If we were to measure the change in concentration of a reactant over time and then plot a graph of our results, we would expect it to look something like this:
Manganate(VII) ions react with ethanedioate ions under acidic conditions to produce carbon dioxide, water and manganese(II) ions.
2MnO4–(aq) + 5C2O42-(aq) + 16H+(aq) ⇾ 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
When we measure the change in concentration of MnO4– ions over time and plot the results on a graph, it takes the shape of a sigmoid curve:
This is an example of an autocatalytic reaction, where one of the products of the reaction acts as a catalyst for the reaction – in this case the catalyst is the Mn2+ ion.
Now the shape of the graph makes more sense …
The rate of the reaction is so slow to begin with because the first step in the mechanism requires two negatively charged ions to react together and they repel each other! The rate increases as the concentration of Mn2+ (the catalyst) increases and then tails off at the end as the reactants are used up and their concentration decreases to zero.
Transition elements such as manganese are commonly used as catalysts, offering an alternative pathway for the reaction with a lower activation energy which makes use of their ability to change oxidation state (and so shuttle electrons around in redox reactions).
Manganate(VII) is an oxidising agent, oxidising carbon from an oxidation state of +3 in the ethanediate ion to +4 in carbon dioxide. Manganese is reduced from an oxidation state of +7 to +2.
The alternative pathway in this reaction comprises two steps with the Mn2+ catalyst being used up in the first step (Mn2+ reduces manganese in MnO4– to Mn3+) and regenerated in the second step (Mn3+ oxidises carbon in C2O42- where it has a +3 oxidation state to an oxidation state of +4 in CO2). The Mn2+ is a homogeneous catalyst in this reaction.
Practice question
When ethanedioate ions are added to an aqueous solution of potassium manganate(VII), the reaction mixture should be heated to begin with.
2MnO4–(aq) + 5C2O42-(aq) + 16H+(aq) ⇾ 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Explain why heating is not needed as the reaction proceeds. Your answer should include an explanation of homogeneous catalysis and make reference to the following electrode potentials.
| E⦵ / V | |
|---|---|
| 2CO2(g) + 2e– ⇌ C2O42–(aq) | + 0.64 |
| Mn3+(aq) + e– ⇌ Mn2+(aq) | + 1.49 |
| MnO4–(aq) + 8H+(aq) + 5e– ⇌ Mn2+(aq) + 4H2O(l) | + 1.51 |
Answer
The reaction needs heating to begin with because the rate is very slow. There is a high activation energy as the ethanedioate ions and manganate ions are both negatively charged and repel each other.
The reaction is catalysed by Mn2+ ions and once the concentration is high enough the rate increases and heating is no longer required. Homogenous catalysts are in the same physical state as the reactants.
The Mn2+ ions provide an alternative pathway for the reaction by firstly reducing the MnO4– ions to Mn3+ ions because the electrode potential for the Mn3+ / Mn2+ half cell is less positive than the electrode potential for the MnO4– / Mn2+ half cell.
Then the Mn3+ ions will oxidise the C2O42- ions because the electrode potential for the Mn3+ / Mn2+ half cell is more positive than the electrode potential for the C2O42- / CO2 half cell.
You can revise standard electrode potentials here 😊.