Question

How much electricity in terms of Faraday is required to produce
(i) $20.0g$ at $Ca$ from molten $CaC{l}_2$
(ii) $40.0g$ of $Al$ from Almolten $A{l}_2{O}_3$.

Answer 1

(a) (w/E) $Eq.$ of $Ca=\frac{i.t}{96500}$

$(C{a}^{2+}+2e\to Ca)$ $\therefore Eq.Wt.$ of $Ca=\frac{40}{2}$

or $\frac{20}{40/2}=\frac{i.t}{96500}$

or $i.t=1\times 96500=1F$

(b) $Eq.$ of $Al=\frac{i.t}{96500}$

$(6e+A{l}_2^{3+}\to 2Al)$ $\therefore Eq.Wt.$ of $Al=\frac{27}{6/2}$

or $\frac{40}{27/3}=\frac{i.t}{96500}$

or $i.t=\frac{120}{27}\times 96500=4.44F$