2CuSO4+4KI (excess) ⟶2 K2SO4+Cu2I2+I2↑ Na2 S2O3+I2⟶Na2 S4O6+2Nal
Excess of KI reacts with CuSO4 solution and then Na2 S2O3 solution is added to it. Which of the statements is incorrect for this reaction?
Held on 30 Apr 2004 · Verified 6 Jul 2026.
Cu2I2 is reduced
Evolved I2 is reduced
Na2 S2O3 is oxidized
Cul2 is formed
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