Edexcel A-Level Chemistry Paper 1, June 2022: Question 8
13 marks · Hard difficulty · Calculations
Calculate the equilibrium constant from cell potential, identify electrode reactions in fuel cells and lead-acid batteries, and determine Kc from titration data.
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Question text
8 This question is about electrode potentials, cells and equilibrium constants.
(a) Chlorine gas can be prepared by the oxidation of chloride ions with
manganate(VII) ions in acid solution.
– – + 2+ d
MnO4(aq) + 5Cl (aq) + 8H (aq) Mn (aq) + 2½Cl2(g) + 4H2O(l) Ecell = +0.15 V
During this reaction, each manganate(VII) ion accepts five electrons.
Calculate the equilibrium constant, K, for this reaction at 298K using
the expression
d
nEcellF
lnK =
RT
where n is the number of electrons transferred in the overall equation,
–1
F is the Faraday constant (96500Cmol ) and
–1 –1
R is the gas constant (8.31Jmol K ).
Units of K are not required.
(2)
(b) A fuel cell produces a voltage from the reaction between a fuel and oxygen.
The reaction occurring at one electrode in a methanol fuel cell is
+ –
CH3OH(g) + H2O(l) → HCOOH(aq) + 4H (aq) + 4e
Which reaction occurs at the other electrode?
(1)
+ –
A 4H (aq) + O2(g) + 4e → 2H2O(l)
– –
B 2H2(g) + 2O2(g) + 4e → 4OH (aq)
– –
C 4OH (aq) → 2H2(g) + 2O2(g) + 4e
+ –
D 2H2O(l) → 4H (aq) + O2(g) + 4e
(c) Lead-acid batteries are used as storage cells in some cars.
The electrolyte is sulfuric acid, one electrode is lead and the other is
lead(IV) oxide, PbO2.
As the cell discharges, the lead and the lead(IV) oxide are both converted to
solid lead(II) sulfate, PbSO4 , and the concentration of the sulfuric acid decreases.
Deduce, using the information given, the two half-equations occurring in the
lead-acid battery.
State symbols are required.
(3)
(d) When solid lead(II) sulfate is added to aqueous sodium iodide, an equilibrium
26 is established.
*P67093RA02632*PbSO4(s)+2I–(aq) PbI2(s)+SO42–(aq)
The expression for the equilibrium constant, Kc , for this reaction is
[SO2− (aq)]
K = 4
c − 2
[I (aq)]
In an experiment, Kc may be determined by adding excess lead(II) sulfate to
3 –3
25.0cm of 0.100moldm sodium iodide.
The volume remains constant at 25.0cm3.
The mixture is left to reach equilibrium at room temperature.
Ice-cold water is added to freeze the position of equilibrium and the mixture is
then titrated with standard silver nitrate solution.
3 –3
The whole mixture requires 12.20cm of 0.0500moldm silver nitrate solution to
react with the aqueous iodide ions at equilibrium.
+ –
Ag (aq) + I (aq) → AgI(s)
Calculate the equilibrium concentrations of the sulfate ions and the iodide ions, 27
and hence the value of*P67093RA02732*Kcat room temperature.
Give your answer to an appropriate number of significant figures and include
units for Kc, if any.
(7)
(Total for Question 8 = 13 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
8(a) Example of calculation (2)
• substitution of values into the equation (1) lnKc = 5 x 0.15 x 96 500
8.31 x 298
• calculation of Kc (1) (lnKc = 29.226)
K = 4.9289 x 1012
c
= 4.9 x 1012 / 4.93 x 1012
TE on their value for lnKc
Ignore SF except 1SF
Correct answer with no working scores (2)
Question Answer Mark
number
8(b) + − (1)
The only correct answer is A (4H (aq) + O2(g) + 4e → 2H2O(l))
B is incorrect because methanol does not react with hydrogen
C is incorrect because this reaction shows an oxidation
D is incorrect because this reaction shows an oxidation
Question
Answer Additional Guidance Mark
Number
8(c) Examples of half-equations (3)
• one half-equation (1) Pb(s) + SO 2−(aq) ⇌ PbSO (s) + 2e−
Allow
Pb(s) + H SO (aq) ⇌ PbSO (s) + 2H+(aq) + 2e−
24 4
• other half-equation (1) PbO (s) + 4H+(aq) + SO 2−(aq) + 2e− ⇌ PbSO (s) + 2H O(l)
24 4 2
Allow
PbO (s) + 2H+(aq) + H SO (aq) + 2e− ⇌ PbSO (s) + 2H O(l)
22 4 4 2
Allow multiples
Allow single headed arrows in the forward direction
Ignore missing charge on electrons
• state symbols (1) Conditional on correct species in one equation that has
scored either M1or M2
Question
Answer Additional Guidance Mark
Number
8(d) Example of calculation (7)
• calculation of initial mol I− (1) initial mol I− = (25.0 x 0.100) ÷1000) = 2.5 x 10−3 / 0.0025 (mol)
• calculation of eqm mol I− (1) eqm mol I− (= mol Ag+) = (12.20 x 0.0500) ÷1000
= 6.1 x 10−4 / 0.00061 (mol)
• calculation of mol I− reacted(1) mol I− reacted = 2.5 x 10−3 – 6.1 x 10−4 = 1.89 x 10−3 / 0.00189 (mol)
• calculation of eqm mol-1 SO 2- eqm mol SO 2−= mol I− reacted / 2 =1.89 x 10−3 ÷ 2
(1) = 9.45 x 10−4 / 0.000945
• calculation of eqm [SO 2−(aq)] eqm [SO 2-] = (9.45 x 10−4 x 1000) ÷ 25 = 0.0378 (mol dm−3)
and and
calculation of eqm [I−(aq)] (1) eqm [I−] = (6.1 x 10−4 x 1000) ÷ 25.0 = 2.44 x 10−2 / 0.0244 (mol dm−3)
• calculation of K K = 0.0378 ÷ 0.02442 = (63.49) = 63 / 63.5
c c
and Do not award unless their numbers are correct or are TE.
