Question

Consider the redox reaction: $2S_2{O}_3^{2-}+I_2\to S_4{O}_6^{2-}+2I^{-}$

• A. $S_4{O}_6^{2-}\text{gets oxidised to}{S}_2{O}_3^{2-}$
• B. $S_2{O}_3^{2-}\text{gets reduced to}{S}_4{O}_6^{2-}$
• C. $I_2\text{gets oxidised to}{I}^{-}$
• D. $I_2\text{gets reduced to}{I}^{-}$

Answer 1

The correct option is D $I_2\text{gets reduced to}{I}^{-}$

The given reaction is:
$2S_2{O}_3^{2-}+I_2\to S_4{O}_6^{2-}+2I^{-}$

Oxidation half-reaction: $\stackrel{+2}{S_2{O}_3^{2-}}\to \stackrel{+2.5}{S_4{O}_6^{2-}}$
Here, $S_2{O}_3^{2-}$ is getting oxidised to $S_4{O}_6^{2-}$ as oxidation number of S is increasing from +2 to +2.5

Reduction half-reaction: ${\stackrel{0}{I}}_2\to \stackrel{-1}{2}{I}^{-}$
here $I_2$ is getting reduced to $2I^{-}$ as oxidation number of S is decreasing from 0 to -1

So, (d) is the correct answer.