CHEM1101
2014-J-13
June 2014
• An aqueous solution of 1 M CuSO4 undergoes electrolysis. At the minimum voltage
necessary for a reaction to proceed, what products form at the anode and the cathode?
Explain your answer.
At the cathode, reduction occurs. The two possible reduction reactions are of
Cu2+(aq) and of H2O(l):
Cu2+(aq) + 2e– → Cu(s)
2H2O + 2e– → H2(g) + 2OH– (aq)
Eo = +0.34 V
Eo = –0.83 V
Reduction of Cu2+(aq) has the more positive (least negative) reduction potential:
it is easier and so this reduction occurs first:
Cu(s) produced at cathode: Cu2+(aq) + 2e– → Cu(s)
At the anode, oxidation occurs. In SO42-, sulfur already has its highest possible
oxidation number of +6 and so it cannot be oxidised. H2O(l) is oxidised at the
anode;
O2(g) produced at anode: 2H2O → O2(g) + 4H+(aq) + 4e–
Write a balanced overall reaction for the electrolytic cell.
2Cu2+(aq) + 2H2O(l) → 2Cu(s) + O2(g) + 4H+(aq)
Assuming no overpotential, what would be the minimum voltage required to drive the
overall reaction at a pH of 0?
For the reduction of Cu2+(aq), Eo(reduction) = +0.34 V. For the oxidation of
H2O(l), Eo(oxidation) = -1.23 V.
For the overall cell, Eo = ((-0.83) + (-1.23)) V = -0.89 V
To drive this reaction to occur, a voltage of greater than 0.89 V needs to be
supplied.
Answer: 0.89 V
At a pH of 7, would a higher or lower voltage be required to drive the reaction?
Explain your answer.
The standard oxidation potential of H2O(l) assumes standard concentrations and
so [H+(aq)] = 1.0 M. At pH = 7, [H+(aq)] is much less, so from Le Chatelier’s
principle the forward reaction will be favoured to produce more H+(aq). Hence
lower voltage would be required.
Marks
6
CHEM1101
2014-N-15
November 2014
• Chlorine is produced by the electrolysis of an aqueous sodium chloride solution using inert electrodes. What products are formed at the anode and cathode? Explain
your answer.
There are two possible oxidation reactions at the anode:
2Cl-(aq) à Cl2(g) + 2e2H2O(l) à O2(g) + 4H+(aq) + 4e-
Eo = -1.36 V
Eo = -1.23 V
Although oxidation of water appears easier, based on its less negative potential,
oxidation of chloride occurs due to the overpotential associated with oxidising
water.
There are two possible reduction reactions at the cathode:
Na+(aq) + e- à Na(s)
H2O(l) + 2e - à H2(g) + 2OH-(aq)
Eo = -2.71 V
Eo = -0.83 V
The reduction of Na+(aq) has a much more negative reduction potential and so
reduction of water occurs, despite the overpotential.
Answer: anode: Cl2(g);
cathode: H2(g) + OH-(aq)
Write a balanced equation for the overall reaction of the electrolytic cell.
2H2O(l) + 2Cl-(aq) à Cl2(g) + H2(g) + 2OH-(aq)
Assuming a [Cl–] of 1.0 M and no overpotential, what would be the minimum voltage
required to drive the overall cell reaction at pH 14? Assume gases are at 1 atm.
At pH =14, [OH-(aq)] = 1.0 M. This corresponds to standard conditions for the
reduction for H2O. [Cl-(aq)] = 1.0 M also corresponds to standard conditions for
the oxidation of Cl-(aq).
Ecello = Eredo + Eoxo = (-0.83 V) + (-1.36) V = -2.19 V
A potential of > 2.19 V is required.
Answer: > 2.19 V
Considering the cell potentials suggest a reason ruthenium oxide electrodes are
employed in this reaction rather than carbon electrodes.
RuO2 acts as a catalyst for the reaction.
Marks
7
CHEM1101
2013-J-12
June 2013
• How many hours does it take to form 10.0 L of O2 measured at 99.8 kPa and 28 °C
from water if a current of 1.3 A passes through the electrolysis cell?
