Question

For titrating certain volume of a reducing substance by $1M$ $KMn{O}_4$, it is found that $20$ mL was used in acidic medium, $33.4$ mL was used in alkaline medium and $100$ mL was used in neutral medium. ${Mn{O}_4}^{-}$ is reduced to $M{n}^{2+}$ in acidic Medium.

To what oxidation state is ${Mn{O}_4}^{-}$ reduced to in alkaline medium?

• A. $+4$
• B. $+1$
• C. $+5$
• D. $+6$

Answer 1

The correct option is D $+6$

Equation for reduction of KMnO4 in acidic medium:

$8H^{+}+Mn{O}_4^{2-}+5e^{-}\to M{n}^{2+}+4H_{2}O$

In neutral medium:

$2H_{2}O+Mn{O}_4^{-}+3e^{-}\to MnO2+4O{H}^{-}$

In basic medium:

$MnO{4}^{-}+e^{-}\to Mn{O}_4^{2-}$

Thus, in an alkaline medium, $Mn{O}_4-$ is reduced to $Mn{O}_4^{2-}$. The oxidation number changes from +7 to +6.

Hence, the correct option is $D$