How does permanganate ion behave in redox reactions?
Answer:
Well, permanganate anion is generally reduced to (colourless) \(M n^{2+}\) in acidic solution……….
Explanation:
\(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}(l)\)
Note that \(M n^{2+}\) 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 \(\mathrm{MnO}_{2}(s)\), a 3 electron reduction:
\(\mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 e^{-} \rightarrow \mathrm{MnO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (but BASIC conditions were specified, so we add \(4 \times H O^{-}\) to both sides…..
\(\mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O}+3 e^{-} \rightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{HO}^{-}\)Of course you need the corresponding oxidation equation. But there is formal \(5 e^{-}\) transfer in acid, and \(3 e^{-}\) transfer in base.