Question

How does permanganate ion behave in redox reactions?

Answer 1

Well, permanganate anion is generally reduced to (colourless) `Mn^(2+)` in acidic solution..........

Explanation:

`MnO_4^(-) +8H^+ +5e^(-) rarr Mn^(2+) + 4H_2O(l)`

Note that `Mn^(2+)` is almost colourless in aqueous solution; it is a `d^5` system, for which electronic transition is spin forbidden. Very concentrated solutions are pale rose.

In basic solution, generally, permanganate is reduced to `MnO_2(s)`, a 3 electron reduction:

`MnO_4^(-) + 4H^+ +3e^(-) rarr MnO_2(s) + 2H_2O(l)` (but BASIC conditions were specified, so we add `4xxHO^-` to both sides.....

`MnO_4^(-) + 2H_2O +3e^(-) rarr MnO_2(s) + 4HO^(-)`

Of course you need the corresponding oxidation equation. But there is formal `5e^-` transfer in acid, and `3e^-` transfer in base.