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(A) The reduction reaction for obtaining $Al$ from $Al^{3+}$ is given by: $Al^{3+} + 3e^{-} 	o Al$ According to the stoichiometry of the reaction,$1

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$3 	imes 96500 \ C$

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(A) The reduction reaction for obtaining $Al$ from $Al^{3+}$ is given by: $Al^{3+} + 3e^{-} 	o Al$ According to the stoichiometry of the reaction,$1 	ext{ mole}$ of $Al^{3+}$ requires $3 	ext{ moles}$ of electrons to be reduced to $1 	ext{ mole}$ of $Al$. Since the charge of $1 	ext{ mole}$ of electrons is $1 	ext{ Faraday} = 96500 \ C$,the total charge required is $3 	imes 96500 \ C$.

The amount of charge required for obtaining one mole of $Al$ from $Al^{3+}$ is:

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