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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.

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Cu2+ ions cannot be oxidized, so only C2O42- will be oxidized by KMnO4 .

5C2O42- + 2MnO4- + 16H+ → 2Mn2+ + 10CO2 + 8H2O

⇒ 2 mmoles of MnO4- ≡ 5 mmoles of C2O42- ions

0.03 x 11.4 mmoles of MnO4- = 5/2 (0.03 x 11.4) m moles of C2O42- ions

or (0.06 x 2.85) mmole of S2O32- = (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

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