Quantitative aspects of electrolysis: We already know that, one Faraday (1F) of the electricity is equal to the charge carried by one mole (6.023 * 1023) of the electrons. Thus in any of the reaction, if one mole of electrons is involved, then that reaction would consume or produce 1F of electricity. As 1F is equal to 96,500 Coulombs, therefore 96,500 Coulombs of the electricity would cause a reaction involving one mole of electrons.
If in any reaction, n moles of electrons are involved, then the total electricity (Q) involved in the reaction is given by the formula stated hear, Q=nF*96500C
Thus, the amount of electricity involved in any reaction is related to,
(i) The number of moles of electrons involved in the reaction,
(ii) The quantity of any substance involved in the reaction.
Therefore, 1 Faraday or 96,500 C or 1 mole of electrons will reduce,
(a) 1 mole of monovalent cation,
(b) 1/2mole of divalent cation,
(c) 1/3 mole of trivalent cation,
(d) 1/n mole of n valent cations.
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