A) \[\text{C}{{\text{u}}_{\text{2}}}{{\text{I}}_{\text{2}}}\] is formed
B) \[\text{Cu}{{\text{I}}_{\text{2}}}\]is formed
C) \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]is oxidized
D) Evolved \[{{I}_{2}}\] is reduced
Correct Answer: B
Solution :
Copper sulphate reacts with KI to give cuprous iodide and iodine. \[2CuS{{O}_{4}}+4KI\to C{{u}_{2}}{{I}_{2}}+2{{K}_{2}}S{{O}_{4}}+{{I}_{2}}\] \[{{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\] In the second reaction oxidation number of \[{{\text{I}}_{\text{2}}}\]changes from zero (in\[{{\text{I}}_{\text{2}}}\]) to -1 (in NaI). Thus, it is reduced. Oxidation number of S changes from +2 \[(in\,N{{a}_{2}}{{S}_{2}}{{O}_{3}})\] to \[2.5(in\,N{{a}_{2}}{{S}_{4}}{{O}_{6}}).\] Thus, \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]is oxidised.
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