Excess of KI reacts with CuSO_4 solution and | vedclass

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(B) The reaction between $KI$ and $CuSO_4$ is: $2CuSO_4 + 4KI 	o Cu_2I_2 + I_2 + 2K_2SO_4$. In this reaction,$Cu^{2+}$ is reduced to $Cu^+$ (forming

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$CuI_2$ is formed

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(B) The reaction between $KI$ and $CuSO_4$ is: $2CuSO_4 + 4KI 	o Cu_2I_2 + I_2 + 2K_2SO_4$. In this reaction,$Cu^{2+}$ is reduced to $Cu^+$ (forming $Cu_2I_2$) and $I^-$ is oxidized to $I_2$. Then,$I_2$ reacts with $Na_2S_2O_3$: $I_2 + 2Na_2S_2O_3 	o Na_2S_4O_6 + 2NaI$. Here,$Na_2S_2O_3$ is oxidized to $Na_2S_4O_6$ and $I_2$ is reduced to $I^-$. $CuI_2$ is not formed because $Cu^{2+}$ oxidizes $I^-$ to $I_2$ immediately,resulting in the formation of $Cu_2I_2$ (cuprous iodide).

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?

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