Question

How do `"MnO"_2` and `"KMnO"_4` behave? Oxidizing or reducing agents?

Answer 1

`"Manganese (IV) oxide"` and `KMnO_4` are `"oxidizing agents."`

Explanation:

`Mn^(2+)` is the reduction product. Because this is a `d^5` high spin metal centre, its electronic transitions are spin-forbidden, and a solution of `Mn^(2+)` is (almost!) completely colourless (very conc. solutions are a pale rose).

So, when we take strongly oxidizing permanganate ion, which is very strongly coloured, the redox reaction has a self-indicating end-point, and decolorizes:

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

We habitually use this reaction in redox titrations.