Electrochemistry

Chapter 18

1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
 Electrochemical processes are oxidation-reduction reactions
in which:
• the energy released by a spontaneous reaction is
converted to electricity or
• electrical energy is used to cause a nonspontaneous
reaction to occur
0 0 2+ 2-
2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-)

O2 + 4e- 2O2- Reduction half-reaction (gain e-)

2
 Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation

number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to
the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
3
4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.

5. Group IA metals are +1, IIA metals are +2 and fluorine is

always –1.

6. The sum of the oxidation numbers of all the atoms in a

molecule or ion is equal to the charge on the
molecule or ion.

4
 Balancing Redox Equations

The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?

1. Write the unbalanced equation for the reaction in ionic form.

Fe2+ + Cr2O72- Fe3+ + Cr3+

2. Separate the equation into two half-reactions.

+2 +3
Oxidation: Fe2+ Fe3+
+6 +3
Reduction: Cr2O7 2- Cr3+

3. Balance the atoms other than O and H in each half-reaction.

Cr2O72- 2Cr3+
5
 Balancing Redox Equations
4. For reactions in acid, add H2O to balance O atoms and H+ to
balance H atoms.
Cr2O72- 2Cr3+ + 7H2O
14H+ + Cr2O72- 2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the
charges on the half-reaction.
Fe2+ Fe3+ + 1e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-
reactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+ 6Fe3+ + 6e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6
 Balancing Redox Equations
7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both
sides must cancel.

Oxidation: 6Fe2+ 6Fe3+ + 6e-

Reduction: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are balanced.

14x1 – 2 + 6 x 2 = 24 = 6 x 3 + 2 x 3

9. For reactions in basic solutions, add OH- to both sides of the

equation for every H+ that appears in the final equation.
7
Example 18.1

Write a balanced ionic equation to represent the oxidation of

iodide ion (I-) by permanganate ion ( MnO-4 ) in basic solution to
yield molecular iodine (I2) and manganese(IV) oxide (MnO2).

8
Example 18.1

Strategy

We follow the preceding procedure for balancing redox

equations. Note that the reaction takes place in a basic
medium.

Solution

Step 1: The unbalanced equation is

MnO-4 + I- MnO2 + I2

9
Example 18.1

Step 2: The two half-reactions are

-1 0
Oxidation: I- I2
+7 +4
Reduction: MnO-4 MnO2

Step 3: We balance each half-reaction for number and type of

atoms and charges. Oxidation half-reaction: We first
balance the I atoms:

2I- I2

10
Example 18.1

To balance charges, we add two electrons to the right-hand

side of the equation:

2I- I2 + 2e-

Reduction half-reaction: To balance the O atoms, we add two

H2O molecules on the right:

MnO-4 MnO2 + 2H2O

To balance the H atoms, we add four H+ ions on the left:

MnO-4 + 4H+ MnO2 + 2H2O

11
Example 18.1
There are three net positive charges on the left, so we add
three electrons to the same side to balance the charges:

MnO-4 + 4H+ + 3e- MnO2 + 2H2O

Step 4: We now add the oxidation and reduction half reactions

to give the overall reaction. In order to equalize the
number of electrons, we need to multiply the oxidation
half-reaction by 3 and the reduction half-reaction by 2 as
follows:

3(2I- I2 + 2e-)
-
2( MnO4 + 4H+ + 3e- MnO2 + 2H2O)
________________________________________
-
6I- + MnO4 + 8H+ + 6e- 3I2 + 2MnO2 + 4H2O + 6e- 12
Example 18.1

The electrons on both sides cancel, and we are left with the
balanced net ionic equation:

-
6I- + 2MnO4 + 8H+ 3I2 + 2MnO2 + 4H2O

This is the balanced equation in an acidic medium. However,

because the reaction is carried out in a basic medium, for every
H+ ion we need to add equal number of OH- ions to both sides
of the equation:
-
6I- + 2 MnO4 + 8H+ + 8OH- 3I2 + 2MnO2 + 4H2O + 8OH-

13
Example 18.1

Finally, combining the H+ and OH- ions to form water, we obtain

6I- + 2MnO-4 + 4H2O 3I2 + 2MnO2 + 8OH-

Step 5: A final check shows that the equation is balanced in

terms of both atoms and charges.

