Chemistry - Redox Reactions Question with Solution | TestHub

The compound is $2{\mathrm{FeC}}_2{\mathrm{O}}_4\cdot {\mathrm{K}}_2{\mathrm{C}}_2{\mathrm{O}}_4\cdot 3{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$.

Oxidation states:

For Fe: ${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}$ (2 moles of Fe, each loses 1 $e^{-}$) = 2 $e^{-}$

For C₂O₄²⁻: ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to 2{\mathrm{CO}}_2$ (each C loses 1 $e^{-}$, so C₂O₄²⁻ loses 2 $e^{-}$).

Total oxalate groups = $2+1+3=6$. So, 6 moles of C₂O₄²⁻ each lose 2 $e^{-}$ = 12 $e^{-}$.

Total electrons lost by the compound (n-factor) = $2+12=14$.

Millimoles of the compound = $40mL\times 0.05M=2mmol$.

Milli-equivalents of the compound = $2mmol\times 14=28meq$.

For ${\mathrm{KMnO}}_4$ in acidic medium, ${\mathrm{Mn}}^{7+}\to {\mathrm{Mn}}^{2+}$, so n-factor = 5.

Normality of ${\mathrm{KMnO}}_4$ = Molarity $\times$ n-factor = $0.4M\times 5=2N$.

Using the equivalence principle: $N_1{V}_1=N_2{V}_2$

$2N\times V_{{\mathrm{KMnO}}_4}=28meq$

$V_{{\mathrm{KMnO}}_4}=\frac{28}{2}=14mL$.