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(C) The chemical reaction for the electrolysis of fused $AlCl_3$ is: $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$. At the cathode: $Al^{3+} + 3e^- \rightarrow

Vedclass · Accepted Answer

$1.12$

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(C) The chemical reaction for the electrolysis of fused $AlCl_3$ is: $AlCl_3 ightarrow Al^{3+} + 3Cl^-$. At the cathode: $Al^{3+} + 3e^- ightarrow Al$. At the anode: $2Cl^- ightarrow Cl_2 + 2e^-$. Given mass of $Al = 0.9 \ g$. Molar mass of $Al = 27 \ g/mol$. Moles of $Al = \frac{0.9}{27} = \frac{1}{30} \ mol$. According to the stoichiometry,$1 \ mol$ of $Al$ requires $3 \ mol$ of electrons,and $2 \ mol$ of $Cl^-$ produce $1 \ mol$ of $Cl_2$ (requiring $2 \ mol$ of electrons). Total electrons passed = $3 	imes 	ext{moles of } Al = 3 	imes \frac{1}{30} = 0.1 \ mol$. Moles of $Cl_2$ produced = $\frac{	ext{Total electrons}}{2} = \frac{0.1}{2} = 0.05 \ mol$. Volume of $Cl_2$ at $STP = 0.05 \ mol 	imes 22.4 \ L/mol = 1.12 \ L$.

During electrolysis of fused aluminium chloride,$0.9 \ g$ of aluminium was deposited on the cathode. The volume of chlorine liberated at the anode will be .............. $litres$ (at $STP$).

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