14
 Galvanic Cells

anode cathode
oxidation reduction

spontaneous
redox reaction

15
 Galvanic Cells
The difference in electrical
potential between the anode
and cathode is called:
• cell voltage
• electromotive force (emf)
• cell potential
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M and [Zn2+] = 1 M

Cell Diagram

phase boundary
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode salt bridge cathode 16
 Standard Reduction Potentials

Standard reduction potential (E0) is the voltage associated

with a reduction reaction at an electrode when all solutes
are 1 M and all gases are at 1 atm.

Reduction Reaction

2e- + 2H+ (1 M) H2 (1 atm)

E0 = 0 V

17
Standard hydrogen electrode (SHE)
 Standard Reduction Potentials

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e-

Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) + 2H+ (1 M) Zn2+ (1 M) + H2 (1 atm)
18
 Standard Reduction Potentials
0 = 0.76 V
Ecell

0 )
Standard emf (Ecell
0 = E0
Ecell 0
cathode - Eanode

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

0 = E 0 + - E 0 2+
Ecell H /H2 Zn /Zn

0.76 V = 0 - EZn0 2+
/Zn
0 2+
EZn /Zn = -0.76 V

Zn2+ (1 M) + 2e- Zn E0 = -0.76 V 19

 Standard Reduction Potentials

0 = 0.34 V
Ecell
0 = E0
Ecell 0
cathode - Eanode

Cu /Cu – EH +/H 2
0 = E 0 2+
Ecell 0

0 2+
0.34 = ECu /Cu - 0
0 2+
ECu /Cu = 0.34 V

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

Anode (oxidation): H2 (1 atm) 2H+ (1 M) + 2e-
Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M) 20
• E0 is for the reaction as
written
• The more positive E0 the
greater the tendency for the
substance to be reduced
• The half-cell reactions are
reversible
• The sign of E0 changes
when the reaction is
reversed
• Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0 21
Example 18.2

Predict what will happen if molecular bromine (Br2) is added to

a solution containing NaCl and NaI at 25°C. Assume all species
are in their standard states.

22
Example 18.2
Strategy To predict what redox reaction(s) will take place, we
need to compare the standard reduction potentials of Cl2, Br2,
and I2 and apply the diagonal rule.

Solution From Table 18.1, we write the standard reduction

potentials as follows:

Cl2(1 atm) + 2e- 2Cl-(1 M) E° = 1.36 V

Br2(l) + 2e- 2Br-(1 M) E° = 1.07 V

I2(s) + 2e- 2I-(1 M) E° = 0.53 V

23
Example 18.2
Applying the diagonal rule we see that Br2 will oxidize I- but will
not oxidize Cl-. Therefore, the only redox reaction that will occur
appreciably under standard-state conditions is

Oxidation: 2I-(1 M) I2(s) + 2e-

Reduction: Br2(l) + 2e- 2Br-(1 M)
______________________________________________
Overall: 2I-(1 M) + Br2(l) I2(s) + 2Br-(1 M)

Check We can confirm our conclusion by calculating E°cell. Try

it. Note that the Na+ ions are inert and do not enter into the
redox reaction.

24
Example 18.3

A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2

solution and a Ag electrode in a 1.0 M AgNO3 solution.
Calculate the standard emf of this cell at 25°C.

25
Example 18.3

Strategy At first it may not be clear how to assign the

electrodes in the galvanic cell. From Table 18.1 we write the
standard reduction potentials of Ag and Mg and apply the
diagonal rule to determine which is the anode and which is the
cathode.

