Consider the reactions :
`2 S_(2)O_(3)^(2-) (aq) I_(2) (s) to S_(4) O_(6)^(2-) (aq) + 2I^(-)(aq)`
`S_(2)O_(3)^(2-) (aq) + 2 Br_(2) (1) + 5 H_(2)O(1) to 2 SO_(4)^(2-) (aq) + 4 Br^(-) (aq) + 10H^(+) (aq)`
Why does the same reductant , 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)`.