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
CHEM1101
2010-J-13
June 2010
• The net reaction discharging the lead acid storage battery is:
PbO2 + Pb + 2H2SO4 → 2PbSO4 + 2H2O
What reaction occurs at the cathode?
PbO2 + 2H2SO4 + 2e– → PbSO4 + 2H2O + SO42–
What reaction occurs at the anode?
Pb + H2SO4 → PbSO4 + 2H+ + 2e–
Why are the cathode and anode not in separate compartments, as in the Cu/Zn
battery?
All the redox active species are solids so can be kept apart without needing
separate compartments.
How does H2SO4 serve as the ‘salt bridge’? Which ions flow in which direction to
maintain electroneutrality?
H2SO4 dissociates to give H+, HSO4– and SO42– ions which carries the current.
The H+ ions migrate to the cathode, whilst the HSO4– and SO42– ions migrate to
the anode.
What is the formula for the equilibrium constant for the discharge reaction above?
Solids do not appear in the equilibrium constant, as their ‘concentration’ is
constant. Water is present as the solvent and is at very high and effectively
constant concentration. Hence, only [H2SO4] varies:
K = [H2SO4]–2
The cell potential for this battery is 2.05 V. If the concentration of the H2SO4 is
4.5 M, what is the standard potential of the cell at 25 °C?
The Nernst equation, Ecell = E° -
𝑹𝑻
𝒏𝑭
lnQ, can be used to work out the standard
potential for this two electron process. As Ecell = 2.05 V when [H2SO4] = 4.5 M:
2.05 V = E° -
𝟖.𝟑𝟏𝟒 𝐉 𝐊 !𝟏 𝐦𝐨𝐥!𝟏 (𝟐𝟗𝟖 𝐊)
𝟐 (𝟗𝟔𝟒𝟖𝟓 𝐂 𝐦𝐨𝐥!𝟏 )
ln(4.5)-2
E° = 2.01 V
Answer: +2.01 V
Marks
7
CHEM1001
2008-J-10
June 2008
• A solar powered light uses a nickel-cadmium battery to store electricity. Calculate the
standard voltage for the battery from the following:
NiO2(s) + Cd(s) + 2H2O(l) → Ni(OH)2(s) + Cd(OH)2
∆Go = –251 kJ mol–1
The reaction is a two electron process (as Cd is being oxidized to Cd(II) and Ni is
being reduced from Ni(IV) to Ni(II)). Hence, n = 2.
The standard electrode potential is linked to Gibbs free energy change by,
∆G° = -nFE°
Hence with ∆G° = -251 kJ mol-1
-251 × 103 J mol-1 = -2 × (96485 C mol-1) × E°
E° = 1.30 V
Answer: 1.30 V
22/01(a)
3
CHEM1001
2007-J-8
June 2007
• A metal-metal hydride battery has the following shorthand notation:
MH(s), M(s) | OH–(aq) || OH–(aq)  NiO(OH)(s), Ni(OH)2(s)
Which component of the battery is the cathode?
NiO(OH)(s)/Ni(OH)2(s)
(the right hand side of the cell
notation)
Give the balanced half equation of the reaction that takes place at the cathode.
NiO(OH)(s) + H2O(l) + e– Ni(OH)2(s) + OH–(aq)
Why is it important that all redox active species are solids in this reaction?
As all redox active species are solids, their concentrations do not change during
the reaction. They therefore do not appear in the equilibrium expression or in the
Nernst equation.
The battery therefore maintains a constant voltage.
22/01(a)
3
CHEM1001
•
2006-J-10
June 2006
A lead-acid battery has the following shorthand notation:
Pb(s), PbSO4(s) | H+(aq), SO42–(aq) || H+(aq), SO42–(aq)  PbO2(s), PbSO4(s)
Which component of the battery is the anode?
Pb(s), PbSO4(s)
Give the balanced half equation of the reaction that takes place at the anode.
Pb(s) + SO42-(aq) PbSO4(s) + 2eWhich component of the battery is the cathode?