Solution The standard reduction potentials are

Ag+(1.0 M) + e- Ag(s) E° = 0.80 V

Mg2+(1.0 M) + 2e- Mg(s) E° = -2.37 V

26
Example 18.3

Applying the diagonal rule, we see that Ag+ will oxidize Mg:

Anode (oxidation): Mg(s) Mg2+(1.0 M) + 2e-

Cathode (reduction): 2Ag+(1.0 M) + 2e- 2Ag(s)
_________________________________________________
Overall: Mg(s) + 2Ag+(1.0 M) Mg2+(1.0 M) + 2Ag(s)

27
Example 18.3
Note that in order to balance the overall equation we multiplied
the reduction of Ag+ by 2. We can do so because, as an
intensive property, E° is not affected by this procedure. We find
the emf of the cell by using Equation (18.1) and Table 18.1:

E°cell = E°cathode - E°anode

= E°Ag+/Ag - E°Mg2+/Mg
= 0.80 V - (-2.37 V)
= 3.17 V

Check The positive value of E° shows that the forward reaction

is favored.

28
 Spontaneity of Redox Reactions
DG = -nFEcell n = number of moles of electrons in reaction
J
DG0 = 0
-nFEcell F = 96,500 = 96,500 C/mol
V • mol
DG0 = -RT ln K = -nFEcell
0

RT (8.314 J/K•mol)(298 K)
0 =
Ecell ln K = ln K
nF n (96,500 J/V•mol)

0 0.0257 V
Ecell = ln K
n
0 0.0592 V
Ecell = log K
n

29
Spontaneity of Redox Reactions

DG0 = -RT ln K = -nFEcell

0

30
Example 18.4

Calculate the equilibrium constant for the following reaction at

25°C:

Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)

31
Example 18.4

Strategy

The relationship between the equilibrium constant K and the

standard emf is given by Equation (18.5):

E°cell = (0.0257 V/n)ln K

Thus, if we can determine the standard emf, we can calculate

the equilibrium constant. We can determine the E°cell of a
hypothetical galvanic cell made up of two couples (Sn2+/Sn and
Cu2+/Cu+) from the standard reduction potentials in Table 18.1.

32
Example 18.4
Solution

The half-cell reactions are

Anode (oxidation): Sn(s) Sn2+(aq) + 2e-

Cathode (reduction): 2Cu2+(aq) + 2e- 2Cu+(aq)

E°cell = E°cathode - E°anode

= E°Cu2+/Cu+ - E°Sn2+/Sn
= 0.15 V - (-0.14 V)
= 0.29 V

33
Example 18.4

Equation (18.5) can be written

o
nE
lnK =
0.0257 V

In the overall reaction we find n = 2. Therefore,

(2)(0.29V)
lnK = = 22.6
0.0257 V

K = e22.6 = 7 ×109
34
Example 18.5

Calculate the standard free-energy change for the following

reaction at 25°C:

2Au(s) + 3Ca2+(1.0 M) 2Au3+(1.0 M) + 3Ca(s)

35
Example 18.5

Strategy

The relationship between the standard free energy change and

the standard emf of the cell is given by Equation (18.3):
ΔG° = -nFE°cell. Thus, if we can determine E°cell, we can
calculate ΔG°. We can determine the E°cell of a hypothetical
galvanic cell made up of two couples (Au3+/Au and Ca2+/Ca)
from the standard reduction potentials in Table 18.1.

36
Example 18.5
Solution

The half-cell reactions are

Anode (oxidation): 2Au(s) 2Au3+(1.0 M) + 6e-

Cathode (reduction): 3Ca2+(1.0 M) + 6e- 3Ca(s)

E°cell = E°cathode - E°anode

= E°Ca2+/Ca - E°Au3+/Au
= -2.87 V - 1.50 V
= -4.37 V

37
Example 18.5
Now we use Equation (18.3):

ΔG° = -nFE°

The overall reaction shows that n = 6, so

G° = -(6) (96,500 J/V · mol) (-4.37 V)

= 2.53 x 106 J/mol
= 2.53 x 103 kJ/mol

Check The large positive value of ΔG° tells us that the reaction
favors the reactants at equilibrium. The result is consistent with
the fact that E° for the galvanic cell is negative.
38
 The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0

-nFE = -nFE0 + RT ln Q

Nernst equation

RT
E = E0 - ln Q
nF

At 298 K

0.0257 V 0.0592 V
E = E0 - ln Q E = E0 - log Q
n n

39
Example 18.6

Predict whether the following reaction would proceed

spontaneously as written at 298 K:

Co(s) + Fe2+(aq) Co2+(aq) + Fe(s)

given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.

