question_answer

Consider the redox reaction : \[2S{{O}_{2}}O_{3}^{2-}+{{I}_{2}}\to {{S}_{4}}O_{6}^{2-}+2{{I}^{-}}\]         

A) \[2S{{O}_{2}}O_{3}^{2-}\] gets oxidised to \[S{{O}_{4}}O_{6}^{2-}\]

B) \[S{{O}_{2}}O_{3}^{2-}\] gets reduced to \[{{S}_{4}}O_{6}^{2-}\]

C) \[{{I}_{2}}\] gets reduced to \[{{I}^{-}}\]

D) \[{{I}_{2}}\] gets oxidised to \[{{I}^{-}}\]

Correct Answer: D

Solution :

[d] \[2{{S}_{2}}O_{3}^{2-}+{{I}_{2}}\to {{S}_{4}}O_{6}^{2-}+2{{I}^{-}}\]

Oxidation half-reaction: \[\overset{+2}{\mathop{{{S}_{2}}}}\,O_{3}^{2-}\to \overset{+4}{\mathop{{{S}_{4}}}}\,O_{6}^{2-}\]

Reduction half-reaction: \[I_{2}^{0}\to 2{{I}^{-}}\]

Hence, \[{{S}_{2}}O_{3}^{2-}\] is getting oxidised to  while \[{{I}_{2}}\] is getting reduced to  \[2{{I}^{-}}\].

So, [d] is the correct answer.