Question

The equivalent mass of potassium permanganate in alkaline medium is:
a.) $\dfrac{{Molar{\text{ mass}}}}{2}$
b.) $\dfrac{{Molar{\text{ mass}}}}{3}$
c.) $\dfrac{{Molar{\text{ mass}}}}{4}$
d.) Molar mass itself

Answer 1

Hint: The equivalent mass of a substance is defined as the Molar mass of substance divided by n-factor. The n-factor is the number of electrons gained by the substance.
Thus, Equivalent Mass= $\dfrac{{Molar{\text{ mass}}}}{{Number{\text{ of electrons gained}}}}$
Alkaline medium means the basic medium.

Complete step by step answer:
We know the equivalent mass of a substance can be obtained using the formula as-
Equivalent mass= $\dfrac{{Molar{\text{ mass}}}}{{n - factor}}$

The term here n-factor is different for acids and bases.
In case of acids, n-factor is the number of replaceable ${H^ + }$ions.
And in case of bases, it is the number of replaceable $O{H^ - }$ions.
In general, we can also be taken as the number of electrons gained.
Now let's first write the reaction of potassium permanganate in an alkaline medium. The reaction can be written as-
$2KMn{O_4} + {H_2}O\xrightarrow{{alkaline}}2Mn{O_2} + 2KOH + 3\left[ O \right]$

It is that two molecules of potassium permanganate react with a molecule of water in an alkaline medium giving two molecules of Manganese (IV) Oxide and two molecules of Potassium hydroxide with evolution of Oxygen gas.
Now, to find the equivalent mass of potassium permanganate in an alkaline medium; we need to know the number of electrons gained by Mn ions. For that lets first find out the oxidation number of Mn in $KMn{O_4}$.

Thus, the oxidation state of Mn in $KMn{O_4}$ is as-
$KMn{O_4} = \left\{ {1 + x + \left( { - 2 \times 4} \right)} \right\} = 0$
$\left\{ {1 + x - 8} \right\} = 0$
$\begin{gathered}
  \left\{ {x - 7} \right\} = 0 \\
  x = 7 \\
\end{gathered} $
Mn is in +7 oxidation state in $KMn{O_4}$.

Now let’s find the oxidation state of Mn in $Mn{O_2}$.
$\begin{gathered}
 Mn{O_2} = \left\{ {x + \left( { - 2 \times 2} \right)} \right\} = 0 \\
  \left\{ {x - 4} \right\} = 0 \\
  x = 4 \\
\end{gathered} $
Mn is in +4 oxidation state in $Mn{O_2}$.

So, we see here a change of +7 oxidation state to +4.
Thus, we have a gain of 3 electrons.

Now, as we have found the number of gained electrons; we can easily find out equivalent mass as-
Equivalent mass= $\dfrac{{{\text{Molar mass}}}}{{Number{\text{ of gained electrons}}}}$
Equivalent mass= $\dfrac{{Molar{\text{ mass}}}}{3}$
So, the correct answer is “Option B”.

Note: Here we were given in question the equivalent mass of potassium permanganate in alkaline medium which means neutral conditions. So, we will have number of electrons gained as 3 but in case we are give highly alkaline medium then the reaction takes place as-
$KMn{O_4} + {H_2}O\xrightarrow{{Highly{\text{ alkaline}}}}{K_3}Mn{O_4} + {H_2}O + \left[ O \right]$
In this reaction, there is a change of oxidation state from +7 to +6 i.e. one electron is gained.
Thus, molar mass itself will be the answer if the condition is given a highly alkaline medium otherwise, we can take neutral/alkaline medium as we did above.