Question

Reaction of $KMnO _{4}$ in neutral or faintly alkaline solution is given by the equation:

• A. $MnO _{4}^{-}+3 e ^{-}+2 H _{2} O \to MnO _{2}+4 OH ^{-}$
• B. $MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \to Mn ^{2+}+4 H _{2} O$
• C. $MnO _{4}^{-}+ e ^{-} \to MnO _{4}^{2-}$
• D. $2 MnO _{4}^{-}+2 OH ^{-} \to 2 MnO _{4}^{2-}+1 / 2 O _{2}+ H _{2} O$

Answer 1

Answer: (A)

In neutral or faintly alkaline medium, $KMnO _{4}$ itself gets changed to $MnO _{2}$.