JEE PREP | APRIL 2026
Last Updated: April 2026
Electrochemistry is one of the most formula-intensive and scoring chapters in JEE Main Chemistry. Expect 4–6 questions in JEE Main 2027. This chapter tests your ability to apply the Nernst equation, calculate EMF, use Faraday’s laws, and understand conductance — all in numerical calculation format. This guide covers every concept with solved problems and a complete formula reference.
Class 12 NCERT Chemistry Chapter 3. JEE Main 2022–2026 average: 4–6 questions. Medium to high difficulty. Numerical-heavy chapter. Key topics: Nernst equation calculations, Faraday’s laws (mass deposited), conductance (Kohlrausch’s law), EMF = E°(cathode) – E°(anode), Gibbs energy relation. Most questions are numerical/application type.
• EMF = E°(cathode) – E°(anode)
• Nernst equation: E = E° – (0.059/n) × log Q at 25°C
• Faraday’s law: mass = (M × I × t) / (n × F), where F = 96500 C/mol
• Kohlrausch’s law: Λ°m(electrolyte) = Σ(λ° of cations) + Σ(λ° of anions)
• ΔG° = -nFE°
• 1 Faraday = 96500 coulombs = charge of 1 mole of electrons
Electrochemical Cells — Galvanic/Voltaic Cells
An electrochemical cell converts chemical energy into electrical energy through spontaneous redox reactions. The Daniell cell (Zn-Cu cell) is the classic example:
- Anode (negative electrode): Oxidation occurs — Zn → Zn²⁺ + 2e⁻
- Cathode (positive electrode): Reduction occurs — Cu²⁺ + 2e⁻ → Cu
- Salt bridge: Maintains electrical neutrality; prevents buildup of charge in half-cells
- External circuit: Electrons flow from anode to cathode
Cell Notation (Cell Representation)
Standard cell notation: Anode | Anode solution || Cathode solution | Cathode
Example: Zn(s) | Zn²⁺(aq, 1M) || Cu²⁺(aq, 1M) | Cu(s)
- Single vertical line (|) = phase boundary (interface between electrode and solution)
- Double vertical line (||) = salt bridge
- Left side = anode (oxidation); Right side = cathode (reduction)
EMF Calculation
E°cell = E°cathode – E°anode
= E°(reduction, cathode) – E°(reduction, anode)
For Zn-Cu cell: E° = E°(Cu²⁺/Cu) – E°(Zn²⁺/Zn) = 0.34 – (-0.76) = +1.10 V
Positive EMF = spontaneous reaction
Standard Electrode Potential — Key Values
| Half-reaction (Reduction) | E° (V) |
|---|---|
| Li⁺ + e⁻ → Li | -3.04 |
| Zn²⁺ + 2e⁻ → Zn | -0.76 |
| Fe²⁺ + 2e⁻ → Fe | -0.44 |
| H⁺ + e⁻ → ½H₂ | 0.00 (reference) |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Au³⁺ + 3e⁻ → Au | +1.50 |
Nernst Equation
The Nernst equation relates EMF to concentration (reaction quotient Q):
E = E° – (RT/nF) × ln Q
At 25°C (298 K): E = E° – (0.0592/n) × log Q
Where: n = number of electrons transferred; Q = reaction quotient; F = 96500 C/mol
Solved Example: For Zn-Cu cell, [Zn²⁺] = 0.1 M, [Cu²⁺] = 0.01 M, E° = 1.10 V, n = 2
Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10; log 10 = 1
E = 1.10 – (0.0592/2) × 1 = 1.10 – 0.0296 = 1.0704 ≈ 1.069 V
Electrochemical Series
The electrochemical series arranges elements in increasing order of their standard reduction potentials. Key applications:
- Predicting reaction feasibility: If E°cell > 0, reaction is spontaneous
- Predicting displacement: A metal higher in activity series (more negative E°) displaces a metal lower in the series from its salt solution
- Hydrogen liberation: Metals with E° < 0 (negative) can displace H₂ from dilute acids (e.g., Zn, Fe, but not Cu, Ag)
- Corrosion: More active metals corrode faster
Gibbs Energy and EMF
ΔG° = -nFE°cell
ΔG = -nFEcell
If E°cell > 0, then ΔG° < 0 → spontaneous reaction
Equilibrium constant: ΔG° = -RT ln K = -nFE°
Therefore: log K = (n × E°) / 0.0592 at 25°C
Example: E° = 1.10 V, n = 2, F = 96500 C/mol
ΔG° = -2 × 96500 × 1.10 = -212300 J = -212.3 kJ
Conductance
Conductance (G) is the reciprocal of resistance: G = 1/R. SI unit: Siemens (S) or ohm⁻¹ (mho).
