Redox Reactions and Electrochemistry

Master oxidation-reduction, electrochemical cells, Nernst equation, and electrolysis for JEE Chemistry.

Electrochemistry deals with the relationship between electrical energy and chemical reactions.

Overview

graph TD
    A[Electrochemistry] --> B[Redox Reactions]
    A --> C[Electrochemical Cells]
    A --> D[Electrolysis]
    C --> C1[Galvanic Cell]
    C --> C2[Electrode Potential]
    C --> C3[Nernst Equation]

Redox Reactions

Oxidation and Reduction

ProcessDefinitionChange
OxidationLoss of electronsIncrease in O.S.
ReductionGain of electronsDecrease in O.S.

Oxidation Number Rules

  1. Free element: 0
  2. Monoatomic ion: equals charge
  3. Oxygen: -2 (except peroxides: -1, OF₂: +2)
  4. Hydrogen: +1 (except hydrides: -1)
  5. Sum = charge on species

Balancing Redox Equations

Ion-Electron Method:

  1. Write half-reactions
  2. Balance atoms (except O, H)
  3. Balance O using H₂O
  4. Balance H using H⁺ (or OH⁻ in basic)
  5. Balance charge using electrons
  6. Multiply and add half-reactions

Electrochemical Cells

Galvanic (Voltaic) Cell

Converts chemical energy to electrical energy.

Components:

  • Two half-cells
  • Salt bridge
  • External circuit

Cell Notation:

$$\text{Anode} | \text{Anode ion} || \text{Cathode ion} | \text{Cathode}$$

Example: Zn | Zn²⁺ || Cu²⁺ | Cu

Electrode Potential

$$E_{cell} = E_{cathode} - E_{anode}$$

Standard Electrode Potential (E°):

  • At 298 K, 1 M concentration, 1 atm pressure
  • SHE is reference (E° = 0)

Electrochemical Series

Arranged in increasing order of reduction potential.

  • Higher E° → Better oxidizing agent
  • Lower E° → Better reducing agent

Nernst Equation

$$\boxed{E = E° - \frac{RT}{nF}\ln Q = E° - \frac{0.059}{n}\log Q}$$

At equilibrium: E = 0 and Q = K

$$E° = \frac{0.059}{n}\log K$$

Relationship with Gibbs Energy

$$\Delta G° = -nFE°$$

For spontaneous reaction: E° > 0, ΔG° < 0

JEE Tip
At 298 K, use the simplified Nernst equation with 0.059/n factor.

Conductance

Resistance and Conductance

$$R = \rho \frac{l}{A}$$ $$G = \frac{1}{R} = \kappa \frac{A}{l}$$

where κ = conductivity (S/m)

Molar Conductivity

$$\Lambda_m = \frac{\kappa \times 1000}{M}$$

where M = molarity

Kohlrausch’s Law

$$\Lambda_m^° = \lambda^°_+ + \lambda^°_-$$

Molar conductivity at infinite dilution = sum of ionic conductivities.

Applications:

  • Calculate Λ°m for weak electrolytes
  • Find degree of dissociation
  • Determine solubility of sparingly soluble salts

Electrolysis

Faraday’s Laws

First Law: Mass deposited ∝ charge passed

$$m = ZQ = Zit$$

Second Law:

$$\frac{m_1}{m_2} = \frac{E_1}{E_2}$$

where E = equivalent weight = M/n

Electrochemical Equivalent

$$Z = \frac{E}{96500} = \frac{M}{nF}$$

1 Faraday = 96500 C = charge of 1 mol electrons

Batteries

Primary Cells

  • Non-rechargeable
  • Example: Dry cell, Mercury cell

Secondary Cells

  • Rechargeable
  • Example: Lead-acid battery, Li-ion

Fuel Cells

  • Continuous supply of reactants
  • Example: H₂-O₂ fuel cell

Practice Problems

  1. Calculate EMF of the cell: Zn | Zn²⁺(0.1M) || Cu²⁺(0.01M) | Cu Given: E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V

  2. How many grams of copper will be deposited by passing 2 A current for 1 hour?

  3. Find the equilibrium constant for the reaction: 2Ag⁺ + Zn → Zn²⁺ + 2Ag

Quick Check
Why is the salt bridge necessary in a galvanic cell?

Further Reading