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Excess of KI reacts with CuSO(4) solutio...

Excess of `KI` reacts with `CuSO_(4)` solution and `Na_(2)SO_(3)` solution is added to it. Which of the following statements in incorrect for the reaction?

A

` Cu_2I_2` is formed

B

` Cu l_2` is formed

C

Evolved `I_2` is reduced,.

D

` Na_2 S_2O_3` is oxidized`.

Text Solution

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The correct Answer is:
B
step 1: Iodide ion is oxdized by copper sulphate
` 2CuSO_4 + 4 KI rarr Cu_2I_2 + I_2 +2K_2SO_4`
Syep 2: Iodine is reduced by sodium thiosulphate
` 2Na_2S_2O_3 +I_2 rarr Na_2S_4O_6 +2 NaI`
thus `, I_2` has been reduced to ` I^(-)` ions while ` Na_2S_2O_3` has been oxidized.
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