8.3 Half-Cell Potentials and Cell Potentials
369
E X A M P L E 8.5
Assuming that ∆H ◦ is constant, find E ◦ at 323.15 K for the Daniell cell. Neglect the liquid
junction potential.
Solution
E ◦ 1.1030 V at 298.15 K. We find from enthalpy changes of formation in Table A.8 of
Appendix A that ∆H ◦ −218.66 kJ mol −1 . From Eq. (8.3-8)
E ◦ (323.15 K)
323.15 K
1.1030 V
298.15 K
−
−218660 J mol −1
(2)(96485 C mol −1 )
1
323.15 K
−
1
298.15 K
3.699 × 10 −3 V K −1 − 2.94 × 10 −4 V K −1
3.405 × 10 −3 V K −1
E ◦ (323.15 K) (3.405 × 10 −3 V K −1 )(323.15 K) 1.100 V
Exercise 8.9
Find E ◦ at 75 ◦ C for the cell with a hydrogen electrode on the left, a silver–silver chloride
electrode on the right, and a solution of HCl. Assume that ∆H ◦ is temperature-independent.
P R O B L E M S
Section 8.3: Half-Cell Potentials and Cell Potentials
8.7 Find the cell voltage of an electrochemical cell with a
Pb/PbSO 4 electrode on the left, a PbO 2 /PbSO 4 electrode
on the right, and a solution of H 2 SO 4 with a molality of
1.00 mol kg −1 at 298.15 K. Assume that the mean ionic
activity coefficient of H 2 SO 4 at this molality is equal
to 0.36.
8.8 a. Using the extrapolation of Eq. (8.2-21), find E ◦
for the cell of Figure 8.2. The following are
(contrived) data for the cell voltage at 298.15 K
with P(H 2 ) P ◦ :
m/mol kg −1 : 0.0100 0.0200 0.0300 0.0400 0.0500
E/volt:
0.4643 0.4305 0.4108 0.3970 0.3863
m/mol kg −1 : 0.0600 0.0700 0.0800 0.0900 0.1000
E/volt:
0.3776 0.3703 0.3639 0.3583 0.3533
b. Using the extrapolation of Eq. (8.2-22), find E ◦ for the
cell of Figure 8.2.
8.9 a. Write the cell symbol for the cell with the half-reactions
Pb
2+
+ 2e
−
− → Pb(s)
AgCl(s) + Cl
−
− → Ag(s) + e
−
Why is a salt bridge needed? What substance would
you not use in the salt bridge?
b. Draw a sketch of the cell of part a.
c. Find the value of E for the cell if the activity of Pb 2+ is
0.100 and that of the Cl − is 0.200 on the molality scale.
State any assumptions.
8.10 Instead of the oxidation half-reaction used in Example 8.1,
use the reverse of the half-reaction
2H 2 O + 2e
−
− → H 2 + 2OH
− E
◦
−0.8277 V
to calculate the reversible cell potential. Since the OH − ion
is not in its standard state, the Nernst equation must be
applied.
8.11 In calculating the E ◦ value for a cell from half-reaction E ◦
values, the half-reaction E ◦ values are added or subtracted
directly, without multiplying by the number of electrons in
369
E X A M P L E 8.5
Assuming that ∆H ◦ is constant, find E ◦ at 323.15 K for the Daniell cell. Neglect the liquid
junction potential.
Solution
E ◦ 1.1030 V at 298.15 K. We find from enthalpy changes of formation in Table A.8 of
Appendix A that ∆H ◦ −218.66 kJ mol −1 . From Eq. (8.3-8)
E ◦ (323.15 K)
323.15 K
1.1030 V
298.15 K
−
−218660 J mol −1
(2)(96485 C mol −1 )
1
323.15 K
−
1
298.15 K
3.699 × 10 −3 V K −1 − 2.94 × 10 −4 V K −1
3.405 × 10 −3 V K −1
E ◦ (323.15 K) (3.405 × 10 −3 V K −1 )(323.15 K) 1.100 V
Exercise 8.9
Find E ◦ at 75 ◦ C for the cell with a hydrogen electrode on the left, a silver–silver chloride
electrode on the right, and a solution of HCl. Assume that ∆H ◦ is temperature-independent.
P R O B L E M S
Section 8.3: Half-Cell Potentials and Cell Potentials
8.7 Find the cell voltage of an electrochemical cell with a
Pb/PbSO 4 electrode on the left, a PbO 2 /PbSO 4 electrode
on the right, and a solution of H 2 SO 4 with a molality of
1.00 mol kg −1 at 298.15 K. Assume that the mean ionic
activity coefficient of H 2 SO 4 at this molality is equal
to 0.36.
8.8 a. Using the extrapolation of Eq. (8.2-21), find E ◦
for the cell of Figure 8.2. The following are
(contrived) data for the cell voltage at 298.15 K
with P(H 2 ) P ◦ :
m/mol kg −1 : 0.0100 0.0200 0.0300 0.0400 0.0500
E/volt:
0.4643 0.4305 0.4108 0.3970 0.3863
m/mol kg −1 : 0.0600 0.0700 0.0800 0.0900 0.1000
E/volt:
0.3776 0.3703 0.3639 0.3583 0.3533
b. Using the extrapolation of Eq. (8.2-22), find E ◦ for the
cell of Figure 8.2.
8.9 a. Write the cell symbol for the cell with the half-reactions
Pb
2+
+ 2e
−
− → Pb(s)
AgCl(s) + Cl
−
− → Ag(s) + e
−
Why is a salt bridge needed? What substance would
you not use in the salt bridge?
b. Draw a sketch of the cell of part a.
c. Find the value of E for the cell if the activity of Pb 2+ is
0.100 and that of the Cl − is 0.200 on the molality scale.
State any assumptions.
8.10 Instead of the oxidation half-reaction used in Example 8.1,
use the reverse of the half-reaction
2H 2 O + 2e
−
− → H 2 + 2OH
− E
◦
−0.8277 V
to calculate the reversible cell potential. Since the OH − ion
is not in its standard state, the Nernst equation must be
applied.
8.11 In calculating the E ◦ value for a cell from half-reaction E ◦
values, the half-reaction E ◦ values are added or subtracted
directly, without multiplying by the number of electrons in