10.0 L corresponds to 0.0100 m3. Using PV = nRT, this corresponds to:
n = PV / RT
= (99.8 × 103 Pa) × (0.0100 m3) / (8.314 Pa m3 mol-1 K-1) × ((28 + 273) K)
= 0.399 mol
O2(g) is formed by electrolysis of water according to the reaction:
2H2O(l) à 4H+(aq) + O2(g) + 4eHence, 4 × 0.399 mol = 1.60 mol of electrons are required. As the number of
moles of electrons passed by a current I in a time t is:
number of moles of electrons = It / F
1.60 mol = (1.3 A)t / (96485 C mol-1)
t = 1.2 × 105 s = (1.2 × 105 / 3600) hours = 33 hours
Answer: 33 hours
Marks
3
CHEM1101
2008-J-14
June 2008
 Impure copper can be purified by electrolysis, with the impure copper as one
electrode and the purified copper as the other. Is the impure copper the cathode or the
anode in the electrolysis cell?
The impure copper is the anode. As oxidation occurs at the anode, the impure
copper is oxidized to give Cu2+, leaving the impurities behind. The Cu2+ is then
reduced at the cathode to pure Cu(s).
If a battery is used as the power source, is the positive terminal of the battery
connected to the impure copper or to the pure copper electrode?
In an electrolytic cell, oxidation at the anode must be forced to occur. This is
achieved by making the anode positive so that it removes electrons from the
reactant (impure copper) at the electrode.
If electrolysis for 1.0 hour with a current of 5.2 A produces 5.9 g of pure copper,
calculate the oxidation number of the copper dissolved in the cell.
As the atomic mass of copper is 63.55 g mol-1, 5.9 g corresponds to:
number of moles of copper =
mass
5.9 g
=
= 0.093 mol
-1
atomic mass
63.55 mol
A current, I, of 5.2 A applied for a time, t, of 1.0 hour corresponds to:
number of moles of electons =
(5.2 A)  (1.0  60  60 s)
It
=
= 0.19 mol
-1
F
(96485 C mol )
0.19
 2.09. As the oxidation number
0.093
is a whole number, this corresponds to an oxidation number of +2.
Each mole of copper therefore requires
Oxidation number: +2 (II)
Explain, with the use of standard reduction potentials, why a silver impurity in the
copper can be recovered from the cell as silver metal, but a nickel impurity is found
dissolved in the electrolyte solution.
The relevant reduction potentials are +0.34 V for Cu2+(aq), +0.80 V for Ag+(aq)
and -0.24 V for Ni2+(aq).
ANSWER CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1101
2008-J-14
June 2008
For formation of silver metal,
Cu(s)  Cu2+(aq) + 2e-(aq)
Ag+(aq) + e-(aq)  2Ag(s)
E° = -0.34 V (reversed for oxidation)
E° = +0.80 V
A voltage of 0.34 V is sufficient to dissolve copper but not to dissolve silver. The
silver thus remains as solid Ag(s). The reaction of Ag(s) with Cu2+(aq) is
unfavourable as the reverse reaction is favourable:
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) E° = ((-0.34) + (+0.80)) V = +0.46 V
As E° > 0, the reaction should occur. Silver metal can thus be formed in the
presence of copper metal.
For formation of nickel metal,
Cu(s)  Cu2+(aq) + 2e-(aq)
Ni2+(aq) + 2e-(aq)  Ni(s)
E° = -0.34 V (reversed for oxidation)
E° = -0.24 V
The overall reaction is:
Cu(s) + Ni2+(aq)  Cu2+(aq) + Ni(s)
E° = ((-0.34) + (-0.24)) V = -0.58 V
As E° < 0, the reaction will not occur. Nickel metal cannot be formed from Ni2+
in the presence of copper metal.
Explain what happens to an iron impurity in the Cu.
The reduction potential for Fe2+(aq) is -0.44 V. Thus, for formation of iron
metal,
Cu(s)  Cu2+(aq) + 2e-(aq)
Fe2+(aq) + 2e-(aq)  Fe(s)
E° = -0.34 V (reversed for oxidation)
E° = -0.44 V
The overall reaction is:
Cu(s) + Fe2+(aq)  Cu2+(aq) + Fe(s)
E° = ((-0.34) + (-0.44)) V = -0.78 V
As E° < 0, the reaction will not occur. Iron metal cannot be formed from Fe2+ in
the presence of copper metal. The iron impurity will stay in the electrolyte
solution as Fe2+.
CHEM1101
2008-J-15
June 2008
 Hydrogen burns or explodes when it reacts with oxygen, but hydrogen and oxygen
can be used to generate electricity safely in a fuel cell.
