Statement 1: When 20 ml of 0.5 M - CuSO₄ solution is treated with excess of KI solution, the liberated I₂ exactly requires 20 ml of 0.5 M - Na₂S₂O₃ solution for titration.
Statement 2: For both the solutions (CuSO₄ and Na₂S₂O₃), their molarity and normality are equal.

Question

Statement 1: When 20 ml of 0.5 M - CuSO₄ solution is treated with excess of KI solution, the liberated I₂ exactly requires 20 ml of 0.5 M - Na₂S₂O₃ solution for titration.

Statement 2: For both the solutions (CuSO₄ and Na₂S₂O₃), their molarity and normality are equal.

TestHub · Accepted Answer

2Cu²⁺ + 4I⁻ → 2CuI + I₂: Cu²⁺ gains one electron, so n = 1 for CuSO₄ and 20 ml x 0.5 M = 10 meq of Cu²⁺ liberate 10 meq of I₂. I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻: thiosulphate loses one electron per ion, n = 1, so 20 ml x 0.5 M = 10 meq of Na₂S₂O₃ - exactly matching, I is correct. II is correct (n = 1 for both, so M = N for each) and is the reason equal volumes of equal molarity react. Hence (a).

Chemistry - Redox Reactions Question with Solution | TestHub

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