In the reactions given below, thiosulphate reacts differently with iodine than with bromine.

$2{\mathrm{S}}_{2}{\mathrm{O}}_{3}^{2-}+{\mathrm{I}}_{2}\to {\mathrm{S}}_{4}{\mathrm{O}}_{6}^{2-}+2{\mathrm{I}}^{-}$

${\mathrm{S}}_{2}{\mathrm{O}}_{3}^{2-}+2{\mathrm{Br}}_{2}+5{\mathrm{H}}_{2}\mathrm{O}\to 2{\mathrm{SO}}_{4}^{2-}+2{\mathrm{Br}}^{-}+10{\mathrm{H}}^{+}$

Which of the following statements best describes the above dual behaviour of thiosulphate?

1. Bromine is a stronger oxidant than iodine

2. Bromine is a weaker oxidant than iodine

3. Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions

4. Bromine undergoes oxidation and iodine undergoes reduction in these reactions

Hint: Gain of electron is reduction and loss of electron is oxidation
The two reactions are as follows:
$2{\stackrel{+2}{\mathrm{S}}}_{2}\stackrel{-2}{{\mathrm{O}}_{3}^{2-}}\left(\mathrm{aq}\right)+{\stackrel{0}{\mathrm{I}}}_{2}\left(\mathrm{s}\right)\to {\stackrel{2.5}{\mathrm{S}}}_{4}{\mathrm{O}}_{6}^{2-}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)$
$\stackrel{+2}{{\mathrm{S}}_{2}}\stackrel{-2}{{\mathrm{O}}_{3}^{2-}}+2\stackrel{0}{\mathrm{B}}{\mathrm{r}}_{2}\left(\mathrm{l}\right)+5{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\to 2\stackrel{+6}{\mathrm{S}}\stackrel{2-}{{\mathrm{O}}_{4}^{2-}}\left(\mathrm{aq}\right)+2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)+10{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$
 Element Change in oxidation state Reaction S +2 ---> 2. 5 oxidation I 0----> -1 reduction S +2-----> +6 Oxidation Br 0 ----> -1 reduction

In both the reactions, halogen acts as an oxidising agent and oxidised ${\mathrm{S}}_{2}{\mathrm{O}}_{3}^{2-}$ and it self gets reduce. The change in oxidation state of sulphur in case of ${\mathrm{Br}}_{2}$ is more than in case of ${\mathrm{I}}_{2}$. Hence, ${\mathrm{Br}}_{2}$ being stronger oxidisng agent than I2