V
O2
H2
What reaction occurs
at the H2 electrode?
Hydrogen is oxidized:
H2(g) + 2OH-(aq)  4OH-(aq) + 2e- (alkali electrolyte)
H2(g)  2H+(aq) + 2e- (acid electrolyte)
What reaction occurs
at the O2 electrode?
Oxygen is reduced:
O2(g) + 2H2O(l) + 4e-  4OH-(aq) (alkali electrolyte)
O2g) + 4H+(aq) + 4e-  2H2O(l) (acid electrolyte)
In which direction do
the electrons flow?
Electrons flow from the H2 electrode to the O2
electrode.
What is conducted through the inert membrane
from the hydrogen to the oxygen electrode?
Mn+(aq) counter ion such as
K+(aq) (alkali electrolyte)
H+(aq) (acid electrolyte)
What is the maximum voltage that this cell can generate under standard conditions?
By definition, the oxidation half cell (H2) has Eo = 0.00 V. The reduction half cell
has Eo = +1.23 V. Thus, Eo = ((0.00) + (+1.23)) V = +1.23 V
How might this voltage be increased by changing the operating conditions?
Standard conditions correspond to 1 atm pressure. The voltage can be increased
by increasing the pressure of H2(g) and O2(g) gas.
Marks
6
CHEM1101
2008-J-15
June 2008
CHEM1101
2008-N-14
November 2008
 In the refining of copper, impure copper electrodes are electrolysed in a manner such
as described in the following figure. The impure Cu electrode contains varying
amounts of metals such as zinc and iron as well as noble metals like gold, silver and
platinum. Indicate in the boxes on the figure, which electrode is the anode and which
is the cathode.
_
+
anode
(oxidation)
cathode
(reduction)
Battery
Impure Cu
electrode
Cu2+
SO422+
Fe
Zn2+
Pure Cu
electrode
Why are noble metals left as a mud on the bottom of the reaction cell?
Noble metals do not undergo oxidation at the voltage used. As the metals do not
form ions, the solid metal just falls to the bottom as the electrode dissolves. Eox
for noble metals is more negative so Cu (and metals with higher Eox) are
preferentially oxidized.
Explain why Zn2+ and Fe2+ are not deposited from solution during this reaction.
Cu2+ is preferentially reduced and hence preferentially deposited as it has the
highest Ered:
Cu2+(aq) + 2e–  Cu(s) Ered = +0.34 V
Zn2+(aq) + 2e–  Zn(s) Ered = –0.76 V
Fe2+(aq) + 2e–  Fe(s)
Ered = –0.44 V
What mass of pure copper (in kg) will be obtained when the electrolytic cell is
operated for 24.0 hours at a constant current of 100.0 A?
The total charge passed during 24.0 hours at a current of 100.0 A is:
charge = (current) × (time in seconds)
= (100.0 A) × (24.0 × 60 × 60 s) = 8.64 × 106 C.
ANSWER CONTINUES ON THE NEXT PAGE
Marks
6
CHEM1101
2008-N-14
November 2008
Using Faraday’s constant, F = 96485 C mol-1, this charge corresponds to:
(8.64×106 C)
charge
moles of e =
=
= 89.5mol
F
(96485 C mol -1 )
-
The reduction of Cu2+ to Cu(s) requires 2e- so the total moles of Cu(s) obtained is
(½ × 89.5) mol = 44.8 mol. As the atomic mass of Cu is 63.55 g mol-1, this
corresponds to:
mass of copper = moles × atomic mass
= (44.8 mol) × (63.55 g mol-1) = 2850 g = 2.85 kg
Answer: 2.85 kg
CHEM1101
2007-J-12
June 2007
• If 1.00 tonne (103 kg) of aluminium metal is produced by the electrolysis of molten
Al2O3, how many tonnes of carbon dioxide are emitted by oxidation of the carbon
electrodes?
Marks
3
The overall reaction is:
2Al2O3(l) + 3C(s) 4Al(s) + 3CO2(g)
3 moles of CO2 are produced for every 4 moles of Al(s). The atomic mass of
aluminium is 26.98 so the amount of aluminium in 103 kg = 106 g is:
amount of aluminium =
mass
(1.00 × 106 )
=
= 37100 mol
atomic mass
(26.98)
Therefore the amount of CO2 produced is (¾ × 37100 mol) = 27800 mol.
