Question

Equivalent weight of Potassium Permanganate in alkaline medium is:

& a)~Molecular\text{ }Weight \\\ & b)\dfrac{Molecular\text{ }Weight}{3} \\\ & c)\dfrac{Molecular\text{ }Weight}{5} \\\ & d)\dfrac{Molecular\text{ }Weight}{6} \\\ \end{aligned}$$

Answer 1

Hint: The equivalent weight of a compound is given by:
$Equivalent\text{ }weight\text{ }=\text{ }\dfrac{molecular\text{ }weight}{n-factor}$
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,
$Equivalent\text{ }weight\text{ }=\text{ }\dfrac{molecular\text{ }weight}{n-factor}$
Therefore, equivalent weight in acidic medium = 158/5 = 31.6 g.
This reaction is given by:
$Mn{{O}_{4}}^{-}+\text{ }8{{H}^{+}}+5{{e}^{-}}\text{ }\xrightarrow{{{H}^{+}}}\text{ }M{{n}^{+2}}\text{ }+\text{ }4{{H}_{2}}O$
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:
$Mn{{O}_{4}}^{-}\text{ }+\text{ }4{{H}_{2}}O\text{ }+3{{e}^{-}}\text{ }\to \text{ }Mn{{O}_{2}}\text{ }+\text{ }4O{{H}^{-}}$

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:
$Mn{{O}_{4}}^{-}\text{ }+\text{ }1{{e}^{-}}\text{ }\xrightarrow{O{{H}^{-}}}\text{ }Mn{{O}_{4}}^{2-}$
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.