40
Example 18.6

Strategy

Because the reaction is not run under standard-state conditions

(concentrations are not 1 M), we need Nernst’s equation
[Equation (18.8)] to calculate the emf (E) of a hypothetical
galvanic cell and determine the spontaneity of the reaction. The
standard emf (E°) can be calculated using the standard
reduction potentials in Table 18.1. Remember that solids do not
appear in the reaction quotient (Q) term in the Nernst equation.
Note that 2 moles of electrons are transferred per mole of
reaction, that is, n = 2.

41
Example 18.6
Solution

The half-cell reactions are

Anode (oxidation): Co(s) Co2+(aq) + 2e-

Cathode (reduction): Fe2+(aq) + 2e- Fe(s)

E°cell = E°cathode - E°anode

= E°Fe2+/Fe - E°Co2+/Co
= -0.44 V – (-0.28 V)
= -0.16 V

42
Example 18.6
From Equation (18.8) we write

o 0.0257 V
E =E - lnQ
n
2+
o 0.0257 V [Co ]
=E - ln
n [Fe 2+ ]
0.0257 V 0.15
= -0.16 V - ln
2 0.68
= -0.16 V + 0.019 V
= -0.14 V

Because E is negative, the reaction is not spontaneous in the

direction written.
43
Example 18.7

Consider the galvanic cell shown in Figure 18.4(a). In a certain

experiment, the emf (E) of the cell is found to be 0.54 V at
25°C. Suppose that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate
the molar concentration of H+.

44
Example 18.7

Strategy

The equation that relates standard emf and nonstandard emf is

the Nernst equation. The overall cell reaction is

Zn(s) + 2H+(? M) Zn2+(1.0 M) + H2(1.0 atm)

Given the emf of the cell (E), we apply the Nernst equation to
solve for [H+]. Note that 2 moles of electrons are transferred per
mole of reaction; that is, n = 2.

45
Example 18.7
Solution As we saw earlier, the standard emf (E°) for the cell is
0.76 V. From Equation (18.8) we write
0.0257 V
E = Eo - lnQ
n
2+
o 0.0257 V [Zn ]PH 2
=E - ln
n [H + ]2
0.0257 V (1.0)(1.0)
0.54 V= 0.76 V - ln
2 [H + ]2
0.0257 V 1
-0.22 V = - ln + 2
2 [H ]
1
17.1 = ln + 2
[H ]
1
e171. =
[H + ]2
1
[H + ]= 7
= 2×10 -4
M 46
3×10
Example 18.7

Check

The fact that the nonstandard-state emf (E) is given in the

problem means that not all the reacting species are in their
standard-state concentrations. Thus, because both Zn2+ ions
and H2 gas are in their standard states, [H+] is not 1 M.

47
 Concentration Cells

Galvanic cell from two half-cells composed of the same

material but differing in ion concentrations.

48
 Batteries

Dry cell

Leclanché cell

Anode: Zn (s) Zn2+ (aq) + 2e-

Cathode: 2NH+4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)

Zn (s) + 2NH4+ (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3(s)

49
 Batteries

Mercury Battery

Anode: Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-

Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)

Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

50
 Batteries

Lead storage
battery

Anode: Pb (s) + SO2-4 (aq) PbSO4 (s) + 2e-

Cathode: PbO2 (s) + 4H+ (aq) + SO2-

4 (aq) + 2e
- PbSO4 (s) + 2H2O (l)

Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO42- (aq) 2PbSO4 (s) + 2H2O (l)

51
 Batteries

52
Solid State Lithium Battery
 Batteries

A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning

Anode: 2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-

Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)

2H2 (g) + O2 (g) 2H2O (l)

53
Chemistry In Action: Bacteria Power
CH3COO- + 2O2 + H+ 2CO2 + 2H2O

54
 Corrosion
Corrosion is the term usually applied to the deterioration of
metals by an electrochemical process.

55
Cathodic Protection of an Iron Storage Tank

56
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.

