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Excess of $KI$ reacts with ${CuSO}_{4}$ solution, and ${Na}_{2} S_{2} O_{3}$ solution is added to it. Which of the following statement is incorrect for the reaction?

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Evolved $I_{2}$ is reduced

${Cu}_{2} I_{2}$ is formed

${Na}_{2} S_{2} O_{3}$ is oxidised

Indicator in the second step is ${(C_{6} H_{5})}_{2} NH$

answer is D.

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Detailed Solution

$2 {CuSO}_{4} + 4 KI \to {Cu}_{2} I_{2} + K_{2} {SO}_{4} + I_{2}; 2 {Na}_{2} S_{2} O_{3} + I_{2} \to {NaS}_{4} O_{6} + 2 NaI;$

This is an example of iodometry.

Indicator used is starch.

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