Question

A solution of 0.2g of a compound containing cupric and oxalate ions on titration with 0.03 M KMnO₄ in presence of H₂SO₄ consumes 11.3 mL of the oxidant. The resulting solution is neutralized with Na₂CO₃ acidified with dil CH₃COOH and treated with excess KI. Iodine liberated  requires 2.85 mL of 0.06 M Na₂S₂O₃ for complete reduction. Find mole ratio of two ions. Also, write balanced redox reaction.

Answer 1

Cu2⁺ ions cannot be oxidized, so only C₂O₄^2- will be oxidized by KMnO₄ .

5C₂O₄^2- + 2MnO₄^- + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

⇒ 2 mmoles of MnO₄^- ≡ 5 mmoles of C₂O₄^2- ions

0.03 x 11.4 mmoles of MnO₄^- = 5/2 (0.03 x 11.4) m moles of C₂O₄^2- ions

or (0.06 x 2.85) mmole of S₂O₃^2- = (0.06 x 2.85) mmole of I₂ 

⇒ 1/2 (0.06 x 2.85) mmole of I₂ = (0.06 x 2.85) mmole of Cu²⁺

The mole ratio is 5 : 1