Question

A solution contains $F{{e}^{2+}}$, $F{{e}^{3+}}$ and ${{I}^{-}}$ ions. This solution was treated with Iodine at ${{35}^{o}}C$. ${{E}^{o}}$ for $F{{e}^{3+}}/F{{e}^{2+}}$ is $+0.77V$ and ${{E}^{o}}$ for ${{I}_{2}}/2{{I}^{-}}=0.536V$. The favourable redox reaction is:
A. ${{I}_{2}}$ will be reduced to ${{I}^{-}}$.
B. There will be no redox reaction.
C. ${{I}^{-}}$ will be oxidised to ${{I}_{2}}$.
D. $F{{e}^{2+}}$ will be oxidised to $F{{e}^{3+}}$.

Answer 1

As per the given question, it is already mentioned that the reaction is a redox reaction and the values for the reduction potential are also given. Remember, greater is the value of reduction potential, greater will be the tendency to get reduced.

Complete step by step answer:
Given that,
A solution contains $F{{e}^{2+}}$, $F{{e}^{3+}}$ and ${{I}^{-}}$ ions. Later, this solution was treated with Iodine at ${{35}^{o}}C$.
The reduction potential for $F{{e}^{3+}}/F{{e}^{2+}}$ is equal to $+0.77V$.
The reduction potential for ${{I}_{2}}/2{{I}^{-}}$ is equal to $0.536V$.
Now coming to reduction potential, it is also called as redox potential, or oxidation/reduction potential and it is denoted by the symbol ${{E}_{h}}$ or ${{E}^{o}}$. It measures the tendency of a chemical species to acquire electrons and thereby be reduced. Generally, it is considered that, the more positive the reduction potential (or higher the value), the greater will the species affinity be reduced.
So, here we can see that the reduction potential for $F{{e}^{3+}}/F{{e}^{2+}}$ is equals to $+0.77V$, while the potential for ${{I}_{2}}/2{{I}^{-}}$ is equals to $0.536V$. So, the reduction potential for iron is greater than that of iodine. So, iron will get reduced when they are reacted in a solution. The balanced chemical reaction with the charge changes indicating oxidation and reduction is shown below;
$2F{{e}^{+3}}+2{{I}^{-}}\to 2F{{e}^{+2}}+{{I}_{2}}$
So, $F{{e}^{3+}}$ is being reduced to $F{{e}^{2+}}$, as it has more reduction potential value than that of ${{I}_{2}}/{{I}^{-}}$, whereas ${{I}^{-}}$ is getting oxidised to ${{I}_{2}}$.
So, the correct answer is “Option C”.

Note: Remember, the reduction potential measures the tendency of a species to acquire or gain electrons and thereby undergoing reduction in its charge. The greater the value of the reduction potential, higher will be its tendency to get reduced.