South Pasadena • AP Chemistry
Name
7 ▪ Electrochemistry
Period
7.2
PROBLEMS
–
Date
NERNST EQUATION
1. Use the Table of Standard Reduction Potentials,
determine whether the following statements are
true or false.
a. Magnesium will react with water to produce
hydrogen gas.
Mg (s) → Mg2+ + 2 e−
E° = +2.37 V
−
−
2 H2O + 2 e → H2(g) + 2 OH E° = −0.83 V
E° = +1.54 V Spontaneous
b. A piece of nickel immersed in a solution of
silver nitrate will become coated with silver.
Ni(s) → Ni2+ + 2 e−
E° = +0.25 V
+
−
Ag + e → Ag(s)
E° = +0.80 V
E° = +1.05 V Spontaneous
c. In basic solution, manganese(IV) oxide will
oxidize mercury to mercury(II) oxide.
MnO2 + 4 H+ + 2 e− → Mn2+ + 2 H2O
E° = +1.21 V
2+
−
Hg → Hg + 2 e
E° = −0.85 V
E° = +0.36 V Spontaneous
d. The iodate ion in acidic solution will oxidize
copper metal.
IO3− + 6 H+ + 5 e− → ½ I2 + 3 H2O
E° = +1.20 V
2+
−
Cu → Cu + 2 e
E° = −0.34 V
E° = +0.86 V Spontaneous
e. Nitric acid will oxidize tin to Sn2+, producing
nitric oxide, NO.
NO3− + 4 H+ + 3 e− → NO + 2 H2O
E° = +0.96 V
2+
−
Sn → Sn + 2 e
E° = +0.14 V
E° = +1.10 V Spontaneous
a. Write the equation for the overall reaction, and
calculate the standard cell potential.
Ox:
2 Al → 2 Al3+ + 6 e−
E° = +1.66 V
2+
−
Red: 3 Zn + 6 e → 3 Zn E° = −0.76 V
2 Al + 3 Zn2+ → 3 Zn + 2 Al3+ E° = +0.90 V
2. Consider the electrochemical cell shown below:
d. Consider the electrochemical cell with
[Zn2+] = 2.0 M and [Al3+] = 0.10 M. Calculate
the value of Ecell at 25°C with these
concentrations.
[Al3+]2 (0.10)2
Q=
=
= 1.25 × 10−3
[Zn2+]3 (2.0)3
R·T
E = E° −
ln Q
ne−·F
= (0.90 V) −
 (8.314 J/mol-K)(298 K)  ln(0.00125)
(6 mol e−)(96500 C/mol e−)
= 0.93 V
b. Calculate the value of the equilibrium
constant.
K = eE°·n·F/R·T
(+0.90 V)(6 mol e−)(96500 C/mol e−)
=e
(8.314 J/mol·K)(298 K)
= 2.1 × 1091
= e210.3
c. For each of the following changes, determine
whether the voltage of the cell would increase,
decrease, or remain the same.
[Al3+]2
i. [Al3+] = 2.0 M
Q=
[Zn2+]3
Q > 1, so Ecell decreases
ii. [Zn2+] = 2.0 M
Q < 1, so Ecell increases
iii. [Al3+] = 0.10 M
Q < 1, so Ecell increases
iv. [Zn2+] = 0.10 M
Q > 1, so Ecell decreases
v. A larger piece of Al° is used.
Remains the same because Q is
unchanged
vi. A larger piece of Zn° is used.
Remains the same because Q is
unchanged
3. Consider the reaction of copper metal with silver
ions in a solution of silver nitrate.
a. Write the oxidation and reduction halfreactions, and the balanced overall net ionic
equation for this reaction.
Cu → Cu2+ + 2 e−
E° = −0.34 V
+
−
2 Ag + 2 e → 2 Ag
E° = +0.80 V
+
2+
Cu + 2 Ag → 2 Ag + Cu E° = +0.46 V
b. Calculate the standard cell potential for this
reaction. Is this process spontaneous?
E°cell = +0.46 V
Process IS spontaneous.
c. Calculate the value of the equilibrium constant
at 25°C.
