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 reductant, thiosulphate, react differently with iodine and bromine?

Explanation

`I_(2)` Oxidises thiosulphate ion to tetrathionate ion i.e, from O.S of +2 for `S (" in " S_(2)O_(3)^(2-) "to" S.O of + 5//2 "for" S ("in" S_(4)O_(6)^(2-) "ion")`
`Br_(2)` oxidises thiosulphate ion to sulphate ion i.e from O.S of + 2 for `S("in "S_(2)O_(3)^(2-))` to O.S of + 6 for `S ("in "SO_(4)^(2-)ion)`
This is because `Br_(2)` is a stronger O.A than `I_(2)(E^(@)_(Br_(2)//2Br-)=1.09V,E^(@)_(I_(2//2I^(-))=0.54V)`