On passing electric current through molten al | vedclass

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Question

(A) The reaction at the anode is: $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$.At $NTP$,$22.4 \ L$ of gas corresponds to $1 \ mol$. Therefore,$11.2 \ L$ of $Cl

Vedclass · Accepted Answer

$9$

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(A) The reaction at the anode is: $2Cl^{-} ightarrow Cl_2 + 2e^{-}$.At $NTP$,$22.4 \ L$ of gas corresponds to $1 \ mol$. Therefore,$11.2 \ L$ of $Cl_2 = \frac{11.2}{22.4} = 0.5 \ mol$ of $Cl_2$.Since $1 \ mol$ of $Cl_2$ is produced by $2 \ mol$ of electrons,$0.5 \ mol$ of $Cl_2$ requires $0.5 	imes 2 = 1 \ mol$ of electrons.The reaction at the cathode is: $Al^{3+} + 3e^{-} ightarrow Al$.This shows that $3 \ mol$ of electrons deposit $1 \ mol$ of $Al$ $(27 \ g)$.Therefore,$1 \ mol$ of electrons will deposit $\frac{1}{3} \ mol$ of $Al$.Mass of $Al = \frac{1}{3} 	imes 27 \ g = 9 \ g$.

On passing electric current through molten aluminium chloride,$11.2 \ L$ of $Cl_2$ is liberated at $NTP$ at the anode. The quantity of aluminium deposited at the cathode is .............. $g$ (atomic weight of $Al = 27$).

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