Question

Excess of $ KI $ reacts with $ CuS{{O}_{4}} $ solution and then $ N{{a}_{2}}{{S}_{2}}{{O}_{3}} $ solution is added to it. Which of the statements is incorrect for this reaction?

• 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

Answer 1

Answer: (B)

Copper sulphate reacts with $K I$ to give cuprous iodide and iodine.
$2 C u S O_{4}+4 K I \rightarrow C u_{2} I_{2}+2 K_{2} S O_{4}+I_{2}$
$I_{2}+2 N a_{2} S_{2} O_{3} \rightarrow N a_{2} S_{4} O_{6}+2 N a I$
In the second reaction oxidation number of $I _{2}$ changes from zero $\left( in I _{2}\right)$ to $-1$ (in $NaI$ ). Thus, it is reduced. Oxidation number of $S$ changes from $+2\left(\right.$ in $\left.Na _{2} S _{2} O _{3}\right)$ to $2.5\left(\right.$ in $\left.Na _{2} S _{4} O _{6}\right)$.
Thus, $N a_{2} S_{2} O_{3}$ is oxidised.