Question

In the reaction, $2{\text{N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} + {{\text{I}}_2} \to {\text{N}}{{\text{a}}_2}{{\text{S}}_4}{{\text{O}}_6} + 2{\text{NaI}}$, ${{\text{I}}_2}$ acts as:
(A) Reducing agent
(B) Oxidising agent
(C) Oxidising as well as reducing agent
(D) None of the above

Answer 1

To solve this we must calculate the oxidation state of iodine in and in . If the oxidation state of iodine increases then iodine is getting oxidised and if the oxidation state decreases it is getting reduced. We will have to identify whether it accepts or releases electrons.

Complete step by step solution:
We are given a reaction,
$2{\text{N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} + {{\text{I}}_2} \to {\text{N}}{{\text{a}}_2}{{\text{S}}_4}{{\text{O}}_6} + 2{\text{NaI}}$
In the reaction, on the reactant side we have ${{\text{I}}_2}$. So, first we will calculate the oxidation state of iodine in ${{\text{I}}_2}$ as follows:
We know that the oxidation state of an element in its free or uncombined state is always zero.
Iodine forms a diatomic molecule to achieve stable outer shell noble gas electronic configuration. Thus, the molecule of iodine is ${{\text{I}}_2}$. Thus, the oxidation state of iodine in ${{\text{I}}_2}$ is zero.
In the reaction, on the product side we have ${\text{NaI}}$. So, now we will calculate the oxidation state of iodine in ${\text{NaI}}$ as follows:
We know that the oxidation state of sodium is +1. The total charge on the compound ${\text{NaI}}$ is zero. Thus,
Oxidation state of iodine $= {\text{Total charge on compound}} - {\text{Oxidation state of Na}}$
Oxidation state of iodine $= 0 - 1$
Oxidation state of iodine $= - 1$
The oxidation state of iodine in the given reaction changes from 0 to $- 1$. Thus, the oxidation state of iodine decreases.
As the oxidation state of iodine decreases, we can say that ${{\text{I}}_2}$ is getting reduced.
We know that the chemical species in the chemical reaction which itself gets reduced and oxidised the other species is known as an oxidising agent.
Thus, in the reaction, $2{\text{N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} + {{\text{I}}_2} \to {\text{N}}{{\text{a}}_2}{{\text{S}}_4}{{\text{O}}_6} + 2{\text{NaI}}$, ${{\text{I}}_2}$ acts as oxidising agent.

Thus, the correct option is (B) oxidising agent.

Note: We must keep in mind that if the oxidation state increases, the species gets oxidised. As the species gets oxidised we can say that it is a reducing agent. We can also identify the oxidizing and reducing agent by identifying whether it has released or accepted the electrons.