Iron(III) ions catalyse the reaction between iodide and persulphate i…

The correct answer is Option (1) → \( \ce{S2O8^2-} \)

Transition metal ions are good catalysts because they can change their oxidation states. Here iron moves between +3 and +2: Fe³⁺ takes electrons from iodide ions, and Fe²⁺ then passes them on to the persulphate ion.

    \( \displaystyle \ce{2Fe^3+ + 2I^- -> 2Fe^2+ + I2} \)

    \( \displaystyle \ce{2Fe^2+ + S2O8^2- -> 2Fe^3+ + 2SO4^2-} \)

Adding the two steps gives the complete reaction, 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻. The Fe³⁺ used in the first step is formed again in the second, so it is not used up; this is why it acts as a catalyst.

Why the other options are incorrect

Option (2) \( \ce{I2} \) — Incorrect. Iodine is the product of the first step, formed when Fe³⁺ oxidises iodide. It remains as a product of the reaction and does not take part in the second step.

Option (3) \( \ce{SO4^2-} \) — Incorrect. Sulphate is formed when persulphate gains electrons in the second step. It is the reduced product, so it does not oxidise Fe²⁺.

Option (4) \( \ce{I^-} \) — Incorrect. Iodide is the reducing agent of the first step; it gives electrons to Fe³⁺. A species that gives electrons cannot oxidise Fe²⁺.

NCERT (2026-27 edition):
4.3.12 Catalytic Properties – “…Also because the transition metal ions can change their oxidation states, they become more effective as catalysts. For example, iron(III) catalyses the reaction between iodide and persulphate ions.

2 I^– + S₂O₈^2– → I₂ + 2 SO₄^2–

An explanation of this catalytic action can be given as: 2 Fe³⁺ + 2 I^– → 2 Fe²⁺ + I₂

2 Fe²⁺ + S₂O₈^2– → 2 Fe³⁺ + 2SO₄^2–”