The molar mass of CO2 is (12.01 (C)) + (2 × 16.00) = 44.01. Hence, this amount
of CO2 corresponds to a mass of:
mass = molar mass × number of moles = (27800) × (44.01) = 1.22 × 106
= 1.22 tonnes
Answer: 1.22 tonnes
• The electrolysis of aqueous sodium chloride produces H2(g) and Cl2(g). Circle the
choice that correctly finishes each of the following statements about this process.
The Cl2(g) is produced at the …………………..
anode
cathode
In the aqueous salt solution, the sodium ions
migrate towards the electrode that produces…...
Cl2(g)
H2(g)
As the electrolysis proceeds the pH of the
aqueous salt solution……………………………
increases
decreases
If the process were being run with a battery, the
positive electrode of the battery would be
connected to the electrode that produces ………
Cl2(g)
H2(g)
The two half cells are:
2H2O(l) + 2e- H2(g) + 2OH-(aq)
2Cl-(aq) Cl2(g) + 2e-
reduction (always occurs at the cathode)
oxidation (always occurs at the anode)
The cathode is negatively charged so Na+(aq) migrate to the cathode where H2O
is preferentially reduced to H2(g).
OH-(aq) is produced at the cathode so the pH increases.
The anode is positively charged with the positive electrode of the battery is
connected to it. Cl2(g) is produced at the anode.
4
CHEM1101
2007-N-9
November 2007
• Indicate the relative entropy of each system in the following pairs of systems.
Use: “>”, “<”, or “=”.
CO2(g)
>
CO2(s)
O2(g) + H2O(l)
>
O2(aq)
hexane, C6H14(g)
>
pentane, C5H12(g)
3O2(g)
>
2O3(aq)
• An electrolytic cell contains a solution of MCl3. A total charge of 3600 C is passed
through the cell, depositing 0.65 g of the metal, M, at the cathode. What is the
identity of the metal, M?
A charge of 3600 C corresponds to:
number of moles of electrons =
charge (C)
3600
mol
Faraday's constant (Cmol ) 96485
-1
=
= 0.037 mol
As the metal, M, has a +3 charge in MCl3, three moles of electrons are required
to reduce each mole of M3+. The number of moles of M deposited is therefore:
number of moles of M =
0.037
mol = 0.012mol
3
As 0.012 mol has a mass of 0.65 g, the atomic mass of M must be:
atomic mass =
mass
0.65 g
=
= 52 g mol -1
number of moles 0.012 mol
This corresponds to chromium.
Answer: Chromium
Marks
2
3
CHEM1101
2006-N-11
November 2006
• A chemical engineer dissolves a mixture of NaBr and MgCl2 in water and decomposes
it in an electrolytic cell. Predict the substance formed at each electrode and write
balanced half reactions and the overall cell reaction.
Marks
4
The possible reduction reactions at the cathode are:
Na+(aq) + e- Na(s)
E0 = -2.71 V
Mg2+(aq) + 2e- Mg(s)
E0 = -2.36 V
2H2O(l) + 2e H2(g) + 2OH (aq) E = -0.42 V (at pH = 7)
The reduction potential for water is the least negative. The overpotential for
water is around 0.6 V so the potential needed to reduce water is ~ -1 V. This is
still smaller than for reduction of the cations. H2(g) is produced at the cathode.
The possible oxidation reactions at the anode are:
E0 = -1.10 V
2Br-(aq) Br2(aq) + 2e2Cl (aq) Cl2(aq) + 2e
E0 = -1.36 V
2H2O(l) O2(g) + 4H+(aq) + 2e- E = -0.82 V (at pH = 7)
The oxidation potential for water is the least negative The overpotential for
water is around 0.6 V so the potential actually needed to oxidise water is ~-1.4 V.
Because of this overpotential, Br2(aq) is actually produced at the anode.
The overall reaction is therefore:
2H2O + 2Br-(aq) H2(g) + 2OH-(aq) + Br2(aq)
• What mass of PbSO4 is reduced at the cathode when a lead-acid storage battery is
charged for 1.5 hours with a constant current of 10.0 A?
The total charge delivered is:
Q = It = (10.0) × (1.5 × 60 × 60) = 54000 A
This is equivalent to
Q 54000
=
= 0.56 moles of electrons.
