Thiosulphate reacts differently with iodine and bromine in the reaction given below: $$ 2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^- $$ $$ S_2O_3^{2-} + 5Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 4Br^- + 10H^+ $$ Which of the following statements justifies the above dual behaviour of thiosulphate?
Thiosulphate with Iodine: In the first reaction, iodine \((I_2)\) is reduced to iodide \((I^-)\), while thiosulphate \((S_2O_3^{2-})\) is oxidized to tetrathionate \((S_4O_6^{2-})\). This indicates that iodine acts as an oxidizing agent.
Thiosulphate with Bromine: In the second reaction, bromine \((Br_2)\) is reduced to bromide \((Br^-)\). Here, the thiosulphate \((S_2O_3^{2-})\) is completely oxidized to sulfate \((SO_4^{2-})\). This shows that bromine acts as a stronger oxidizing agent compared to iodine.
Conclusion:
The difference in behavior is due to the relative strengths of iodine and bromine as oxidizing agents. Bromine is a stronger oxidant than iodine, which is why it can oxidize thiosulphate to a greater extent, all the way to sulfate.
Based on this analysis, the correct statement that justifies the dual behavior of thiosulphate is:
Bromine is a stronger oxidant than iodine.
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Approach Solution -2
In the reaction with I$_2$, the oxidation state of sulphur changes from $+2$ to $+2.5$. In the reaction with Br$_2$, the oxidation state of sulphur changes from $+2$ to $+6$. Thus, both I$_2$ and Br$_2$ are oxidants, but Br$_2$ is stronger as it increases the oxidation state of sulphur further. Final Answer: Bromine is a stronger oxidant than iodine.
