Handout ElectroChemistry BY S.K
Handout ElectroChemistry BY S.K
Handout ElectroChemistry BY S.K
CHAPTER: - 2
ELECTROCHEMISTRY
CHAPTER:- 4
ELECTROCHEMISTRY
ELECTROCHEMISTRY
Electrochemistry is the area of chemistry dealing with the interconversion of electrical energy
and chemical energy. There are many applications of this in everyday life. Batteries, control of
corrosion, metallurgy and electrolysis are just a few examples of the applications of
electrochemistry. This handout will look at the basic principles of electrochemistry and show
some of its applications.
In general, metals tend to make good reducing agents because they can only be oxidized.
The reducing ability of the metal is given by the activity series. The more active metal is able to
reduce the less active metal cation. This activity series is:
Li>K>Ca>Na>Mg>Al>Zn>Cr>Fe>Ni>Sn>Pb>H2>Cu>Hg>Ag>Pt>Au
most active least active
So for example, magnesium metal is able to reduce copper(II) ions in solution to form
magnesium ions and copper metal: Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
Chemistry for Grade 12
PREFERENTIAL DISCHARGE
The preferential discharge of ions is affected by:
1) the nature of the electrodes,
2) the positions of the ions in the electrochemical series and
3) the concentration of the ions in the electrolyte.
The schematic shows how a zinc anode connected by a salt bridge to a copper cathode will
convert chemical energy to electrical energy and allow the light bulb to light up. Lets look at
the electrochemistry involved. For any electrochemical cell, oxidation always occurs at the
anode and reduction always occurs at the cathode. An easy way to remember this is to keep the
vowels together and the consonants together (oxidation at the anode; reduction at the cathode).
Also, when drawing a galvanic cell, most people use the convention where the anode is on the
left and the cathode is on the right. Use the mnemonic:ABC to remember
(Anode/Bridge/Cathode).
1. What is salt bridge? Give examples of such electrolytes that are used as salt bridge?
2. What are the purposes (uses) of salt bridge? (List at least three)
The standard potential is: Ecell = Ered (Ag+ / Ag) – Ered ( Ni+2/ Ni)
= 0.8V – (-0.28V) = +1.08 V (This reaction is spontaneous & the forward rxn is
favored)
An E˚ > 0 means the cell will be spontaneous (produce a voltage). An E˚ < 0 means the cell must
be supplied that voltage from an external power supply in order for the cell reaction to run in
the direction it is written. If anything, that cell would want to run in the reverse direction.
Often these nonspontaneous cells are called electrolytic cells.
Exercise-1:
1. For each of the following galvanic cell, write the anode and cathode half-cell reactions, the
overall cell reaction, and calculate the standard cell potential. Use the standard reduction half-
cell potentials
(a) Zn|Zn2+(aq, 1 M)||Fe3+,Fe2+(aq, 1 M)|Pt;
(b) Zn|Zn2+(aq, 1 M)||Br2,Br-(aq, 1 M)|Pt;
2. Write the cell notation and calculate the standard cell potential for galvanic cells in which the
following reactions occur:
(a) Pb(s) + 2 Fe3+(aq) Pb2+(aq) + 2Fe2+(aq)
(b) 2Li(s) + I3-(aq) 2Li+(aq) + 3I-(aq)
We can show that : G = -nFEcell , where G is the change in free energy, n is the
number of moles of electrons transferred , F is Faradays constant and E is the emf of the
cell.
1F = 96,500 Coulomb/mol= 96,500 J/ V-mol
Since n and F are positive, if G is negative, then Ecell must be positive and the reaction
will be spontaneous
Ecell = E0cell - 0.0592V/n (log Q) for Zn(s) + Cu+2 Zn+2 + Cu(s) we get the following:
Ecell = 1.10V - 0.0592V/2 log [Zn+2] / [Cu+2] in this case n=2 since 2 electrons are being
transferred. Remember solids = 1 in an equilibrium expression.
