Question

When $KMn{O}_4$ is reduced with oxalic acid in acidic solution , the oxidation number of Mn changes from:

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

Answer 1

The correct option is C +7 to +2

Reaction involved:

$KMn{O}_4+H_2{C}_2{O}_4+H_{2}S{O}_4\to K_{2}S{O}_4+MnS{O}_4+C{O}_2+H_{2}O$

Assign the oxidation number to Mn

Let the oxidation number of Mn be 'x'.

Oxidation number of $S{O}_4^{2-}=O=-2,K=+1$.

In $KMn{O}_4$: $1+x+4\times (-2)=0$

$x+1-8=0$ or $x=+7$

In $MnS{O}_4$: $x+(-2)=0$

$x=+2$in the reaction thus:

$K\stackrel{+7}{Mn}{O}_4+H_2{C}_2{O}_4+H_{2}S{O}_4\to K_{2}S{O}_4+\stackrel{+2}{M}nS{O}_4+C{O}_2+8H_{2}O$
When $KMn{O}_4$ is reduced with oxalic acid in acidic solution , the oxidation number of Mn changes from +7 to +2.