JEE MAIN - Chemistry (2004 - No. 46)
Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction?
Cu2I2 is reduced
Evolved I2 is reduced
Na2S2O3 is oxidized
CuI2 is formed
Explanation
$$4\mathop {KI}\limits^{ - 1} + 2CuS{O_4} \to \mathop {{I_2}}\limits^0 + C{u_2}{I_2} + 2{K_2}S{O_4}$$
$$\mathop {{I_2}}\limits^0 + 2N{a_2}\mathop {{S_2}}\limits^{2 + } {O_3} \to N{a_2}\mathop {{S_4}}\limits^{ + 2.5} {O_6} + \mathop {2NaI}\limits^{ - 1} $$
In this $$Cu{I_2}$$ is not formed
$$\mathop {{I_2}}\limits^0 + 2N{a_2}\mathop {{S_2}}\limits^{2 + } {O_3} \to N{a_2}\mathop {{S_4}}\limits^{ + 2.5} {O_6} + \mathop {2NaI}\limits^{ - 1} $$
In this $$Cu{I_2}$$ is not formed
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