In the reaction between peroxodisulphate ions and iodide ions, the iodide ions are oxidised to iodine whilst the peroxodisulphate ions are reduced to sulphate(VI) ions.
This is confirmed by the standard electrode potential data for the two half reactions …
S2O82-(aq) + 2e- ⇌ 2SO42-(aq) E⦵ = + 2.01V
I2(aq) + 2e- ⇌ 2I–(aq) E⦵ = + 0.54V
The S2O82-(aq) / 2SO42-(aq) half cell has the more positive standard electrode potential so we can see that it will be the reduction half cell and the I2(aq) / 2I–(aq) half cell will be the oxidation half cell.
Although the reaction is energetically favourable (Ecell = + 1.47V), the reaction is very slow indeed. This is because it requires two negatively charged ions to come together. We need a catalyst and this reaction is catalysed by aqueous Fe2+ ions.
The catalyst and the reactants are in the same phase (they are all ions in solution) which is the definition of a homogeneous catalyst.
A catalyst provides an alternative pathway for a reaction with a lower activation energy:
catalyst + S2O82- ⇾ intermediates ; intermediate + I– ⇾ products + catalyst
In a redox reaction, the job of the catalyst is to transfer electrons between the reactants and products and a transition metal ion can do this by changing between two different oxidation states.
If we consider the standard electrode potential for a Fe2+(aq) / Fe3+(aq) half cell, we see it falls between the standard electrode potentials of the two half cells that make up this reaction (this is always a good indicator when picking a suitable catalyst).
Fe3+(aq) + e- ⇌ Fe2+(aq) E⦵ = + 0.77V
Fe2+ can reduce the peroxodisulphate ion, itself being oxidised to Fe3+ and then the Fe3+ can oxidise the iodide ion, itself being reduced back to Fe2+. Our catalyst has been regenerated.
Step 1: 2Fe2+(aq) + S2O82-(aq) ⇾ 2Fe3+(aq) + 2SO42-(aq)
Step 2: 2Fe3+(aq) + 2I–(aq) ⇾ 2Fe2+(aq) + I2(aq)
We can show this on a reaction profile:
Of course, we could choose to use Fe3+ as our catalyst in which case the two steps in our reaction mechanism would be reversed.
Step 1: 2Fe3+(aq) + 2I–(aq) ⇾ 2Fe2+(aq) + I2(aq)
Step 2: 2Fe2+(aq) + S2O82-(aq) ⇾ 2Fe3+(aq) + 2SO42-(aq)
Transition metal compounds are also widely used as heterogeneous catalysts with the catalyst being solid and the reactants being in the gas phase, for example in catalytic converters.
Vanadium(V) oxide catalyses the reaction between sulphur dioxide and oxygen in the Contact process (the final industrial product is sulphuric acid). The mechanism is thought to be similar to the homogeneous catalysis seen above – the variable oxidation states of vanadium enabling electron transfer in each step of the reaction.
SO2(g) + V2O5(s) ⇾ SO3(g) + V2O4(s)
2V2O4(s) + O2(g) ⇾ 2V2O5(s)
The V2O5 catalyst is regenerated and the overall equation becomes:
2SO2(g) + O2(g) ⇾ 2SO3(g)
However, with heterogeneous catalysis there are other processes happening which will increase the rate of the reaction by lowering the overall activation energy. The reactant gases bond to the surface of the solid catalyst in a process known as chemisorption. Bonds within the reactant gases are stretched and weakened or broken enabling reactions between them to proceed more quickly. In addition, the concentration of reactant gases near the surface of the catalyst is increased.
Practice questions
- The oxidation of sulfate(IV) ions to sulfate(VI) ions is catalysed by Co2+ ions in acidic solution.
SO32- + ½O2 ⇾ SO42-
(a) What property of cobalt allows it to act as a catalyst?
(b) Explain how Co2+ ions catalyse this reaction in a two step mechanism. You should include relevant ionic equations in your answer.
2. Explain why chromium commonly shows the oxidation states of +2, +3 and +6 in its compounds in terms of ionisation energies.
Answers
- (a) Cobalt can change between oxidation states enabling electron transfer between reactants.
(b) In the first step Co2+ is oxidised to Co3+ and O2 is reduced.
Co2+ + ½O2 + 2H+ ⇾ Co3+ + H2O
In the second step Co3+ its reduced back to Co2+ and SO32- is oxidised to SO42-. The catalyst has been regenerated and both steps have a combined lower activation energy than the reaction without a catalyst.
Co3+ + SO32- + H2O ⇾ Co2+ + SO42- + 2H+
The Co2+, Co3+, H+ and H2O cancel out when we add these two equations together to give:
SO32- + ½O2 ⇾ SO42-
Exam tip: this is very hard question and it relies on you knowing how to write half and ionic equations in acidic conditions (this was the clue that H+ and H2O would be needed in each step) – if you need to revise the rules for writing half equations you can do it here!
2. The successive ionisation energies of chromium are very close together so there is only a gradual increase from one oxidation state to the next.