Consider the reactions:
$2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$
$\left.\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \mathrm{aq}\right)+2 \mathrm{Br}_{2}(\mathrm{l})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+4 \mathrm{Br}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})$
Why does the same reductant, thiosulphate react differently with iodine and bromine?
The average oxidation number (O.N.) of $S$ in $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is $+2$. Being a stronger oxidising agent than $\mathrm{I}_{2}, \mathrm{Br}_{2}$ oxidises $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ to $\mathrm{SO}_{4}^{2-}$, in which the $\mathrm{O} . \mathrm{N}$. of $\mathrm{S}$ is $+6$. However, $\mathrm{I}_{2}$ is a weak oxidising agent.
Therefore, it oxidises $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ to $\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$, in which the average $\mathrm{O} . \mathrm{N} .$ of $\mathrm{S}$ is only $+2.5 . \mathrm{As}$ a result, $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ reacts differently with iodine and bromine.
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