GATE Electrochemistry — Nernst Equation Applications
The Nernst equation is one line of algebra, and it is also the bridge between thermodynamics and every measurement an electrochemist makes. GATE Chemistry does not usually ask you to state it. It asks you to use it — to extract an equilibrium constant, a solubility product, a pH, or the EMF of a cell whose two halves differ only in concentration. This article covers the five applications that recur, each worked out completely.
The equation and what each symbol means
E is the cell potential at the actual composition, E° the standard cell potential, n the number of electrons transferred in the balanced cell reaction, F the Faraday constant (96 485 C mol⁻¹) and Q the reaction quotient written for the cell reaction as you have balanced it. The 0.0592 V comes from (RT/F) × ln 10 at 298.15 K: (8.314 × 298.15 / 96 485) × 2.3026 = 0.0592 V. Recompute it if a question specifies a different temperature — that substitution is itself a favourite GATE question.
Two companion relations complete the toolkit:
E°cell = E°cathode − E°anode (both as reduction potentials)
Application 1 — EMF at non-standard concentrations
Worked example 1. For the cell Zn | Zn²⁺ (0.10 M) ‖ Cu²⁺ (0.010 M) | Cu, with E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, find E at 298 K.
Cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2
E° = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V
Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 0.010 = 10 (solids do not appear in Q)
E = 1.10 − (0.0592/2) log 10 = 1.10 − (0.0296)(1) = +1.07 V
The potential dropped because the product ion is in excess relative to the reactant ion — exactly the direction Le Chatelier predicts.
Application 2 — Equilibrium constant from E°
Worked example 2. Find K for the Daniell cell reaction at 298 K.
log K = nE° / 0.0592 = (2 × 1.10) / 0.0592 = 2.20 / 0.0592 = 37.16
K = 1037.16 ≈ 1.4 × 10³⁷
Also ΔG° = −nFE° = −(2)(96 485)(1.10) = −212 267 J mol⁻¹ ≈ −212 kJ mol⁻¹. A K of 10³⁷ is the quantitative statement that this reaction goes essentially to completion.
Application 3 — Concentration cells
A concentration cell has identical electrodes in identical chemistry, differing only in ion concentration. Therefore E° = 0 and the entire potential comes from the log term. The more dilute half-cell is always the anode, because the spontaneous direction is the one that reduces the concentration difference.
Worked example 3. Ag | Ag⁺ (0.0010 M) ‖ Ag⁺ (0.10 M) | Ag at 298 K.
n = 1, E° = 0
E = (0.0592/1) log (0.10 / 0.0010) = 0.0592 × log(100) = 0.0592 × 2 = +0.118 V
A hundredfold concentration ratio buys you 118 mV for a one-electron couple. That 59.2 mV per decade per electron is worth memorising as a number, not a formula.
Application 4 — pH measurement
For the hydrogen electrode 2H⁺ + 2e⁻ → H₂ at unit H₂ pressure, E° = 0 by definition, and the Nernst equation collapses to a straight line in pH:
Worked example 4. A hydrogen electrode dipped in an unknown solution and measured against the standard hydrogen electrode gives E = −0.413 V. Find the pH.
pH = −E / 0.0592 = 0.413 / 0.0592 = 6.98
This is the principle behind every glass pH electrode: the instrument is a voltmeter that has been calibrated to display the log term instead of the millivolts.
Application 5 — Solubility product from standard potentials
This is the most satisfying use of the Nernst framework and a recurring GATE question, because it extracts a quantity that seems impossible to measure electrochemically.
Worked example 5. Given E°(AgCl/Ag, Cl⁻) = +0.222 V and E°(Ag⁺/Ag) = +0.799 V, find Ksp of AgCl at 298 K.
Construct the dissolution as a cell reaction:
Cathode (reduction): AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq) E° = +0.222 V
Anode (oxidation): Ag(s) → Ag⁺(aq) + e⁻ (reduction E° = +0.799 V)
Overall: AgCl(s) → Ag⁺(aq) + Cl⁻(aq), n = 1
E°cell = 0.222 − 0.799 = −0.577 V
log Ksp = nE°/0.0592 = (1)(−0.577)/0.0592 = −9.75
Ksp = 10−9.75 ≈ 1.8 × 10⁻¹⁰
ΔG° = −nFE° = −(1)(96 485)(−0.577) = +55.7 kJ mol⁻¹ — strongly positive, which is precisely why AgCl is insoluble.
The rest of the electrochemistry syllabus
Nernst is the centre of gravity, but GATE also draws on conductance. Keep these in working memory: molar conductivity Λm = κ/c; Kohlrausch's law of independent migration Λm° = ν₊λ₊° + ν₋λ₋°, which is how you obtain Λm° for a weak electrolyte that can never be extrapolated directly; the degree of dissociation α = Λm/Λm°; and the Debye–Hückel limiting law for activity coefficients in dilute solution. Electrode kinetics (the Butler–Volmer equation, overpotential, exchange current density) sits at the boundary of the syllabus and is worth a conceptual reading rather than heavy drilling.
Quick reference
| You are asked for | Use | Key check |
|---|---|---|
| E at given concentrations | E = E° − (0.0592/n) log Q | Solids and pure liquids excluded from Q |
| K of the cell reaction | log K = nE°/0.0592 | Positive E° must give K > 1 |
| ΔG or ΔG° | ΔG = −nFE | Answer in J; F = 96 485 C mol⁻¹ |
| EMF of a concentration cell | E = (0.0592/n) log(cconc/cdil) | E° = 0; dilute side is the anode |
| pH from EMF | E = −0.0592 pH | Requires PH₂ = 1 bar |
| Ksp from E° values | Build the dissolution as a cell, then log K = nE°/0.0592 | n = 1 for a 1:1 salt |
Errors that repeatedly appear in electrochemistry answers
- E° is not multiplied when you scale a half-reaction. Doubling Cu²⁺ + 2e⁻ → Cu does not double E°. Potential is an intensive property. ΔG° is extensive, which is why the correct route for combining half-reactions is through ΔG° = −nFE°, not by adding potentials directly.
- Inverting Q. Q must be written for the cell reaction exactly as you balanced it, products over reactants. Writing it upside down flips the sign of the correction term and typically turns a 1.07 V answer into 1.13 V.
- Getting n wrong. n is the number of electrons in the balanced overall reaction after the half-equations have been made to cancel, not the number in one half-reaction as written in a data table.
- Using 0.0592 at a temperature other than 298 K. That constant is temperature-specific. At 310 K it is (8.314 × 310 / 96 485) × 2.3026 = 0.0615 V.
- Including solids or the solvent in Q. Pure solids and pure liquids have unit activity; putting a concentration in for Zn(s) is a guaranteed wrong answer.
Verify your cell potentials in seconds. The Nernst equation calculator takes E°, n, temperature and the reaction quotient and returns E with the substitution shown, so you can find precisely which step of a hand calculation went astray.
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