A solution of 0.2 g of a compound containing `Cu^(2+) "and" C_(2)O_(4)^(2-)` ions on titration with 0.02 M `KMnO_(4)` in presence of `H_(2)SO_(4)` consumes 22.6 mL of the oxidant. The resultant solution is neutralized with `Na_(2)CO_(3)` acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.5 M `Na_(2)S_(2)O_(3)` solution for complete reduction. Find out the mole ratio of `Cu^(2+) "to" C_(2)O_(4)^(2-)` in the compound. Write down the balanced redox reactions involved in the above titrations.

Explanation

Let 'a' moles of `Cu^(2+)` and 'b' moles of `C_(2)O_(4)^(2-)` be present in solution
(i) The solution is oxidised by `KMnO_(4)` with only `C_(2)O_(4)^(2-)`
`Mn^(7+) + 5 e^(-) rarr Mn^(2+)`
`C_(2)^(6+) rarr 2C^(4+) + 2e^(-)`
Meq. of `C_(2)O_(4)^(2-)` = Meq. of `KMnO_(4)`
`b xx 2 xx 1000 = 0.02 xx 5 xx 2.26 `
`b = 1.13 xx 10^(-3)`
(ii) After oxidation of `C_(2)O_(4)^(2-)` the resulting solution is neutralized by `Na_(2)CO_(3)` acidified with dilute `CH_(3)COOH` and then treated with excess of KI. The liberated `I_(2)` required `Na_(2)S_(2)O_(3)` for its neutralization,
i.e. `Cu^(2+) overset("KI")to I_(2) overset(Na_(2)S_(2)O_(3))to Na_(2)S_(4)O_(6) + I^(-)`
Meq. of `Cu^(2+)` = Meq. of `I_(2)` liberated = Meq. of `Na_(2)S_(2)O_(3)` used
Meq. of `Cu^(2+)` = Meq. of `Na_(2)S_(2)O_(3)` used
`a xx 1 xx 1000 = 11.3 xx 0.5 xx 1`
`a = 5.65 xx 10^(-3)`
Molar ratio of `Cu^(2+)//C_(2)O_(4)^(2-) = a/b implies 5.65 xx 10^(-3)/(1.13 xx 10^(-3)) = 5/1 `
Hence, molar ratio of `Cu^(2+) , C_(2)O_(4)^(2-) = 5 : 1`
Involved reactions are : `2 MnO_(4)^(-) + 5 C_(2)O_(4)^(2-) + 16H^(+) rarr 2Mn^(2+) + 10 CO_(2) + 8H_(2)O`
`2Cu^(2+) + 4I^(-) rarr Cu_(2)I_(2) + I_(2)`
`I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) + S_(4)O_(6)^(2-)`