A current of 15.0 amperes is passed through a solution of ${\mathrm{CrCl}}_2$ for 45 minutes. The volume of ${\mathrm{Cl}}_2(\mathrm{in} L)$ obtained at the anode at 1 atm and 273 K is around          $\left(\mathrm{lF}=96500{\mathrm{Cmol}}^{-1}\text{, At. wt. of }\mathrm{Cl}=35.5,\mathrm{R}=0.082 \mathrm{L}.atm.mo{l}^{-1}{K}^{-1}\right)$
 

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{electrons}: \\ 15 \times 45\times 60=40500 \mathrm{C} \\ \mathrm{number} \mathrm{of} \mathrm{F}=\frac{40500 \mathrm{C}}{96500 \mathrm{C}}\times 1 \mathrm{F}=0.42 \mathrm{F} \\ \mathrm{mole} {\mathrm{e}}^{-}=0.42 \mathrm{F}\times \frac{1 \mathrm{mole} {\mathrm{e}}^{-}}{1 \mathrm{F}}=0.42 \mathrm{mol} {\mathrm{e}}^{-} \end{gathered}$$

Moles of chlorine gas produced:

$$\begin{gathered} 0.42\times \frac{1 \mathrm{mole} {\mathrm{Cl}}_2}{2 \mathrm{mole} {\mathrm{e}}^{-}}=0.21 \mathrm{mole} {\mathrm{Cl}}_2 \\ \mathrm{volume}=\frac{0.21\times 0.082\times 273}{1 \mathrm{atm}}=4.7 \mathrm{L} \end{gathered}$$