How much electricity in terms of Faraday is r | vedclass

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Question

(A) The reduction reaction for the production of aluminum is: $	ext{Al}^{3+} + 3e^- 	o 	ext{Al}$.
According to the stoichiometry, $1 	ext{ mole}$

Vedclass · Accepted Answer

$4.44$

Vedclass · Answer

(A) The reduction reaction for the production of aluminum is: $	ext{Al}^{3+} + 3e^- 	o 	ext{Al}$.
According to the stoichiometry, $1 	ext{ mole}$ of $	ext{Al}$ $(27 	ext{ g})$ requires $3 	ext{ Faradays}$ $(3 	ext{ F})$ of electricity.
To produce $40.0 	ext{ g}$ of $	ext{Al}$, the charge required is calculated as:
$	ext{Charge} = \frac{3 	ext{ F}}{27 	ext{ g}} 	imes 40.0 	ext{ g} = \frac{120}{27} 	ext{ F} \approx 4.44 	ext{ F}$.
Thus, the amount of electricity required is $4.44 	ext{ F}$.

How much electricity in terms of Faraday is required to produce $40.0 \text{ g}$ of $\text{Al}$ from molten $\text{Al}_2\text{O}_3$ (in $\text{ F}$)?

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