Q.(a) Transition metals can act as catalysts because these can change their oxidation state. How does Fe(III) catalyse the reaction between iodide and persulphate ions?

The key idea here is that transition metals like Fe(III) can cycle between oxidation states, providing an alternate reaction pathway with lower activation energy.

(a) Fe(III) catalyses the reaction between iodide ($I^-$) and persulphate ($S_2O_8^{2-}$) by undergoing reduction and re-oxidation in two steps:

1. Fe(III) is first reduced to Fe(II) by iodide:

2. Fe(II) is then re-oxidised back to Fe(III) by persulphate:

The net reaction is $2I^- + S_2O_8^{2-} \rightarrow I_2 + 2SO_4^{2-}$, with Fe(III) regenerated — hence it acts as a true catalyst.

(b) Three processes where transition metals act as catalysts:

• Haber's process: Iron (Fe) catalyses the synthesis of ammonia from $N_2$ and $H_2$. …

Why this works — the idea of a redox shuttle

A catalyst works by offering an alternative reaction path with a lower activation energy. For transition metals, this often comes from their ability to exist in multiple oxidation states with small energy differences. Fe(III) and Fe(II) are both stable in aqueous solution, and the energy barrier to switch between them is modest. That makes iron an ideal electron ferry.

In the uncatalysed reaction, iodide and persulphate ions must collide directly and transfer two electrons in one go — a slow, high‑energy step. With Fe(III) present, the electron transfer is broken into two easier steps, each involving only one electron. The catalyst is consumed in the first step and regenerated in the second, so it is not used up.

Step‑by‑step mechanism

1. Fe(III) oxidises iodide Fe³⁺ accepts one electron from I⁻, forming Fe²⁺ and iodine radical (or, in net terms, half an I₂ molecule).

This step is fast because Fe³⁺ is a good oxidising agent (standard reduction potential $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}$) and I⁻ is a moderate reducing agent.

2. Fe(II) reduces persulphate The Fe²⁺ produced in step 1 now donates an electron to S₂O₈²⁻, regenerating Fe³⁺ and forming sulphate radicals (which quickly become sulphate ions).

This step is also fast because Fe²⁺ is a good reducing agent and S₂O₈²⁻ is a strong oxidiser ($E^\circ(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}) = +2.01\ \text{V}$).

3. Net reaction is unchanged Adding the two steps cancels the Fe³⁺ and Fe²⁺:

The iron ions appear on both sides of the mechanism and therefore do not appear in the overall equation. They are true catalysts — consumed in one step, regenerated in the next.

Solution Method: Catalytic Mechanism Analysis via Oxidation State Change

(a) Fe(III) Catalysis of Iodide–Persulphate Reaction

Method: Two-step redox cycle (alternating oxidation states)

Steps:

1. Identify the uncatalysed reaction The direct reaction between iodide ($\text{I}^-$) and persulphate ($\text{S}_2\text{O}_8^{2-}$) is:

This is slow because both reactants are negatively charged — repulsion makes collision unfavourable.

2. Introduce the catalyst — Fe(III)

   Fe(III) can accept an electron (get reduced) and later donate it (get reoxidised). This provides a lower-energy pathway.

3. Step 1: Reduction of Fe(III)

   Fe(III) oxidises iodide:

4. Step 2: Reoxidation of Fe(II) The Fe(II) formed is then oxidised back by persulphate:

5. Net result Adding the two steps gives the overall reaction. Fe(III) is regenerated — it is not consumed, only cycled between +3 and +2 states.

Key insight: The catalyst works because Fe(III)/Fe(II) has a variable oxidation state with a redox potential intermediate between the two reactants, allowing electron transfer in two easy steps instead of one difficult step.

(b) Three Processes Where Transition Metals Act as Catalysts

| Process | Catalyst | Role |

|---------|----------|------| …

Here is a breakdown of the common mistakes and how to avoid them, structured for exam success.

(a) How does Fe(III) catalyse the reaction between iodide and persulphate ions?

This is a classic example of auto-oxidation (or redox catalysis). The key is that the catalyst shuttles between two oxidation states.

The Correct Mechanism (Step-by-Step):

1. The Slow Step (Catalyst is Reduced): The catalyst, $\text{Fe}^{3+}$ (Iron III), oxidises the iodide ion ($\text{I}^-$).

2. The Fast Step (Catalyst is Regenerated): The newly formed $\text{Fe}^{2+}$ (Iron II) is then oxidised back to $\text{Fe}^{3+}$ by the persulphate ion ($\text{S}_2\text{O}_8^{2-}$).

Overall Reaction (Net):

The $\text{Fe}^{3+}$ is regenerated at the end, so it is not consumed. It simply provides a lower-energy pathway (lower activation energy) for the reaction.

✗ Common Mistake #1: Writing the wrong products for the first step.

• The Error: Students write $\text{Fe}^{3+} + \text{I}^- \rightarrow \text{Fe}^{2+} + \text{I}$ (a single iodine atom). This is incorrect because iodine atoms are highly unstable and immediately pair up.
• How to Avoid: Remember that iodine exists as a diatomic molecule ($\text{I}_2$). You must balance the equation. For every 2 electrons gained by $\text{Fe}^{3+}$, you need 2 $\text{I}^-$ ions to lose 2 electrons to form one $\text{I}_2$ molecule. Always balance the charge and the atoms.

✗ Common Mistake #2: Confusing the roles of the ions.

• The Error: Saying "Fe(III) oxidises persulphate" or "Fe(II) reduces iodide." This reverses the entire mechanism.
• How to Avoid: Use the Oxidation Number trick.
  • In step 1: $\text{Fe}^{3+} \rightarrow \text{Fe}^{2+}$ (gain of electron = reduction). Therefore, $\text{Fe}^{3+}$ is the oxidising agent for $\text{I}^-$.
  • In step 2: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$ (loss of electron = oxidation). Therefore, $\text{Fe}^{2+}$ is the reducing agent for $\text{S}_2\text{O}_8^{2-}$.
  • Mnemonic: "LEO the lion says GER" (Loss of Electrons = Oxidation; Gain of Electrons = Reduction).

✗ Common Mistake #3: Forgetting to mention the regeneration of the catalyst.

• The Error: Only writing the first step and stopping. The examiner wants to see that you understand the catalytic cycle.
• How to Avoid: Always write both steps. The definition of a catalyst is that it is chemically unchanged at the end. You must show how it gets back to its original state ($\text{Fe}^{3+}$).

(b) Mention any three processes where transition metals act as catalysts.

The key here is to be specific. Don't just say "Haber process." Say which metal is the catalyst.

✗ Common Mistake #4: Giving vague or incorrect examples.

• The Error: Saying "Iron is used in the Contact process" (it's Vanadium Pentoxide) or "Nickel is used in the Haber process" (it's Iron).
• How to Avoid: Memorise the Metal + Process + Role triplet.

Here are three correct, high-scoring examples:

1. Haber's Process (Ammonia synthesis):
   • Catalyst: Finely divided Iron ($\text{Fe}$) with promoters like $\text{Mo}$ or $\text{K}_2\text{O}$. …