Question

A solution of \[\ce{KMnO4}\] 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?

Answer 1

 \[\ce{KMnO4}\] acts as an oxidising agent in alkaline, neutral and acidic mediums, i.e., oxidising behaviour of  \[\ce{KMnO4}\] depends on pH of the solution.

In alkaline medium (pH > 7),

\[\ce{MnO^{-}4 + e^{-} -> MnO^{2-}4}\]

i.e., permanganate is changed to manganate which gives green solution.

\[\ce{2KMnO4 + 2KOH -> \underset{Green}{2K2MnO4} + H2O + [O]}\]

Reducing agent + \[\ce{[O] -> Product}\].

In neutral solution (pH = 7), permanganate is changed to manganese dioxide (brown ppt).

\[\ce{MnO^{-}4 + 2H2O ->[+3e^{-}] \underset{Brown ppt}{MnO2} + 4OH^{-}}\]

\[\ce{2KMnO4 + H2O -> 2KOH + 2MnO2 + 3[O]}\]

In acidic medium (pH < 7), permanganate is changed to manganous ion (Colourless).

\[\ce{MnO^{-}4 + 8H^{+}5e^{-} -> Mm^{2+} + 4H2O}\]

\[\ce{2KMnO4 + 3H2SO4 -> \underset{\underset{solution}{Colourless}}{2MnSo4} + K2SO4 + 3H2O + 5[O]}\]