Types of Conductance
- Specific conductance (κ): Conductance of a solution with unit cross-section and unit length. Unit: S·m⁻¹ or S·cm⁻¹
- Molar conductance (Λm): Conductance of solution containing 1 mole of electrolyte between electrodes of large area separated by unit distance. Λm = (κ × 1000)/M (in S·cm²/mol)
- Equivalent conductance: Conductance of solution containing 1 gram equivalent of electrolyte
Kohlrausch’s Law
At infinite dilution, the molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its individual ions:
Λ°m = v₊λ°₊ + v₋λ°₋
Where v₊ and v₋ are stoichiometric coefficients
Application: Calculate Λ°m of weak electrolytes (CH₃COOH) from strong electrolytes
Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) – Λ°m(NaCl)
Electrolysis
Electrolysis uses electrical energy to drive non-spontaneous chemical reactions. Unlike galvanic cells, electrical energy is converted to chemical energy.
Faraday’s Laws of Electrolysis
m = (M × Q) / (n × F) = (M × I × t) / (n × F)
Where: M = molar mass, I = current (A), t = time (s), n = valency/no. of electrons, F = 96500 C/mol
Second Law: When same charge is passed through different electrolytic cells in series, the masses of substances deposited are proportional to their equivalent weights (M/n).
Example: Mass of Ag deposited when 2 F of charge passes through AgNO₃:
m = (108 × 2 × 96500) / (1 × 96500) = 108 × 2 = 216 g
Products of Electrolysis — Common Cases
| Electrolyte | Cathode Product | Anode Product |
|---|---|---|
| Dilute H₂SO₄ | H₂ gas | O₂ gas |
| CuSO₄ solution | Cu metal deposited | O₂ gas (inert anode); Cu dissolves (Cu anode) |
| AgNO₃ solution | Ag metal deposited | O₂ gas |
| Molten NaCl | Na metal | Cl₂ gas |
Battery Types
| Battery | Electrolyte | Voltage | Key Feature |
|---|---|---|---|
| Leclanche (dry cell) | NH₄Cl paste + MnO₂ | 1.5 V | Non-rechargeable; Zn anode, C cathode |
| Lead storage battery | H₂SO₄ (38%) | 2 V/cell (12V total) | Rechargeable; used in cars; Pb anode, PbO₂ cathode |
| Nickel-Cadmium (NiCd) | KOH solution | 1.2 V | Rechargeable; Cd anode, NiO(OH) cathode |
| Hydrogen fuel cell | KOH solution | ~1.23 V | H₂ + O₂ → H₂O; no pollution; used in space crafts |
Important Formulas — Quick Reference
• E°cell = E°cathode – E°anode
• E = E° – (0.0592/n) × log Q [at 25°C]
• ΔG° = -nFE° [F = 96500 C/mol]
• log K = nE°/0.0592 [at 25°C]
• m = MIt/(nF) [Faraday’s 1st law]
• Λm = (κ × 1000)/M [molar conductance]
• Λ°m = v₊λ°₊ + v₋λ°₋ [Kohlrausch’s law]
• κ = G × (l/A) = G × cell constant
10 Solved JEE Problems
Q1. For the cell Ag|Ag⁺(0.001M)||Cu²⁺(0.01M)|Cu, E°(Ag⁺/Ag)=0.80V, E°(Cu²⁺/Cu)=0.34V. Calculate E°cell and predict if forward cell reaction is spontaneous.