(+0.46 V)(2 mol e−)(96500 C/mol e−)
b. Find the values of K and ∆G°.
∆G° = −ne−·F·E°
2 mol e−  96500 C +0.98 J
= −
1 mol rxn  mol e−   C 
= −189,000 J/mol = −189 kJ/mol
K = eE°·n·F/R·T
K = eE°·n·F/R·T
=e
(8.314 J/mol·K)(298 K)
= 3.6 × 1015
4. A galvanic cell is based on the following half
reactions at 25°C:
Ag+ + e– → Ag
H2O2 + 2 H+ + 2e– → 2 H2O
a. Write the overall net ionic equation for this
reaction. Find the value of E°cell.
H2O2 + 2 H+ + 2e– → 2 H2O E° = +1.78 V
2 Ag → 2 Ag+ + 2 e−
E° = −0.80 V
+
H2O2 + 2 H + 2 Ag → 2 H2O + 2 Ag+
E° = +0.98 V
(+0.98 V)(2 mol e−)(96500 C/mol e−)
= e35.8
d. Calculate the value of ∆G° for this reaction.
∆G° = −ne−·F·E°
2 mol e−  96500 C +0.46 J
= −
1 mol rxn  mol e−   C 
= −89,000 J/mol = −89 kJ/mol
=e
(8.314 J/mol·K)(298 K)
= 1.4 × 1033
= e76.3
c. Predict whether Ecell is larger or smaller than
E°cell for each of these cases. Then calculate
Ecell.
i. [Ag+] = 1.0 M, [H2O2] = 2.0 M,
[H+] = 2.0 M
[Ag+]2
(1.0)2
Q=
= 0.125
+ 2=
[H2O2][H ] (2.0)(2.0)2
Since Q < 1, E > E°.
R·T
E = E° −
ln Q
ne−·F
= (0.98 V) −
(8.314 J/mol-K)(298 K)
ln(0.125)
(2 mol e−)(96500 C/mol e−)
= +1.01 V
ii. [Ag+] = 2.0 M, [H2O2] = 1.0 M,
[H+] = 1.0 × 10–7 M
[Ag+]2
(2.0)2
Q=
+ 2=
[H2O2][H ] (1.0)(1.0 × 10−7)2
= 4.0 × 1014
Since Q > 1, E < E°.
R·T
E = E° −
ln Q
ne−·F
= (0.98 V) −
(8.314 J/mol-K)(298 K)
ln(4×1014)
(2 mol e−)(96500 C/mol e−)
= +0.55 V
AP Chemistry 2006B #2
Answer the following questions about voltaic cells.
(a) A voltaic cell is set up using Al/Al3+ as one half-cell and Sn/Sn2+ as the other half-cell. The half-cells contain
equal volumes of solutions and are at standard conditions.
i. Write the balanced net-ionic equation for the spontaneous cell reaction.
3 Sn2+ + 6 e− → 3 Sn
E°red = −0.14 V
2 Al → 2 Al3+ + 6 e− E°ox = +1.66 V
2 Al + 3 Sn2+ → 3 Sn + 2 Al3+
ii. Determine the value, in volts, of the standard potential, E°, for the spontaneous cell reaction.
E°cell = (−0.14 V) + (+1.66 V) = +1.52 V
iii. Calculate the value of the standard free-energy change, ∆G°, for the spontaneous cell reaction. Include
units with your answer.
6 mol e−  96500 C +1.52 J
∆G° = −ne−·F·E° = −
1 mol rxn  mol e−   C  = −880,000 J/mol = −880 kJ/mol
(b) In another voltaic cell with Al/Al3+ and Sn/Sn2+ half cells, [Sn2+] is 0.010 M and [Al3+] is 1.00 M. Calculate
the value, in volts, of the cell potential, Ecell, at 25°C.
[Al3+]2 (1.00)2
Q=
=
= 1.0 × 106
[Sn2+]3 (0.010)3
R·T
(8.314 J/mol-K)(298 K)
E = E° −
ln Q = (+1.52 V) −
ln(1.0 × 106) = +1.46 V
ne−·F
(6 mol e−)(96500 C/mol e−)