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(C) The reduction reaction is: $Al^{3+} + 3e^{-} \longrightarrow Al$This equation shows that $3 \ moles$ of electrons are required to produce $1 \ mol

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The reduction reaction is: $Al^{3+} + 3e^{-} \longrightarrow Al$This equation shows that $3 \ moles$ of electrons are required to produce $1 \ mole$ of $Al$,which is equivalent to $27 \ g$ of $Al$.Thus,$3 \ F$ of electricity is required to produce $27 \ g$ of $Al$.Therefore,the number of faradays of electricity required to produce $45 \ g$ of $Al$ is calculated as:$	ext{Faradays} = \frac{3 \ F}{27 \ g} 	imes 45 \ g = 5 \ F$.

Find the number of faradays of electricity required to produce $45 \ g$ of $Al$ from molten $Al_2O_3$. (in $F$)

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