Iron (III) catalyses the reaction between iodide and persulphate ions, in which
A. Fe³⁺ oxidises the iodide ion
B. Fe³⁺ oxidises the persulphate ion
C. Fe²⁺ reduces the iodide ion
D. Fe²⁺ reduces the persulphate ion
Choose the most appropriate answer from the options given below:

Solution

Chemistry ma’am · 4 steps

A catalyst must come back unchanged, so iron goes Fe³⁺ → Fe²⁺ → Fe³⁺ in a loop.

1. Step 1. Fe³⁺ takes an electron from iodide, so it oxidises I⁻ to iodine and becomes Fe²⁺: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
2. Step 2. Fe²⁺ gives that electron to persulphate, so it reduces S₂O₈²⁻ to sulphate and returns to Fe³⁺: 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻.

   2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻

3. Check the loop. Add the two steps and the iron cancels, leaving only the overall reaction above. Iron is regenerated, which is what a catalyst does. The other statements fail: iodide at -1 is already in its lowest state and cannot be reduced, and persulphate is the oxidant, so it cannot be oxidised by Fe³⁺.
4. So. Statements A and D are correct, which is option (c).

The idea to keep. For a redox catalyst, write the two half-steps: the ion oxidises one reactant, then its reduced form reduces the other and comes back. If the catalyst is not regenerated, the mechanism is wrong.

Answer from the NTA final answer key. Solution by PrepWiser.