South Pasadena • AP Chemistry
Name
7 ▪ Electrochemistry
Period
Date
–
ELECTROLYSIS
7.3
PROBLEMS
1. Possible anode reactions in the electrolysis of
seawater are:
2Cl−(aq)  Cl2(g) + 2e−
Eox = −1.36 V
[Ered = +1.36 V]
+
−
2H2O  4H (aq) + O2(g) + 4e E = −1.229 V
[Ered = +1.229 V]
Possible cathode reactions in the electrolysis of
sea water are:
Na+(aq) + e−  Na(s)
E = −2.714 V
−
−
2H2O + 2e  2OH (aq) + H2(g) E = −0.8277 V
(a) Which two half-reactions will occur when
seawater is electrolyzed?
2H2O  4H+(aq) + O2(g) + 4e− E = −1.229 V
2H2O + 2e−  2OH−(aq) + H2(g) E = −0.8277 V
(b) What is the minimum voltage required for an
external power source to drive this electrolysis
process?
E° = −2.057 V
Power source > +2.057 V
2. Silver plating occurs when electrolysis of a
Ag2SO4 solution is used.
(a) At which electrode is silver metal formed?
(cathode/anode)
Ag+ + e− → Ag
Cathode
(b) Write the equation that occurs at this
electrode.
(c) If a cell is run for 200. s with a current of
0.250 A, how many grams of Ag will be
deposited?
0.25 C1 mol e−1 mol Ag107.9 g Ag
200 s
 s 96485 C 1 mol e−  1 mol Ag 
= 0.0559 g
3. A current of 10.0 amperes flows for 2.00 hours
through an electrolytic cell containing a molten
salt of metal X. This results in the decomposition
of 0.250 mole of metal X at the cathode. What is
the oxidation state of X in the molten salt?
(X+, X2+, X3+, X4+)
3600 s10.0 C1 mol e−
−
2 h
 1 h  s 96485 C = 0.75 mol e
0.75 mol e−
= 3 mol e−/mol X 
X3+
0.25 mol X
4. Solutions of Ag+, Cu2+, Fe3+ and Ti4+ are
electrolyzed with a constant current until 0.10 mol
of metal is deposited. Which will require the
greatest length of time?
Ti4+ will require the greatest time because more
e− need to be gained than the other metals,
when the e− are transferred at the same rate.
5. Electrolysis of a molten metal chloride (MCl3)
using a current of 6.50 A for 1397 seconds
deposits 1.41 g of the metal at the cathode. What
is the metal?
6.50 C1 mol e−1 mol M
1397 s
 s 96485 C 3 mol e− 
= 0.0314 mol M
m
1.41 g M
MM = =
= 44.9 g/mol
n 0.0314 mol M
M = Sc
6. What volume of F2 gas, at 25°C and 1.00 atm is
produced when molten KF is electrolyzed by a
current of 10.0 A for 2.00 hours? What mass of
potassium metal is produced? At which electrode
does the reaction occur?
Anode: 2 F− → F2 + 2 e− Cathode: K+ + e− → K
3600 s10.0 C1 mol e−
−
2 h
 1 h  s 96485 C = 0.746 mol e
1 mol F2
0.746 mol e−
 2 mol e−  = 0.373 mol F2
nRT (0.373 mol)(0.0821)(298 K)
V=
=
= 9.12 L F2
P
(1.00 atm)
1 mol K 39.1 g K
0.746 mol e− 
1 mol e− 1 mol K  = 29.2 g K
AP Chemistry 2013 #2
Answer the following questions involving the stoichiometry and thermodynamics of reactions containing
aluminum species.
2 Al2O3(l) + 3 C(s) → 4 Al(l) + 3 CO2(g)
An electrolytic cell produces 235 g of Al(l) according to the equation above.
(a) Calculate the number of moles of electrons that must be transferred in the cell to produce the 235 g of Al(l).
In this process, Al is going from oxidation state of +3 to 0, so each Al needs to gain 3 e−.
1 mol Al  3 mol e− 
−
235 g Al 
26.98 g Al1 mol Al = 26.1 mol e
(b) A steady current of 152 amp was used during the process. Determine the amount of time, in seconds, that was
needed to produce the Al(l) .
96485 C 1 s 
26.1 mol e−
1 mol e−152 C = 16,600 s
(c) Calculate the volume of CO2(g) , measured at 301 K and 0.952 atm, that is produced in the process.
1 mol Al 3 mol CO2
152 g Al 
26.98 g Al 4 mol Al  = 4.23 mol CO2
nRT (4.23 mol)(0.0821)(301 K)
V=
=
= 110. L
P
(0.952 atm)
(d) For the electrolytic cell to operate, the Al2O3 must be in the liquid state rather than in the solid state. Explain.
In a solid state, ions cannot migrate toward the electrodes to be reduced/oxidized.
When Al(s) is placed in a concentrated solution of KOH at 25°C, the reaction represented below occurs.
2 Al(s) + 2 OH−(aq) + 6 H2O(l) → 2[Al(OH)4]−(aq) + 3 H2(g)
Half-Reaction
E° (V)
[Al(OH)4]−(aq) + 3 e− → Al(s) + 4 OH−(aq)
−2.35
2 H2O(l) + 2 e− → H2(g) + 2 OH−(aq)
−0.83
(d) Using the table of standard reduction potentials shown above, calculate the following.
(i) E°, in volts, for the formation of [Al(OH)4]−(aq) and H2(g) at 25°C
2 Al(s) + 8 OH−(aq) → 2 [Al(OH)4]−(aq) + 6 e−
+2.35 V
6 H2O(l) + 6 e− → 3 H2(g) + 6 OH−(aq)
−0.83 V
−
−
2 Al(s) + 2 OH (aq) + 6 H2O(l) → 2[Al(OH)4] (aq) + 3 H2(g)
+1.52 V
(ii) ∆G°, in kJ/molrxn, for the formation of [Al(OH)4]−(aq) and H2(g) at 25°C
6 mol e− 96485 C+1.52 J
∆G° = −ne−⋅F⋅E° = −
1 mol rxn1 mol e− C  = −880,000 J/molrxn = −880 kJ/molrxn