Question

In an alkaline medium, $KMn{O}_4$ reacts as follows (At wt. of $K=39.09$, $Mn=54.94$, $O=16.0$):

$2KMn{O}_4+2KOH\to 2K_2{MnO}_4+H_{2}O+\left[O\right]$

i) The equivalent weight of $KMn{O}_4$ in alkaline medium is 158.
ii) The number of electrons gained by one molecule of $KMn{O}_4$ in the alkaline medium is 1.
iii) In the alkaline medium, $KMn{O}_4$ acts as a reducing agent.

Select the correct option.

• A. Only i is correct
• B. i and ii are correct
• C. Onle iii is correct
• D. Only ii is correct

Answer 1

The correct option is B i and ii are correct

$KMn{O}_4\to M{n}^{+7}$

$K_{2}Mn{O}_4\to M{n}^{+6}$

So, one $e^{-}$ is gained by a single molecule of $KMn{O}_4$ in the alkaline medium and thus, the valency factor is 1.

So, $equivalentweight=molecularweight$

$=39\times 1+54\times 94+4\times 16$

$=158$

So, i and ii are correct statement.

Oxidation number is reduced and therefore, potassium has reduced itself which makes it an oxidizing agent.