A solution of 0.2 g of a compound containing Cu2+ and C2O_4^(2-) ions on titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solutions neutralized with Na2CO3, acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 M Na2S2O3 solution for complete reduction. Find out the mole ratio of Cu2+ to C2O_4^(2-) in the compound. Write down the balanced redox reactions involved in the above titrations.

Sol. With KMnO2, oxalate ion is oxidized only as : 5C2O_4^- + 2MnO_4^- + 16H+ → 2Mn2+ + 10CO2 + 8H2O Let, in the given mass of compound, x millimol of C2O_4^(2-) ion is present, then Meq of C2O_4^(2-) = Meq of MnO_4^- ⇒ 2x = 0.02 × 5 × 22.6 ⇒ x = 1.13 At the later stage, with I-, Cu2+ is reduced as : 2Cu2+ + 41- → 2CuI + I2 and I2 + 2S2O_3^(2-) → 2I- + S4O_6^(2-) Let there be x millimol of Cu2+. ⇒ Meq of Cu2+ = Meq of I2 = meq of hypo ⇒ x = 11.3 × 0.05 = 0.565 ⇒ Moles of Cu2+ : moles of C2O_4^(2-) = 0.565 : 1.13 = 1 : 2