(i) $\mathrm{CaCl}_2 \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{Cl}^{-}$ or, $\mathrm{Ca}^{2+} + \underset{2\text{ mol}}{2e^{-}} \rightarrow \underset{1\text{ mol (40 g)}}{\mathrm{Ca}}$ ∵ 40 g of calcium needs = 2 mole of electrons = 2 × 96500 coulombs ∴ 20 g of calcium needs = 96500 coulombs (1F). $\mathrm{Al}_2\mathrm{O}_3 \rightarrow 2\mathrm{Al}^{3+} + 3\mathrm{O}^{2-}$ or, $\mathrm{Al}^{3+} + \underset{3\text{ mole}}{3e^{-}} \rightarrow \underset{1\text{ mole (27 g)}}{\mathrm{Al}}$ ∵ 27 g of aluminium needs = 3 mole of electrons = 3 × 96500 coulombs∴ 40.0 g of aluminium needs =$\frac{3 \times 96500 \times 40.0}{27}$ coulombs= 4.28888 × 10$^{5}$coulombs (4.44 F)