Chemistry: Atoms First 2e · Electrochemistry

Potential, Free Energy, and Equilibrium

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On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

This topic connects electrochemistry to the thermodynamics of Chapter 12. A cell potential measures the driving force per electron, and multiplying by the number of electrons and the charge they carry gives the Gibbs free energy: ΔG° = −nFE°. What thermodynamics calls a negative ΔG° (spontaneous) now appears as a positive E°cell — two sides of one coin. Because free energy also fixes the equilibrium constant K, every cell potential implies one, and the extends the prediction to nonstandard concentrations. A battery's voltage is not fixed; it drifts as reactants are consumed — the Nernst equation tells you exactly how.

Why this matters

These equations make electrochemistry quantitative. E° alone tells you whether a reaction can run; adding n and F tells you how much work it can do (ΔG°), how far it will go (K), and what voltage a real cell produces at nonstandard concentrations (Nernst). These calculations underlie battery design, fuel-cell efficiency, and analytical tools such as pH meters — essentially concentration cells. On exams, expect at least one ΔG°, K, or Nernst problem.

The college version

Core Concepts

Free energy and cell potential

For a redox reaction transferring n moles of electrons, the standard free-energy change is:

$Δ G ° = -nFE °$

where n is the moles of electrons transferred per mole of reaction, F is the Faraday constant, F = 96,485 C/mol (the charge on one mole of electrons), and E° is in volts. The units work because a volt is a joule per coulomb: C × V = J. A positive E° gives a negative ΔG° — spontaneous — matching the thermodynamic criterion, and the free energy sets the maximum electrical work available from the cell.

Equilibrium constants from cell potentials

Because ΔG° = −RT ln K as well (Chapter 12), the two expressions can be equated:

$E ° = RT nF ln K$

At 25 °C, converting to base-10 logs and inserting R = 8.314 J/(mol·K), T = 298 K, and F = 96,485 C/mol gives the practical form:

$log 10 K = nE ° 0.0592 V$

A large positive E° implies a huge K — the reaction essentially runs to completion, which is why a charged battery's reactants are almost entirely consumed.

The Nernst equation

Real cells rarely run at 1 M concentrations, and voltage depends on concentrations through the reaction quotient Q. The Nernst equation corrects E° to actual conditions:

$E cell = E ° cell - RT nF ln Q$

At 25 °C this becomes:

$E cell = E ° cell - 0.0592 V n log 10 Q$

As the cell discharges, Q grows (products accumulate, reactants deplete) and E drops; at equilibrium Q = K and E = 0 — a dead battery in thermodynamic language. Q follows equilibrium rules: solids and pure liquids are omitted, and each concentration is raised to its stoichiometric coefficient. Identical half-reactions at different concentrations form a whose entire voltage comes from Q.

How It Works / Step-by-Step Process

Balance the overall redox reaction and count n, the moles of electrons transferred.
Compute E°cell from the table (Topic 3): E°cell = E°cathode − E°anode.
Apply ΔG° = −nFE°, substituting values with units, and convert J to kJ if desired.
For K: use log₁₀ K = nE°/0.0592 V (25 °C) or the general E° = (RT/nF) ln K; for nonstandard concentrations, build Q and apply the Nernst equation (at equilibrium E = 0).

Common Confusions

Do Not Confuse	With	Difference
ΔG° and E° signs	Spontaneity direction	E° > 0 means ΔG° < 0 (spontaneous); they always have opposite signs.
n in ΔG° = −nFE°	n in the Nernst equation	Same n: moles of electrons per mole of reaction as written — count them from the balanced half-reactions.
Q and K	Which one changes as the cell runs	Q changes continuously toward K; K is fixed at a given temperature.
Concentration cell	Cell with different metals	A concentration cell uses the same couple at two concentrations; all voltage comes from the ratio.
Multiplying the reaction by 2	Doubling E°	Doubling coefficients doubles n and ΔG°, but E° is unchanged (intensive).

Eli explains

The same idea, in plain words

Explain it like I’m 10

A battery's voltage is like the height of a water slide: the taller the slide, the more energy each drop can deliver. The free-energy equation multiplies that "height" by how many drops (electrons) go down. As the water level at the top falls and the pool below fills, the slide gets less exciting — that's the Nernst equation, explaining why batteries run down as they work.

