Question

Potassium permanganate acts as an oxidant in alkaline and acidic medium. The final products formed from $KMn{O}_4$ in the two conditions are respectively

• A. $Mn{O}^{2-}$ and $M{n}^{3+}$
• B. $M{n}^{3+}$ and $M{n}^{2+}$
• C. $M{n}^{2+}$ and $M{n}^{3+}$
• D. $Mn{O}_2$ and $M{n}^{2+}$

Answer 1

Answer: (D)

In dilute alkaline solution, $Mn{O}_2$ is formed. $\stackrel{+7}{\mathrm{Mn}}{O}_4^{-}+2H_{2}O+3e^{-}\to \stackrel{+4}{\mathrm{Mn}}{O}_2+4O{H}^{-}$
But in very strong base and with an excess of however manganate ion $(Mn{O}_4^{2-})$ is formed.
$Mn{O}_4^{2-}+e^{-}\to Mn{O}_4^{2-}$
In the acidic medium MnS04 (i.e., $M{n}^{2+}$ )is formed. $2\stackrel{+7}{\mathrm{KM}}n{O}_4+3H_{2}S{O}_4\to K_{2}S{O}_4+\stackrel{+2}{2Mn}S{O}_4$
$+3H_{2}O+5(O)$