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. |
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