Question

Consider the reactions:
$2S_{2}O_{3}^{2-}(aq)+I_{2}(s)\longrightarrow S_{4}O_{6}^{2-}(aq)+2I^{-}(aq)$
$S_{2}O_{3}^{2-}(aq)+2Br_{2}(l)+5H_{2}O(l)\longrightarrow 2SO_{4}^{2-}(aq)+4Br^{-}(aq)+10H^{+}(aq)$
Why does the same reductant, thiosulphate react differently with iodine and bromine?

Answer 1

In both equations, ${ S }_{ 2 }{ O }_{ 3 }^{ 2- }$ ions are reducing agents while ${ I }_{ 2 }$ and ${ Br }_{ 2 }$ are oxidizing\quad agents.

${ Br }_{ 2 }$ oxidizes more than ${ I }_{ 2 }$. A stronger oxidizing agent is capable of bringing about the higher oxidation state.

$\begin{matrix} { S }_{ 2 }{ O }_{ 3 }^{ 2- } \\ +2 \end{matrix}\quad \xrightarrow { { I }_{ 2 } } \quad \begin{matrix} { S }_{ 4 }{ O }_{ 6 }^{ 2- } \\ 2.5 \end{matrix}$

$\begin{matrix} -2{ S }_{ 2 }{ O }_{ 3 }^{ 2- } \\ +2 \end{matrix}\quad \xrightarrow { Br_{ 2 } } \quad \begin{matrix} { S }{ O }_{ 4 }^{ 2- } \\ +6 \end{matrix}$

Therefore, ${ Br }_{ 2 }$ and ${ I }_{ 2 }$ react with ${ S }_{ 2 }{ O }_{ 3 }^{ 2- }$ differently.