CHEM1001
2014-J-10
June 2014
• Rechargeable nickel-cadmium batteries normally operate (discharge) with the
following oxidation and reduction half-cell reactions.
Cd(OH)2(s) + 2e– → Cd(s) + 2OH–(aq)
E° (1) = –0.82 V
AlF63–(aq) + 3e– → Al(s) + 6F–(aq)
E° (2) = –2.07 V
Write out a balanced overall cell reaction.
The reduction potential is more negative for the second reaction so this is
reversed. To balance the electrons, the first reaction is multiplied by 3 and the
second reaction is multiplied by 2.
3Cd(OH)2(s) + 2Al(s) + 12F–(aq) → 3Cd(s) + 6OH–(aq) + 2AlF63–(aq)
Calculate the overall cell potential under standard conditions.
With the second reaction reversed, E°(oxidation) = +2.07 V. Hence:
E° = E°(reduction) + E°(oxidation) = ((-0.82) + (+2.07)) V = 1.75 V
Answer: + 1.25 V
A constant current of 3.15 A is measured during the operation of this cell. What
would be the change in mass of the aluminium electrode after 10.0 minutes?
The number of electrons passed by a 3.15 A current in 10.0 minutes is:
number of moles of e- = It / F = (3.15 A) × (10.0 × 60.00 s) / (96485 C mol-1)
= 0.0196 mol
Formation of AlF63- from Al(s) requires 3e- per mol so the number of moles
formed is:
number of moles of Al(s) lost = 0.0196 / 3 mol = 0.00653 mol
Hence the mass of aluminium lost is:
mass of aluminium = number of moles × molar mass
= 0.00653 mol × 26.98 g mol-1 = 0.176 g
Answer: 0.176 g is lost
ANSWER CONTINUES ON THE NEXT PAGE
22/01(a)
Marks
8
CHEM1001
2014-J-10
June 2014
Write out the overall cell reaction that would occur spontaneously if half-cell (1) were
coupled to a standard hydrogen electrode (SHE).
By definition, the standard hydrogen electron has Eo = 0.00 V. As the reduction
half cell (1) has a more negative potential, it is reversed and becomes the
oxidation half cell:
Cd(s) + 2OH–(aq) → Cd(OH)2(s) + 2e–
This is then coupled with the SHE, 2H+(aq) + 2e- → H2(g), to give the overall
reaction:
Cd(s) + 2H2O(l) → Cd(OH)2(s) + H2(g)
What would be the cell potential for this new cell?
The SHE has, by definition, E° (reduction) = 0.00 V. As cell (1) is reversed, E°
(oxidation) = +0.82 V. Hence:
Eo = Eo(reduction) + Eo(oxidation) = (0.00 + 0.82) V = 0.82 V
Answer: 0.82 V
22/01(a)
CHEM1001
2013-J-8
June 2013
• Rechargeable nickel-cadmium batteries normally operate (discharge) with the
following oxidation and reduction half-cell reactions.
Cd(s) + 2OH–(aq) → Cd(OH)2(s) + 2e–
E° = 0.82 V
NiO(OH)(s) + H2O(l) + e– → Ni(OH)2(s) + OH–(aq)
E° = 0.60 V
Write out a balanced overall cell reaction.
Cd(s) + 2NiO(OH)(s) + 2H2O(l) à Cd(OH)2(s) + 2Ni(OH)2(s)
Calculate the overall cell potential.
E° cell = E° oxidation + E° reduction = (0.82 + 0.60) V = +1.42 V
Answer: +1.42 V
Using your balanced cell reaction, briefly explain why the cell potential does not
change as the battery discharges itself.
Cell potentials depend on concentrations (as quantified using the Nernst
equation). The overall cell reaction only involves solids and a pure liquid. The
concentration of solids and pure liquids are constant during the reaction (until
they are used up). As the concentrations do not change, the cell potential remains
constant (until the battery is used up.)
Write out the balanced overall reaction that occurs when this battery is being
recharged.
Cd(OH)2(s) + 2Ni(OH)2(s) à Cd(s) + 2NiO(OH)(s) + 2H2O(l)
A current of 2.75 A is measured during recharging with an external potential of 2.0 V.
