The reduction half equation of the following reaction is: $Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}$

Analyze the question: You need the reduction half equation for the reaction between zinc and copper(II) sulphate.

Concept: Reduction is gain of electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain). In a redox reaction, the species that gains electrons is reduced, and its half equation has the electrons on the left side.

Solution:

1. Write the ionic equation. Zn and Cu are the species that change; the SO₄²⁻ ions are spectators: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

2. Check the oxidation numbers: • Zn: 0 → +2 (loses 2 electrons, so it is oxidised) • Cu: +2 → 0 (gains 2 electrons, so it is reduced)

3. Write the half equations: • Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻ (Option A) • Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s) (Option C)

Checking the other options:

• A is the oxidation half equation, not the reduction one.
• B is just the dissolving (dissociation) of copper(II) sulphate, not a redox step.
• D has the wrong number of electrons. Cu²⁺ going to Cu needs 2 electrons, and the charge would not balance with 1 electron (+2 − 1 = +1 on the left, but 0 on the right).

Answer: C

Tip: In a reduction half equation the electrons appear on the left (reactants side). Also check that the charges balance on both sides: Cu²⁺ + 2e⁻ gives (+2) + (−2) = 0, which matches Cu(s). A quick way to remember: "Reduction = Reactant side electrons."