Electrochemistry Class 12 Notes — CBSE Chemistry Chapter 3

Chapter 3 of Class 12 Chemistry — Electrochemistry — connects chemistry with electricity. This chapter explains how chemical reactions can produce electric current (galvanic cells) and how electricity can drive non-spontaneous reactions (electrolytic cells). With 7-8 marks in Board exams, this is one of the highest-weightage chapters in Chemistry. Master the Nernst equation numericals and you’ll ace this!

Key Concepts

Electrochemical Cells

FeatureGalvanic (Voltaic) CellElectrolytic Cell
Energy conversionChemical → ElectricalElectrical → Chemical
ΔGNegative (spontaneous)Positive (non-spontaneous)
E_cellPositiveNegative (external EMF applied)
AnodeNegative terminalPositive terminal
CathodePositive terminalNegative terminal
ExampleDaniel cellElectrolysis of NaCl
Memory Aid: In BOTH cells: Oxidation at Anode, Reduction at Cathode (think “An Ox” and “Red Cat”). The sign of the electrode just flips!

Daniel Cell — The Classic Galvanic Cell

Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu

  • Anode (oxidation): Zn → Zn²⁺ + 2e⁻
  • Cathode (reduction): Cu²⁺ + 2e⁻ → Cu
  • Salt bridge: Maintains electrical neutrality (usually KCl in agar-agar)
  • E⁰_cell = E⁰_cathode − E⁰_anode = +0.34 − (−0.76) = +1.10 V

Standard Electrode Potential (E⁰)

Measured against the Standard Hydrogen Electrode (SHE) which is assigned E⁰ = 0.00 V.

E⁰_cell = E⁰_cathode − E⁰_anode
If E⁰_cell > 0 → reaction is spontaneous
If E⁰_cell < 0 → reaction is non-spontaneous

Electrochemical Series (selected values):

ElectrodeE⁰ (V)Tendency
Li⁺/Li−3.05Strongest reducing agent
K⁺/K−2.93Strong reducing agent
Zn²⁺/Zn−0.76Good reducing agent
H⁺/H₂ (SHE)0.00Reference
Cu²⁺/Cu+0.34Weak oxidising agent
Ag⁺/Ag+0.80Good oxidising agent
F₂/F⁻+2.87Strongest oxidising agent

Nernst Equation

E_cell = E⁰_cell − (RT/nF) ln Q

At 298 K: E_cell = E⁰_cell − (0.0591/n) log Q

Where: n = number of electrons transferred, Q = reaction quotient
At equilibrium: E_cell = 0, so E⁰_cell = (0.0591/n) log K_c

Relationship Between ΔG and E_cell

ΔG⁰ = −nFE⁰_cell
Where: n = moles of electrons, F = 96485 C/mol (Faraday constant)

If E⁰_cell > 0 → ΔG⁰ < 0 → spontaneous
If E⁰_cell < 0 → ΔG⁰ > 0 → non-spontaneous

Conductance of Electrolytic Solutions

Conductance (G) = 1/R (unit: S or Ω⁻¹ or mho)
Specific conductance (κ) = 1/ρ = G × (l/A) — unit: S/cm
Molar conductivity (Λ_m) = κ × 1000/c — unit: S·cm²/mol
Where c = concentration in mol/L

Variation of Conductivity with Concentration

  • κ (specific conductance) decreases with dilution — fewer ions per unit volume
  • Λ_m (molar conductivity) increases with dilution — more dissociation, less inter-ionic attraction

Kohlrausch’s Law of Independent Migration

Λ⁰_m = ν₊λ⁰₊ + ν₋λ⁰₋
Where ν = number of ions, λ⁰ = limiting molar conductivity of each ion

Example: Λ⁰_m(NaCl) = λ⁰(Na⁺) + λ⁰(Cl⁻)

Applications:

  • Calculate Λ⁰_m of weak electrolytes (like CH₃COOH) which can’t be measured directly
  • Λ⁰_m(CH₃COOH) = Λ⁰_m(CH₃COONa) + Λ⁰_m(HCl) − Λ⁰_m(NaCl)
  • Determine degree of dissociation: α = Λ_m / Λ⁰_m

Electrolysis — Faraday’s Laws

First Law: m = ZIt = (M × I × t) / (n × F)
Where: m = mass deposited, Z = electrochemical equivalent, I = current, t = time

Second Law: m₁/m₂ = E₁/E₂ (equivalent weights)
When same charge passes through different electrolytes

Batteries

BatteryTypeAnodeCathodeE_cell
Dry Cell (Leclanché)PrimaryZnMnO₂ + C1.5 V
Mercury CellPrimaryZn-HgHgO1.35 V
Lead-acid (car)SecondaryPbPbO₂2 V/cell
Ni-CdSecondaryCdNiO(OH)1.2 V
H₂-O₂ Fuel CellFuelH₂O₂1.23 V

Corrosion

Corrosion is an electrochemical process. Rusting of iron:

  • Anode: Fe → Fe²⁺ + 2e⁻ (iron dissolves)
  • Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻
  • Fe²⁺ + 2OH⁻ → Fe(OH)₂ → further oxidises to Fe₂O₃·xH₂O (rust)

