts in some ores of lead.
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.
21-54
Sample Problem 21.5
21-55
Cell Potential and Concentration
21-56
Sample Problem 21.6 Using the Nernst Equation to Calculate Ecell
21-57
Sample Problem 21.6
21-58
Figure 21.11A The relation between Ecell and log Q for the zinc-
copper cell.
If the reaction starts with [Zn2+] < [Cu2+] (Q < 1), Ecell is higher than the
standard cell potential.
As the reaction proceeds, [Zn2+] decreases and [Cu2+] increases, so
Ecell drops. Eventually the system reaches equilibrium and the cell can
no longer do work.
21-59
Figure 21.11B The relation between Ecell and log Q for the zinc-
copper cell.
21-60
Concentration Cells
21-61
Figure 21.12 A concentration cell based on the Cu/Cu2+ half-reaction.
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Oxidation half-reaction
Cu(s) Cu2+(aq, 0.1 M) + 2e- Ecell > 0 as long as the half-cell
concentrations are different.
Reduction half-reaction
Cu2+(aq, 1.0 M) + 2e- Cu(s) The cell is no longer able to do work
once the concentrations are equal.
Overall (cell) reaction
Cu2+(aq,1.0 M) Cu2+(aq, 0.1 M)
21-62
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, the electrolyte is 0.0100 M AgNO3; in half-cell
B, it is 4.0x10-4 M AgNO3. What is the cell potential at
298.15 K?
21-63
Sample Problem 21.7
21-64
Figure 21.13 Laboratory measurement of pH.
21-65
Table 21.3 Some Ions Measured with Ion-Specific Electrodes
21-66
Electrochemical Processes in Batteries
21-67
Figure 21.15 Alkaline battery.
21-68
Figure 21.16 Silver button battery.
21-69
Figure 21.17 Lithium battery.
Anode (oxidation):
3.5Li(s) 3.5Li+ + 3.5e-
Cathode (reduction):
AgV2O5.5 + 3.5Li- + 3.5e- Li3.5V2O5.5
Overall (cell) reaction:
AgV2O5.5 + 3.5Li(s) Li3.5V2O5.5
21-70
The reactions in a lead-acid battery:
The cell generates electrical energy when it discharges as a voltaic cell.
Anode (oxidation): Pb(s) + HSO4-(aq) PbSO4(s) + H+(aq) + 2e-
Cathode (reduction):
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- PbSO4(s) + 2H2O(l)
Overall (cell) reaction (discharge):
PbO2(s) + Pb(s) + H2SO4(aq) 2PbSO4(s) + 2H2O(l) Ecell = 2.1 V
21-71
Figure 21.19 Nickel-metal hydride battery
21-72
Figure 21.20 Lithium-ion battery.
Anode (oxidation):
LixC6(s) xLi+ + xe- + C6(s)
Cathode (reduction):
Li1-xMn2O4(s) + xLi+ + xe- LiMn2O4(s)
Overall (cell) reaction:
LixC6(s) + Li1-xMn2O4(s) LiMn2O4(s)
Ecell = 3.7 V
21-73
Figure 21.21 Hydrogen fuel cell.
21-74
Fuel Cells
21-75
Corrosion: an Environmental Voltaic Cell
21-76
The Rusting of Iron
The loss of iron:
Fe(s) Fe2+(aq) + 2e- [anodic region; oxidation]
O2(g) + 4H+(aq) + 4e- 2H2O(l) [cathodic region; reduction]
2Fe(s) + O2(g) + 4H+(aq) 2Fe2+(aq) + 2H2O(l) [overall]
H+ ions are consumed in the first step, so lowering the pH increases the
overall rate of the process. H+ ions act as a catalyst, since they are
regenerated in the second part of the process.
21-77
Figure 21.22 The corrosion of iron.
21-78
Figure 21.23 Enhanced corrosion at sea.
21-79
Figure 21.24 The effect of metal-metal contact on the corrosion
of iron.
