Question

A solution of $KMnO_{4}$ is reduced to various products depending upon its pH. At pH $< 7$ it is reduced to a colourless solution (1), at pH = 7 it forms a brown precipitate (2) and at pH > 7 it gives a green solution (3). (1), (2) and (3) are

(1) (2) (3)

• A. $Mn^{2 +}$ $MnO_{2}$ $MnO_{4}^{2 -}$
• B. $MnO_{2}$ $Mn^{2 +}$ $MnO_{4}^{2 -}$
• C. $Mn^{2 +}$ $MnO_{4}^{2 -}$ $MnO_{2}$
• D. $MnO_{4}^{2 -}$ $Mn^{2 +}$ $MnO_{2}$

Answer 1

Answer: (A)

$Mn^{2 +}$ $MnO_{2}$ $MnO_{4}^{2 -}$

Answer 2

At pH <7, in acidic medium

$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow \underset{(Colourless)}{Mn^{2 +}} + 4H_{2}O$

At pH = 7 in neutral medium

$MnO_{4}^{-} + 2H_{2}O + 3e^{-}\overset{\quad\quad}{\rightarrow}\underset{(Brwonprecipitate)}{MnO_{2}} + 4OH^{-}$

At pH > 7, in alkaline medium

$MnO_{4}^{-} + e^{-}\overset{\quad\quad}{\rightarrow}\underset{(green)}{MnO_{4}^{2 -}}$