After 5.00 minutes charging, how many moles of Cd(s) will be redeposited?
The number of moles of electrons used is given by:
number of moles of e– = It / F
= (2.75 A) × (5.00 × 60.00 s) / (96485 C mol-1)
= 0.00855 mol
From the half-cell reaction, each mole of Cd(s) requires 2 mol of electrons. The
number of moles of Cd(s) redeposited is therefore:
number of moles of Cd(s) = ½ × 0.00855 mol = 0.00428 mol
Answer: 0.00428 mol
22/01(a)
Marks
9
CHEM1001
2012-J-9
June 2012
• H+ is reduced to H2 in an electrochemical cell. What is the total charge transferred
when a current of 2 A is passed through the cell for 20 minutes?
Total charge = current × time = (2 A) × (20 × 60 s) = 2400 C = 2000 C (to 1 s.f.)
Answer: 2000 C
What amount of H2 (in mol) is produced under these conditions?
The equation for the reduction of H+ to H2 is:
2H+(aq) + 2e- à H2(g)
The number of moles of electrons in 2000 C is:
moles of electrons = total charge / Faraday’s constant = 2000 C / 96485 C mol-1
= 0.02 mol
Two moles of electrons is required for each mole of H2. Hence the number of
moles of H2 is half of this:
number of moles of H2 = (1/2) × 0.02 mol = 0.01 mol
Answer: 0.01 mol
What volume would this gas occupy at 25 °C and 90 kPa?
As 1 atm = 101.3 kPa, 90 kPa corresponds to:
pressure = (90 / 101.3) atm = 0.89 atm
Using the ideal gas equation, PV = nRT:
V = nRT / P
= (0.01 mol) × (0.08206 atm L K-1 mol-1) × ((25 + 273) K)/ (0.89 atm)
= 0.3 L
Answer: 0.3 L
22/01(a)
5
CHEM1001
2010-J-11
June 2010
• Write the two half equations and hence balance the equation for the following redox
reaction:
22/01(a)
Marks
3
MnO2 + NaCl + H2SO4 → MnSO4 + H2O + Cl2 + Na2SO4
Working
The Mn is being reduced since it begins with oxidation number +4 (in MnO2) and
ends with oxidation number +2 (in MnSO4). The reduction reaction is:
MnO2 + 4H+ + 2e– → Mn2+ + 2H2O
The Cl is being oxidised since it begins with oxidation number -1 (in Cl-) and
ends with oxidation number 0 (in Cl2). The oxidation reaction is:
2Cl– → Cl2 + 2e–
These are then summed together to give the overall reaction.
Balanced equation:
MnO2 + 2Cl– + 4H+ → Mn2+ + 2H2O + Cl2
Which species is oxidised?
Cl– is oxidised
• In the electro-refining of Pt, what mass of Pt is deposited from a solution of PtCl62– in
1.00 hour, by a current of 1.62 A?
PtCl62- contains Pt(IV) and so 4 moles of electrons are required to reduced 1 mol.
The number of moles of electrons delivered by a current of 1.62 A in 1.00 hour is:
number of moles of electrons = It / F
= (1.62 C s-1)(1.00 × 60 × 60 s) / (96485 C mol-1)
= 0.0604 mol
This will deposit:
number of moles of Pt deposit = ¼ × 0.0604 mol = 0.0151 mol
As the molar mass of Pt is 195.09 g mol-1, this corresponds to:
mass of Pt = number of moles × molar mass = (0.0151 mol) × (195.09 g mol-1)
= 2.95 g
Answer: 2.95 g
2
CHEM1001
2009-J-8
June 2009
• Adiponitrile, a key intermediate in the manufacture of nylon, is prepared by the
reduction of acrylonitrile.
Anode:
2H2O → O2 + 4H+ + 4e–
Cathode:
2CH2=CHCN + 2H+ + 2e– → NC(CH2)4CN
Write a balanced equation for the overall electrochemical reaction.
4CH2=CHCN + 2H2O → 2NC(CH2)4CN + O2
What mass of adiponitrile (in kg) is produced in 10.0 hours in a cell that has a constant
current of 3.00 × 103 A?