Prevention: Galvanisation (Zn coating), electroplating, cathodic protection, painting, alloying

Important Definitions

TermDefinition
Galvanic CellDevice that converts chemical energy to electrical energy via spontaneous redox reaction
Electrolytic CellDevice that uses electrical energy to drive non-spontaneous chemical reactions
Standard Electrode PotentialPotential of an electrode measured against SHE under standard conditions (1M, 1 atm, 25°C)
Salt BridgeU-tube with electrolyte that maintains electrical neutrality in a galvanic cell
Molar ConductivityConductance of a solution containing 1 mole of electrolyte between electrodes 1 cm apart
Faraday ConstantCharge on 1 mole of electrons = 96485 C/mol
CorrosionElectrochemical destruction of metals by reaction with the environment

Solved Examples — NCERT Based

Example 1: EMF of Cell using Nernst Equation

Q: Calculate the EMF of the cell: Zn | Zn²⁺(0.001 M) || Cu²⁺(0.1 M) | Cu. Given E⁰(Zn²⁺/Zn) = −0.76 V, E⁰(Cu²⁺/Cu) = +0.34 V

Solution:

E⁰_cell = 0.34 − (−0.76) = 1.10 V, n = 2

Q = [Zn²⁺]/[Cu²⁺] = 0.001/0.1 = 0.01

E_cell = 1.10 − (0.0591/2) × log(0.01)

E_cell = 1.10 − 0.02955 × (−2) = 1.10 + 0.0591 = 1.159 V

Example 2: ΔG from E⁰_cell

Q: Calculate ΔG⁰ for the reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. Given E⁰(Fe³⁺/Fe²⁺) = +0.77 V, E⁰(I₂/I⁻) = +0.54 V

Solution:

E⁰_cell = E⁰_cathode − E⁰_anode = 0.77 − 0.54 = 0.23 V

n = 2 (2 electrons transferred)

ΔG⁰ = −nFE⁰ = −2 × 96485 × 0.23 = −44,383 J = −44.4 kJ

Example 3: Faraday’s Law — Mass Deposited

Q: How much copper is deposited on the cathode if a current of 0.5 A is passed through CuSO₄ solution for 2 hours? (M of Cu = 63.5, n = 2)

Solution:

t = 2 × 3600 = 7200 s

m = (M × I × t) / (n × F) = (63.5 × 0.5 × 7200) / (2 × 96485)

m = 228600 / 192970 = 1.185 g

Example 4: Kohlrausch’s Law

Q: Calculate Λ⁰_m of CH₃COOH using: Λ⁰_m(CH₃COONa) = 91.0, Λ⁰_m(HCl) = 426.2, Λ⁰_m(NaCl) = 126.5 (all in S·cm²/mol)

Solution:

Λ⁰_m(CH₃COOH) = Λ⁰_m(CH₃COONa) + Λ⁰_m(HCl) − Λ⁰_m(NaCl)

= 91.0 + 426.2 − 126.5 = 390.7 S·cm²/mol

Important Questions for Board Exams

1 Mark Questions

  1. What is the SI unit of molar conductivity?
  2. How does molar conductivity change with dilution?
  3. What is the role of salt bridge in a galvanic cell?
  4. State Faraday’s first law of electrolysis.
  5. Why is E⁰ of SHE taken as zero?

2 Mark Questions

  1. Differentiate between galvanic and electrolytic cells.
  2. Write the Nernst equation and define each term.
  3. Explain why Λ_m of a weak electrolyte increases sharply at very low concentrations.
  4. State Kohlrausch’s law. Give one application.

3 Mark Questions

  1. Describe the construction and working of the Daniel cell with a diagram.
  2. How is ΔG related to EMF? Calculate ΔG⁰ for a cell with E⁰ = 1.10 V and n = 2.
  3. Explain corrosion of iron as an electrochemical process. How can it be prevented?
  4. Calculate the EMF of a cell at 298 K using the Nernst equation given specific conditions.

5 Mark Questions

  1. What is a fuel cell? Explain the construction and working of the hydrogen-oxygen fuel cell. What are its advantages?
  2. State and explain Faraday’s laws of electrolysis. Calculate the mass of aluminium deposited when 5 A current is passed through Al₂O₃ for 1 hour.

Quick Revision Points

  • Anode = Oxidation, Cathode = Reduction (in ALL cells)
  • E⁰_cell = E⁰_cathode − E⁰_anode; positive → spontaneous
  • Nernst equation at 298K: E = E⁰ − (0.0591/n) log Q
  • At equilibrium: E = 0, E⁰ = (0.0591/n) log K
  • ΔG⁰ = −nFE⁰ (link between thermodynamics and electrochemistry)
  • κ decreases with dilution; Λ_m increases with dilution
  • Kohlrausch: Λ⁰_m = ν₊λ⁰₊ + ν₋λ⁰₋
  • Faraday: m = MIt/nF
  • Lead-acid: rechargeable, 6 cells × 2V = 12V car battery
  • Fuel cells: high efficiency (~70%), clean energy, produce only water
  • Corrosion = electrochemical; prevent by galvanisation, cathodic protection

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