Question

When $KMn{O}_4$ acts as an oxidising agent, it can form $Mn{O}_4^{2-},Mn{O}_2,{Mn}_2{O}_3$ or ${Mn}^{2+}$, the numbers of electrons transferred in each case respectively, are :

• A. 4, 3, 1, 5
• B. 1, 5, 3, 7
• C. 1, 3, 4, 5
• D. 3, 5, 7, 1

Answer 1

Answer: (C)

1, 3, 4, 5

The half cell reactions are as follows:
$e^{-}+Mn{O}_4^{\ominus }⟶Mn{O}_4^{2-}$
$3e^{-}+Mn{O}_4^{\ominus }⟶Mn{O}_2$
$4e^{-}+2Mn{O}_4^{\ominus }⟶{Mn}_2{O}_3$
$5e^{-}+Mn{O}_4^{\ominus }⟶{Mn}^{2+}$
Hence, the number of electrons transferred are $1,3,4$ and $5$ respectively in the formation of $Mn{O}_4^{2-},Mn{O}_2,{Mn}_2{O}_3$ and ${Mn}^{2+}$ from $KMn{O}_4$.