AQA A-Level Chemistry Paper 1, June 2024: Question 6

10 marks · Medium difficulty · State/Explain/Numerical

Use standard electrode potentials to explain vanadium reduction reactions, construct an electrochemical cell representation and calculate EMF, and determine the percentage purity of ammonium vanadate(V) by redox titration.

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Question text

06 This question is about vanadium ions.

Table 3 shows some standard electrode potential values.

Table 3

Eo / V

O (g) + 4 H+(aq) + 4 e− → 2 H O(l) +1.23

VO +(aq) + 2 H+(aq) + e− → VO2+(aq) + H O(l) +1.00

VO2+(aq) + 2 H+(aq) + e− → V3+(aq) + H O(l) +0.34

V3+(aq) + e− → V2+(aq) –0.26

Fe2+(aq) + 2 e− → Fe(s) –0.44

Zn2+(aq) + 2 e− → Zn(s) –0.76

V2+(aq) + 2 e− → V(s) –1.20

Mg2+(aq) + 2 e− → Mg(s) –2.38

06.1 Use the data in Table 3 to explain why Zn reduces an aqueous solution of VO + ions

to V2+ ions, but does not reduce it any further.

[2 marks]

06.2 Identify the species in Table 3 that can reduce an aqueous solution of VO + to V

[1 mark]

06.3 Two half-cells Fe2+(aq) / Fe(s) and VO2+(aq) / V3+(aq) are connected.

Calculate the EMF of this cell.

Give the conventional representation for this cell.

Give a half-equation for the reaction that occurs at the negative electrode.

[3 marks]

EMF

Cell representation

Half-equation

06.4 0.151 g of impure NH4VO3 is added to dilute sulfuric acid to form a solution containing

aqueous VO + ions.

All the VO – ions are converted to VO + ions.

These VO + ions are reduced to aqueous V2+ ions by reaction with an excess of zinc.

2 VO +(aq) + 8 H+(aq) + 3 Zn(s) → 3 Zn2+(aq) + 2 V2+(aq) + 4 H O(l)

The excess of zinc is removed by filtration and washed.

The filtrate, containing the V2+ ions, is titrated with a 0.0200 mol dm–3 solution of

acidified KMnO4

29.43 cm3 of KMnO solution are needed to oxidise all the V2+ ions to VO + ions.

The ionic equation for the reaction of MnO – ions with V2+ ions is

3 MnO –(aq) + 5 V2+(aq) + 4 H+(aq) → 2 H O(l) + 3 Mn2+(aq) + 5 VO +(aq)

42 2

Calculate the percentage purity of the NH4VO3

Give your answer to 3 significant figures.

[4 marks]

Percentage purity

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

M1 Eo V3+ (/V2+) > Eo Zn2+ (/ Zn)

Or EMF of reaction between V3+ and Zn = (+)0.50 V

06.1 o 2+ o 2+

M2 E V (/V) < E Zn (/ Zn) (2 x AO3)

Or EMF of reaction between V2+ and Zn = – 0.44 V

only 1

06.2 Mg

(1 x AO3)

EMF = (+) 0.78 (V)

Fe(s) l Fe2+(aq) ll VO2+(aq), H+(aq), V3+(aq) l Pt(s)

Allow Fe(s) l Fe2+(aq) ll VO2+(aq), V3+(aq) l Pt(s)

Ignore state symbols 3

06.3 (1 × AO1,

2 × AO2)

Fe(s) → Fe2+(aq) + 2e– Ignore state symbols

M1 n MnO – = 29.43 × 10–3 × 0.0200 = 5.89 × 10–4 mol

2+ –4 5 –4 M2 = M1 ×

M2 n V = 5.89 x 10 × = 9.81 × 10 mol 3

–4 M3 = M2 × 116.9 4

06.4 M3 mass NH4VO3 = 9.81 × 10 × 116.9 = 0.1147 g

(4 x AO2)

