Question

$1$ mole of $Fe{C}_2{O}_4$ is oxidised by $x$ moles of ${Cr}_2{O}_7^{2-}$ in acidic medium, $x$ is:

• A. $3$
• B. $1.5$
• C. $0.5$
• D. $1.0$

Answer 1

Answer: (C)

$0.5$

The balanced reaction is as follows:

$Fe{C}_2{O}_4\rightleftharpoons {Fe}^{2+}+C_2{O}_4^{2-}$

${Fe}^{2+}$ is oxidised to ${Fe}^{3+}$ 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,
$6{Fe}^{2+}\equiv 1{Cr}_2{O}_7^{2-}$

$1{Fe}^{2+}\equiv \frac{1}{6}{Cr}_2{O}_7^{2-}$

$(C_2{O}_4^{2-}⟶2C{O}_2\times 2e^{-})\times 3$

${Cr}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2{Cr}^{3+}+7H_{2}O$

Thus,
$3C_2{O}_4^{2-}\equiv 1{Cr}_2{O}_7^{2-}$

$1C_2{O}_4^{2-}=\frac{1}{3}{Cr}_2{O}_7^{2-}$

or, $\frac{1}{6}+\frac{1}{3}=\frac{1}{2}=0.5mol$

Hence, the correct option is $C$