Read the passage carefully and answer the questions.

The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron, and nickel (in Catalytic Hydrogenation) are some of the examples. Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowering). Also because the transition metal ions can change their oxidation states, they become more effective as catalysts.

Reaction between iodide and persulphate ions takes place as follows:

$\mathrm{2I^{-}+S_2O_8^{2-}\rightarrow I_2+2SO}_4^{2-}$

Which of the following catalyses the above reaction?

$\mathrm{Fe}^{3+}$

The correct answer is Option (2) → $\mathrm{Fe}^{3+}$

Redox catalysis — A catalyst provides an alternative reaction pathway by undergoing temporary oxidation–reduction and getting regenerated at the end.

This reaction is slow due to electrostatic repulsion between negatively charged ions.

Fe³⁺ acts as a catalyst through a redox cycle.

Step 1: Fe³⁺ + I⁻ → Fe²⁺ + ½ I₂

Fe³⁺ oxidises iodide.

Step 2: Fe²⁺ + S₂O₈²⁻ → Fe³⁺ + 2SO₄²⁻

Fe²⁺ reduces persulphate and regenerates Fe³⁺.

Thus: • Fe³⁺ participates in reaction • Gets regenerated • Provides faster pathway

Hence acts as catalyst.

Option 1: Fe²⁺ Intermediate formed, not the catalyst.

Option 3: Mn²⁺ Does not form effective redox catalytic cycle.

Option 4: Sn²⁺ Acts as reducing agent, not catalyst.