0.1147×100 M4 = M3 ×

M4 % purity = = 76.0 % 0.151

0.151 allow 75.9 or 76.2 %

answer to 3 significant figures

How to answer it

Vanadium Electrochemistry, Cell Notations & Redox Titration

WHAT THIS QUESTION TESTS

This question assesses your mastery of Transition Metals (Vanadium) and Electrochemical Cells:

Using standard electrode potentials ( E° ) to predict feasible redox reactions and stepwise reduction limits.
Selecting appropriate reducing agents from standard reduction potentials.
Calculating cell EMF, constructing IUPAC conventional cell representations, and identifying negative half-cell reactions.
Multistep redox back-titration calculations involving permanganate ( MnO₄⁻ ) and vanadyl species to determine percentage purity.

QUESTION 06.1 • 2 MARKS

Explaining Stepwise Reduction of VO₂⁺ with Zinc

Predicting reaction boundaries using electrode potentials

✅ Correct Answer

Mark 1: E°(V³⁺/V²⁺) > E°(Zn²⁺/Zn)
OR: EMF for V³⁺ + Zn → V²⁺ + Zn²⁺ = +0.50 V (> 0, feasible)
Mark 2: E°(V²⁺/V) < E°(Zn²⁺/Zn)
OR: EMF for V²⁺ + Zn → V + Zn²⁺ = -0.44 V (< 0, non-feasible)

🧠 Exam Technique

For a reaction to occur spontaneously under standard conditions:

EMF = E°(reduction) - E°(oxidation) > 0

Alternatively, the oxidising agent must have a more positive electrode potential than the reducing agent's couple.

💡 Key Knowledge

VO₂⁺ → VO²⁺ (+1.00 V vs -0.76 V → EMF = +1.76 V > 0)
VO²⁺ → V³⁺ (+0.34 V vs -0.76 V → EMF = +1.10 V > 0)
V³⁺ → V²⁺ (-0.26 V vs -0.76 V → EMF = +0.50 V > 0, reduces!)
V²⁺ → V (-1.20 V vs -0.76 V → EMF = -0.44 V < 0, does not reduce!)

❌ Common Errors

Stating vague phrases like "Zn is more reactive" without citing specific E° values.
Referring to the wrong vanadium couples (e.g., comparing Zn to VO₂⁺ or VO²⁺ instead of explaining why reduction stops at V²⁺ ).

Mark Scheme Note: Award 1 mark for justifying the reduction of V³⁺ to V²⁺, and 1 mark for justifying why V²⁺ cannot be reduced to V(s).

QUESTION 06.2 • 1 MARK

Identifying a Strong Enough Reducing Agent

Reducing VO₂⁺ all the way to metallic Vanadium (V)

✅ Correct Answer

Mg (or Magnesium)

Must be given as the element/metal, not the ion.

💡 Key Knowledge

To reduce V²⁺ all the way to V (which requires E° = -1.20 V ), the reducing agent must have an E° value more negative than -1.20 V .

From Table 3, only the Mg²⁺/Mg couple ( -2.38 V ) is more negative than -1.20 V .

❌ Common Errors

Writing Mg²⁺ instead of Mg . A cation cannot act as a reducing agent!
Selecting Zinc or Iron, neither of which has an E° low enough to overcome -1.20 V .

Mark Scheme Note: Only Mg is accepted. "Mg²⁺" scores 0.

