Question

How many electrons are involved in the oxidationby $KMnO_4$ in basic medium ?

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (A)

1

Answer 2

Oxidation of $KMnO _{4}$ takes place in all the three medium acidic, basic and the three medium
acidic, basic and oxidation number is different .
In acidic medium :
$2 KMnO _{4}+3 H _{2} SO _{4} \rightarrow K _{2} SO _{4}+2 MnO _{4}+3 H _{2} O +50$
The net reaction is :
$MnO _{4}^{-} \rightarrow Mn ^{2+}$
change in oxidation number $=7-2=+5$
So electrons involved $=5 e ^{-}$
In basic medium :
$2 KMnO _{4}+2 KOH \rightarrow 2 K _{2} MnO _{4}+ H _{2} O +0$
Net reaction is
$\overset{+7}{MnO _{4}^{-}} \rightarrow \overset{+6}{MnO _{4}^{-2}}$
Change in oxidation number $=7-6=+1$
So electron involved is $=1 e ^{-}$
In Neutral medium:
$2 KMnO _{4}+ H _{2} O \rightarrow 2 KOH +2 MnO _{2}+30$
$Net$ reaction
$\overset{+7}{MnO _{4}^{-}} \rightarrow \overset{+4}{MnO _{2}}$
Change in oxidation no. $=7-4=+3$
So electrons involved : $3 e ^{-}$