In the electrolysis of $CuCl_2$,a current of $2.0 \ A$ is passed through the solution at $1 \ bar$ pressure and $300 \ K$ temperature for $1 \ hour$. Calculate the amount of product obtained at the anode and cathode. $[Cu = 63.5, \ Cl = 35.5, \ R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}]$

(N/A) $(i)$ Calculation for quantity of electricity: $Q = It = 2.0 \ A \times (1 \times 60 \times 60) \ s = 7200 \ C = \frac{7200}{96500} \ F \approx 0.0746 \ F$.
Ionization: $CuCl_2 \rightarrow Cu^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$.
Reduction reaction at cathode: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$.
According to the reaction,$2 \ mol$ of electrons produce $1 \ mol$ of $Cu$.
So,$2 \ F$ of electricity produces $63.5 \ g$ of $Cu$.
Therefore,mass of $Cu$ produced by $\frac{72}{965} \ F = \frac{63.5}{2} \times \frac{72}{965} \approx 2.368 \ g$ of $Cu$ at the cathode.
Oxidation at anode: $2Cl^{-}_{(aq)} \rightarrow Cl_{2(g)} + 2e^{-}$.
$2 \ mol$ of electrons $(2 \ F)$ produce $1 \ mol$ of $Cl_2$ gas.
Therefore,moles of $Cl_2$ gas produced by $\frac{72}{965} \ F = \frac{72}{965} \times \frac{1}{2} = \frac{36}{965} \approx 0.0373 \ mol$ of $Cl_2$.
The mass of $Cl_2 = 0.0373 \times 71 \approx 2.648 \ g$.
Using the ideal gas law $PV = nRT$,the volume of $Cl_2$ is $V = \frac{nRT}{P} = \frac{0.0373 \times 0.08314 \times 300}{1} \approx 0.9305 \ L$ of $Cl_2$ gas at the anode.