QUESTION 06.3 • 3 MARKS

Electrochemical Cell: Fe²⁺/Fe and VO²⁺/V³⁺

EMF, IUPAC cell convention, and negative electrode equation

✅ Correct Answers

EMF: (+) 0.78 V
Cell representation:
Fe(s) | Fe²⁺(aq) || VO²⁺(aq), H⁺(aq), V³⁺(aq) | Pt(s)
(Also allowed without H⁺: Fe(s) | Fe²⁺(aq) || VO²⁺(aq), V³⁺(aq) | Pt(s) )
Half-equation at negative electrode:
Fe(s) → Fe²⁺(aq) + 2e⁻ (or Fe → Fe²⁺ + 2e⁻ )

🧠 Exam Technique & Conventions

EMF: E°(RHS) - E°(LHS) = (+0.34) - (-0.44) = +0.78 V .
Negative electrode: Has the more negative standard electrode potential ( Fe²⁺/Fe = -0.44 V vs +0.34 V ). Oxidation occurs here, so it is placed on the left.
Inert electrode: The vanadium half-cell contains only aqueous species ( VO²⁺ and V³⁺ ), so an inert platinum electrode Pt(s) must be included on the outside right.

❌ Common Errors

Forgetting the inert Pt electrode on the right.
Writing a phase boundary ( | ) between aqueous ions in the same solution instead of a comma ( , ).
Writing the reduction reaction for the negative electrode ( Fe²⁺ + 2e⁻ → Fe ) instead of oxidation ( Fe → Fe²⁺ + 2e⁻ ).

Mark Scheme Note: Ignore state symbols. Accept species in any order within each aqueous compartment as long as phase boundaries and salt bridge ( || ) are correct.

QUESTION 06.4 • 4 MARKS

Redox Titration & Percentage Purity Calculation

Determining the purity of an ammonium metavanadate (NH₄VO₃) sample

📐 Step-by-Step Calculation

Step 1: Calculate moles of KMnO₄ (titrant) used

n(MnO₄⁻) = concentration × volume = 0.0200 × (29.43 / 1000)
n(MnO₄⁻) = 5.886 × 10⁻⁴ mol (award Mark 1)

Step 2: Use stoichiometric ratio to find moles of V²⁺ (and VO₂⁺)

Equation: 3MnO₄⁻(aq) + 5V²⁺(aq) + 4H⁺(aq) → 2H₂O(l) + 3Mn²⁺(aq) + 5VO₂⁺(aq)
Mole ratio: n(V²⁺) = n(MnO₄⁻) × (5 / 3)
n(V²⁺) = 5.886 × 10⁻⁴ × (5 / 3) = 9.810 × 10⁻⁴ mol (award Mark 2)

Since 1 mol of NH₄VO₃ → 1 mol VO₃⁻ → 1 mol VO₂⁺ → 1 mol V²⁺ , n(NH₄VO₃) = 9.810 × 10⁻⁴ mol .

Step 3: Calculate the pure mass of NH₄VO₃

M_r(NH₄VO₃) = 14.0 + (4 × 1.0) + 50.9 + (3 × 16.0) = 116.9 g mol⁻¹
mass = n × M_r = 9.810 × 10⁻⁴ × 116.9 = 0.1147 g (award Mark 3)

Step 4: Calculate percentage purity (to 3 significant figures)

% purity = (pure mass / impure mass) × 100
% purity = (0.1147 / 0.151) × 100 = 76.0% (award Mark 4)

❌ Common Errors & Pitfalls

Inverting the ratio: Multiplying by 3/5 instead of 5/3. Always check: 3 moles of MnO₄⁻ react with 5 moles of V²⁺, so V²⁺ must be greater in number than MnO₄⁻!
Volume conversion: Forgetting to divide 29.43 cm³ by 1000.
Significant figures: Writing 75.95% or 76% instead of exactly 3 sig figs (76.0%) as explicitly demanded by the question.

🧠 Examiner Insights

The prompt specifies: "Give your answer to 3 significant figures." Failure to write 76.0% (or 75.9% / 76.2% depending on intermediate rounding) loses the final mark.
Follow-through marks (consequential marking) apply if M1 or M2 contains an arithmetic slip, provided the method is sound.

Mark Scheme Note: Allow 75.9% or 76.2% depending on intermediate rounding. 76.0% is the preferred unrounded answer.

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.11 Electrode Potentials · 3.1.2 Amount of Substance · 3.2.5 Transition Metals

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.