Excess of KI reacts with CuSO4 solution ...

Excess of KI reacts with `CuSO_4` solution and then `Na_2S_2O_3` solution is added to it. Which of the statement is incorrect in this reaction?

Evolved I, is reduced

` C u l_2` is formed

`Na_2 S_2 O_3`is oxidised

`Cu_2 l_2 `is formed

Text Solution

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The correct Answer is:

`2 CuSO_4 + 4Ki to Cu_2 I_2 + 2K_2 SO_4 +I_2`
` I_2 + 2Na_2 S_2 O_3 to Na_2 S_4 O_6 + 2Nal`
this liberated `I_2` is reduced and `NaSO_4` is oxidised.

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