Excess of $\mathrm{KI}$ reacts with $\mathrm{CuSO}_4$ solution and then $\mathrm{Na}_2 \mathrm{~S}_2…

Excess of $\mathrm{KI}$ reacts with $\mathrm{CuSO}_4$ solution and then $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution is added to it. Which of the statements is incorrect for this reaction?
  1. $\mathrm{Cu}_2 \mathrm{I}_2$ is reduced
  2. Evolved $\mathrm{I}_2$ is reduced
  3. $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ is oxidized
  4. $\mathrm{Cul}_2$ is formed

Solution

$2 \mathrm{CuSO}_4+4 \mathrm{KI}$ (excess) $\longrightarrow 2 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{Cu}_2 \mathrm{I}_2+\mathrm{I}_2 \uparrow$ $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{Nal}$

Asked in: JEE Main 2004

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