answer to 2 / 3 SF (1) Allow TE throughout. Correct answer with working gains 7 marks
• units (1) dm3 mol−1 (standalone mark)
Allow dm3 mol− / mol−1 dm3 / mol− dm3
(Total for Question 8 = 13 mark)
How to answer it
Electrode Potentials, Cells and Equilibrium Constants
What this question tests
This comprehensive multi-part question assesses your mastery of redox equilibria, electrochemical cells, fuel cell half-equations, constructing complex storage cell half-equations with state symbols, and carrying out advanced multi-step equilibrium constant ($K_c$) calculations using titration data.
Calculating Equilibrium Constant ($K$) from Cell Potential
✅ Correct Answer
ln K = 29.226
K = 4.9 × 10¹² (or 4.93 × 10¹² )
💡 Key Knowledge
- Relating thermodynamics, cell potentials, and equilibrium constants via ln K = (n × E_cell × F) / (R × T) .
- Identifying n = 5 from the given text ("each manganate(VII) ion accepts five electrons").
📐 Step-by-Step Calculation
- Identify values: n = 5 , E_cell = +0.15 V , F = 96500 C mol⁻¹ , R = 8.31 J mol⁻¹ K⁻¹ , T = 298 K .
- Substitute into rearranged expression:
ln K = (5 × 0.15 × 96500) / (8.31 × 298) - Evaluate ln K: 29.228... (or 29.226 depending on rounding of terms).
- Inverse log (exp): K = e^(29.226) = 4.93 × 10¹² .
❌ Common Errors
- Forgetting to determine or misidentifying the number of electrons transferred ( n ).
- Mathematical errors when handling multi-term fractions in the denominator. Always use brackets around the denominator in your calculator!
Identifying Counter-Electrode Reactions in Fuel Cells
✅ Correct Answer
A: 4H⁺(aq) + O₂(g) + 4e⁻ → 2H₂O(l)
🧠 Exam Technique & Examiner Insight
A fuel cell consists of an oxidation half-cell (fuel) and a reduction half-cell (oxygen). Since the given methanol reaction generates electrons (oxidation, loss of electrons), the other electrode must be a reduction process where electrons are accepted.
Options C and D show electrons on the product side (oxidation). Option B introduces hydrogen gas, which is not present in an oxygen/air cathode feed. Therefore, A is the only viable oxygen reduction half-equation in acidic medium.
Lead-Acid Battery Half-Equations
✅ Correct Answer
Half-equation 1 (Lead electrode):
Pb(s) + SO₄²⁻(aq) ⇌ PbSO₄(s) + 2e⁻
*(also accepts Pb(s) + H₂SO₄(aq) ⇌ PbSO₄(s) + 2H⁺(aq) + 2e⁻ )*
Half-equation 2 (Lead(IV) oxide electrode):
PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ ⇌ PbSO₄(s) + 2H₂O(l)
💡 Key Knowledge
Discharge converts both Pb and PbO₂ into solid PbSO₄ inside a sulfuric acid electrolyte. State symbols ( s , aq , l ) are strictly required.
❌ Common Errors
Students frequently omit sulfate ions ( SO₄²⁻ ) or hydrogen ions ( H⁺ ) required to balance the mass and charge in storage cell environments, forgetting that the electrolyte actively participates in the reaction.
Complex Equilibrium Titration Calculation
✅ Correct Answer
Equilibrium [SO₄²⁻]: 0.0378 mol dm⁻³
Equilibrium [I⁻]: 0.0244 mol dm⁻³
Value of K_c: 63 or 63.5
Units of K_c: dm³ mol⁻¹ (or L mol⁻¹ )
📐 Step-by-Step Calculation
- Initial moles of I⁻:
(25.0 × 0.100) / 1000 = 2.50 × 10⁻³ mol - Equilibrium moles of I⁻ (from titration with Ag⁺):
n(I⁻) = n(Ag⁺) = (12.20 × 0.0500) / 1000 = 6.10 × 10⁻⁴ mol - Moles of I⁻ reacted:
2.50 × 10⁻³ - 6.10 × 10⁻⁴ = 1.89 × 10⁻³ mol - Equilibrium moles of SO₄²⁻ formed:
From stoichiometry ( 2I⁻ : 1SO₄²⁻ ), divide reacted moles by 2:
1.89 × 10⁻³ / 2 = 9.45 × 10⁻⁴ mol - Equilibrium concentrations (in 25.0 cm³ total volume):
- [SO₄²⁻] = (9.45 × 10⁻⁴ × 1000) / 25.0 = 0.0378 mol dm⁻³
- [I⁻] = (6.10 × 10⁻⁴ × 1000) / 25.0 = 0.0244 mol dm⁻³ - Calculate K_c:
K_c = [SO₄²⁻] / [I⁻]² = 0.0378 / (0.0244)² = 63.5 dm³ mol⁻¹ (to 2 or 3 SF).
❌ Common Calculation Traps
- Forgetting to square the iodide concentration term in the denominator when calculating K_c .
- Using initial volume instead of considering mole ratios properly via stoichiometry.
- Failing to derive the correct unit:
K_c = (mol dm⁻³) / (mol dm⁻³)² = 1 / (mol dm⁻³) = dm³ mol⁻¹ .
Topics
Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 11: Equilibrium II · Topic 14: Redox II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.