Excess of KI reacts with $\mathrm{CuSO}_{4}$ solution and then $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ solution is added to it. Which of the statements is incorrect for this reaction?

(D) $\mathrm{CuI}_{2}$ is formed

(A) $\mathrm{Cu}_{2} \mathrm{I}_{2}$ is reduced

(B) Evolved $\mathrm{I}_{2}$ is reduced

(C) $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ is oxidized

JEE Main · 2004 · 24 Apr 2004 · Offline · Redox Reactions and Electrochemistry

Excess of KI reacts with

CuSO_{4}

solution and then

Na_{2} S_{2} O_{3}

solution is added to it. Which of the statements is incorrect for this reaction?

Answer: (D)

Excess potassium iodide reduces copper(II) to copper(I):

2 CuSO X_{4} + 4 KI 2 K X_{2} SO X_{4} + Cu X_{2} I X_{2} + I X_{2}

, so

Cu X^{2 +}

becomes the cuprous iodide

Cu X_{2} I X_{2}

while iodide is oxidised to iodine. Added sodium thiosulphate then reduces that liberated

I X_{2}

back to iodide and is itself oxidised to tetrathionate. The statement that fails is the one saying

CuI X_{2}

forms; copper(II) iodide is unstable, and it is the copper(I) salt

Cu X_{2} I X_{2}

that actually appears.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option