Question

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

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

Answer 1

Answer: (A), (C)

$I_2\text{gets reduced to}{I}^{-}$
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 :
${\stackrel{0}{I}}_2\to \stackrel{-1}{I^{-}}$
Here $I_2$ is getting reduced to $I^{-}$ since oxidation number of iodine changes from $0$ to $-1$