Electrolysis of molten NaCl

57
Electrolysis of Water

58
Example 18.8
An aqueous Na2SO4 solution is electrolyzed, using the
apparatus shown in Figure 18.18. If the products formed at the
anode and cathode are oxygen gas and hydrogen gas,
respectively, describe the electrolysis in terms of the reactions
at the electrodes.

59
Example 18.8

Strategy

Before we look at the electrode reactions, we should consider

the following facts: (1) Because Na2SO4 does not hydrolyze, the
pH of the solution is close to 7. (2) The Na+ ions are not
reduced at the cathode and the SO2- 4 ions are not oxidized at
the anode. These conclusions are drawn from the electrolysis of
water in the presence of sulfuric acid and in aqueous sodium
chloride solution, as discussed earlier. Therefore, both the
oxidation and reduction reactions involve only water molecules.

60
Example 18.8
Solution

The electrode reactions are

Anode: 2H2O(l) O2(g) + 4H+(aq) + 4e-

Cathode: 2H2O(l) + 2e- H2(g) + 2OH-(aq)

The overall reaction, obtained by doubling the cathode reaction

coefficients and adding the result to the anode reaction, is

6H2O(l) 2H2(g) + O2(g) + 4H+(aq) + 4OH-(aq)

61
Example 18.8

If the H+ and OH- ions are allowed to mix, then

4H+(aq) + 4OH-(aq) 4H2O(l)

and the overall reaction becomes

2H2O(l) 2H2(g) + O2(g)

62
Electrolysis and Mass Changes

charge (C) = current (A) x time (s)

1 mol e- = 96,500 C

63
 Chemistry In Action: Dental Filling Discomfort

2+
Hg2 /Ag2Hg3 0.85 V
2+
Sn /Ag3Sn -0.05 V
2+
Sn /Ag3Sn -0.05 V

64
Example 18.9

A current of 1.26 A is passed through an electrolytic cell

containing a dilute sulfuric acid solution for 7.44 h. Write the
half-cell reactions and calculate the volume of gases generated
at STP.

65
Example 18.9

Strategy

Earlier we saw that the half-cell reactions for the process are

Anode (oxidation): 2H2O(l) O2(g) + 4H+(aq) + 4e-

+ -
1
Cathode (reduction): 4[H (aq) + e H (g)]
2 2
__________________________________________________
Overall: 2H2O(l) 2H2(g) + O2(g)

66
Example 18.9

According to Figure 18.20, we carry out the following

conversion steps to calculate the quantity of O2 in moles:

current x time → coulombs → moles of e- → moles of O2

Then, using the ideal gas equation we can calculate the volume
of O2 in liters at STP. A similar procedure can be used for H2.

67
Example 18.9
Solution

First we calculate the number of coulombs of electricity that

pass through the cell:
3600 s 1C
? C = 1.26 A × 7.44 h × × = 3.37 × 10 4 C
1h 1 A s

Next, we convert number of coulombs to number of moles of

electrons

4 1 mol e-
3.37 × 10 C × = 0.349 mol e-
96,500 C

68
Example 18.9
From the oxidation half-reaction we see that 1 mol O2 =
4 mol e-. Therefore, the number of moles of O2 generated is

-1 mol O2
0.349 mol e × -
= 0.0873 mol O2
4 mol e

The volume of 0.0873 mol O2 at STP is given by

nRT
V =
P
(0.0873 mol)(0.0821 L  atm/K  mol)(273 K)
= = 1.96 L
1 atm

69
Example 18.9

The procedure for hydrogen is similar. To simplify, we combine

the first two steps to calculate the number of moles of H2
generated:
1 mol e- 1 mol H2
4
3.37 × 10 C ×  -
= 0.175 mol H2
96,500 C 2 mol e

The volume of 0.175 mol H2 at STP is given by

nRT
V =
P
(0.175 mol)(0.0821 L  atm/K  mol)(273 K)
= = 3.92 L
1 atm
70
Example 18.9

Check

Note that the volume of H2 is twice that of O2 (see Figure

18.18), which is what we would expect based on Avogadro’s
law (at the same temperature and pressure, volume is directly
proportional to the number of moles of gases).

71