Consider the following reaction equation: SO42-(aq) + 2H+(aq) + ye- $\to$ SO42-(aq) + H2O(1). The value of y in the equation is

Analyze the question

The equation as printed has SO4²⁻ on both sides, which cannot be right. The original WASSCE question is almost certainly: SO4²⁻(aq) + 2H⁺(aq) + ye⁻ → SO3²⁻(aq) + H2O(l). This is a reduction of sulphate to sulphite. I'll solve it in that form.

Concept

In a balanced half-equation, both atoms and charge must balance. We use charge balance to find the number of electrons.

Solution

Left side charge:

• SO4²⁻ gives −2
• 2H⁺ gives +2
• ye⁻ gives −y

Total = −2 + 2 − y = −y

Right side charge:

• SO3²⁻ gives −2
• H2O gives 0

Total = −2

Equating: −y = −2, so y = 2.

Check using oxidation numbers: S in SO4²⁻ is +6, and S in SO3²⁻ is +4. The change is 6 → 4, so 2 electrons are gained (reduction). This agrees.

Atom check:

• O: 4 on the left, 3 + 1 = 4 on the right ✓
• H: 2 on the left, 2 on the right ✓

Note on the typo: If the equation really had SO4²⁻ on both sides, the charge balance would give −y = −2 as well (with H⁺ cancelling the charge), so y = 2 either way. The oxygen count wouldn't balance, though, which confirms the product is meant to be SO3²⁻.

Answer: A (y = 2)

Exam tip: For half-equations, balance the charge first: total charge on the left must equal total charge on the right. Then confirm with the change in oxidation number, since electrons gained equals the drop in oxidation state for a reduction.