Homogeneous and heterogeneous catalysts — practice questions

The uncatalysed reaction is slow because it involves a collision between two negatively charged ions (S₂O₈²⁻ and I⁻), leading to strong electrostatic repulsion which results in a high activation energy.

Iron(II) ions act as a homogeneous catalyst because they are in the same aqueous phase as the reactants. The catalysis occurs in two steps:

Step 1: The Fe²⁺ ions are oxidised to Fe³⁺ ions by the peroxodisulfate(VI) ions. The S₂O₈²⁻ is reduced to SO₄²⁻. $S_{2}O_{8}^{2}⁻(aq) + 2Fe^{2}⁺(aq) \to 2SO_{4}^{2}⁻(aq) + 2Fe^{3}⁺(aq)$ This step is faster because it involves reaction between oppositely charged ions (S₂O₈²⁻ and Fe²⁺), reducing electrostatic repulsion.

Step 2: The Fe³⁺ ions formed then oxidise the iodide ions to iodine, regenerating the Fe²⁺ catalyst. $2Fe^{3}⁺(aq) + 2I⁻(aq) \to 2Fe^{2}⁺(aq) + I_{2}(aq)$ This step is also fast due to the attraction between oppositely charged ions (Fe³⁺ and I⁻).

Overall: The Fe²⁺ is regenerated at the end, fulfilling its role as a catalyst. The two steps provide an alternative pathway with lower activation energies than the direct reaction.

Vanadium(V) oxide is a solid, while the reactants are gases, so this is heterogeneous catalysis. The catalyst works by being reduced and then re-oxidised.

Step 1: The vanadium(V) oxide oxidises sulfur dioxide to sulfur trioxide. In this process, the vanadium is reduced from the +5 oxidation state to the +4 state (in the form of V₂O₄ or VO₂). $V_{2}O_{5}(s) + SO_{2}(g) \to V_{2}O_{4}(s) + SO_{3}(g)$ (Or using VO₂: $2VO_{2}(s) + SO_{2}(g) \to V_{2}O_{5}(s) + SO_{3}(g)$ is incorrect, the correct first step is reduction of V(V): $V_{2}O_{5}(s) + SO_{2}(g) \to 2VO_{2}(s) + SO_{3}(g)$ is also a valid representation, but the first one is simpler. Let's stick to the V₂O₄ representation for clarity). Correction: A better representation is showing the oxidation state change clearly. $V_{2}O_{5}(s) + SO_{2}(g) \to 2VO_{2}(s) + SO_{3}(g)$ is not balanced. Let's use the standard representation: $SO_{2}(g) + V_{2}O_{5}(s) \to SO_{3}(g) + V_{2}O_{4}(s)$ - This is not balanced either. Let's try again: $SO_{2}(g) + 2VO_{2}⁺ \to SO_{3}(g) + 2VO^{2}⁺$ is not syllabus standard. Let's stick to the oxides.

Correct balanced equations: Step 1: Vanadium(V) oxide is reduced by sulfur dioxide. $V_{2}O_{5}(s) + SO_{2}(g) \to V_{2}O_{4}(s) + SO_{3}(g)$ This is balanced. V is +5 in V₂O₅ and +4 in V₂O₄.

Step 2: The reduced form of the catalyst, vanadium(IV) oxide, is re-oxidised back to vanadium(V) oxide by oxygen. $V_{2}O_{4}(s) + \frac{1}{2}O_{2}(g) \to V_{2}O_{5}(s)$ To avoid fractions, we can double everything: $2V_{2}O_{4}(s) + O_{2}(g) \to 2V_{2}O_{5}(s)$

Summary: The V₂O₅ is regenerated, ready to catalyse another cycle. The variable oxidation states of vanadium (+5 and +4) are key to its catalytic activity.