Question

How many electrons are involved in oxidation of $ KMnO_4 $ in basic medium ?

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (A)

Oxidation of $KMnO_{4}$ takes place in all the three medium acidic, basic and neutral. In all three medium the oxidation number is different . In acidic medium
$2KMnO_{4}+3H_{2}SO_{4} \to K_{2}SO_{4} +2MnSO_{4}+3H_{2}O+5O$
The net. reaction is $MnO^{-}_{4} \to Mn^{2+}$
Change in oxidation numbers $=7-2=5$
So, electrons involved $=5e^{-}$
In basic medium
$2KMnO_{4}+2KOH \to 2K_{2}MnO_{4}+H_{2}O+O$
The net reaction is
$\overset{+7}{Mn}O_4^- \rightarrow \overset{+6}{Mn}O_4^{-2}$
Change in oxidation numbers $=7-6=+1$
So, electrons involved $=1e^{-}$
In neutral medium
$2KMnO_{4}+H_{2}O \to 2kOH+2MnO_{2}+3O$ net reaction is
$MnO^{-}_{4} \to MnO_{2}$
Change in oxidation numbers $= 7 - 4 = + 3$
So, electrons involved $=3e^{-}$