Last Updated: April 2026
Electrochemistry is a high-yield chapter in JEE Main 2027 Chemistry, consistently contributing 2-4 questions per year. The chapter (NCERT Class 12 Chemistry Chapter 3) covers electrochemical cells, electrode potentials, Nernst equation, electrolysis (Faraday’s laws), conductance, and practical applications like batteries and corrosion. This guide covers all concepts with solved problems for JEE Main 2027.
JEE Main Electrochemistry — Weightage
| Topic | JEE Questions/Year | Type |
|---|---|---|
| EMF of cell, Standard electrode potential | 1-2 | Calculation |
| Nernst Equation | 1 | Calculation |
| Faraday’s Laws of Electrolysis | 1 | Calculation |
| Conductance (molar, specific, equivalent) | 0-1 | Concept/Calculation |
| Batteries and Corrosion | 0-1 | Concept |
1. Electrochemical Cells
Types of Cells
| Feature | Galvanic (Voltaic) Cell | Electrolytic Cell |
|---|---|---|
| Energy conversion | Chemical → Electrical | Electrical → Chemical |
| Spontaneity | Spontaneous (ΔG < 0) | Non-spontaneous (ΔG > 0) |
| Anode | Negative (oxidation) | Positive (oxidation) |
| Cathode | Positive (reduction) | Negative (reduction) |
| Example | Daniel cell, batteries | Electroplating, refining |
Daniel Cell
- Zn(s) | ZnSO₄(aq) || CuSO₄(aq) | Cu(s)
- Anode (negative): Zn → Zn²⁺ + 2e⁻ (oxidation)
- Cathode (positive): Cu²⁺ + 2e⁻ → Cu (reduction)
- Overall: Zn + Cu²⁺ → Zn²⁺ + Cu
- E°cell = E°cathode − E°anode = +0.34 − (−0.76) = +1.10 V
2. Standard Electrode Potential and EMF
Standard Hydrogen Electrode (SHE)
Reference electrode: 2H⁺(aq, 1M) + 2e⁻ → H₂(g, 1 atm), E° = 0.00 V
All standard electrode potentials are measured against SHE at 25°C, 1 atm, 1M concentration.
EMF Calculation
E°cell = E°reduction (cathode) − E°reduction (anode)
OR: E°cell = E°cathode − E°anode (using reduction potentials for both)
- If E°cell > 0 → spontaneous cell reaction (ΔG < 0)
- If E°cell < 0 → non-spontaneous (requires external energy)
Relationship between EMF and Gibbs Free Energy
ΔG° = −nFE°cell
- n = number of electrons transferred in balanced equation
- F = Faraday constant = 96,500 C/mol (or 96,485 C/mol)
- ΔG° = −RT ln K (equilibrium constant relationship)
- Therefore: E°cell = (RT/nF) ln K = (0.0591/n) log K (at 25°C)
3. Nernst Equation
EMF varies with concentration. At non-standard conditions:
E_cell = E°_cell − (0.0591/n) log Q (at 25°C)
- Q = reaction quotient
- n = number of electrons transferred
- At equilibrium: E_cell = 0, Q = K
Applications of Nernst Equation
- Calculation of EMF at non-standard concentrations
- Finding equilibrium constant: E°cell = (0.0591/n) log K
- pH measurement using hydrogen electrode (E = −0.0591 × pH)
- Concentration cells (both electrodes are identical but different concentrations)
4. Conductance
Terminology
| Term | Formula | Unit |
|---|---|---|
| Resistance (R) | R = ρl/A | Ohm (Ω) |
| Conductance (G) | G = 1/R | Siemens (S) |
| Resistivity (ρ) | ρ = RA/l | Ω·m |
| Conductivity (κ) | κ = 1/ρ = G × l/A | S/m or S/cm |
| Molar Conductivity (Λm) | Λm = κ × 1000/M (M = molarity) | S·cm²/mol |
Kohlrausch’s Law
At infinite dilution, molar conductivity of an electrolyte equals sum of contributions from individual ions:
Λ°m = ν₊λ°₊ + ν₋λ°₋
- Used to calculate Λ°m of weak electrolytes (which cannot be measured directly)
- Example: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl)
5. Faraday’s Laws of Electrolysis
First Law
Mass of substance deposited is directly proportional to charge passed: m = ZQ = ZIt
- Z = electrochemical equivalent (grams per coulomb)
- I = current (amperes), t = time (seconds)
Second Law
When the same charge is passed through different electrolytes, masses deposited are proportional to equivalent weights (M/n):
m₁/m₂ = (M₁/n₁)/(M₂/n₂)
Important: 1 Faraday (96,500 C) deposits 1 gram equivalent of substance at an electrode.
6. Solved Problems
Problem 1 (EMF Calculation)
E° for Zn²⁺/Zn = −0.76V and Cu²⁺/Cu = +0.34V. Find E°cell for the Daniel cell.
Solution: E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V
Problem 2 (Nernst Equation)
For the cell Zn|Zn²⁺(0.1M)||Cu²⁺(0.01M)|Cu, find E_cell. (E°cell = 1.10V, n = 2)
Solution: Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10
E_cell = 1.10 − (0.0591/2)log(10) = 1.10 − 0.02955 = 1.07 V
Problem 3 (Faraday’s Law)
A current of 2A flows for 30 minutes through CuSO₄ solution. Mass of Cu deposited? (M of Cu = 64, n = 2)
Solution: Charge = I × t = 2 × 30 × 60 = 3600 C
Equivalents deposited = 3600/96500 = 0.03731
Mass = equivalents × equivalent weight = 0.03731 × 32 = 1.19 g
Frequently Asked Questions (FAQ)
What is the most common mistake students make in Electrochemistry for JEE?
The most common mistake is confusing anode and cathode signs in galvanic vs electrolytic cells. In a galvanic cell, anode is NEGATIVE (electrons flow out, oxidation occurs). In an electrolytic cell, anode is POSITIVE (connected to + terminal of external battery). Another frequent error is using the wrong sign in the Nernst equation — remember it’s MINUS (0.0591/n) × log Q, not plus.
How many electrons are transferred in typical JEE electrochemistry problems?
n (number of electrons transferred) depends on the balanced half-reactions. For common systems: Cu²⁺/Cu → n=2; Ag⁺/Ag → n=1; Fe³⁺/Fe → n=3; Al³⁺/Al → n=3; Zn²⁺/Zn → n=2; H⁺/H₂ → n=2 (per mole H₂). Always balance the half-reactions first to determine n before using the Nernst equation or ΔG = -nFE calculations.
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