A current of 0.5 A is passed through molten | vedclass

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(D) The total charge passed is $Q = I 	imes t = 0.5 \ A 	imes 96.5 \ s = 48.25 \ C$. For the reaction at cathode: $Al^{3+} + 3e^- ightarrow Al(s)$

Vedclass · Accepted Answer

$4.5, 5.6$

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(D) The total charge passed is $Q = I 	imes t = 0.5 \ A 	imes 96.5 \ s = 48.25 \ C$. For the reaction at cathode: $Al^{3+} + 3e^- ightarrow Al(s)$. Moles of electrons = $\frac{48.25}{96500} = 5 	imes 10^{-4} \ mol$. Moles of $Al$ deposited = $\frac{5 	imes 10^{-4}}{3} \ mol$. Mass of $Al$ $(x)$ = $\frac{5 	imes 10^{-4}}{3} 	imes 27 \ g = 4.5 	imes 10^{-3} \ g = 4.5 \ mg$. For the reaction at anode: $2Cl^- ightarrow Cl_2(g) + 2e^-$. Moles of $Cl_2$ liberated = $\frac{5 	imes 10^{-4}}{2} = 2.5 	imes 10^{-4} \ mol$. Volume of $Cl_2$ $(y)$ = $2.5 	imes 10^{-4} \ mol 	imes 22400 \ mL/mol = 5.6 \ mL$. Thus,$x = 4.5$ and $y = 5.6$.

$A$ current of $0.5 \ A$ is passed through molten $AlCl_3$ for $96.5 \ s$. The mass of aluminium deposited at cathode is $x \ mg$ and volume of chlorine liberated (at $STP$) at anode is $y \ mL$. $x$ and $y$ are respectively

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