Consider the reaction:
`2S_(2)O_(3)^(2-)(aq)+I_(2)(s) rarr S_(4)O_(6)^(2-)(aq) + 2I^(Θ)(aq)`
`2S_(2)O_(3)^(2-)(aq) + 2Br_(2)(l) + 5H_(2)O(l) rarr 2SO_(4)^(2-)(aq) + 4Br^(Θ)(aq)+10H^(o+)(aq)`
Why does the same reducatnt, thiosulphate, react differently with iodine and bromine?

Explanation

The average oxidation number of `S` in `S_(2)O_(3^(2-)` is `+2` while in `S_(4)O_(6)^(2-)` it is `+2.5`. The oxidation number of `S` in `SO_(4)^(2-)` is `+6`. since `Br_(2)` is a stronger oxidising agent that `I_(2)`. It oxidies `S` to `S_(2)O_(3)^(2-)` to a higher oxidation state of `+6` and hence forms `SO_(4)^(2-)` ion. `I_(2)`, however, being a weaker oxidising agent oxidises `S` of `S_(2)O_(3)^(2-)` ion to a lower oxidation of `+2.5` in `S_(4)O_(6)^(2-)` ion. It is because of this reason thai thiosulphate reacts differently with `Br_(2)` and `I_(2)`.