Question

When $KMn{O_4}$is titrated against ferrous ammonium sulphate in acidic medium then the equivalent mass of $KMn{O_4}$will be:
A. $\dfrac{{Molecular{\text{ mass}}}}{{10}}$
B. $\dfrac{{Molecular{\text{ mass}}}}{5}$
C. $\dfrac{{Molecular{\text{ mass}}}}{3}$
D. $\dfrac{{Molecular{\text{ mass}}}}{2}$

Answer 1

Equivalent mass of a substance (oxidant or reductant) is equal to the molecular mass divided by number of electrons lost or gained by one molecule of the substance in redox reaction.

Complete step by step answer:
Potassium permanganate is a strong oxidizing agent in acidic medium. Ferrous ammonium sulphate is a double salt known as Mohr’s salt.
In this titration, potassium permanganate acts as an oxidizing agent and Mohr’s salt acts as a reducing agent. So, the reaction between potassium permanganate and Mohr’s salt is a redox reaction. In this redox reaction, ferrous ions from Mohr’s salt get oxidised. Manganese present in potassium permanganate( pink coloured ) which is in$+ 7$oxidation state gets reduced to colourless $M{n^{ + 2}}$state.
The chemical reaction and the molecular chemical equation is given below:
Reduction half reaction: Potassium permanganate ($KMn{O_4}$) reacts with sulphuric acid and produces nascent oxygen.
$2KMn{O_4} + 3{H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5\left[ O \right]$
In ionic form the reaction can be written as,
${\text{MnO}}_{\text{4}}^{{\text{2 - }}}$+${\text{8}}{{\text{H}}^{\text{ + }}} + {\text{5}}{{\text{e}}^{\text{ - }}}$ $\to$ ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$+ ${\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$
Oxidation half reaction is written as,
$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$
The overall reaction in ionic form can be written as,
$Mn{O_4}^ - + 8{H^ + } + 5F{e^{ + 2}}\xrightarrow{{}}M{n^{ + 2}} + 5F{e^{ + 3}} + 4{H_2}O$
The reaction in normal form with all the compounds oxidised and reduced is written as follows,
$2KMN{O_4} + 10FeS{O_4}{\left( {N{H_4}} \right)_2}S{O_4} \cdot 6{H_2}O \to {K_2}S{O_4} + 2MnS{O_4} + 5F{e_2}{\left( {S{O_4}} \right)_3} + 10{\left( {N{H_4}} \right)_2}S{O_4} + 68{H_2}O$
The above reaction is studied in volumetric analysis called Titration.
The n – factor of $KMn{O_4}$is $5$because it gains $5$ electrons in the reaction.
By using the formula,
Equivalent weight of oxidizing agent $= \dfrac{{Molecular{\text{ mass}}}}{{n - factor}}$
$\therefore$Equivalent weight of $KMn{O_4}\left( {O.A} \right) = \dfrac{{Molecular{\text{ mass}}}}{5}$

Hence, the correct option is (B).

Additional information: This titration is based upon oxidation – reduction titrations. When potassium permanganate solution is titrated against Ferrous ammonium sulphate in the presence of acidic medium then $KMn{O_4}$is oxidised $F{e^{ + 2}}$to $F{e^{ + 3}}$and itself reduces $M{n^{ + 7}}$to $M{n^{ + 2}}$

Note:
Potassium permanganate $\left( {KMn{O_4}} \right)$acts as an oxidizing agent in acidic, alkaline and neutral medium. In the above reaction, acidic medium is necessary in order to prevent precipitation of manganese oxide. $KMn{O_4}$acts as a self-indicator and the above titration is called permanganate titration.