21-80
Figure 21.25 The use of sacrificial anodes to prevent iron corrosion.
21-81
Electrolytic Cells
21-82
Figure 21.26 The tin-copper reaction as the basis of a voltaic and
an electrolytic cell.
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
21-83
Figure 21.27 The processes occurring during the discharge and
recharge of a lead-acid battery.
VOLTAIC (discharge)
Switch
ELECTROLYTIC (recharge)
21-84
Table 21.4 Comparison of Voltaic and Electrolytic Cells
Electrode
Cell Type G Ecell Name Process Sign
21-85
Products of Electrolysis
21-86
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 need to determine which metal and nonmetal will form more
easily at the electrodes. We list the ions as oxidizing or reducing
agents.
If a metal holds its electrons more tightly than another, it has a
higher ionization energy (IE). Its cation will gain electrons more
easily, and it will be the stronger oxidizing agent.
If a nonmetal holds its electrons less tightly than another, it has a
lower electronegativity (EN). Its anion will lose electrons more
easily, and it will be the reducing agent.
21-87
Sample Problem 21.8
SOLUTION:
Possible oxidizing agents: Na+, Mg2+
Possible reducing agents: Br-, Cl-
Mg is to the right of Na in Period 3. IE increases from left to right across
the period, so Mg has the higher IE and gives up its electrons less
easily. The Mg2+ ion has a greater attraction for e- than the Na+ ion.
21-88
Figure 21.28 The electrolysis of water.
21-89
Electrolysis of Aqueous Salt Solutions
When an aqueous salt solution is electrolyzed
- The strongest oxidizing agent (most positive electrode potential) is
reduced, and
- The strongest reducing agent (most negative electrode potential) is
oxidized.
21-90
Summary of the Electrolysis of Aqueous Salt Solutions
21-91
Sample Problem 21.9 Predicting the Electrolysis Products of
Aqueous Salt Solutions
PROBLEM: What products form at which electrode during electrolysis of
aqueous solution 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 0.6 V overvoltage
into account. The reduction half-reaction with the less negative
E occurs at the cathode, while the oxidation half-reaction with
the less positive E occurs at the anode.
SOLUTION:
(a) KBr K+(aq) + e- K(s) E = -2.93
2H2O(l) + 2e- H2(g) + 2OH-(aq) E = -0.42V
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.
21-92
Sample Problem 21.9
21-93
Sample Problem 21.9
21-94
Stoichiometry of Electrolysis
21-95
Figure 21.29 A summary diagram for the stoichiometry of
electrolysis.
MASS (g)
of substance CURRENT
oxidized or (A)
reduced
AMOUNT (mol)
AMOUNT (mol) CHARGE
of substance
of electrons (C)
oxidized or
transferred
reduced balanced Faraday
half-reaction constant
(C/mol e-)
21-96
Sample Problem 21.10 Applying the Relationship Among Current,
Time, and Amount of Substance
PROBLEM: A technician plates a faucet with 0.86 g of Cr metal by
electrolysis of aqueous Cr2(SO4)3. If 12.5 min is allowed for
the plating, what current is needed?
PLAN: To find the current, we divide charge by time, so we need to find
the charge. We write the half-reaction for Cr3+ reduction to get
the amount (mol) of e- transferred per mole of Cr. We convert
mass of Cr needed to amount (mol) of Cr. We can then use the
Faraday constant to find charge and current.
mass (g) of Cr needed
divide by M
mol of Cr
3 mol e- = 1 mol Cr
mol e- transferred Charge (C) current (A)
1 mol e- = 9.65x104 C divide by time in s
21-97
Sample Problem 21.10
SOLUTION:
Cr3+(aq) + 3e- Cr(s)
1 mol e-
21-98
Chemical Connections
Figure B21.1 The mitochondrion
21-99
Chemical Connections
Figure B21.2 The main energy-yielding steps in the electron-
transport chain (ETC).
21-
Chemical Connections
Figure B21.3 Coupling electron transport to proton transport to
ATP synthesis.
21-