# How does permanganate ion behave in redox reactions?

Apr 25, 2017

Well, permanganate anion is generally reduced to (colourless) $M {n}^{2 +}$ in acidic solution..........

#### Explanation:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O \left(l\right)$

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 $M n {O}_{2} \left(s\right)$, a 3 electron reduction:

$M n {O}_{4}^{-} + 4 {H}^{+} + 3 {e}^{-} \rightarrow M n {O}_{2} \left(s\right) + 2 {H}_{2} O \left(l\right)$ (but BASIC conditions were specified, so we add $4 \times H {O}^{-}$ to both sides.....

$M n {O}_{4}^{-} + 2 {H}_{2} O + 3 {e}^{-} \rightarrow M n {O}_{2} \left(s\right) + 4 H {O}^{-}$

Of course you need the corresponding oxidation equation. But there is formal $5 {e}^{-}$ transfer in acid, and $3 {e}^{-}$ transfer in base.