Question

Equivalent weight of Potassium Permanganate in alkaline medium is:
$$\begin{array}{rl}& a)MolecularWeight\\ & b){\displaystyle \frac{MolecularWeight}{3}}\\ & c){\displaystyle \frac{MolecularWeight}{5}}\\ & d){\displaystyle \frac{MolecularWeight}{6}}\end{array}$$

Answer 1

Hint: The equivalent weight of a compound is given by:
$$Equivalentweight={\displaystyle \frac{molecularweight}{n-factor}}$$
With this formula in mind, try to work out what the answer to this question could possibly be. Also, try calculating its n-factor by figuring out the action of Permanganate ion in an alkaline medium.
Complete answer:
Let us first try and calculate the oxidation state of Potassium Permanganate before moving on towards the solution of this question.
The Mn in $$KMn{O}_4$$ exists in +7 state.
Now, we know that in an acidic/alkaline medium, the n-factor is given by the number of electrons gained or lost in the required reaction.
In Acidic medium, This $$M{n}^{+7}$$goes to $$M{n}^{+2}$$ state and hence there is a net gain of 5 electrons.
Now,
$$Equivalentweight={\displaystyle \frac{molecularweight}{n-factor}}$$
Therefore, equivalent weight in acidic medium = 158/5 = 31.6 g.
This reaction is given by:
$$Mn{O_4}^{-}+8H^{+}+5e^{-}\stackrel{H^{+}}{\to }M{n}^{+2}+4H_{2}O$$
Now, let us analyse the transfer of electrons in similar reactions in a neutral medium.
$$M{n}^{+7}$$changes into $$M{n}^{+4}$$ therefore, gain of 3 electrons Therefore, the n-factor in this circumstance is 3.
Therefore, equivalent weight in neutral medium = 158/3 = 52.67g
The reaction for the same is given by:
$$Mn{O_4}^{-}+4H_{2}O+3e^{-}\to Mn{O}_2+4O{H}^{-}$$

Finally, let us analyse the transfer of electrons in similar reactions in a strongly alkaline medium.
$$M{n}^{+7}$$changes into $$M{n}^{+6}$$. Therefore, there is a gain of 1 electron making the n-factor in this circumstance 1
Hence, equivalent weight in strongly alkaline medium = 158/1 = 158g
And the reaction in question is:
$$Mn{O_4}^{-}+1e^{-}\stackrel{O{H}^{-}}{\to }Mn{O_4}^{2-}$$
Thus, we can safely conclude that the answer to this question is a)


Note:
For acid-base reactions, the equivalent weight of an acid or base is the mass which supplies or reacts with one mole of hydrogen cations ($$H^{+}$$). For redox reactions, the equivalent weight of each reactant supplies or reacts with one mole of electrons ($$e^{-}$$) in a redox reaction.