Question

Find the n-factor in the following chemical changes :

$KMn{O}_4\stackrel{H^{+}}{\to }M{n}^{2+}$

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

The correct option is D 5

In acidic medium, $M{n}^{7+}$ goes to $M{n}^{2+}$ state and hence there is a net gain of 5 electrons.

In $KMn{O}_4$ the oxidation state of Mn is $+7$. In an acidic medium, the reaction is:

$Mn{O}_4^{¯}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

As you can see, $M{n}^{7+}$ gains 5 electrons. (Making it an oxidizing agent.)

So the 'n factor' is 5 for $KMn{O}_4$