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:
  $\frac{z<em>1}{z</em>2} = \frac{E<em>1}{E</em>2}$

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+:
  $\text{Reaction: } \text{Mn}<em>4 + 8H^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4H</em>2O$
  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.
   • $\frac{W<em>A}{W</em>B} = \frac{E<em>A}{E</em>B}$
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 = $c$:
  $\text{Volume} = a \times b \times c$
  $\text{Mass} = (a \times b \times c) \times d$
  $\text{Using: } (a \times b \times c) \times d = \frac{I \times t \times E}{96500}$