A current of 15.0 A is passed through a solu | vedclass

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(A) The anodic reaction is: $2Cl^{-} ightarrow Cl_2 + 2e^{-}$The total charge passed is $Q = I 	imes t = 15.0 \ A 	imes (45 	imes 60) \ s = 40500

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(A) The anodic reaction is: $2Cl^{-} ightarrow Cl_2 + 2e^{-}$The total charge passed is $Q = I 	imes t = 15.0 \ A 	imes (45 	imes 60) \ s = 40500 \ C$.The number of moles of electrons passed is $n_{e^-} = \frac{Q}{F} = \frac{40500}{96500} \approx 0.4197 \ mol$.From the stoichiometry of the reaction,$2 \ mol$ of $e^-$ produce $1 \ mol$ of $Cl_2$. Therefore,$n_{Cl_2} = \frac{n_{e^-}}{2} = \frac{0.4197}{2} \approx 0.20985 \ mol$.Using the ideal gas law $PV = nRT$ at $1 \ atm$ and $273 \ K$ ($STP$ conditions),the volume is $V = \frac{nRT}{P} = \frac{0.20985 	imes 0.082 	imes 273}{1} \approx 4.7 \ L$.

$A$ current of $15.0 \ A$ is passed through a solution of $CrCl_2$ for $45 \ min$. The volume of $Cl_2$ (in $L$) obtained at the anode at $1 \ atm$ and $273 \ K$ is around ($1 \ F = 96500 \ C \ mol^{-1}$,At. wt. of $Cl=35.5$,$R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$) (in $.7$)

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