Concentration Cells
Chemistry for Grade 12
A cell whose EMF is generated solely by a difference in concentration
Consider a cell with 2 Ni(s) electrodes but a difference in the concentration of Ni+2 . One
cell has [Ni+2] = 1M and the other [Ni+2] = 0.001M
The standard cell potential is obviously zero. However, the cell is operating under
nonstandard conditions
The driving force is the difference in concentration
Using the Nernst equation we calculate a EMF of + 0.0888V
Cell EMF and Chemical Equilibrium
For the voltaic cell, Cu|Cu2+(aq)||Ag+(aq)|Ag, the Nernst equation is expressed as:
Ecell = (EoAg+|Ag + EoCu|Cu2+) - (0.0591 V/2) log ([Cu2+]/[Ag+]2)
The cell reaction will occur spontaneously until it reaches equilibrium, at which point Q = K
(the equilibrium constant) and
Ecell = Eocell - (RT/nF) ln(K) = 0 ;
ln(K) = nFEocell/RT; K = exp(nFEocell/RT)
Note also that at equilibrium, G = Go – RT ln(K) = 0 (no current flows through the circuit)
Exercise-3:
1. Write an overall net ionic equation and calculate the cell potential at 25 oC for:
Zn|Zn2+(aq, 0.0050 M)||Cu2+(aq,1.0 M)|Cu
2. Write an overall net ionic equation and calculate the cell potential at 25 oC for:
Cu|Cu2+(aq, 0.0050 M)||Ag+(aq, 0.50 M)|Ag
3. The standard cell potential (Eocell) for the following galvanic cell is 1.10 V
Zn|Zn2+(aq, 1M)||Cu2+(aq, 1M)|Cu
What is the K value for the reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) ?
Concentration Cells
A concentration cell is an electrochemical cell in which both half-cells are of the same type,
but with different electrolyte concentrations. The following cell notations are examples of
concentration cells:
Cu|Cu2+(aq, 0.0010 M)||Cu2+(aq, 1.0 M)|Cu
Ag|Ag+(aq, 0.0010 M) ||Ag+(aq, 0.10 M)|Ag
In concentration cells, the half-cell with the lower electrolyte concentration serves as an anode
half-cell and one with the higher electrolyte concentration is the cathode half-cell. At the anode
half-cell, oxidation reaction occurs to increase the electrolyte concentration and at the cathode
half-cell, a reduction reaction occurs to decrease its electrolyte concentration. Oxidation-
reduction reaction will continue until the electrolyte concentrations in both half-cells become
equal.
BATTERIES
Batteries are galvanic cells or a group of galvanic cells connected in series, where the total battery
potential is equal to the sum of the potentials of the individual cells. There are three types of batteries
– primary batteries, secondary batteries, and the fuel cell. Primary batteries are not re-chargeable,
where as secondary batteries are re-chargeable. Fuel cell will last as long as there is an ample supply
of fuel to provide the energy.
Chemistry for Grade 12
Dry Cells:- The normal (acidic) dry batteries, alkaline batteries, and the mercury batteries are
example of primary batteries.
1. LeClanchè cells The “acidic” dry battery consists of Zinc casing as container, anode and as the
reducing agent; graphite rod as (inert) cathode; aqueous NH 4Cl paste as electrolyte, and MnO2
powder as the oxidizing agent.
Zn|Zn2+,NH4+,NH3(aq)||Mn2O3,MnO2|C(s)
2. Alkaline batteries also use zinc (as the reducing agent) and MnO2 (as oxidizing agent), but aqueous
paste containing KOH, instead of NH4Cl, is used as the electrolyte. The anode and cathode reactions
are as follows:
At anode: Zn(s) + 2OH-(aq) ZnO(s) + H2O(l) + 2e-;
At cathode: 2MnO2(s) + H2O(l) + 2e- Mn2O3(s) + 2OH-(aq);
Net reaction: Zn(s) + 2MnO2(s) ZnO(s) + Mn2O3(s)
Since all reactants involved are in the solid form, alkaline batteries can deliver a fairly constant
voltage until the limiting reactant is completely used up. They also last longer because zinc metal
corrodes more slowly under basic conditions.
3. mercury batteries Batteries used in calculators and watches are mercury batteries. The following
reactions occur at the anode and cathode sections of the cell:
Anode reaction: Zn(s) + 2OH-(aq) ZnO(s) + H2O + 2e-;
Cathode reaction: HgO(s) + H2O + 2e- Hg(l) + 2OH-(aq)
Net cell reaction: Zn(s) + HgO(s) ZnO(s) + Hg(l)
7, Fuel Cells:
A fuel cell is a galvanic cell that uses hydrogen (as fuel), which reacts with oxygen, and a large
amount of energy from the reaction is available to produce electricity. The fuel is supplied
continuously from an external tank. The hydrogen-oxygen fuel cells are used in the space shuttle
modules.
