One of the reasons for the difference in the oxidation states of sulphur in both reactions is due to difference in reactivity of Bromine and Iodine, as Iodine is a bigger molecule thus it has more tendency to share electrons than to donate as compared to bromine. Whereas bromine with higher ionic character tends to donate electron rather than sharing. So the more readily an atom donates electron the more easily it gives out energy in the form of electric potential due to ions formed.

Another reason is, as Br (Br₂/Br^- = 1.09) has higher reduction potential as compared to I₂ (I₂/I^- = 0.54) it more readily undergoes reduction itself which in turn increases the driving and thus acting as a stronger oxidizing agent than I₂.