Faraday's Law for Gaseous Electrolytic Product
For the gases, we use
where, Volume of gas evolved at S.T.P. at an electrode
Equivalent volume = Volume of gas evolved at an electrode at S.T.P. by 1 Faraday charge
Quantitative aspects of electrolysis:
We know that, one Faraday (1F) of electricity is equal to the charge carried by one mole of electrons. So, in any reaction, if one mole of electrons are involved, then that reaction would consume or produce 1F of electricity. Since 1F is equal to 96,500 Coulombs, hence 96,500 Coulombs of electricity would cause a reaction involving one mole of electrons.
If in any reaction, n moles of electrons are involved, then the total electricity involved in the reaction is given by,
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 amount 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.