Q. How many electrons are involved in oxidation of ${KMnO}_{4}$ in basic medium?
A) 1
B) 2
C) 5
D) 3
Explanation
(a)
In basic medium
$2{KMnO}_{4} + 2KOH \longrightarrow 2K_{2}{MnO}_{4} + H_{2}O + O$
Net reaction is _{+7 +6}
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {MnO}_{4}^{-} \longrightarrow {MnO}_{4}^{- 2}$
Change in oxidation number
=7$- 6 = + 1$
So, electrons involved =$1e^{-}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.