AQA A-Level Chemistry Paper 3, June 2022: Question 5

✅ Correct Answers

Equation:

C₁₄H₃₀ → C₆H₁₂ + C₈H₁₈

(Also allowed: C₁₄H₃₀ → 2 C₃H₆ + C₈H₁₈ )

Heterogeneous:

The catalyst is in a different phase (or physical state) to the reactants.

💡 Key Knowledge

• Cycloalkanes have the general formula C_nH_2n (same as straight-chain alkenes, due to one ring closing).
• Tetradecane = 14 carbons; Octane = 8 carbons. Remaining carbons = 14 − 8 = 6, with 30 − 18 = 12 hydrogens → C₆H₁₂ (cyclohexane).
• Phase vs State: "Phase" is preferred. In heterogeneous catalysis with zeolites, the zeolite is solid while reactants are gases/liquids.

🧠 Exam Technique

• Always check the wording: the question specifically asked for a cycloalkane, not an alkene! While both have the formula C_nH_2n, double-check that your hydrogen count balances.
• When defining "heterogeneous", reference the reactants, not the products.

❌ Common Errors

• Writing "catalyst is in a different phase to the products" (loses the mark).
• Writing an equation that produces an alkene instead of balancing specifically for octane + cycloalkane.

✅ Correct Answer Breakdown

• Autocatalyst: A product of the reaction acts as a catalyst for the reaction.
• Why slow at first: Both reactant ions (MnO₄⁻ and C₂O₄²⁻) are negatively charged and therefore repel each other.
• This electrostatic repulsion causes a high activation energy (Eₐ).
• Why it speeds up: Mn²⁺ acts as a catalyst; there is an attraction between oppositely charged ions (negative reactant ions and positive catalyst Mn²⁺/Mn³⁺).
• Catalytic Step 1:
  4 Mn²⁺ + MnO₄⁻ + 8 H⁺ → 5 Mn³⁺ + 4 H₂O
• Catalytic Step 2:
  2 Mn³⁺ + C₂O₄²⁻ → 2 Mn²⁺ + 2 CO₂

💡 The Reaction Mechanism Steps

The overall titration reaction is:

2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O

• Mn²⁺ is not initially present, so the uncatalysed collision of two anions (MnO₄⁻ and C₂O₄²⁻) must occur first.
• Once formed, Mn²⁺ reduces MnO₄⁻ to form Mn³⁺ intermediates.
• Mn³⁺ then oxidises ethanedioate C₂O₄²⁻ to CO₂, regenerating Mn²⁺.

🧠 Exam Technique: Securing all 6 Marks

• Mark 1: Define autocatalyst clearly using the word product.
• Marks 2 & 3: State the problem: "Negative ions repel" + "high activation energy".
• Mark 4: State why the catalyst works: "Attraction between oppositely charged ions" (cations attract anions).
• Marks 5 & 6: Write both individual steps accurately. Ensure charges balance on both sides!

❌ Common Errors

• Calling an autocatalyst a "reactant that speeds up the reaction".
• Forgetting to mention the charges on the ions (e.g. just saying "the molecules repel").
• Unbalanced charges in the two equations: in Step 1, LHS charge is 4(+2) − 1 + 8(+1) = +15; RHS is 5(+3) = +15.

✅ Correct Answer Model

• Mark 1 (Variable oxidation state): Co²⁺ is oxidised to Co³⁺ and then reduced back to Co²⁺ (or changes between Co²⁺ and Co³⁺).
• Mark 2 (First step feasibility):
  E°(S₂O₈²⁻ / SO₄²⁻) > E°(Co³⁺ / Co²⁺) (i.e. +2.01 V > +1.82 V), so S₂O₈²⁻ oxidises Co²⁺ to Co³⁺.
  2 Co²⁺ + S₂O₈²⁻ → 2 Co³⁺ + 2 SO₄²⁻ (E°_cell = +0.19 V)
• Mark 3 (Second step feasibility):
  E°(Co³⁺ / Co²⁺) > E°(I₂ / I⁻) (i.e. +1.82 V > +0.54 V), so Co³⁺ oxidises I⁻ to I₂.
  2 Co³⁺ + 2 I⁻ → 2 Co²⁺ + I₂ (E°_cell = +1.28 V)

📐 Step-by-Step Feasibility Rule

For any redox reaction to be feasible under standard conditions:

E°_cell = E°(reduction) − E°(oxidation) > 0

1. Step 1: S₂O₈²⁻ reduced (+2.01 V), Co²⁺ oxidised (+1.82 V):
   E°_cell = +2.01 − (+1.82) = +0.19 V (> 0, feasible)
2. Step 2: Co³⁺ reduced (+1.82 V), I⁻ oxidised (+0.54 V):
   E°_cell = +1.82 − (+0.54) = +1.28 V (> 0, feasible)

🧠 Exam Technique

• The prompt asks you to use the electrode potential data. Simply stating "Co oxidises and reduces" without quoting the values or stating which E° is more positive will cap your marks.
• Remember: the species with the more positive E° is the stronger oxidising agent and will gain electrons (undergo reduction).

❌ Common Errors

• Attempting an alternative route where Co²⁺ is first reduced to Co (not possible in aqueous solution and not given in table).
• Comparing numbers incorrectly: stating that +0.54 V oxidises +1.82 V.
• Failing to state that the catalyst is regenerated back to its original state (Co²⁺).