Transition metals as catalystsEdexcel A-Level Chemistry: Revision notes

Section 1

Heterogeneous and homogeneous catalysts

A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy, and is regenerated at the end. Transition metals and their compounds are good catalysts because they have variable oxidation states (they can accept and donate electrons) and can adsorb molecules on their surface.

A heterogeneous catalyst is in a different phase from the reactants (usually a solid with gases or liquids). The reaction occurs at its surface, so a large surface area helps.
A homogeneous catalyst is in the same phase as the reactants (usually aqueous). The catalysed reaction proceeds via an intermediate species.

Key termscatalystheterogeneoushomogeneous

Common mistake

Writing that a catalyst 'takes no part' in the reaction. It takes part, but is regenerated.

Section 2

V₂O₅ in the Contact process

Solid $\mathrm{V_2O_5}$ is a heterogeneous catalyst for $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$ (gases, about 450 °C). Vanadium changes oxidation number:

$\mathrm{SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4}$ (V: +5 to +4)

$\mathrm{V_2O_4 + \tfrac{1}{2}O_2 \rightarrow V_2O_5}$ (V: +4 to +5)

The catalyst is regenerated, and adding the two equations gives $\mathrm{SO_2 + \tfrac{1}{2}O_2 \rightarrow SO_3}$ .

Key termscontact process

Section 3

Catalytic converters

A catalytic converter has platinum, palladium or rhodium on a ceramic honeycomb, giving a large surface area for a small mass of metal. It removes CO and NO from exhaust gases:

$\mathrm{2CO + 2NO \rightarrow 2CO_2 + N_2}$

The steps are:

Adsorption of CO and NO onto the surface.
Weakening of the bonds in the adsorbed molecules.
Reaction between the adsorbed species.
Desorption of the products CO₂ and N₂.

Key termsadsorptiondesorption

Exam tip

Name the four steps in order: adsorption, bond weakening, reaction, desorption.

Section 4

Fe²⁺ catalysing I⁻ and S₂O₈²⁻

$\mathrm{S_2O_8^{2-} + 2I^- \rightarrow 2SO_4^{2-} + I_2}$ is slow because two negative ions repel. Fe²⁺ is a homogeneous catalyst that avoids this:

$\mathrm{2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}}$ , $E^\ominus_{cell} = 2.01 - 0.77 = +1.24$ V

$\mathrm{2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2}$ , $E^\ominus_{cell} = 0.77 - 0.54 = +0.23$ V

Both steps involve oppositely charged ions, and both are feasible. The iron cycles between +2 and +3 (the intermediate is Fe³⁺).

Key termsintermediate

Section 5

Autocatalysis: Mn²⁺ with MnO₄⁻ and C₂O₄²⁻

In acid, $\mathrm{2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2}$ . It starts slowly (two negative ions repel), then speeds up because the product Mn²⁺ is the catalyst: autocatalysis.

$\mathrm{4Mn^{2+} + MnO_4^- + 8H^+ \rightarrow 5Mn^{3+} + 4H_2O}$

$\mathrm{2Mn^{3+} + C_2O_4^{2-} \rightarrow 2Mn^{2+} + 2CO_2}$

The purple colour fades slowly at first, then quickly as [Mn²⁺] increases. The concentration of MnO₄⁻ against time has an S-shaped (sigmoid) curve.

Key termsautocatalysis

Common mistake

Saying the reaction speeds up because of temperature or because concentration of reactants rises. The reactant concentrations fall; the rate increases because the catalyst Mn²⁺ is being made.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Transition metals as catalysts

In the Contact process, sulfur dioxide gas is oxidised to sulfur trioxide gas by oxygen in the presence of a solid vanadium(V) oxide, V₂O₅, catalyst at about 450 °C: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Write two equations to show how V₂O₅ acts as a catalyst for the oxidation of SO₂ and is regenerated.2 marks
A car's catalytic converter contains platinum and rhodium spread as a thin layer over a ceramic honeycomb. It converts carbon monoxide and nitrogen monoxide in exhaust gases into carbon dioxide and nitrogen.
Suggest why the platinum and rhodium are used as a thin layer on a ceramic honeycomb rather than as solid blocks of metal.2 marks
The reaction between iodide ions and peroxodisulfate ions, S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq), is very slow, but it is catalysed by a few drops of iron(II) sulfate solution. Standard electrode potentials: S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻, E° = +2.01 V; Fe³⁺ + e⁻ ⇌ Fe²⁺, E° = +0.77 V; I₂ + 2e⁻ ⇌ 2I⁻, E° = +0.54 V.
Explain why the uncatalysed reaction is slow, and why iron(II) ions speed it up. State the type of catalysis involved.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).