Galvanic Cells & Standard Electrode Potentials: JEE

Electrochemistry contributes three to five questions per JEE Main session, and galvanic cells with electrode potentials form the conceptual core. Students who cannot read a cell notation, calculate EMF from standard potentials, or apply the Nernst equation with confidence drop marks in every electrochemistry question. This guide builds these skills systematically.

Cell Notation and the Daniel Cell

Standard cell notation: anode (oxidation, left) | anode electrolyte || cathode electrolyte | cathode (reduction, right). Vertical bars = phase boundary; double vertical bars = salt bridge. The Daniel cell: Zn | ZnSO₄ || CuSO₄ | Cu. At the anode (left): Zn → Zn²⁺ + 2e⁻ (oxidation, zinc dissolves). At the cathode (right): Cu²⁺ + 2e⁻ → Cu (reduction, copper deposits). Net reaction: Zn + Cu²⁺ → Zn²⁺ + Cu. E°_cell = E°_cathode − E°_anode = (+0.34 V) − (−0.76 V) = +1.10 V. Positive E°_cell means the reaction is spontaneous (ΔG° = −nFE°_cell < 0).

Standard electrode potential E° (reduction potential): measured relative to the Standard Hydrogen Electrode (SHE, E° = 0 V by definition). Species with E° > 0 (F₂: +2.87V, Au³⁺: +1.50V, Ag⁺: +0.80V, Cu²⁺: +0.34V) are stronger oxidising agents — they accept electrons readily. Species with E° < 0 (Zn²⁺: −0.76V, Fe²⁺: −0.44V, Pb²⁺: −0.13V) are stronger reducing agents — their metals readily give up electrons. JEE uses the electrochemical series to predict: (1) which metal displaces which from solution (higher in activity series displaces lower); (2) which half-reaction acts as anode and which as cathode in a given cell. Take a free electrochemistry mock. See our electrochemistry guide.

Nernst Equation: EMF at Non-Standard Conditions

E_cell = E°_cell − (RT/nF) × ln Q = E°_cell − (0.0592/n) × log₁₀ Q (at 298 K). Where Q = reaction quotient = [products]^(stoich)/ [reactants]^(stoich). n = electrons transferred in the balanced cell reaction. F = 96500 C/mol. Practical form: E_cell = E°_cell − (0.0592/n) log Q. At equilibrium, E_cell = 0, Q = K: E°_cell = (0.0592/n) log K. This gives the relationship between standard EMF and equilibrium constant — a common JEE question type (find K given E°_cell, or find E°_cell given K).

JEE concentration-cell question: a concentration cell has the same electrode material in different concentrations. E°_cell = 0 (same electrode). E_cell = −(0.0592/n) log ([dilute]/[concentrated]). The cell operates until concentrations equalise (E_cell = 0). For the concentration cell Zn | Zn²⁺(0.1 M) || Zn²⁺(1 M) | Zn: E_cell = −(0.0592/2) log(0.1/1) = +(0.0296)(1) = +0.0296 V. Current flows from dilute to concentrated solution side (through external circuit).

Gibbs Energy and Equilibrium Constant from EMF

ΔG° = −nFE°_cell. At 298 K: ΔG° = −n × 96500 × E°_cell (in joules). Relationship with K: ΔG° = −RT lnK = −2.303 RT log K. Therefore: E°_cell = (RT/nF) ln K = (0.0592/n) log K at 298 K. Example: if E°_cell = 0.59 V and n = 2, then log K = (2 × 0.59)/0.0592 = 19.93 → K ≈ 10²⁰. Very large K means the reaction goes essentially to completion under standard conditions. JEE frequently asks: given E°_cell, find ΔG° and K; or given K, find E°_cell.

Electrode Potential and Spontaneity Tests

For any galvanic cell: E°_cell > 0 → ΔG° < 0 → spontaneous. E°_cell < 0 → ΔG° > 0 → non-spontaneous (electrolysis required). To calculate whether Fe can displace Pb from PbCl₂ solution: E°(Pb²⁺/Pb) = −0.13 V (cathode), E°(Fe²⁺/Fe) = −0.44 V (anode). E°_cell = −0.13 − (−0.44) = +0.31 V (positive, spontaneous). Yes, Fe can displace Pb. This "can X displace Y" question type appears directly in JEE. The rule: a metal with more negative E° (higher in the activity series) displaces one with less negative E° from its salt solution. For Faraday's laws see our electrolysis guide and for the Nernst equation in depth see our Nernst equation guide.