Electrochemistry

NCERT Class 12 Chemistry Chapter 2

Electrochemistry for Class 12 Chemistry, JEE and NEET: galvanic cells, electrode potentials, Nernst equation, conductance, electrolysis and batteries.

8 topics Updated 8 Nov 2026

For: Class 12, JEE and NEET, unless a topic says otherwise

In the syllabus: JEE Class 12 syllabus, NEET Class 12 syllabus

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NCERT Chemistry Part I (Class 12), Chapter 2 “Electrochemistry”, reprint 2026-27.

Open the official NCERT chapter (PDF, 4.4 MB, ncert.nic.in) (opens in a new tab)

Checked on 19 Oct 2026.

Learn in this order

Electrochemical cells
The Daniell cell, galvanic vs electrolytic cells, anode and cathode signs, a diagram and solved examples.

NCERT Section 2.1 12 min read 4 solved examples 9 practice questions
Galvanic cells
Half-cells, salt bridge, cell notation, emf, the standard hydrogen electrode and standard electrode potentials.

NCERT Section 2.2 23 min read 5 solved examples 10 practice questions
Nernst equation
The formula at 298 K, its derivation, K and ΔG from the emf, and solved numericals from NCERT.

NCERT Section 2.3 17 min read 6 solved examples 11 practice questions
Conductance of electrolytic solutions
Unit, cell constant, strong vs weak electrolytes, Kohlrausch’s law, examples.

NCERT Section 2.4 23 min read 7 solved examples 11 practice questions
Electrolytic cells and electrolysis
Electrolytic cells, Faraday’s laws, the faraday, mass deposited, and the products of electrolysis of brine.

NCERT Section 2.5 16 min read 6 solved examples 10 practice questions
Batteries
Primary vs secondary cells, dry and mercury cells, lead storage battery reactions and diagram, nickel-cadmium cell.

NCERT Section 2.6 15 min read 5 solved examples 10 practice questions
Fuel cells
The hydrogen–oxygen fuel cell, its reactions and diagram, efficiency against thermal plants, the hydrogen economy.

NCERT Section 2.7 13 min read 4 solved examples 9 practice questions
Corrosion
Rusting of iron as an electrochemical cell, its electrode reactions, and prevention by paint, galvanising or sacrificial metals.

Class 12 only NCERT Section 2.8 12 min read 4 solved examples 10 practice questions

Key formulas in this chapter

Electrochemical cells

Daniell cell: Zn + Cu^²⁺ → Zn^²⁺ + Cu, 1.1 V at 1 mol dm^⁻³
E_ext < 1.1 V: galvanic, electrons Zn to Cu
E_ext = 1.1 V: no current, no net reaction
E_ext > 1.1 V: electrolytic, the reaction runs in reverse
Galvanic: anode negative, cathode positive
Electrolytic: anode positive, cathode negative

Galvanic cells

E(cell) = E(right) − E(left) = E(cathode) − E(anode)
E^⊖(H⁺/H₂) = 0.00 V at all temperatures
Cell notation: anode | anode solution || cathode solution | cathode
Daniell cell: E^⊖(cell) = 0.34 − (−0.76) = 1.10 V
E^⊖ > 0: oxidised form reduced more easily than H⁺; E^⊖ < 0: reduced form oxidised by H⁺

Nernst equation

E = E^⊖ − (RT/nF) ln (1/[Mⁿ⁺])
E(cell) = E^⊖(cell) − (RT/nF) ln Q
At 298 K: E(cell) = E^⊖(cell) − (0.059/n) log Q
E^⊖(cell) = (2.303RT/nF) log K_c
Δ_rG^⊖ = −nFE^⊖(cell) = −RT ln K

Conductance of electrolytic solutions

R = ρl/A; G = 1/R; κ = 1/ρ
Cell constant = l/A = Rκ; κ = (cell constant)/R
Λ_ₘ = κ/c; Λ_ₘ = 1000κ/M (S cm^² mol^⁻¹)
Strong electrolytes: Λ_ₘ = Λ°_ₘ − A√c
Kohlrausch: Λ°_ₘ = ν_₊λ°_₊ + ν_₋λ°_₋
Weak electrolytes (dilute approximation): α ≈ Λ_ₘ/Λ°_ₘ; K_ₐ ≈ cα^²/(1 − α)

Electrolytic cells and electrolysis

Q = It
F = 96 487 C mol^⁻¹ ≈ 96 500 C mol^⁻¹
m = QM/(nF) = ItM/(nF)
Second law: w_₁/w_₂ = (M_₁/n_₁)/(M_₂/n_₂)
Brine: cathode H_₂O + e^⁻ → ½H_₂ + OH^⁻; anode Cl^⁻ → ½Cl_₂ + e^⁻

Batteries

Dry cell: Zn → Zn^²⁺ + 2e^⁻; MnO_₂ + NH_₄^⁺ + e^⁻ → MnO(OH) + NH_₃
Mercury cell: Zn(Hg) + HgO → ZnO + Hg(l), about 1.35 V
Lead storage (discharge): Pb + PbO_₂ + 2H_₂SO_₄ → 2PbSO_₄ + 2H_₂O
Nickel-cadmium (NCERT): Cd + 2Ni(OH)_₃ → CdO + 2Ni(OH)_₂ + H_₂O

Fuel cells

Anode: 2H_₂ + 4OH^⁻ → 4H_₂O + 4e^⁻
Cathode: O_₂ + 2H_₂O + 4e^⁻ → 4OH^⁻
Overall: 2H_₂ + O_₂ → 2H_₂O
E^⊖(cell) = 1.23 − 0.00 = 1.23 V
Ideal efficiency at 298 K = Δ_rG^⊖/Δ_rH^⊖ ≈ 83% (H₂O liquid, HHV)

Corrosion

Anode: 2Fe → 2Fe^²⁺ + 4e^⁻, E⊖(Fe^²⁺/Fe) = −0.44 V
Cathode: O_₂ + 4H^⁺ + 4e^⁻ → 2H_₂O, E⊖ = 1.23 V
Overall: 2Fe + O_₂ + 4H^⁺ → 2Fe^²⁺ + 2H_₂O, E⊖(cell) = 1.67 V
Rust: Fe_₂O_₃·xH_₂O

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Key formulas in this chapter

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