Transition metals as catalysts

Heterogeneous catalysis (spec 17.26 to 17.29)

A heterogeneous catalyst is in a different phase from the reactants; the reaction happens on its surface.

• Vanadium(V) oxide in the contact process, by changing oxidation number:

SO2 + V2O5 -> SO3 + V2O4 (V: +5 to +4)

V2O4 + ½O2 -> V2O5 (V: +4 to +5)

• Catalytic converters (platinum, rhodium, palladium on a ceramic honeycomb): CO and NO are adsorbed onto the metal surface, bonds weaken and they react, then CO2 and N2 desorb: 2CO + 2NO -> 2CO2 + N2.

Homogeneous catalysis (spec 17.30, 17.31)

A homogeneous catalyst is in the same phase as the reactants and works through an intermediate species.

Peroxydisulfate and iodide react slowly because both are negative ions that repel: S2O8²- + 2I- -> 2SO4²- + I2. Fe2+ catalyses it:

S2O8²- + 2Fe2+ -> 2SO4²- + 2Fe3+

2Fe3+ + 2I- -> 2Fe2+ + I2

Each step involves oppositely charged ions, so each has a lower activation energy. Fe3+ works equally well, because it starts at the second step.

Autocatalysis (spec 17.32)

In autocatalysis a product catalyses the reaction. Manganate(VII) oxidising ethanedioate:

2MnO4- + 16H+ + 5C2O4²- -> 2Mn2+ + 8H2O + 10CO2

is catalysed by Mn2+, a product. The reaction starts slowly, speeds up as Mn2+ builds up, then slows as the reactants run out, so a graph of [MnO4-] against time is an inverted S shape (a slow fall at first, a steep middle, then levelling off). This is why the first drops of manganate(VII) in the titration decolourise slowly.

Worked example

Explain, using oxidation numbers, how vanadium(V) oxide catalyses 2SO2 + O2 -> 2SO3.

1. V2O5 oxidises SO2 to SO3 and is reduced: V +5 to +4.
2. V2O4 is reoxidised by oxygen: V +4 to +5.
3. V2O5 is regenerated, so it is a catalyst.

Answer: SO2 + V2O5 -> SO3 + V2O4; V2O4 + ½O2 -> V2O5

Check yourself

1. How does the oxidation number of vanadium change as V2O5 catalyses the oxidation of SO2?
   • +5 to +4, then back to +5
   • +5 to +3, then back to +5
   • +4 to +5, then back to +4
   • It stays at +5
   Show answer

   +5 to +4, then back to +5. V2O5 is reduced to V2O4 by SO2, then reoxidised by O2.

2. In a catalytic converter, what happens to CO and NO at the catalyst surface?
   • They dissolve in the catalyst
   • They are absorbed into the ceramic permanently
   • They are adsorbed, react, and the products CO2 and N2 desorb
   • They react in the gas phase away from the surface
   Show answer

   They are adsorbed, react, and the products CO2 and N2 desorb. Adsorption weakens bonds; products leave the surface.

3. What is a homogeneous catalyst?
   • A catalyst in the same phase as the reactants, which works via an intermediate
   • A solid catalyst for gas reactions
   • A catalyst that is used up
   • A catalyst that changes the equilibrium position
   Show answer

   A catalyst in the same phase as the reactants, which works via an intermediate. Same phase; it forms an intermediate and is regenerated.

4. Why is the uncatalysed reaction between S2O8²- and I- slow?
   • The reaction is endothermic
   • Iodine is a catalyst poison
   • Peroxydisulfate is a reducing agent
   • The two reactants are negative ions, so they repel each other
   Show answer

   The two reactants are negative ions, so they repel each other. Repulsion means a high activation energy.

5. Which equation is the first step when Fe2+ catalyses the reaction of S2O8²- with I-?
   • 2Fe3+ + 2I- -> 2Fe2+ + I2
   • S2O8²- + 2Fe2+ -> 2SO4²- + 2Fe3+
   • Fe2+ + I- -> FeI
   • S2O8²- + 2I- -> 2SO4²- + I2
   Show answer

   S2O8²- + 2Fe2+ -> 2SO4²- + 2Fe3+. Fe2+ is oxidised by S2O8²-; then Fe3+ oxidises I- in the second step.

6. What is autocatalysis?
   • A catalyst that works only in cars
   • Catalysis by a solid surface
   • A reaction that needs no activation energy
   • Catalysis of a reaction by one of its products
   Show answer

   Catalysis of a reaction by one of its products. Mn2+, formed from MnO4-, catalyses the MnO4-/C2O4²- reaction.

7. How does the rate of the reaction of manganate(VII) with ethanedioate change over time at constant temperature?
   • Fastest at the start, then slower throughout
   • Slow at first, then faster as Mn2+ forms, then slower as reactants are used up
   • Constant throughout
   • Zero until all the Mn2+ has gone
   Show answer

   Slow at first, then faster as Mn2+ forms, then slower as reactants are used up. Autocatalysis gives an inverted S-shaped [MnO4-]-time curve: slow, then fast, then slow.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.