Question

$2$ moles of $Fe{C}_2{O}_4$ is oxidised by $x$ moles of ${Cr}_2{O}_7^{2-}$ in acidic medium. The value of $x$ is :

Answer 1

The reactions are as follows:
$Fe{C}_2{O}_4\rightleftharpoons {Fe}^{2+}+C_2{O}_4^{2-}$
${Fe}^{2+}$ is oxidised to ${Fe}^{2+}$ and $C_2{O}_4^{2-}$ to $C{O}_2$ by ${Cr}_2{O}_7^{2-}$.
${Fe}^{2+}⟶{Fe}^{3+}+e^{-}\times 6$
${Cr}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2{Cr}^{3+}+7H_{2}O$
Thus,
$6{Fe}^{2+}\equiv 1{Cr}_2{O}_7^{2-}$
$1{Fe}^{2+}\equiv \frac{1}{6}{Cr}_2{O}_7^{2-}$
${Cr}_2{O}_7^{2-}⟶2C{O}_2\times 2e^{-}\times 3$
${Cr}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2{Cr}^{3+}+7H_{2}O$
$3C_2{O}_4^{2-}={Cr}_2{O}_7^{2-}$
$1C_2{O}_4^{2-}=\frac{1}{3}{Cr}_2{O}_7^{2-}$
$\frac{1}{6}+\frac{1}{3}=\frac{1}{2}=0.5mol$.
Hence, the value of $x$ is $\frac{1}{6}$.