Question

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

Answer 1

The average oxidation number (O.N.) of \(S\) in \(S_2O_3^{2-}\)is \(+2\). Being a stronger oxidising agent than \(I_2\), \(Br_2\) oxidises \(S_2O_3^{2-}\) to \(SO_4^{2-}\) , in which the O.N. of \(S\) is \(+6\). 
However, \(I_2\) is a weak oxidising agent. 
Therefore, it oxidises \(S_2O_3^{2-}\) to \(S_4O_6^{2-}\), in which the average O.N. of \(S\) is only \(+2.5\). 
As a result, \(S_2O_3^{2-}\) reacts differently with iodine and bromine.