A student electrolysed dilute sulfuric acid using inert platinum electrodes. Which half-equation correctly represents the reaction at the anode?

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Question

A student electrolysed dilute sulfuric acid using inert platinum electrodes. Which half-equation correctly represents the reaction at the anode?

Accepted Answer

4OH⁻ → O₂ + 2H₂O + 4e⁻ — Award 1 mark for the correct oxidation half-equation at the anode. A is incorrect — this is the reduction half-equation at the cathode. C is incorrect — sulfate ions are not discharged during electrolysis of dilute sulfuric acid. D is incorrect — this shows reduction, not oxidation, and the direction of electron flow is wrong for an anode reaction.

Answer

O₂ + 4e⁻ → 2O²⁻

Answer

2H⁺ + 2e⁻ → H₂

Answer

4OH⁻ → O₂ + 2H₂O + 4e⁻

Answer

SO₄²⁻ → SO₄ + 2e⁻

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