Question

The equivalent weight of $Fe{C}_2{O}_4$ in the reaction, $Fe{C}_2{O}_4⟶{Fe}^{3+}+C{O}_2$ is :

• A. $\frac{M}{3}$
• B. $\frac{M}{6}$
• C. $\frac{M}{2}$
• D. $\frac{M}{1}$

Answer 1

The correct option is A $\frac{M}{3}$

Total number of electrons involved per molecule oxidation of $Fe{C}_2{O}_4$ to $F{e}^{3+}$ and $C{O}_2$ is 3.
The oxidation of ferrous ions to ferric ions involves 1 electron.
$F{e}^{2+}\to F{e}^{3+}+e^{-}$
The oxidation of oxalate ions to carbon dioxide involves 2 electrons.
$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$
$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{total number of electrons involved per molecule}}$
Hence, $E=\frac{M}{3}$