Faraday's Laws of Electrolysis and Applications

Faraday’s Laws of Electrolysis

Overview

  • Faraday’s laws govern the relationship between electricity and chemical changes during electrolysis.

Faraday’s First Law of Electrolysis

  • Statement: When an electric current passes through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed.
  • Relationship:
    • If W is the mass of the substance deposited and Q is the charge:
      W \propto Q
  • Charge Definition:
  • 1 coulomb is the charge when 1 ampere of current is passed for 1 second:
    Q = I \times t
  • Revising the Mass Equation:
    W \propto I \times t
    W = z \times I \times t
    Where z is the electrochemical equivalent (depends on the substance).

Faraday’s Second Law of Electrolysis

  • Statement: When the same quantity of charge passes through different electrolytes, the masses deposited are in the ratio of their equivalent masses.
  • Relationship:
    W = Z \times Q
  • For Q = 96500 coulombs,
    E = Z \times 96500
    Z = \frac{E}{96500}
  • Ratio of electrochemical equivalents:
    \frac{z1}{z2} = \frac{E1}{E2}

Fundamental Unit of Charge

  • A g-equivalent of an ion is liberated by 96500 coulombs.
  • Charge Carried by 1 g-equivalent:
    • If valency of an ion is n:
      \text{Charge} = \frac{nF}{6.02 \times 10^{23}}
  • For monovalent ions (n = 1):
    \text{Fundamental unit of charge} = \frac{F}{6.02 \times 10^{23}} = \frac{96500}{6.02 \times 10^{23}} = 1.6 \times 10^{-19} \text{ coulombs}
  • Therefore, 1 coulomb is equivalent to 6.24 \times 10^{18} electrons.

Charge Calculations for Ions

  • Charge on 1 g-ion of N3-:
    \text{Charge} = 3 \times 1.6 \times 10^{-19} \text{ coulombs}
    \text{Charge on one g-ion} = 3 \times 1.6 \times 10^{-19} \times 6.02 \times 10^{23} = 2.89 \times 10^{5} \text{ coulombs}

Charge Requirement for Reduction/Oxidation

  • To reduce 1 mole of Al3+ to Al:
    \text{Reaction: } \text{Al}^{3+} + 3e^{-} \rightarrow \text{Al}
    Charge required:
    Q = 3F = 3 \times 96500 = 289500 \text{ coulombs}
  • To reduce 1 mole of Mn4- to Mn2+:
    \text{Reaction: } \text{Mn}4 + 8H^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4H2O
    Charge required:
    Q = 5F = 5 \times 96500 = 485000 \text{ coulombs}

Applications of Faraday’s Laws

  1. Determination of Equivalent Masses:
    • By comparing the masses of metals deposited in different electrolytes.
    • \frac{WA}{WB} = \frac{EA}{EB}
  2. Electron Metallurgy:
    • Alkaline and alkaline earth metals are obtained via electrolysis from fused salts.
  3. Manufacture of Non-Metals:
    • Hydrogen, chlorine, etc., via electrolysis.
  4. Electro-Refining of Metals:
    • Metals like copper, gold refined through electrolysis.
  5. Manufacture of Compounds:
    • Example: NaOH, KOH through electrolysis.
  6. Electroplating:
    • Coating inferior metal with a superior layer for protection and aesthetics.

Electroplating Requirements

  • Clean surface and roughness to ensure proper adhesion.
  • Controlled concentration of electrolytes and consistent current density.

Calculating Coating Thickness

  • Given dimensions: Area = a \times b and thickness = c:
    \text{Volume} = a \times b \times c
    \text{Mass} = (a \times b \times c) \times d
    \text{Using: } (a \times b \times c) \times d = \frac{I \times t \times E}{96500}