Question

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

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

Answer 1

The correct option is A $7$ to $4$

$Mn{O}_4^{-}+8H^{+}+5e\to 2M{n}^{+2}+4H_{2}O$
So the oxidation state of $Mn{O}_4^{-}$ is changed from +7 to +2 when it is converted to $M{n}^{+2}$
Hence option C is correct.