Worked example 1
For Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), E°cell = +1.10 V. Calculate ΔG°.
Try it first: Find n from the balanced equation before substituting.
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Electrochemistry & Applications of ThermodynamicsElectrochemistryContent level: Core 28 min
Cell Potential, Free Energy, and the Equilibrium Constant
What you'll be able to do: Convert freely among E°cell, ΔG°, and K, and explain why all three describe the same thermodynamic fact.
Best after: Standard Reduction Potentials and Cell Potential
Introduction
E°cell, ΔG°, and K look like three separate topics, but they are three readings of one instrument. Learn the two equations that connect them and a whole class of problems collapses into arithmetic.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Relate the signs of E°cell, ΔG°, and K to spontaneity
- Calculate ΔG° from a standard cell potential
- Calculate K from E°cell at 25 °C
- Explain why ΔG° is extensive while E° is intensive
Lesson
Electrical work and free energy
The maximum useful work a spontaneous reaction can deliver is its free energy change. In a cell, that work is electrical: charge nF moved through a potential difference E. Setting the two equal, with a minus sign because work done by the system lowers its free energy, gives ΔG = -nFE.
ΔG° = -n F E°cell
What n and F mean
n is the moles of electrons transferred per mole of reaction as the equation is balanced, found by looking at how many electrons cancelled when the half-reactions were combined. F is the charge on one mole of electrons, 96485 C/mol. With E in volts, the product nFE comes out in joules, so divide by 1000 to report kilojoules.
A volt is a joule per coulomb, which is why C x V gives J directly.
Bringing in K
Thermodynamics already gives ΔG° = -RT ln K. Setting the two expressions for ΔG° equal and solving for E° gives the bridge between voltage and equilibrium. At 298 K, substituting R = 8.314 J/(mol K) and F = 96485 C/mol and converting to base-10 logs produces the 0.0592 shortcut.
E°cell = (0.0592 / n) log K (at 298 K)
The three-way sign table
Product-favored: E°cell positive, ΔG° negative, K greater than 1. Reactant-favored: E°cell negative, ΔG° positive, K less than 1. At equilibrium: E = 0, ΔG = 0, Q = K. These cannot disagree, so a mismatch in your answers is a signal that a sign was dropped.
A dead battery is a cell that has reached equilibrium: E = 0, not E° = 0.
Extensive versus intensive
Doubling every coefficient in the overall reaction doubles n, so ΔG° doubles, while E° is unchanged. K, meanwhile, is squared. All three behaviors are consistent with the equations, and testing them is a favorite exam move.
ΔG°(2x) = 2 ΔG°, E°(2x) = E°, K(2x) = K²
Why cell voltages imply enormous K values
Rearranging gives K = 10nE°/0.0592. With n = 2 and E° = 1.10 V, the exponent is about 37. Small changes in voltage therefore mean astronomically different equilibrium positions, which is why voltage is such a sensitive measure of driving force.
Scaling changes ΔG and K but not E
If every coefficient in a cell reaction is doubled, the transferred electron count n doubles and so does ΔG°=-nFE°. The potential E° remains the same because it is energy per unit charge. Since ΔG°=-RT ln K, the new equilibrium constant is K². State the reaction as written before comparing values.
E is intensive; ΔG is extensive; K follows the chosen reaction stoichiometry.
Key ideas
Rule
Free energy from potential
ΔG° = -nFE°cell, with F = 96485 C/mol e⁻.
Rule
Free energy from K
ΔG° = -RT ln K.
Rule
Potential from K
E°cell = (0.0592/n) log K at 298 K.
Definition
n
Moles of electrons transferred per mole of reaction as balanced.
Rule
Scaling
ΔG° and ln K scale with the reaction; E° does not.
Equation
Free energy and potential
ΔG° = -n F E°cell
- n = moles of electrons transferred
- F = 96485 C per mole of electrons
- E°cell = standard cell potential in volts
Equation
Potential and K
E°cell = (0.0592 / n) log K
- K = equilibrium constant at 298 K
Equation
Free energy and K
ΔG° = -R T ln K
- R = 8.314 J/(mol K)
- T = absolute temperature in K
Worked examples
Worked example 2
Find K at 25 °C for that same reaction.
Try it first: Use the 0.0592 shortcut rather than going through ΔG°.
0 of 3 steps revealed.
Worked example 3
A cell has ΔG° = -145 kJ/mol with n = 3. Find E°cell.
Try it first: Watch the units: joules, not kilojoules.
0 of 3 steps revealed.
Worked example 4
A cell reaction as written has E°=0.20 V, n=1, and equilibrium constant K. What happens if every coefficient is doubled?
Try it first: The physical cell potential does not depend on how the equation is scaled.
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Common mistakes
Leaving ΔG° in kJ when solving for E°.
Why it's wrong: F carries coulombs, so joules are required; using kJ gives an answer 1000 times too small.
Check instead: Convert to joules before dividing.
Dropping the negative sign in ΔG° = -nFE°.
Why it's wrong: A positive potential must give a negative free energy change.
Check instead: Check that the signs of E° and ΔG° end up opposite.
Taking n as the charge of one ion.
Why it's wrong: n is the electrons transferred in the balanced overall reaction, which may differ after scaling.
Check instead: Count the electrons that cancelled when the half-reactions were combined.
Multiplying E° when doubling the reaction to keep ΔG° consistent.
Why it's wrong: ΔG° doubles through n, not through E°.
Check instead: Scale n only; leave E° alone.
Saying E° = 0 at equilibrium.
Why it's wrong: E° is a fixed constant for the reaction; it is the actual E that falls to zero.
Check instead: Distinguish E from E°, just as you distinguish Q from K.
After the lesson · Question bank
Independent practice
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What you should now know
The standard cell potential, the standard free energy change, and the equilibrium constant are three ways of stating how far a redox reaction lies from equilibrium under standard conditions. They are linked by ΔG° = -nFE°cell and ΔG° = -RT ln K, which combine to give E°cell = (RT/nF) ln K, or at 25 °C the convenient form E°cell = (0.0592/n) log K. Here n is the number of moles of electrons transferred in the balanced overall reaction and F is the Faraday constant, 96485 C per mole of electrons. A positive E°cell corresponds to a negative ΔG° and to K greater than 1, all three signalling a product-favored reaction. Unlike E°, ΔG° is extensive: doubling the reaction doubles ΔG° because n doubles while E° stays fixed. Because the exponent in K = 10nE°/0.0592 is large, even a modest cell voltage of about 1 V with n = 2 corresponds to an equilibrium constant near 10³⁴, which is why a fresh battery drains rather than reaching a visible equilibrium.
- ΔG° = -nFE°cell links voltage to free energy
- ΔG° = -RT ln K links free energy to equilibrium
- E°cell = (0.0592/n) log K at 298 K
- Positive E°, negative ΔG°, and K > 1 always travel together
- ΔG° and K respond to scaling; E° does not
Sources and further reading
- Chemistry 2e, Section 17.4: Potential, Free Energy, and EquilibriumOpenStax · Rice University · Chapter 17.4View source
Access for free at openstax.org License