Question

In the reduction of $KMn{O}_4$, by warm acidified oxalic acid, 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 $+2$

$[Mn{O}_4^{-}+8H^{+}+5e^{-}⟶{Mn}^{2+}+4H_{2}O]\times 2$
$[C_2{O}_4^{-}⟶2{CO}_2+2e^{-}]\times 5$
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$2{Mn}^{2+}{O}_4^{-}+5C_2{O}_4^{-}+16H^{+}⟶2{Mn}^{2+}+10{CO}_2+8H_{2}O$
So, oxidation number of $Mn$ changes in the above reaction from $+7$ to $+2$.