AQA A-Level Chemistry AS Paper 1, June 2022: Question 6

✅ Correct Answer

Electron acceptor OR gains electrons

❌ Common Errors & Pitfalls

• "Electron pair acceptor": 0 marks! That is the definition of an electrophile or a Lewis acid, not an oxidising agent.
• "Gain of electrons": 0 marks. That defines reduction (the process), not the oxidising agent (the species performing the action). An agent is the entity that accepts/gains electrons.
• Confusing agents: Saying "loses electrons" describes a reducing agent.

✅ Correct Answers

Oxidation half-equation:

2I⁻ → I₂ + 2e⁻

Reduction half-equation:

2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O

📐 Step-by-Step Half-Equation Logic

1. Oxidation Half-Equation: Iodide (oxidation state -1) is oxidised to iodine (oxidation state 0).
   Balance atoms: 2I⁻ → I₂
   Balance charge: add 2e⁻ to the right → 2I⁻ → I₂ + 2e⁻ .
2. Reduction Half-Equation: Iodate(V), IO₃⁻ (oxidation state +5), is reduced to I₂ (0).
   Balance iodine atoms: 2IO₃⁻ → I₂
   Balance oxygen using H₂O: 2IO₃⁻ → I₂ + 6H₂O
   Balance hydrogen using H⁺: 2IO₃⁻ + 12H⁺ → I₂ + 6H₂O
   Balance charges: Left is +10, right is 0, so add 10e⁻ to left:
   2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O

🧠 Exam Technique

Check that multiplying the oxidation half-equation by 5 gives 10e⁻:
10I⁻ → 5I₂ + 10e⁻
Adding to the reduction half-equation gives:
2IO₃⁻ + 10I⁻ + 12H⁺ → 6I₂ + 6H₂O
Dividing through by 2 reproduces the overall given equation:
IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O . This confirms both half-equations are completely correct!

✅ Correct Answers

Equation for yellow solid:

6I⁻ + 6H⁺ + H₂SO₄ → S + 3I₂ + 4H₂O

Also accepted:
• 6HI + H₂SO₄ → S + 3I₂ + 4H₂O
• 6I⁻ + 8H⁺ + SO₄²⁻ → S + 3I₂ + 4H₂O

Identity of foul-smelling gas:

H₂S OR hydrogen sulfide (or hydrogen sulphide)

💡 Key Knowledge: Reduction Products of H₂SO₄

Iodide is a powerful reducing agent and reduces sulfur in H₂SO₄ (+6) in three distinct stages:

• SO₂ (oxidation state +4): Colourless, choking gas.
• S (oxidation state 0): Yellow solid.
• H₂S (oxidation state -2): Foul-smelling gas (bad eggs).

📐 How to Construct the Equation for Sulfur Formation

1. Reduction of sulfuric acid to sulfur:
   H₂SO₄ + 6H⁺ + 6e⁻ → S + 4H₂O (S changes from +6 to 0, requiring 6e⁻).
2. Oxidation of iodide:
   2I⁻ → I₂ + 2e⁻ .
3. Multiply oxidation by 3 to balance electrons:
   6I⁻ → 3I₂ + 6e⁻ .
4. Combine the two equations:
   6I⁻ + 6H⁺ + H₂SO₄ → S + 3I₂ + 4H₂O .

❌ Common Errors

• Writing the wrong gas: Writing SO₂ instead of H₂S for the foul-smelling gas. (SO₂ has a choking/sharp smell; H₂S is the notoriously foul "rotten egg" gas).
• Unbalanced water or protons: Writing 6H⁺ + 6I⁻ but forgetting the extra 2H atoms in H₂SO₄, resulting in unbalanced oxygen/hydrogen atoms on the right. Always check that the 4 oxygen atoms in H₂SO₄ balance into 4H₂O.