Question

If $KMn{O}_4$ is reduced by oxalic acid in an acidic medium, then the oxidation number of $Mn$ changes from:

• A. $+4$ to $+2$
• B. $+6$ to $+4$
• C. $+7$ to $+2$
• D. $+7$ to $+4$

Answer 1

The correct option is C $+7$ to $+2$

In an acidic medium, oxalic acid reduces $KMn{O}_4$ to $M{n}^{2+}$.

The redox reaction is as follows:

$2Mn{O}_4^{-}+5H_2{C}_2{O}_4+6H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

The oxidation state of $Mn$ in $KMn{O}_4$ and $M{n}^{2+}$ are $+7$ and $+2$ respectively.

Hence, the oxidation state of $Mn$ changes from $+7$ to $+2$.

Option C is correct.