Question

$KMn{O_4}$ $\left( {mol.wt. = 158} \right)$ oxidizes oxalic acid in acidic medium to $C{O_2}$ and water as follows:
$5{C_2}O_4^{2 - } + 2MnO_4^ - + 16{H^ + }\xrightarrow{{}}10C{O_2} + 2M{n^{2 + }} + 8{H_2}O$
What is the equivalent weight of$KMn{O_4}$?
A: $158$
B: $31.6$
C: $39.5$
D: $79$

Answer 1

Hint: Equivalent weight of a substance is defined as molecular weight of substance divided by its valency or change in oxidation number. Valency of an atom is charged on that atom the compound is ionic.
Formula used: equivalent weight$ = \dfrac{M}{{O.S.}}$ where, $M$ is molecular weight and $O.S.$ is change in oxidation state.

Complete step by step solution:
Oxalic acid is also called carboxylic acid. Molecular formula of oxalic acid is ${C_2}{H_2}{O_4}$ .
$KMn{O_4}$ is potassium permanganate. It was first discovered as a disinfectant. It is widely used to treat skin problems. It has different oxidation states in different mediums.
Acidic conditions represent a solution which contains excess of ${H^ + }$ ion concentration, hence making the solution acidic.
To find an equivalent weight of $KMn{O_4}$ we have to find its change in oxidation state. Reaction of $KMn{O_4}$ is as follows:
$MnO_4^ - \xrightarrow[{}]{}M{n^{2 + }}$
Before reaction, charge on whole compound$ = - 1$
We know, oxidation state of oxygen atom$ = - 2$
Let oxidation state of manganese be $x$
Then, $ - 2 \times 4 + x = - 1$
                      $x = + 7$
Therefore oxidation state of manganese before reaction is $ + 7$
Oxidation state of manganese after reaction is $ + 2$
Change in oxidation state is $5$
Here we are considering the oxidation state of manganese only because the oxidation state of oxygen is not changing. Change in oxidation state of compounds is only due to manganese.
Hence equivalent weight of $KMn{O_4}$ $ = \dfrac{M}{{O.S.}}$
Molecular weight is given that is $158$
Change in oxidation state is $5$ (calculated above)
So equivalent weight$ = \dfrac{{158}}{5} = 31.6$
Answer to this question is option B that is$31.6$.

Note: $KMn{O_4}$ shows different oxidation states in different mediums. Oxidation state of $KMn{O_4}$ is reduced to $ + 2$ in acidic medium. In a neutral medium its oxidation state is reduced to $ + 4$ . In basic medium $KMn{O_4}$ is reduced to$Mn{O_2}$ .