Excess of KI reacts with (CuSO\_4) solution and then (Na\_2S... | Chemistry | SolveeIt | Solveeit

Today, August 18

Question

Excess of KI reacts with $CuSO_4$ solution and then $Na_2S_2O_3$ solution is added to it. Which of the statements is incorrect for this reaction?

A

$Cu_2I_2$ is reduced

B

Evolved $I_2$ is reduced

C

$Na_2S_2O_3$ is oxidized

D

$CuI_2$ is formed

Answer

$CuI_2$ is formed

Explanation

Solution

${2 \,CuSO4 + 4KI\, (excess) -> 2K2SO4 + Cu2 I2 + I2 ?}$ ${Na2S2O3 + I2 -> Na2S4O6 + 2NaI}$