Question

Thiosulphate reacts differently with iodine and bromine in the reaction given below.
$2S_2{O}_3^{2-}+I_2\to S_4{O}_6^{2-}+2I^{-}$
$S_2{O}_3^{2-}+2B{r}_2+5H_{2}O\to 2S{O}_4^{2-}+2B{r}^{-}+10H^{+}$
Which of the following statement justifies the above dual behaviour of thiosulphate?

• A. Bromine is stronger oxidant than iodine
• B. Bromine is weaker oxidant than iodine
• C. Thiosulphate undergoes oxidation by bromine and reduction by iodine in above reactions
• D. Bromine undergoes oxidation and iodine undergoes reduction in above reactions

Answer 1

The correct option is A Bromine is stronger oxidant than iodine

Thiosulphate undergoes oxidation by bromine and iodine. For the reaction iodine and bromine undergo reduction by accepting electrons.

In the thiosulphate reaction with $I_2$ the oxidation state of $S$ in $S_2{O}_3^{2-}$ is $+2$ while the ON of $S$ in $S_4{O}_6^{2-}$ is $+2.5$.

In the thiosulphate reaction with $B{r}_2$ the oxidation state of $S$ in $S_2{O}_3^{2-}$ is $+2$ changing to $+6$ in $S{O}_4^{2-}$.

$B{r}_2$ is a stronger oxidizing agent than $I_2$, therefore it oxidizes sulphur from lower oxidation state to higher oxidation state. Iodine is a weaker oxidizing agent and is not able to oxidize Sulphur to higher oxidation

state. That’s why $S_2{O}_3^{2-}$ on reacting with $B{r}_2$ forms Sulphate , while it forms $S_4{O}_6^{2-}$ when reacts with $I_2$.