Question

How many electrons are involved in oxidation of $KMn{O}_4$ in basic medium ?

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (A)

Oxidation of $KMn{O}_4$ takes place in all the three medium acidic, basic and neutral. In all three medium the oxidation number is different . In acidic medium
$2KMn{O}_4+3H_{2}S{O}_4\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5O$
The net. reaction is $Mn{O}_4^{-}\to M{n}^{2+}$
Change in oxidation numbers $=7-2=5$
So, electrons involved $=5e^{-}$
In basic medium
$2KMn{O}_4+2KOH\to 2K_{2}Mn{O}_4+H_{2}O+O$
The net reaction is
$\stackrel{+7}{\mathrm{Mn}}{O}_4^{-}\to \stackrel{+6}{\mathrm{Mn}}{O}_4^{-2}$
Change in oxidation numbers $=7-6=+1$
So, electrons involved $=1e^{-}$
In neutral medium
$2KMn{O}_4+H_{2}O\to 2kOH+2Mn{O}_2+3O$ net reaction is
$Mn{O}_4^{-}\to Mn{O}_2$
Change in oxidation numbers $=7-4=+3$
So, electrons involved $=3e^{-}$