PbO2(s), PbSO4(s)
Give the balanced half equation of the reaction that takes place at the cathode.
PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- PbSO4(s) + 2H2O(l)
22/01(a)
4
CHEM1001
2005-J-9
June 2005
 The half reactions describing the discharge of a silver-zinc cell are:
Zn(s) + 2OH–(aq)  ZnO(s) + H2O(l) + 2e–
Ag2O(s) + H2O(l) + 2e–  2Ag(s) + 2OH–(aq)
List the chemical species that will be consumed as the battery discharges.
Zn(s) and Ag2O(s)
Why is a saturated solution of KOH used in the battery?
It acts like a salt bridge allowing the migration of ions away from and towards
the electrodes. OH–(aq) ions are produced at the cathode and consumed at the
anode.
Why is the voltage in the silver-zinc cell constant during discharge?
The overall cell reaction is:
Zn(s) + Ag2O(s)  2Ag(s) + ZnO(s).
No ions with constantly changing concentrations appear, so the cell produces a
constant voltage until one of the reactants is exhausted and it stops functioning.
3
CHEM1001
2004-J-8
June 2004
 Explain the following features of the lead acid storage battery.
Marks
3
It has a relatively constant voltage.
The reagents are solids or very concentrated solutions.
It needs no salt bridge.
Both the anode and cathode are solids so can be placed in the same electrolyte.
It can be recharged.
The products are solids so remain trapped in the same electrode. External
voltage can reverse the reaction and regenerate the cathode and anode.
3
 Consider the following cell reaction.
Pb(s) + Sn2+(aq)
Pb2+(aq) + Sn(s)
Use the Nernst equation to calculate the ratio of cation concentrations for which the
cell potential, E = 0. (See the data sheet for a table of standard reduction potentials.)
The half reactions and potentials are:
Pb(s)  Pb2+(aq)+ 2eSn2+(aq) + 2e-  Sn(s)
E° = 0.13 V (reversed for oxidation required)
E° = 0.14 V
Hence, E° = {(+0.13) + (0.14)} V = 0.01 V
The Nernst equation gives Ecell = E° - 2.303 ×
E° = 2.303 ×
𝑹𝑻
𝒏𝑭
logQ = 2.303 ×
𝑹𝑻
𝒏𝑭
𝑹𝑻
𝒏𝑭
logQ. Ecell = 0 when:
logK
The ratio of cation concentrations is just Q =
[Pb 2+ (aq)]
. The process involves
[Sn 2+ (aq)]
2e- so n = 2 and Q can be obtained using the E° = 0.01 V:
Kc = 𝟏𝟎𝒏𝑭𝑬°/𝟐.𝟑𝟎𝟑𝑹𝑻= 0.5
Answer: 0.5
CHEM1001
2004-J-9
June 2004
 The standard dry cell (battery) has the following shorthand notation:
Zn(s) | Zn2+(aq) || MnO2(s), Mn2O3(s) | graphite(s)
Which component of the battery is the anode?
Zn is the anode
Give the balanced half equation that takes place at the anode.
Zn(s)  Zn2+(aq) + 2e–
Which component of the battery is the cathode?
The graphite electrode
Give the balanced half equation that takes place at the cathode.
2MnO2(s) + 2NH4+(aq) + 2e–  Mn2O3(s) + H2O(l) + 2NH3(aq)
Marks
4
CHEM1001/CHEM1101 combined
2003-N-9
November 2003
 What is meant by “cathodic protection”? Which of the following metals can provide
cathodic protection to iron and why?
Zn, Ni, Al, Sn
To protect a metal such as iron, it is connected to a more active metal. This
ensures that in any electrochemical reaction, the iron will be the cathode and the
more active metal the anode. The more active metal will therefore undergo any
oxidation reaction that occurs and be sacrificed to protect the iron.
Efor oxidation of Fe to Fe2+ is 0.44 V. Zn and Al have more positive Eox values
than Fe and hence can provide cathodic protection of iron. Ni and Sn have less
positive Eox values so cannot protect the iron.
2