Electrolysis

Ionic Theory

The ionic theory assumes that a solid electrolyte is made up of two types of charged particles: positively charged ions (cations) and negatively charged ions (anions). These particles are held together by electrostatic forces of attraction.

When such a solid is dissolved in a solvent, these forces are weakened, allowing the electrolyte to dissociate into free-moving ions. This process is known as ion solvation.

The ionic theory helps explain electrolysis. In an electrolyte, ions are free to move in all directions. When an electric current is passed through, negative ions (anions) are attracted to the positive electrode (anode), while positive ions (cations) move toward the negative electrode (cathode).

Electrolytes and Non-Electrolytes

Electrolytes

Electrolytes are molten ionic compounds or aqueous solutions that contain ions capable of conducting electricity. During conduction, these ions are decomposed.

Strong Electrolytes

Strong electrolytes are substances that dissociate almost completely into ions in aqueous solution, even at moderate dilutions. Their dissociation reactions are typically irreversible.

Examples:

Weak Electrolytes

Weak electrolytes are substances that dissociate only slightly in aqueous solution, meaning only a small fraction of the molecules form ions.

Examples:

Non-Electrolytes

Non-electrolytes are substances that do not conduct electricity in their aqueous or molten states because they do not form ions.

Examples:

Electrolysis

Electrolysis is the chemical decomposition that occurs when an electric current is passed through a liquid or solution containing ions.

Mechanism of Electrolysis

Before the current flows, ions in the electrolyte move randomly. Once electrolysis begins, positively charged ions (cations) are attracted to the negatively charged cathode, where they gain electrons and are reduced. Negatively charged ions (anions) are drawn to the positively charged anode, where they lose electrons and are oxidized. Both ions become electrically neutral and are released (discharged).

Electrodes

Electrodes are conductive materials—such as wires, rods, or plates—through which electric current enters or exits the electrolyte.

Factors Affecting Preferential Discharge of Ions

  1. Position of ions in the electrochemical series
    Ions that are lower in the electrochemical series are discharged more readily than those higher up.
  2. Concentration of ions in the electrolyte
    When other conditions are constant, increasing the concentration of a specific ion increases its chances of being discharged. This factor is especially significant when the competing ions are close in the electrochemical series.
  3. Nature of the electrode
    The type of electrode (active or inert) can influence which ion is discharged during electrolysis.

Electrolysis of Concentrated NaCl

Ions Present: Na+, H+, OH, Cl

Reaction at the Anode

Chloride ions (Cl) lose electrons to form chlorine gas, even though OH is easier to discharge:

2Cl(aq) → Cl2(g) + 2e

Reaction at the Cathode

Hydrogen ions gain electrons to form hydrogen gas:

2H+(aq) + 2e → H2(g)

Overall Reaction

2NaCl(aq) + 2H2O(l) → H2(g) + Cl2(g) + 2NaOH(aq)

Note: The remaining Na+ and OH form sodium hydroxide.


Electrolysis of Dilute H2SO4

Ions Present: H+, OH, SO42−

Reaction at the Anode

OH ions lose electrons to form oxygen gas and water:

4OH(aq) → O2(g) + 2H2O(l) + 4e

Reaction at the Cathode

Hydrogen ions are reduced to hydrogen gas:

2H+(aq) + 2e → H2(g)

Note: Only water is electrolyzed, so the H2SO4 solution becomes more concentrated.

Types of Electrodes

Inert Electrodes: Do not react with the electrolyte or products (e.g., platinum, graphite).

Active Electrodes: React with the electrolyte or products, affecting electrolysis (e.g., copper).

Electrolysis of CuSO4 Using Inert Electrodes

Ions Present: Cu2+, H+, OH, SO42−

Reaction at the Anode

OH is discharged to produce oxygen:

4OH(aq) → O2(g) + 2H2O(l) + 4e

Reaction at the Cathode

Cu2+ is reduced to form copper metal:

Cu2+(aq) + 2e → Cu(s)

Note: The blue color fades as Cu2+ ions are used up.


Electrolysis of CuSO4 Using Active Electrodes

Ions Present: Cu2+, H+, OH, SO42−

Reaction at the Anode

Copper electrode loses electrons:

Cu(s) → Cu2+(aq) + 2e

Reaction at the Cathode

Cu2+ from the anode is reduced to metallic copper:

Cu2+(aq) + 2e → Cu(s)

Note: The anode decreases in size while the cathode increases.

Electroplating

Electroplating involves coating an object with a thin layer of metal using electrolysis. The object is the cathode, and the metal to be plated is the anode. The electrolyte contains the plating metal ion.

Example: Plating Iron with Nickel

Reaction at the Anode

Nickel loses electrons:

Ni(s) → Ni2+(aq) + 2e

Reaction at the Cathode

Ni2+ gains electrons and is deposited:

Ni2+(aq) + 2e → Ni(s)

Overall Effect: The iron object gets a nickel coating, and the solution concentration remains unchanged.

Uses of Electroplating

Faraday’s Laws of Electrolysis

Faraday’s First Law of Electrolysis

The mass (m) of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity (Q) passed through the electrolyte.

Mathematical expression:

\( Q = I \times t \)

\( m = \frac{Q \times \text{RAM}}{F \times Z} \)

Where:

Example:

Problem: What mass of copper is deposited when a current of 10.0 A is passed through a CuSO₄ solution for 1 hour?

Given:

  • \( I = 10.0 \, \text{A} \)
  • \( t = 3600 \, \text{s} \)
  • \( \text{RAM of Cu} = 63.5 \)
  • \( F = 96,500 \, \text{C/mol} \)
  • \( Z = 2 \)

Solution:

\( Q = I \times t = 10.0 \times 3600 = 36,000 \, \text{C} \)

\( m = \frac{36,000 \times 63.5}{96,500 \times 2} \approx 11.8 \, \text{g} \)

Answer: 11.8 g of copper is deposited.

Faraday’s Second Law of Electrolysis

When the same quantity of electricity is passed through different electrolytes, the number of moles of elements deposited is inversely proportional to the charges on their ions.

Mathematical expression:

\( \frac{m_1}{\text{RAM}_1 \times Z_1} = \frac{m_2}{\text{RAM}_2 \times Z_2} \)

Where: