Quantitative Aspect of Electrolysis

We have already studied that during electrolysis anodic reaction involve oxidation and cathodic reaction involves reduction. For example, cathodic reaction in electrolysis of copper(II) tetraoxosulphate(VI) is

Cu2+ + 2e à Cu

This means that for the production of one mole of copper,

2 moles of electrons are needed. If the number of electrons available to copper ions are less than 2 moles, the amount of copper deposited at the cathode will also be less than 2 moles. Thus, the amount of product formed at the electrode depends on the number of electrons, which in turn depends on the quantity of electricity flowing through the electrolytic cell.

Now charge on one electron has been found to be

1.602 x 10-19 coulombs.

Charge on one mole of electrons

= 1.602 c 10-19 x 6.02 c 1023 = 96488 C mol -l

This magnitude of charge (= 96488 C moi-1) is called Faraday constant and is denoted by F. It is in honour of Michael Faraday who discovered laws of electrolysis.

F = 96488 C mol -1 or = 96500 C mol -1

Charge on n moles of electrons (Q) is given by

Q=nF

Thus, for n moles of electrons, the charge equal to nF, i.e., n x 96500 C of electricity has to be passed through the electrolyte. For example, during the electrolysis of aluminium chloride 3 moles of electrons are required for producing 1 mole of aluminium.

Al 3+ + 3e à Al

Therefore, charge equal to 3 F or 3 x 96500 C has to be passed through the electrolyte.

Now, quantity of charge flowing through the electrolyte can be calculated from the current strength and time for which the current is passed according to following relation.

where,

Q =I x t

Q = Quantity of charge (coulombs)

I = Current in amperes

t = Time in seconds.

Thus, from the above considerations it is possible to calculate the amount of a substance deposited or evolved (if gas) when a definite amount of current is passed for a definite period of time. For illustrations refer to solved examples.