Sol: E°cell = 0.34 – 0.80 = -0.46 V. Since E°cell < 0, forward reaction is NOT spontaneous. The reverse reaction (Cu dissolving, Ag depositing) is spontaneous.
Q2. How long (in seconds) must a current of 2A pass through CuSO₄ solution to deposit 0.635 g of Cu? (M=63.5, n=2, F=96500)
Sol: t = mNF/(MI) = (0.635 × 2 × 96500)/(63.5 × 2) = 122555/127 ≈ 965 s
Q3. The molar conductivity of KCl at infinite dilution is 149.9 S·cm²/mol, λ°(K⁺) = 73.5 S·cm²/mol. Find λ°(Cl⁻).
Sol: λ°(Cl⁻) = Λ°m – λ°(K⁺) = 149.9 – 73.5 = 76.4 S·cm²/mol
Q4. Calculate ΔG° for the Zn-Cu cell (E° = 1.10 V, n = 2, F = 96500).
Sol: ΔG° = -nFE° = -2 × 96500 × 1.10 = -212,300 J = -212.3 kJ
Q5. Using Nernst equation, find EMF of Zn-Cu cell when [Zn²⁺]=0.001M, [Cu²⁺]=1M.
Sol: Q = 0.001/1 = 10⁻³; log Q = -3; E = 1.10 – (0.0592/2)×(-3) = 1.10 + 0.0888 = 1.189 V
Practice MCQs — Electrochemistry
Test your JEE preparation with these 10 MCQs on Electrochemistry concepts and calculations:
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Frequently Asked Questions
What is the Nernst equation and how is it used in JEE?
The Nernst equation relates the EMF of a cell to concentration: E = E° – (0.0592/n) × log Q at 25°C, where n is the number of electrons transferred and Q is the reaction quotient. In JEE, you’re typically asked to calculate the EMF when concentrations are not standard (1M). Steps: (1) Identify the cell reaction and count electrons (n); (2) Write Q = [products]/[reactants]; (3) Substitute into the equation. At equilibrium, E = 0, so E° = (0.0592/n) × log K, which lets you calculate the equilibrium constant K.
How to solve Faraday’s law problems in JEE?
Use the formula: m = MIt/(nF), where m = mass deposited (g), M = molar mass, I = current (amperes), t = time (seconds), n = number of electrons (valency), F = 96500 C/mol. Key steps: (1) Identify the ion being deposited and its charge (n); (2) Calculate total charge Q = I × t; (3) Calculate moles of electrons = Q/F; (4) Calculate moles of metal = (Q/F)/n; (5) Mass = moles × M. For problems with cells in series, the same charge passes through all cells, so compare using equivalent weights (M/n).
What is the relationship between EMF and Gibbs energy?
The relationship is: ΔG° = -nFE°cell, where n is moles of electrons, F is Faraday’s constant (96500 C/mol), and E° is standard cell EMF. If E°cell is positive, ΔG° is negative (spontaneous reaction). At equilibrium, ΔG = 0 and E = 0. The three quantities are also related to equilibrium constant K by: ΔG° = -RT ln K = -nFE°. At 25°C: E° = (0.0592/n) × log K. This three-way relationship between E°, ΔG°, and K is a very common JEE question type.
What is Kohlrausch’s law and how is it applied?
Kohlrausch’s law states that at infinite dilution, the limiting molar conductivity of an electrolyte equals the sum of limiting molar conductivities of its constituent ions: Λ°m = v₊λ°₊ + v₋λ°₋. Key application in JEE: calculating Λ°m of weak electrolytes (like CH₃COOH, NH₄OH) which cannot be measured directly by extrapolation because their conductance curve doesn’t plateau linearly. For example: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) – Λ°m(NaCl). This is a standard numerical type in JEE Main.