Question

In the reaction,
$\begin{matrix}
   {{\text{I}}_{\text{2}}} & \text{+} & \text{2S2O}_{\text{3}}^{\text{2-}} & \to & \text{2}{{\text{I}}^{\text{-}}} & \text{+} & \text{S4O}_{\text{6}}^{\text{2-}} \\
\end{matrix}$
The equivalent weight of iodine will be equal to
A) $\dfrac{4}{6}$ Of molecular weight
B) Molecular weight
C) $\dfrac{2}{9}$ Of molecular weight
D) Twice the molecular weight

Answer 1

Hint: The equivalent weight is obtained by dividing the molecular weight of species by the valence factor. For the redox reaction, the valence factor is the number equal to the total number of electrons gained or lost by the species. Therefore the equivalent weight for the species can be given as:
\[\text{Equivalent weight of atom or molecule =}\dfrac{\text{Molecular weight of atom or molecule}}{\text{No of electron gain or lose during redox reaction}}\]

Complete answer:
The equivalent weight of the compound is defined as the molecular weight of the compound per number of equivalent molecules of it. Equivalent moles are also called the valence factor which depends upon the chemical properties of the compound.
Therefore, the equivalent weight of the compound or species can be given as:
\[\text{Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}\]
In a redox reaction, one species gains the electron and gets reduced while the others lose the electrons and get oxidised. During the reaction, the electrons are transferred from one species to the other. These electrons in the reaction are considered as the valence factor for the redox reaction.
The number of equivalent moles is also called the valence factor.
We are given the following reaction:
$\begin{matrix}
   {{\text{I}}_{\text{2}}} & \text{+} & \text{2S2O}_{\text{3}}^{\text{2-}} & \to & \text{2}{{\text{I}}^{\text{-}}} & \text{+} & \text{S4O}_{\text{6}}^{\text{2-}} \\
\end{matrix}$
This is the redox reaction between iodine and thiosulfate. The iodine undergoes the reduction reaction. However, the thiosulphates undergo the oxidation reaction to form sulfate.
The reduction reaction of the iodine is as shown below:
$\text{ }{{\text{I}}_{\text{2}}}\text{ + 2}{{\text{e}}^{\text{-}}}\text{ }\to \text{ 2}{{\text{I}}^{\text{-}}}$
The iodine $\text{(}{{\text{I}}_{\text{2}}}\text{)}$ gains $\text{2}{{\text{e}}^{\text{-}}}$ and forms the two iodide ions$\text{(}{{\text{I}}^{\text{-}}}\text{)}$. Therefore each iodine gains an electron and the oxidation state changes from $\text{0}\to \text{ (-1)}$ .thus valence factor for iodine is equalled to 1.
 Let consider the molecular weight of iodine as M.Then the equivalent weight of the iodine can be given as,
\[\begin{align}
  & \text{ Equivalent weight of iodine (I) =}\dfrac{\text{Molecular weight of iodine}}{\text{No of electron gain by }{{\text{I}}_{\text{2}}}} \\
 & \text{i}\text{.e}\text{. Equivalent weight of iodine (I) =}\dfrac{\text{M}}{\text{1}} \\
 & \\
 & \Rightarrow \text{ Equivalent weight of iodine = Molecular weight of Iodine} \\
\end{align}\]

Hence, (B) is the correct option.

Note:
The valence factor varies with the substance of interest. For example, for acids, the valence factor is the basicity of acid. For metals, it is the valency of metal, etc.