Question

Question #1bdf3

Answer 1

The two roles of `KMnO_4` are as an oxidising agent and as a provider of a catalyst.

I am assuming you are referring to the oxidation of ethandioate ions by `"Mn(VII)"`.

The half equations are:

`MnO_4^(-)+8H^++5erarrMn^(2+)+4H_2O`

`C_2O_4^(2-)rarr2CO_2+2e`

You can see that to get the electrons to balance we need to X the top equation by 2 and the bottom equation by 5, then add. This gives:

`2MnO_4^(-)+16H^(+)+5C_2O_4^(2-)rarr2Mn^(2+)+8H_2O+10CO_2`

This reaction is quite slow at the start since it involves collision between two negatively charged species.

The `Mn^(2+)` ions catalyse the reaction by reacting with `MnO_4^-` ions to give `Mn^(3+)`:

`4Mn^(2+)+MnO_4^(-)+8H^(+)rarr5Mn^(3+)+4H_2O`

These can then oxidise the ethandioate ions, regenerating `Mn^(2+)`.

`2Mn^(3+)+C_2O_4^(2-)rarr2CO_2+Mn^(2+)`

This is known as "autocatalysis".