Question

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 neutralised with $Na _{2} CO _{3}$, acidified with dilute acetic acid and treated with excess $KI$. The liberated iodine requires $11.3 \, mL$ of $0.05 \,M \, Na$ ${ }_{2} S _{2} O _{3}$, solution for complete reduction. Find out the mole ratio of $Cu -+$ to $C _{2} O _{4}^{2-}$ in the compound. Write down the balanced redox reactions involved in the above titrations.

Answer 1

With $KMnO _{4}$, oxalate ion is oxidised only as :
$5C _{2} O _{4}^{2-}+2 MnO _{4}^{-}+16 H ^{+} \longrightarrow 2 Mn ^{2+}+10 CO _{2}+8 H _{2} O$
Let, in the given mass of compound, $x$ millimol of $C _{2} O _{4}^{2-}$ ion is present, then
Meq of $ C _{2} O _{4}^{2-} =$ Meq of $ MnO _{4}^{-} $
$\Rightarrow 2 x =0.02 \times 5 \times 22.6$
$\Rightarrow x =1.13$
At the later stage, with $I ^{-}, Cu ^{2+}$ is reduced as :
$2 Cu ^{2+}+4 I ^{-} \longrightarrow 2 CuI + I _{2}$
and $I _{2}+2 S _{2} O _{3}^{2-} \longrightarrow 2 I ^{-}+ S _{4} O _{6}^{2-}$
Let there be $x$ millimol of $Cu ^{2+}$.
$\Rightarrow Meq of Cu ^{2+}= Meq$ of $I _{2}=$ meq of hypo
$\Rightarrow x=11.3 \times 0.05=0.565$
$\Rightarrow$ Moles of $Cu ^{2+}: $ moles of $C _{2} O _{4}^{2-}=0.565: 1.13=1: 2$