Worked example

Example 1: Free energy of the Daniell cell

For Zn + Cu²⁺ → Zn²⁺ + Cu, E°cell = 1.10 V and n = 2. Write the formula, then substitute:

$Δ G ° = -nFE ° = -(2)(96,485 C mol)(1.10 V) = -2.12 \times 105 J mol$

Convert to kilojoules (the C·V cancels to J):

$Δ G ° = -2.12 \times 105 J mol \times 1 kJ 1000 J = -212 kJ mol$

The negative value confirms the reaction is spontaneous, and the magnitude is the maximum work available per mole of reaction — the energy budget a Daniell cell can deliver.

Example 2: Equilibrium constant of the Daniell cell

Use the same cell to find K at 25 °C. Formula first, then substitution:

$log 10 K = nE ° 0.0592 V = (2)(1.10 V) 0.0592 V = 37.2$

So:

$K = 10 37.2 \approx 1.6 \times 1037$

An enormous equilibrium constant: the reaction runs essentially to completion, which is why a fresh cell delivers close to its theoretical voltage until the zinc is nearly gone.

Example 3: The Nernst equation at nonstandard concentrations

What is E if [Cu²⁺] = 2.0 M and [Zn²⁺] = 0.10 M at 25 °C? First build Q — solids are omitted:

$Q = [Zn 2+] [Cu 2+] = 0.10 2.0 = 0.050$

Then apply the Nernst equation with n = 2:

$E = E ° - 0.0592 V n log 10 Q = 1.10 V - 0.0592 V 2 log 10 (0.050)$

Since log₁₀(0.050) = −1.30:

$E = 1.10 V - (0.0296 V)(-1.30) = 1.10 V + 0.038 V = 1.14 V$

The voltage rises slightly above E° because the product (Zn²⁺) is depleted and the reactant (Cu²⁺) is enriched, pushing the reaction away from equilibrium. As it runs, Q climbs toward K and E falls toward zero.

Key takeaways

ΔG° = −nFE° with F = 96,485 C/mol; units: C × V = J, so ΔG° comes out in joules per mole.
E° > 0 ⟺ ΔG° < 0 ⟺ K > 1: all mean the reaction is product-favored.
log₁₀ K = nE°/0.0592 V at 25 °C; E° = (RT/nF) ln K is the general form.
Nernst at 25 °C: E = E° − (0.0592 V/n) log₁₀ Q.
Q uses concentrations of aqueous ions and gas pressures; solids and pure liquids are omitted.
At equilibrium, Q = K and E = 0 — the cell is "dead."
A concentration cell's E°cell is zero; all its voltage comes from the concentration ratio.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

For a cell with E°cell = +0.45 V and n = 2, calculate ΔG° in kJ/mol.
Show answer
ΔG° = −nFE° = −(2)(96,485 C/mol)(0.45 V) = −8.7 × 10⁴ J/mol = −87 kJ/mol.
A reaction has K = 1.0 × 10⁻⁵ at 25 °C. Is E°cell positive or negative? What about ΔG°?
Show answer
K < 1 means the reaction is reactant-favored, so E°cell is negative and ΔG° is positive.
Write the Nernst equation at 25 °C and identify what happens to E as a cell discharges.
Show answer
E = E° − (0.0592 V/n) log₁₀ Q. As the cell discharges, Q increases, the log term grows, and E falls toward zero.
Why is the voltage of a concentration cell nonzero even though both half-reactions are identical?
Show answer
The two half-reactions have identical E° values, so E°cell = 0, but the concentrations differ; the Nernst term (Q ≠ 1) generates the voltage.
A fuel cell has E°cell = 1.23 V and n = 4. What is ΔG° per mole of O₂ reduced? (This connects to Topic 5.)
Show answer
ΔG° = −(4)(96,485 C/mol)(1.23 V) = −4.75 × 10⁵ J/mol ≈ −475 kJ per mole of O₂ (about −237 kJ per mole of H₂).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Faraday constant (F): The charge carried by one mole of electrons, 96,485 C/mol.
Gibbs free energy (ΔG°): The maximum useful work available from a reaction under standard conditions.
reaction quotient (Q): The concentration ratio of products to reactants at any moment.
equilibrium constant (K): The value of Q when the reaction is at equilibrium.
Nernst equation: E = E° − (RT/nF) ln Q: voltage as a function of concentrations.
concentration cell: A cell with identical half-reactions at different concentrations.
Faraday constant F: Charge carried by one mole of electrons: 96,485 C mol^-1

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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