How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl _{2} ?
(ii) 40.0 g of Al from molten Al _{2} O _{3} ?
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl _{2} ?
(ii) 40.0 g of Al from molten Al _{2} O _{3} ?

Solution:

(i) From given data, we write the equation of cell,

    \[Ca ^{2+}+2 e^{-} \rightarrow Ca\]

Evaluating the value of electricity in terms of Faraday we have,

Electricity required to produce 40 g of calcium =2 F

Therefore, electricity required to produce 20 g of calcium =(2 \times 20) / 40 F

    \[=1 F\]

(ii) From given data, we write the equation of cell

    \[A l^{3+}+3 e^{-} \rightarrow Al\]

Evaluating the value of electricity in terms of Faraday we have,

Electricity required to produce 27 g of Al =3 F

Therefore, electricity required to produce 40 g of Al =(3 \times 40) / 27 F

    \[=4.44 F\]