Question

In an oxidation-reduction reaction, \[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\] the ion is converted to $\text{M}{{\text{n}}^{\text{2+}}}$. What is the number of equivalents of $\text{KMn}{{\text{O}}_{\text{4}}}$(\[\text{mol}\text{.wt}\text{. =158}\] ) present in \[\text{250 mL}\] of \[\text{ 0}\text{.04 N }\] $\text{KMn}{{\text{O}}_{\text{4}}}$?
(A) \[~~0.02\]
(B) \[~~0.05\]
(C) \[~~0.04\]
(D) \[~~0.07\]

Answer 1

Hint: The normality (N) is a measure of concentration. It is the ratio of equivalent weight per unit volume in$\text{ d}{{\text{m}}^{\text{3}}}$. Here, the number of moles is equal to the product of normality and volume. Then calculate the number of equivalents weigh from the following relation,
$\text{ No}\text{. of equivalent weight = valence factor }\!\!\times\!\!\text{ number of moles }$
For the redox reaction, the valence factor is the number equal to the total number of electrons gained or lost by the species. The $\text{Mn}$ in $\text{MnO}_{\text{4}}^{-}$ gains $5{{e}^{-}}$ and the oxidation state changes from$+7\to +2$.

Complete step by step solution:
The equivalent weight of the compound is defined as the molecular weight of the compound per number of equivalent molecules of it. For the redox reaction, the valence factor is equal to the number of electron gain or loss by the species.
The equivalent weight is related to the normality (N) of the solution. The normality is given as,
\[\text{Normality (N) = }\dfrac{\text{Equivalent weight }}{\text{Volume of solution in liter}}\text{ = }\dfrac{\text{Eq}\text{.}}{\text{V}}\]
Where w is the mass of the compound, E is the equivalent weight of the compound, M is the molar mass and n is the valence factor.
We have given the following information.
The molecular weight of $\text{ KMn}{{\text{O}}_{\text{4}}}$ is$158\text{ g mo}{{\text{l}}^{-1}}$.
The normality of $\text{ KMn}{{\text{O}}_{\text{4}}}$ the solution is $\text{0}\text{.04 N}$ and the volume is $250\text{ mL}$. We have to find the equivalent weight of$\text{ KMn}{{\text{O}}_{\text{4}}}$.
The $\text{ KMn}{{\text{O}}_{\text{4}}}$ or the $\text{MnO}_{\text{4}}^{-}$ is converted to the $\text{ M}{{\text{n}}^{\text{2+}}}$. In acidic medium the $\text{MnO}_{\text{4}}^{-}$ gains $5{{e}^{-}}$ and reduced to$\text{ M}{{\text{n}}^{\text{2+}}}$. The reaction is as follows,
$\text{MnO}{}_{\text{4}}^{-}\text{+8}{{\text{H}}^{\text{+}}}\text{+5}{{\text{e}}^{\text{-}}}\to \text{M}{{\text{n}}^{\text{2+}}}\text{+4}{{\text{H}}_{\text{2}}}\text{O}$
Therefore, the valence factor, $\text{n = 5}$.
Let us find out the number of moles of $\text{ KMn}{{\text{O}}_{\text{4}}}$ the solution.
\[\begin{align}
& \text{The number of moles in }\!\!~\!\!\text{ solution = N }\!\!\times\!\!\text{ V } \\
& \text{ = 0}\text{.04 N }\!\!\times\!\!\text{ }\dfrac{\text{250}}{\text{1000}}\text{ } \\
& \text{ = 0}\text{.01 mol} \\
& \therefore \text{Number of moles (n) = 0}\text{.01 mol} \\
\end{align}\]
The number of moles can be related to the no. of equivalent. The number of equivalent weight is,
$\begin{align}
& \text{No}\text{.of equivalent of KMn}{{\text{O}}_{\text{4}}}\text{ = valence factor }\!\!\times\!\!\text{ moles of KMn}{{\text{O}}_{\text{4}}} \\
& \text{ = 5 }\times \text{ 0}\text{.01 } \\
& \text{ = 0}\text{.05 } \\
& \text{ } \\
\end{align}$
Therefore, the number of the equivalent of $\text{ KMn}{{\text{O}}_{\text{4}}}$ is $\text{ 0}\text{.05 }$.

Hence, (A) is the correct option.

Note: In this reaction, the $\text{KMn}{{\text{O}}_{\text{4}}}$ acts as an oxidizing agent. Because here the $\text{KMn}{{\text{O}}_{\text{4}}}$ has got reduced to +2 oxidation state from the +7 oxidation state. Remember that, the number of equivalent weights in terms of moles. Similarly, the molarity is also explained in terms of normality.
$\begin{align}
& \text{N = z }\!\!\times\!\!\text{ M} \\
& \text{eq = z }\times \text{ n} \\
\end{align}$
Where the z is valence factor, M is molarity, and eq. is no. of equivalent .