Faraday laws of electrolysis

Faraday’s laws helps to calculate the amount of products deposited or liberated at the respective electrodes during electrolysis. Charge carried by 1 mole of electrons is known as Faraday constant (F).   Charge on one electron is 1.6 x 10^-19 C. Therefore, Charge carried by 1 mole of electrons:  C = 96,487 C. Therefore, 1F is approximately 96,500C

Faraday’s 1st law of electrolysis:

Amount of substance deposited or liberated at respective electrodes is directly proportional to the amount of charge passed through the electrolyte.

$$w\propto Q$$

$$w = Z \times Q \quad \text{Where} \quad Q = I \times t$$

Where, Z = Electrochemical equivalent which is the amount of substance deposited or liberated at respective electrodes when 1 coulomb (1 ampere per second) charge passed through the solution. If Q = 96,500C, then amount of substance deposited is equal to the equivalent mass(E) of the substance.

$$E = \frac{\text{Atomic mass}}{\text{Charge}}$$

Let Mⁿ⁺ is deposited at the cathode by taking ‘n’ moles of electrons (i.e., nF Faraday charge = n x 96,500C )

$$M^{n+} + ne^- \rightarrow M(s)$$

n x 96,500C  of charge deposits 1 mole of M (Let atomic mass of M is m g/mol, 1 mole of M = m g). Now we will calculate the amount of M deposited when Q(I x t) Coulomb charge passed through electrolyte.

$$n\times 96500 C ……………….m g$$

$$Q \quad \text{coulomb}………………. ?$$

Therefore, Mass of M deposited when Q coulomb charge passed through electrolyte

$$ = \frac{Q\times m}{n\times 96500}$$

$$\text{In the above equation} \quad \frac{m}{n} = \frac{\text{Atomic mass}}{\text{Charge}} = \text{Equivalent mass} (E) \quad \text{and} \quad Q = I \times t$$

$$\text{Therefore,} W = \frac{E \times I \times t}{96500}$$