Question

When $KMn{O}_4$ act as an oxidising agent and ultimately form $Mn{O}_4^{2-},Mn{O}_2,{Mn}_2{O}_3$ and ${Mn}^{2+}$, then 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

The correct option is C 1, 3, 4, 5

$e^{-}+Mn{O}_4^{\ominus }⟶Mn{O}_4^{2-}$

$3e^{-}+Mn{O}_4^{\ominus }⟶Mn{O}_2$

$4e^{-}+2Mn{O}_4^{\ominus }⟶{Mn}_2{O}_3$

$5e^{-}+Mn{O}_4^{\ominus }⟶{Mn}^{2+}$