Anode reaction: 2H2(g) + 4OH-(aq) 4H2O(l) + 4e-;
Cathode reaction: 2H2O(l) + O2(g) + 4e- 4OH-(aq)
Net cell reaction: 2H2(g) + O2(g) 2H2O(l) + Energy
CORROSION CONTROL
Corrosion is a spontaneous oxidation of metals by atmospheric oxygen – a process that has
great economic impact. Consider the rusting process of a steel pipe. The process is accelerated by the
presence of water droplets on the metal surface. At the edge of the droplet, the part of iron exposed to
the air acts as the cathode for the electrochemical reaction. At the center of the droplet, the metal acts
as anode and get oxidized (corroded).
For example, if magnesium is used as sacrificial anode, the following reactions will occur:
2Mg(s) 2Mg2+(aq) + 4e-; (at anode)
and, O2(g) + 2H2O(l) + 4 e- 4OH-(aq) (at cathode)
Net reaction: 2Mg(s) + O2(g) + 2H2O 2Mg(OH)2(s);
This method of corrosion prevention using active metals such as magnesium as sacrificial anode is
called a cathodic protection, because as a cathode the pipe will not oxidized.
Electrochemistry Worksheet
1. Assign oxidation numbers to each atom in the following:
a. P4O6 b. BiO3 − c. N2H4 d. Mg(BrO4)2 e. MnSO4 f. Mn(SO4)2
2. For each of the reactions below identify the oxidizing agent and the reducing agent.
A. 2 KCl + MnO2 + H2SO4 → K2SO4 + MnSO4 + Cl2 + H2O
B. SiCl4 + 2 Mg(s) → 2 MgCl2 + Si
3. Use the half-reaction method to balanceeach of the following oxidation-reduction reactions.
Identify the oxidizing agent and the reducing agent.
+ 2+
A. Cu(s) + Ag (aq) → Ag(s) + Cu
B. Al(s) + I2(s) → AlI3 (s)
3+ 2+
C. Pb(s) + Fe (aq) → Pb (aq) + Fe(s)
4.Balance each of the following oxidation-reduction reactions. Identify the oxidizing agent and the
reducing agent.
2− − − 2−
A. S2O3 + OCl → Cl + S4O6 (in acid )
−
B. CH3OH + MnO4 → HCOO + MnO2 (in base )
C. NO3 − + Zn → NH4+ + Zn2+ (in acid)
D. Br2 → Br + BrO3
− −
(in base)
5. Write the cell notation for the voltaic cell that incorporates the following redox reaction.
Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s)
Chemistry for Grade 12
2+ 2+
6. Drawa voltaic cell that is constructed with a Mn/Mn electrode and a Cd/Cd electrode. Use the
emf table in the textbook to determine which electrode will be the cathode and which will be the
anode. Your drawing should include all of the following components:
2+ 2+
a. Label the location of each substance (Mn, Mn , Cd, and Cd )
b. Label the cathode and the anode
c. Label the direction of electron flow
d. Label which electrode is positively charged and which is negatively charged.
e. Include a salt bridge with NaNO3. Label the direction that each ion flows
f. Write the cell notation for this voltaic cell.
g. Calculate the cell potential of this cell.
7. Given the following half-reactions and half-cell potentials, write the balanced overall
electrochemical reaction that would occur and calculate the cell potential of a voltaic cell
incorporating these two half reactions.
O2(g) + 2 H2O(l) + 4 e− ––→ 4 OH−(aq) E° = +0.40 V
Cr3+ (aq) + 3 e− ––––→ Cr(s) E° = −0.74 V
8. Balance the following skeleton reaction, calculate E°cell, and determine whether the reaction
would be spontaneous as written. You will need to use the emf table in the textbook.
Cr3+(aq) + Cu(s) → Cr(s) + Cu2+(aq)
2+ − 3+
9.Answer the question below regarding the reaction, Cl2(g) + Fe (aq) –––→ Cl (aq) + Fe (aq).
a. Balance the reaction using the half-reaction method.
b. calculate the E°cell and determine whether this reaction would occur in a voltaic cell or an
electrolytic cell.
c. Use E°cell to calculate Kc for this reaction at 25 °C.
d. Use E°cell to calculate ΔG° for this reaction
2+ +
10.Answer the question below regarding the reaction, Pb (aq) + Ag(s) →Pb(s) + Ag (aq).
a. Balance the reaction using the half-reaction method.
b. calculate the E°cell and determine whether this reaction would occur in a voltaic cell or an
electrolytic cell.
c. Use E°cell to calculate Kc for this reaction at 25 °C.
d. Use E°cell to calculate ΔG° for this reaction
11. Calculate the cell potential for a voltaic cell with Pt/Pt2+ and Ag/Ag+ half-cells and the initial
concentrations
[Pt2+] = 0.90 M and [Ag+] = 0.20 M.