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.
0 of 3 steps revealed.
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.
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.
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.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
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