For approximate calculations we use 1F ≃ 96500 C mol–1 For the electrode reactions:
Mg2+(l) + 2e– ¾® Mg(s)
(2 31)
Al3+(l) + 3e– ¾® Al(s)
(2 32)
It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons
(2F) and 3 mol of electrons (3F) respectively