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The molecular mass of \[KMn{O_4}\] is\[M.\]Its equivalent mass in acidic medium will be.
A. $M$
B. $\dfrac{M}{2}$
C. $\dfrac{M}{5}$

Last updated date: 13th Jun 2024
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Hint:Molar mass is the mass of one mole of a substance. Equivalent weight of a substance (oxidizing and reducing agents) is equal to the molar mass divided by the number of gain or loss by one molecule of the substance in the redox reaction.
Formula used: Equivalent weight of an oxidizing agent $ = \dfrac{{Molar\,mass}}{{number\,of\,electrons\,\,gain\,\,{\text{b}}y\,\,metal}}$.

Complete answer:
Given: Molecular mass of $KMn{O_4}$is \[M.\]
$KMn{O_4}$act as an oxidizing agent in acidic, basic and neutral medium.
To find equivalent mass of $KMn{O_4}$,
In an acidic medium, when potassium permanganate reacts with acid then it will give manganous sulphate and nascent oxygen. Then the reaction is
\[2KMn{O_4} + 3{H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5[O]\]
\[2K\mathop M\limits^{ + 7} n{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2\mathop {Mn}\limits^{ + 2} S{O_4} + 3{H_2}O + 5[O]\]
In ionic form: $Mn{O_4}^{ - 1} + 8{H^ + } + 5{e^ - }\xrightarrow{{}}M{n^{2 + }} + 4{H_2}O$ (oxidation state of $Mn$in $KMn{O_4}$ is $ + 7$)
Oxidation number assigned to an element in a chemical combination which represents the number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound.
Oxidation state is a positive or negative number that represents the effective charge of an atom or element and that indicates the extent or possibility of its oxidation.
Therefore, equivalent weight of $KMn{O_4}$ in acidic medium \[ = \dfrac{{{\text{Molar mass of KMn}}{{\text{O}}_4}}}{{{\text{number of electrons gain by metal}}}}{\text{ }}\]
Then, equivalent weight $ = \dfrac{M}{5}$

Hence, the correct answer is option (C).

Note: In acidic medium, oxidation state of \[Mn\] is $ + 7$ to $ + 2$ state, hence there is a net gain of five electron then equivalent weight of $KMn{O_4}$ is $\dfrac{M}{5}$. In basic medium, oxidation state of\[Mn\]is $ + 7$ to $ + 6$ state that gain of one electron then equivalent weight of \[KMn{O_4}\] is $\dfrac{M}{1}$ and in neutral medium, oxidation state of \[Mn\] is $ + 7$ to $ + 4$ that gain three electron then the equivalent weight of \[KMn{O_4}\] is $\dfrac{M}{3}$. Oxidation state of a pure element is always zero. The oxidation state for a pure ion is equivalent to its ionic charge. In general, hydrogen has an oxidation state of \[ + 1,\] while oxygen has an oxidation state of \[ - 2.\] The sum of the oxidation states for all atoms of a neutral molecule must add up to zero.