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Electrochemistry, thermodynamics and equilibrium are taught as three chapters, but for a cell reaction they are one idea. Measure the standard EMF of a cell and you know how much useful work the reaction can do and how far it goes before it stops. This post puts the three quantities on one line, shows how to move between them in either direction, works four examples using NCERT's electrode potentials, and points out the trap that catches students when an equation is halved.
Where is this in NCERT and the syllabus?
The current NCERT Class 12 Chemistry Part I chapter Electrochemistry (2026-27 reprint) gives the equilibrium form E°cell = (2.303RT/nF) log Kc (Eq. 2.14) in Section 2.3.1, and ΔrG° = −nFE°cell (Eq. 2.16) with ΔrG° = −RT ln K in Section 2.3.2. It stresses that E°cell is an intensive property while ΔrG depends on n. The JEE (Main) 2026 syllabus lists the "relationship between cell potential and Gibbs' energy change", and the NEET (UG) 2026 syllabus lists the same. Check your exam year's syllabus.
This post assumes you can already find E°cell from a table. If not, start with our guide to the Nernst equation and cell EMF.
How are E°, ΔG° and K connected?
Put the two NCERT equations side by side:
−nFE° = ΔG° = −RT ln K
Here n is the number of electrons transferred in the equation as written and F ≈ 96,485 C/mol. Rearranging gives ln K = nFE°/RT. At 298 K, 2.303RT/F = 2.303 × 8.314 × 298 ÷ 96,485 ≈ 0.059 V, so
log K = nE°/0.059
Three sign rules follow. A positive E° means a negative ΔG° and K greater than 1, so products are favoured. A negative E° means a positive ΔG° and K less than 1. And E° = 0 means K = 1.
Worked example 1: a cell with almost no driving force
Fe²⁺ + Ag⁺ → Fe³⁺ + Ag, with NCERT's values E°(Ag⁺/Ag) = 0.80 V and E°(Fe³⁺/Fe²⁺) = 0.77 V.
E°cell = E°cathode − E°anode = 0.80 − 0.77 = 0.03 V. Here n = 1.
ΔG° = −1 × 96,485 × 0.03 = −2,895 J/mol ≈ −2.9 kJ/mol.
log K = 1 × 0.03/0.059 = 0.51, so K ≈ 3.2.
A K of about 3 means a real mixture of reactants and products at equilibrium. Small E° values matter: a question can ask whether a reaction "goes to completion", and here it clearly does not.
Worked example 2: chlorine and bromide
Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, with E°(Cl₂/Cl⁻) = 1.36 V and E°(Br₂/Br⁻) = 1.09 V.
E°cell = 1.36 − 1.09 = 0.27 V, and n = 2.
ΔG° = −2 × 96,485 × 0.27 = −52,100 J/mol ≈ −52.1 kJ/mol.
log K = 2 × 0.27/0.059 = 9.15, so K ≈ 1.4 × 10⁹.
A cell EMF only nine times larger than in example 1, with twice the electrons, raises K by a factor of over 10⁸. That is why chlorine water displaces bromine almost completely.
Worked example 3: what happens if you halve the equation?
Write the same reaction as ½Cl₂ + Br⁻ → Cl⁻ + ½Br₂. Now n = 1.
E° is unchanged at 0.27 V, because it is intensive.
ΔG° = −1 × 96,485 × 0.27 ≈ −26.1 kJ/mol, half the value in example 2.
log K = 0.27/0.059 = 4.58, so K ≈ 3.8 × 10⁴, which is the square root of 1.4 × 10⁹.
NCERT makes the same point with the Daniell cell: doubling the equation doubles n and ΔrG while E° stays the same. K, in turn, is squared.
Worked example 4: going backwards to a solubility product
The method also runs the other way. NCERT's table gives E°(AgCl/Ag, Cl⁻) = 0.22 V and E°(Ag⁺/Ag) = 0.80 V. Combine them to get AgCl(s) → Ag⁺ + Cl⁻:
E° = 0.22 − 0.80 = −0.58 V, with n = 1.
log Ksp = −0.58/0.059 = −9.83, so Ksp ≈ 1.5 × 10⁻¹⁰.
ΔG° = +56.0 kJ/mol. Positive, so AgCl dissolves only very slightly.
A cell measurement has given a solubility product without weighing any precipitate.
Which mistakes cost marks?
Multiplying E° by coefficients. Never. Multiply ΔG° and n instead.
Using the wrong n. n is the electrons in the balanced equation you are using, not in one half-reaction you happened to look at.
Mixing kJ and J. F × E° gives joules. Convert at the end.
Using 0.059 at other temperatures. It holds only at 298 K. Elsewhere, use ln K = nFE°/RT with the actual T.
Confusing E with E°. ΔG° and K come from standard values. Non-standard concentrations need the Nernst equation first.
When a chain of conversions goes wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. For the thermodynamics side of the same equation, read our post on Gibbs energy and spontaneity.
A practice routine
Build a three-column table of your own for E°, ΔG° and log K, filling in one column from each of five cell reactions.
Find K for Cu + 2Ag⁺ → Cu²⁺ + 2Ag (E° = 0.46 V) and compare with NCERT's Example 2.2.
Redo example 4 for AgBr using E°(AgBr/Ag) = 0.10 V. Check: Ksp ≈ 1.4 × 10⁻¹².
Write one line explaining why halving an equation leaves E° unchanged.
Tonight, redo example 2 from a blank page and confirm K ≈ 1.4 × 10⁹.
