Faraday’s Law of Electrolysis

The relationship between the quantity of electric charge passed through an electrolyte, and the amount of substance deposited at the electrodes was given by Faraday in 1834, in the form of the law of electrolysis.

Faraday’s First Law

When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte.

If W is the mass of the substance deposited by passing Q coulomb of charge, then according to this law:

W ∝ Q

Now, Q = I ✕ t

W ∝ I ✕ t

W = z ✕ I ✕ t

Where Z is a constant, known as electrochemical equivalent and is characteristic of a substance deposited.

Faraday’s constant (F) – It is the charge possessed by 1 mole of electrons, and it is equal to 96500 coulombs (approx.). In terms of Faraday’s constant, the number of gram equivalent of electrolyte discharged at an electrode is equal to Faraday’s passed,

W = E X Q / 96500

Faraday’s Second Law

When the same quantity of charge is passed through different electrolytes, then the mass of different substances deposited at the respective electrodes will be in the ratio of their equivalent masses.

Mathematically, it is represented as,

W₁ / W₂ = Z₁ / Z₂

Where, W₁ and W₂ are the weight of two substances which are deposited at their respective electrodes, and Z₁ and Z₂ are their respective equivalent weight.

Solved Problems

1. Given that,

Fe²⁺/ Fe = -0.44 v, Ni⁺²/ Ni = 0.25 v

Ag⁺ / Ag = 0.80 v, Cu⁺²/ Cu = 0.34 v

Which of the following reactions under standard conditions will not take place in the specified direction?

(a) Ni⁺²(aq) + Cu(s) → Ni(s) + Cu⁺²(aq)

(b) Cu(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag(s)

(c) Cu(s) + 2H⁺(aq)→ Cu⁺² + H₂ (g)

(d) Fe(s) + 2H⁺(aq)→ Fe⁺² (aq) + 3 H₂

Solution:

Options a and c.

Ni⁺²(aq) + Cu(s) → Ni(s) + Cu⁺²(aq)

E^o_cell= 0.25 – 0.34

= -0.11v

The reaction is not feasible (negative E^o_cell)

Cu(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag(s)

E^o_cell =0.80 – 0.34 = 0.46 v

The reaction is feasible (positive E^o_cell)

Cu(s) + 2H⁺(aq)→ Cu⁺² + H₂ (g)

E^o_cell = 0 – 0.34 = – 0.34 v

The reaction is not feasible (negative E^o_cell)

Fe(s) + 2H⁺(aq) → Fe⁺² (aq) + 3 H₂

E^o_cell = 0 – (-0.44) = 0.44 v

The reaction is feasible (positive E^o_cell)

2. The emf of a cell corresponding to the reaction, Zn + 2H⁺ (aq) → Zn⁺² (0.1M) + H₂(g) 1 atm is 0.30 v at 25 ⁰ C. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. (E⁰_cell = – 0.76 v )

Solution :

E^o_cell = 0 – (-0.76) = 0.76 v

Applying the Nernst equation,

E_cell = E^o_cell – 0.0591 / 2 log [zN⁺²][H2] / [H]⁺

0.30 = 0.76 – 0.0591 / 2 log 0.1 x 1 / [H⁺]²

log 0.1 / [H⁺]² = 2 x 0.46 / 0.0591

log 0.1 – log [H⁺]² = 15.56

2 pH = 15.56 – log 0.1

pH = 16.56 / 2 = 8.28

3. The solution of metal of atomic mass X was electrolysed for 1 hour with a current of 0.25 ampere. The mass of the metal deposited was 0.295 g. Find the metal X if its valency is 2.

Solution:

Given, I = 0.25 ampere, t= 1 hr = 60 x 60 = 3600s

Q = I ✕ t

Q = 0.25 ✕ 3600

= 900 coulombs

Therefore, 900 coulombs of electricity deposit = 0.295

96500 coulomb of electricity deposit = 0.295 X 96500 / 900 = 31.63g

Valency of metal = atomic mass / equivalent mass

Atomic mass of metal X = 31.63 X 2 = 63.26g

Therefore, the metal X is copper.