Quantitative aspects of electrolysis

a. The mass of reactant consumed or the mass of product formed at an electrode during electrolysis can be calculated by knowing the stoichiometry of the half reaction at the electrode. The calculation involves four steps.

i. Calculation of quantity of electricity passed : To calculate the quantity of electricity ($Q$) passed during electrolysis, the amount of current $I$ passed through the cell is measured, and the time for which the current is passed is noted:

ii. Calculation of moles of electrons passed — the total charge passed is $Q$ coulombs, and the charge of one mole of electrons is 96500 coulombs (C), referred to as one faraday (1 F). Hence,

iii. Calculation of moles of product formed — the balanced equation for the half reaction occurring at the electrode is devised; the stoichiometry of the half reaction indicates the moles of electrons passed and the moles of product formed. For the reaction $\mathrm{Cu^{2+}\,(aq) + 2e^- \longrightarrow Cu\,(s)}$, two moles of electrons are required for the production of one mole of Cu. To simplify further we introduce the entity mole ratio, given by

For the reaction of Cu, mole ratio = $\frac{1}{2}$. Therefore,

iv. Calculation of mass of product :

b. Suppose two cells containing different electrolytes are connected in series; the same quantity of electricity is passed through them. The masses of the substances liberated at the electrodes of the two cells are then related. The mass of the substance produced at the electrode of the first cell is given by

$W_1 = \frac{Q\,(\mathrm{C})}{96500\ (\mathrm{C/mol}\ e^-)} \times (\text{mole ratio})_1 \times M_1, \quad\text{hence}\quad \frac{Q\,(\mathrm{C})}{96500\ (\mathrm{C/mol}\ e^-)} = \frac{W_1}{(\text{mole ratio})_1 \times M_1}$ …