JEE Main 2024 — Redox Reactions Question with Solution

In the given reactions:

$$\begin{array}{rl}& 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+{\mathrm{I}}_2\to {\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{I}}^{-}\\ & \\ & {\mathrm{S}}_2{\mathrm{O}}_3^{2-}+5{\mathrm{Br}}_2+5{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{SO}}_4^{2-}+4{\mathrm{Br}}^{-}+10{\mathrm{H}}^{+}\end{array}$$

In the first reaction with iodine ($I_2$), thiosulfate ($S_2{O}_3^{2-}$) acts as a reducing agent, converting $I_2$ to $I^{-}$, while it is itself oxidized to $S_4{O}_6^{2-}$. This indicates that iodine is acting as an oxidant (oxidizing agent), getting reduced in the process.

In the second reaction with bromine ($B{r}_2$), thiosulfate is oxidized more extensively, being converted all the way to sulfate ($S{O}_4^{2-}$), with hydrogen ions ($H^{+}$) and bromide ions ($B{r}^{-}$) also produced. This shows that bromine acts as a stronger oxidant compared to iodine, as it promotes a complete oxidation of $S_2{O}_3^{2-}$ to $S{O}_4^{2-}$.

Given these observations, the correct option that justifies the dual behavior of thiosulfate reacting with iodine and bromine differently is:

Option C: Bromine is a stronger oxidant than iodine

This statement correctly explains why thiosulfate is oxidized to $S_4{O}_6^{2-}$ in the presence of iodine, and to $S{O}_4^{2-}$ in the presence of bromine. Bromine's stronger oxidizing ability leads to a more complete oxidation of thiosulfate compared to the oxidation by iodine.