F 96485
The reduction reaction during recharging is a 2e- process:
PbSO4 + H+ + 2e- Pb + HSO40.56
= 0.28 moles of PbSO4 are reduced. The molar mass of PbSO4 is
2
(207.2 (Pb)) + (32.07 (S)) + (4 × 16.00 (O)) = 303.27. Hence, the mass reduced is
given by:
Hence,
mass of PbSO4 reduced = 0.28 × 303.27 = 85 g
3
CHEM1101
2005-J-10
June 2005
 Calculate the mass of aluminium metal that would be produced by the
electroreduction of Al3+ by a current of 2.5 105 A for a period of 1.0 hour.
Marks
4
The total charge passed during 1.0 hours at a current of 2.5 × 105 A is:
charge = (current) × (time in seconds)
= (2.5 × 105 A) × (1.0 × 60 × 60 s) = 9.0 × 108 C.
Using Faraday’s constant, F = 96485 C mol-1, this charge corresponds to:
charge
9.0×109 C
=
= 9300mol
moles of e =
F
96485C mol -1
The reduction of Al3+ to Al(s) requires 3e- so the total moles of Al(s) obtained is
(⅓ × 9300) mol = 3100 mol. As the atomic mass of Al is 26.98 g mol-1, this
corresponds to:
mass of aluminium = moles × atomic mass
= (3100 mol) × (26.98 g mol-1) = 84 kg
Answer: 84 kg
Explain why, in the Hall-Heroult process, a molten mixture of Al2O3 and Na3AlF6 is
electrolysed, rather than either an aqueous solution of Al3+ or molten Al2O3.
Water is reduced rather than aluminium ions due to the relevant reduction
potentials, so electrolysis of aqueous solution of Al3+ ions will not produce Al(s).
2H2O + 2e–  2OH–(aq) + H2(g)
E° = -0.83 V
Al3+(aq) + 3e–  Al(s)
E° = -1.68 V
Al2O3 has a very high melting point. Adding cryolite (Na3AlF6) produces a
mixture of much lower melting point (an important energy consideration in
industrial processes) and provides a source of ions, Na+ and [AlF6]3–, so that the
melt conducts a current.
 In the chlor-alkali process OH–(aq) and Cl2(g) are produced from the electrolysis of a
saturated solution of sodium chloride. Write the half-reactions for the production of
each of these.
OH–
2H2O + 2e–  2OH–(aq) + H2(g)
Cl2
2Cl–(aq)  Cl2(g) + 2e–
ANSWER CONTINUES ON THE NEXT PAGE
4
Compare the oxidation potential of Cl– to that of water and explain why Cl– is
oxidised preferentially.
E°ox for Cl– ions (above reaction) is –1.36 V.
The E°ox for water is –1.23 V:
(2H2O  O2(g) + 4H+(aq) + 2e–)
Water should be oxidised preferentially based on E°ox values, but there is an
overpotential of about 0.6 V associated with production of O2(g). This means
that in practice, Cl2(g) is formed instead.
CHEM1101
2003-N-13
November 2003
 In the refining of copper, impure copper electrodes are electrolysed in a manner such
as described in the following figure. Indicate in the boxes on the figure, which
electrode is the anode and which is the cathode.
cathode
anode
Impure Cu
electrode
Pure Cu
electrode
Mud from noble metals
Why are noble metals left as a mud on the bottom of the reaction cell?
Noble metals do not undergo oxidation at the voltage used. As the metals do not
form ions, the solid metal just falls to the bottom as the electrode dissolves.
Explain why Zn2+ and Fe2+ are not deposited from solution during this reaction.
Cu2+ has a higher electrode reduction potential so is more readily reduced.
How many kilograms of pure copper will be obtained when the electrolytic cell is
operated for 24.0 hours at a constant current of 100.0 A?
The total charge passed during 24.0 hours at a current of 100.0 A is:
charge = (current) × (time in seconds)
= (100.0 A) × (24.0 × 60 × 60 s) = 8.64 × 106 C.
Using Faraday’s constant, F = 96485 C mol-1, this charge corresponds to:
moles of e- =
(8.64×106 C)
charge
=
= 89.5mol
F
(96485 C mol -1 )
The reduction of Cu2+ to Cu(s) requires 2e- so the total moles of Cu(s) obtained is
(½ × 89.5) mol = 44.8 mol. As the atomic mass of Cu is 63.55 g mol-1, this
corresponds to:
mass of copper = moles × atomic mass
= (44.8 mol) × (63.55 g mol-1) = 2850 g = 2.85 kg
Answer: 2850 g or 2.85 kg
Marks
6