A solution of `KMnO_(4)` on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

Explanation

Oxidising behaviour of `KMnO_4` depends on the pH of the solution.
In acidic medium (pH < 7), it gives colourless `Mn^(2+)` ions:
In alkaline medium (pH > 7), green coloured `MnO_4^(2-)` ions are formed :
`MnO_4^(-) + e^(-) to underset("green")(MnO_4^(2-))`
In neutral medium(pH = 7), brown coloured `MnO_3` formed :
`MnO_4^(-) + 2H_2O + 3e^(-) to underset("brown precipitate")(MnO_2) + 4OH^(-) `