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

(N/A) In $S_2O_3^{2-}$,the average oxidation number of $S$ is $+2$. In $S_4O_6^{2-}$,the average oxidation number of $S$ is $+2.5$. In $SO_4^{2-}$,the oxidation number of $S$ is $+6$.
$Br_2$ is a stronger oxidizing agent than $I_2$. Therefore,$Br_2$ is capable of oxidizing $S_2O_3^{2-}$ (where $S$ is $+2$) to $SO_4^{2-}$ (where $S$ is $+6$).
$I_2$ is a weaker oxidizing agent. Therefore,it only oxidizes $S_2O_3^{2-}$ $(S=+2)$ to $S_4O_6^{2-}$ $(S=+2.5)$.
Thus,the difference in the oxidizing strength of the halogens leads to different products.