ElectrochemistryCourse level: AP Chemistry

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

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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

Electrical work and free energy

At constant temperature and pressure, the maximum non-expansion work deliverable by a spontaneous process has magnitude −ΔG; its signed reversible electrical work on the system is ΔG. 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.

Count n from the balanced equation, not from one ion's charge alone. For 2 Al + 3 Cu²⁺ → 2 Al³⁺ + 3 Cu, six electrons cancel, so n = 6. With E° = 2.00 V, ΔrG° = −(6)(96485)(2.00) = −1.16 × 10⁶ J/mol reaction = −1.16 × 10³ kJ/mol reaction. A half-sized equation transfers n = 3 and has half the ΔrG°, while its potential remains 2.00 V. The “per mole” basis is the equation as written.

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.

Use ln K = nFE°/(RT) at any stated T, or log₁₀K ≈ nE°/0.0592 at 25 °C. With n = 1 and E° = 0.1184 V, log K ≈ 2.00 and K ≈ 100. Exponentiate only after computing the logarithm, and keep guard digits: rounding log K can multiply the final K by a substantial factor. Potentials quoted to a few hundredths of a volt often support only an order-of-magnitude K estimate. No concentration units belong inside a logarithm when K is written rigorously in activities.

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.

SituationEΔrGRelation of Q to K
Forward direction favored for actual mixturePositiveNegativeQ < K
Equilibrium of the actual cell reactionZeroZeroQ = K
Reverse direction favored for actual mixtureNegativePositiveQ > K

For the corresponding standard quantities, E° > 0 means ΔrG° < 0 and K > 1. E° = 0 means K = 1; it does not describe every possible equilibrium mixture. A cell with large positive E° can reach actual equilibrium at a very large Q while retaining that same E°.

Connect standard voltage, free energy and K. Read this visual alongside the section; the text alternative describes its scientific content and assumptions.
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 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.

    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.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Leaving ΔG° in kJ when solving for E°.F carries coulombs, so joules are required; using kJ gives an answer 1000 times too small.Convert to joules before dividing.
          Dropping the negative sign in ΔG° = -nFE°.A positive potential must give a negative free energy change.Check that the signs of E° and ΔG° end up opposite.
          Taking n as the charge of one ion.n is the electrons transferred in the balanced overall reaction, which may differ after scaling.Count the electrons that cancelled when the half-reactions were combined.
          Multiplying E° when doubling the reaction to keep ΔG° consistent.ΔG° doubles through n, not through E°.Scale n only; leave E° alone.
          Saying E° = 0 at equilibrium.E° is a fixed constant for the reaction; it is the actual E that falls to zero.Distinguish E from E°, just as you distinguish Q from K.

          Scroll horizontally if needed to read all columns.

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          Independent practice

          Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.

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          What you should now know

          For the balanced cell reaction, ΔrG° = −nFE° and ΔrG° = −RT ln K connect standard voltage, reaction Gibbs energy and equilibrium. Positive E° corresponds to negative ΔrG° and K > 1. Actual equilibrium instead has E = 0, ΔrG = 0 and Q = K; E° need not vanish. Scaling the equation scales n and ΔrG°, raises K to the same power, and leaves E° unchanged. Preserve energy units and logarithmic precision.

          • 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
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          Sources and further reading

          • Chemistry 2e, Section 17.4: Potential, Free Energy, and Equilibrium
            OpenStax · Rice University · Chapter 17.4
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            Recorded attribution / license claim (applicability unverified): Access for free at openstax.org

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