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Question

(C) The reduction reactions at the cathodes are:$Al^{3+} + 3e^- 	o Al$ (requires $3 \ mol$ of electrons for $1 \ mol$ of $Al$)$Cu^{2+} + 2e^- 	o Cu$

Vedclass · Accepted Answer

$1 \ mol : 1.5 \ mol : 3 \ mol$

Vedclass · Answer

(C) The reduction reactions at the cathodes are:$Al^{3+} + 3e^- 	o Al$ (requires $3 \ mol$ of electrons for $1 \ mol$ of $Al$)$Cu^{2+} + 2e^- 	o Cu$ (requires $2 \ mol$ of electrons for $1 \ mol$ of $Cu$)$Na^+ + e^- 	o Na$ (requires $1 \ mol$ of electrons for $1 \ mol$ of $Na$)Given $3 \ F$ ($3 \ mol$ of electrons) are passed through each cell:Moles of $Al$ deposited $= \frac{3 \ F}{3 \ F/mol} = 1 \ mol$Moles of $Cu$ deposited $= \frac{3 \ F}{2 \ F/mol} = 1.5 \ mol$Moles of $Na$ deposited $= \frac{3 \ F}{1 \ F/mol} = 3 \ mol$Therefore,the ratio is $1 \ mol : 1.5 \ mol : 3 \ mol$.

Three faradays of electricity are passed through molten $Al_2O_3$,aqueous solution of $CuSO_4$ and molten $NaCl$ taken in different electrolytic cells. The amount of $Al$,$Cu$ and $Na$ deposited at the cathodes will be in the ratio of

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