Question

Consider the reactions:
$2S_2O_3^{2-}(aq) + I_2(s) \rightarrow S_4O_6^{2-}(aq) + 2I ^- (aq)$
$S_2O_3^{2-}(aq) + 2Br_2(l) + 5 H_2O(l) \rightarrow 2SO_4^{2-}(aq) + 4Br ^- (aq) + 10H ^+ (aq)$
Why does the same reductant, thiosulphate react differently with iodine and bromine?

Answer 1

The average oxidation number (O.N.) of $S$ in $S_2O_3^{2-}$is $+2$. Being a stronger oxidising agent than $I_2$, $Br_2$ oxidises $S_2O_3^{2-}$ to $SO_4^{2-}$ , in which the O.N. of $S$ is $+6$.
However, $I_2$ is a weak oxidising agent.
Therefore, it oxidises $S_2O_3^{2-}$ to $S_4O_6^{2-}$, in which the average O.N. of $S$ is only $+2.5$.
As a result, $S_2O_3^{2-}$ reacts differently with iodine and bromine.