8.3 Half-Cell Potentials and Cell Potentials
361
An alternate extrapolation method can also be used. For small values of m the
(m/m ◦ ) 1/2 term negligible compared to unity and the bm term is negligible compared
to the αm 1/2 term. If these terms are omitted, Eq. (8.2-21) becomes
E +
2RT
F
ln
m
m ◦
E
◦
+
2RT
F
αm
1/2
(8.2-22)
If the left-hand side of this equation is plotted as a function of m 1/2 the extrapolation
to m 0 should be nearly linear for small values of m 1/2 , giving the value of E ◦ from
the intercept.
P R O B L E M S
Section 8.2: Electrochemical Cells
8.4 a. Find the reversible cell voltage for the cell of Figure 8.2
at 298.15 K if the partial pressure of hydrogen is equal to
712 torr and the molality of the HCl solution is
0.100 mol kg −1 . Assume the hydrogen gas to be ideal
and use the Davies equation to estimate the activity
coefficient of the HCl.
b. Repeat the calculation of part a using the value
of the activity coefficient from Table A.11 of
Appendix A.
8.5 Find the reversible voltage of the cell of Figure 8.2 at
298.15 K if the HCl solution has a molality of 1.00 mol kg −1
and the pressure of the hydrogen gas is 755 torr. Use the
value of the activity coefficient from Table A.11 of
Appendix A.
8.6 The cell
Cd|CdCl 2 (6.62 m)|Cl 2 (1.000 atm)|Pt
has a cell voltage equal to 1.8111 V at 25 ◦ C. The
6.62 mol kg −1 solution of CdCl 2 is saturated and is at
equilibrium with CdCl 2 ·
5
2
H 2 O and water vapor at 16.5 torr.
The activity coefficient of CdCl 2 at this molality is equal to
0.025. The vapor pressure of pure water at this temperature
is 23.8 torr. Find E ◦ for the cell and ∆G ◦ for the reaction
Cd(s) + Cl 2 (g) +
5
2
H 2 O(l) − → CdCl 2 ·
5
2
H 2 O(s)
8.3
Half-Cell Potentials and Cell Potentials
Figure 8.3 depicts a cell that has a hydrogen electrode on the left side and a calomel
electrode on the right side. The calomel electrode contains liquid mercury in contact
with solid calomel (Hg 2 Cl 2 ). A platinum wire extends from the pool of mercury and
acts as a terminal. The solution in this cell is an aqueous solution of HCl, just as in the
cell of Figure 8.2. The reduction half-reaction at the calomel electrode is
Hg 2 Cl 2 (s) + 2e
−
− → 2Hg(l) + 2Cl
−
(8.3-1)
The oxidation half-reaction of the hydrogen electrode is the same as in Eq. (8.2-1). The
cell reaction equation is
H 2 (g) + Hg 2 Cl 2 (s) − → 2Hg(l) + 2H
+
+ 2Cl
−
The standard-state cell voltage is equal to 0.268 V for this cell.
Exercise 8.5
a. Write the Nernst equation for the cell of Figure 8.3.
b. Write an equation analogous to Eq. (8.2-21) that could be used to determine the value of
E ◦ for the cell of Figure 8.3.
361
An alternate extrapolation method can also be used. For small values of m the
(m/m ◦ ) 1/2 term negligible compared to unity and the bm term is negligible compared
to the αm 1/2 term. If these terms are omitted, Eq. (8.2-21) becomes
E +
2RT
F
ln
m
m ◦
E
◦
+
2RT
F
αm
1/2
(8.2-22)
If the left-hand side of this equation is plotted as a function of m 1/2 the extrapolation
to m 0 should be nearly linear for small values of m 1/2 , giving the value of E ◦ from
the intercept.
P R O B L E M S
Section 8.2: Electrochemical Cells
8.4 a. Find the reversible cell voltage for the cell of Figure 8.2
at 298.15 K if the partial pressure of hydrogen is equal to
712 torr and the molality of the HCl solution is
0.100 mol kg −1 . Assume the hydrogen gas to be ideal
and use the Davies equation to estimate the activity
coefficient of the HCl.
b. Repeat the calculation of part a using the value
of the activity coefficient from Table A.11 of
Appendix A.
8.5 Find the reversible voltage of the cell of Figure 8.2 at
298.15 K if the HCl solution has a molality of 1.00 mol kg −1
and the pressure of the hydrogen gas is 755 torr. Use the
value of the activity coefficient from Table A.11 of
Appendix A.
8.6 The cell
Cd|CdCl 2 (6.62 m)|Cl 2 (1.000 atm)|Pt
has a cell voltage equal to 1.8111 V at 25 ◦ C. The
6.62 mol kg −1 solution of CdCl 2 is saturated and is at
equilibrium with CdCl 2 ·
5
2
H 2 O and water vapor at 16.5 torr.
The activity coefficient of CdCl 2 at this molality is equal to
0.025. The vapor pressure of pure water at this temperature
is 23.8 torr. Find E ◦ for the cell and ∆G ◦ for the reaction
Cd(s) + Cl 2 (g) +
5
2
H 2 O(l) − → CdCl 2 ·
5
2
H 2 O(s)
8.3
Half-Cell Potentials and Cell Potentials
Figure 8.3 depicts a cell that has a hydrogen electrode on the left side and a calomel
electrode on the right side. The calomel electrode contains liquid mercury in contact
with solid calomel (Hg 2 Cl 2 ). A platinum wire extends from the pool of mercury and
acts as a terminal. The solution in this cell is an aqueous solution of HCl, just as in the
cell of Figure 8.2. The reduction half-reaction at the calomel electrode is
Hg 2 Cl 2 (s) + 2e
−
− → 2Hg(l) + 2Cl
−
(8.3-1)
The oxidation half-reaction of the hydrogen electrode is the same as in Eq. (8.2-1). The
cell reaction equation is
H 2 (g) + Hg 2 Cl 2 (s) − → 2Hg(l) + 2H
+
+ 2Cl
−
The standard-state cell voltage is equal to 0.268 V for this cell.
Exercise 8.5
a. Write the Nernst equation for the cell of Figure 8.3.
b. Write an equation analogous to Eq. (8.2-21) that could be used to determine the value of
E ◦ for the cell of Figure 8.3.