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

(N/A) The oxidising behaviour of $KMnO_{4}$ depends on the $pH$ of the solution.
$1$. In acidic medium $(pH < 7)$:
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$
Here,$Mn$ is reduced from $+7$ to $+2$ oxidation state,resulting in a colourless solution.
$2$. In alkaline medium $(pH > 7)$:
$MnO_{4}^{-} + e^{-} \rightarrow MnO_{4}^{2-}$ (Green solution)
Here,$Mn$ is reduced from $+7$ to $+6$ oxidation state.
$3$. In neutral or weakly alkaline medium $(pH \approx 7)$:
$MnO_{4}^{-} + 2H_{2}O + 3e^{-} \rightarrow MnO_{2} + 4OH^{-}$ (Brown precipitate)
Here,$Mn$ is reduced from $+7$ to $+4$ oxidation state.