(i) Ca²⁺ + 2e^- → Ca

⇒ Here, 1 mole of Ca, i.e., 40g of Ca requires = 2 F electricity (F if Faraday)

∴ 20g of Ca requires =

= 1 F of electricity

Electricity in terms of Faraday required to produce 20.0 g of Ca from molten CaCl₂ is 1 F of electricity.

(ii) Al^{3 +} + 3e ^- → Al

⇒ 1 mole of Al, i.e., 27g of Al requires = 3 F electricity (F if Faraday)

∴ 40.0 g of Al will require =

= 4.44 F of electricity

Electricity in terms of Faraday required to produce 40.0 g of Al from molten Al₂O₃ is 4.44 F of electricity