A current of 3.00 × 103 A passed for 10.0 hours corresponds to:
number of moles of electrons =
. . = 1120 mol
From the cathode half cell reaction, 2 mol of electrons are needed to produce 1
mol of adiponitrile. Therefore, 1120 mol of electrons will produce 560. mol of this
product.
The molar mass of NC(CH2)4CN is (2 × 14.01 (N) + 6 × 12.01 (C) + 8 × 1.008 (H))
g mol-1 = 108.44 g mol-1.
560 mol therefore corresponds to (560. mol) × (108.44 g mol-1) = 60500 g = 60.5
kg.
Answer: 60.5 kg
22/01(a)
Marks
3
CHEM1001
2008-J-10
June 2008
• How many minutes will be required for a 1.50 A current to electroplate 1.97 g of gold
from a solution containing AuCl4– ions?
1.97 g of gold corresponds to:
amount of gold =
=
. . = 0.0100 mol
The oxidation number of gold is +3 (as AuCl4- = Au3+ 4Cl-). Hence, 3 moles of
electrons are required to reduce 1 mole of AuCl4- to gold. Thus, 0.0300 mol of
electrons are required to reduce 0.0100 mol.
1.97 g of gold corresponds to:
amount of gold =
=
. . = 0.0100 mol
The oxidation number of gold is +3 (as AuCl4- = Au3+ 4Cl-). Hence, 3 moles of
electrons are required to reduce 1 mole of AuCl4- to gold. Thus, 0.0300 mol of
electrons are required to reduce 0.0100 mol.
As,
moles of electrons =
The time, t, required with current I = 1.50 A to deliver 0.0300 mol is:
t=
. . = 1930 s = 32.2 minutes
Answer: 32.2 minutes
22/01(a)
2
CHEM1001
2007-J-8
June 2007
• 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?
The total number of electrons passed is given by:
It Q
3600
number of moles of electrons =
= =
= 0.037 mol.
F F 96485
As the solution contains MCl3, these electrons are reducing M3+(aq) ions, each of
which requires 3e-. Hence, the number of moles of metal, M, formed at the
cathode is:
number of moles of M = ⅓ × 0.037 = 0.012
This amount of M has a mass of 0.65 g. The atomic mass of M is therefore:
atomic mass =
=
. . = 52 g mol-1
This atomic mass corresponds to that of chromium (Cr).
Answer: Chromium (Cr)
22/01(a)
Marks
4
CHEM1001
2006-J-10
June 2006
• Calculate the mass of silver nitrate, AgNO3, required to make 500 mL of 0.200 M
aqueous solution.
The number of moles of AgNO3 in 500 mL of a 0.200 M solution is:
number of moles = concentration × volume = 0.200 ×
500
= 0.100 mol
1000
The formula mass of AgNO3 is:
formula mass = (107.87 (Ag)) + (14.01 (N)) + (3 × 16.00 (O)) = 169.88
The mass of 0.100 mol is therefore 0.100 × 169.88 = 17.0 g
Answer: 17.0 g
Calculate the time required (in minutes) to deposit 7.0 g of silver from a 0.200 M
silver nitrate solution using a current of 4.5 A.
mass
7.0
=
= 0.065 mol
atomic mass 107.87
To deposit silver requires reduction of Ag+(aq), requiring 1 mole of electrons per
mole of silver. Hence, 0.065 mol of electrons is required. This corresponds to a
charge of Q = nF = 0.065 × F = 0.065 × 96485 = 6300 C.
The number of moles of silver in 7.0 g is
As Q = I × t, the time taken to deliver this charge at a current of 4.5 A is:
t=
Q 6300
=
=1400 s = 23 minutes
I
4.5
Answer: 23 minutes
22/01(a)
Marks
4
CHEM1001/CHEM1101 combined
2003-N-10
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:
8.64 × 106 C`
charge
=
= 89.5mol
moles of e- =
F
96485C mol -1
The reduction of Cu2+ to Cu(s) requires 2e- so the total moles of Cu(s) obtained is
½ × 89.5 = 44.8 mol. As the atomic mass of Cu is 63.55, 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