Electrolysis and laws of electrolysis

Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction. This process occurs in an electrolytic cell, which is the opposite of a galvanic cell. Electrolysis is crucial for many industrial processes, including the production of metals, purification of metals, and electroplating. For JEE, understanding the principles of electrolysis and the quantitative laws governing it is essential.

1.0Electrolytic Cell

An electrolytic cell is an apparatus used to carry out electrolysis. It consists of:

• Electrolyte: A substance (molten or in an aqueous solution) that contains free ions and conducts electricity.
• Electrodes: Two metallic conductors dipped in the electrolyte.
• External Power Source: A battery or power supply that provides the electrical energy to drive the reaction.

In an electrolytic cell, the positive terminal of the battery is connected to the anode (where oxidation occurs), and the negative terminal is connected to the cathode (where reduction occurs).

2.0Mechanism of Electrolysis

When an electric current is passed through the electrolyte, the ions migrate towards the oppositely charged electrodes.

• At the Cathode (Negative Electrode): Cations (positive ions) move towards the cathode and undergo reduction (gain of electrons).
• At the Anode (Positive Electrode): Anions (negative ions) move towards the anode and undergo oxidation (loss of electrons).

Example: Electrolysis of molten sodium chloride (NaCl)

• Electrolyte: Molten NaCl (contains $N{a}^{+}andC{l}^{-}$ ions).
• At Cathode: $N{a}^{+}$ ions get reduced: $N{a}^{+}(l)+e^{-}\to Na(s)$
• At Anode: $C{l}^{-}$ ions get oxidized: $2C{l}^{-}(l)\to C{l}_2(g)+2e^{-}$
• Overall Reaction: $2NaCl(l)\stackrel{\text{electrolysis}}{\to }2Na(s)+C{l}_2(g)$

3.0Faraday's Law of Electrolysis

Michael Faraday established a quantitative relationship between the amount of electricity passed through an electrolyte and the amount of substance deposited or liberated at the electrodes.

Faraday's First Law of Electrolysis

This law states that the mass of a substance (m) deposited or liberated at an electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte.

• Expression: $m\propto Q$
• Since Q=I×t (Current × Time), we can write: m=ZIt
• • m = mass of substance deposited (in grams)
  • Q = quantity of electricity passed (in Coulombs)
  • I = current (in Amperes)
  • t = time (in seconds)
  • Z = Electrochemical Equivalent of the substance.

The Electrochemical Equivalent (Z) is the mass of a substance deposited by one Coulomb of charge. Its value is given by:

$Z=\frac{\text{Molar Mass}}{n\times F}$

• n = number of electrons transferred per mole of the substance.
• F = Faraday constant (96485 C ), which is the charge of one mole of electrons.

Faraday's Second Law of Electrolysis

This law states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances liberated at the electrodes are directly proportional to their chemical equivalent weights.

• Expression:

$\frac{m_1}{m_2}=\frac{{\text{Equivalent Weight}}_1}{{\text{Equivalent Weight}}_2}$

• Equivalent Weight: The equivalent weight is the molar mass divided by the number of electrons transferred per mole (n-factor).

$\text{Equivalent Weight}=\frac{\text{Molar Mass}}{n}$

Quantitative Aspects of Electrolysis

The fundamental unit of charge in electrolysis is the Faraday (F).

• 1 Faraday = 1 mole of electrons = 96485 Coulombs (C).
• The charge required to deposit 1 mole of a substance is equal to n×F.

General Formula for Quantitative Calculations:

$\text{Moles of substance}=\frac{I\times t}{n\times F}$

$\text{Mass of substance}=\frac{I\times t\times \text{Molar Mass}}{n\times F}$

Example: How much copper is deposited by passing 2 A of current for 10 minutes through a CuSO4​ solution?

• I=2 A, t=10 min=600 s
• Reaction: $C{u}^{2+}+2e^{-}\to Cu(so,n=2)$.
• Molar Mass of Cu = 63.5 g/mol
• Mass of Cu = $\frac{2\times 600\times 63.5}{2\times 96485}\approx 0.395g$

4.0Factors Affecting the Products of Electrolysis

When an aqueous solution is electrolysed, water can also be oxidised or reduced, competing with the electrolyte ions. The product formed depends on the standard reduction potentials of the species present.

• At the Cathode: The species with the higher standard reduction potential will be reduced.
• At the Anode: The species with the lower standard reduction potential will be oxidized.

Example: Electrolysis of aqueous NaCl solution using inert electrodes.

• Species Present: $N{a}^{+},C{l}^{-},H_{2}O$
• At Cathode (Reduction):

$N{a}^{+}+e^{-}\to Na(s)(E^{\circ }=-2.71\text{ V})$

$2H_{2}O+2e^{-}\to H_2(g)+2O{H}^{-}(aq)(E^{\circ }=-0.83\text{ V})$

• Since water has a higher reduction potential, it gets reduced. Product: H_{2}(g)
• At Anode (Oxidation):

$2C{l}^{-}\to C{l}_2(g)+2e^{-}(E^{\circ }=-1.36\text{ V})$

$2H_{2}O\to O_2(g)+4H^{+}+4e^{-}(E^{\circ }=-1.23\text{ V})$

Despite water having a lower oxidation potential, Cl− is often oxidized due to the concept of overpotential. Product: $C{l}_2(g)$