How to Use the Nernst Equation

Real cells aren’t standard

Standard cell potentials (E°) assume every species is at 1 M, every gas at 1 atm, and the temperature at 25 °C. Real cells almost never look like that. A battery starts strong because the reactants are concentrated and the products absent; it dies as Q climbs toward K and the cell potential approaches zero. The Nernst equation is what tells you the actual voltage at any moment in that journey.

E = E° − (RT/nF) · ln(Q)

• E: actual cell potential (V)
• E°: standard cell potential (V)
• R: 8.314 J/(mol·K), T: temperature in K, F: 96,485 C/mol
• n: electrons transferred in the balanced overall reaction
• Q: reaction quotient (same form as K, but with current concentrations)

At 25 °C, the prefactor RT/F evaluates to 0.02569 V. Convert from ln to log₁₀ (factor of 2.303) and you get the form most courses use:

E = E° − (0.05916/n) · log₁₀(Q) (at 25 °C)

How to use it

1. Write the balanced overall reaction. Combine the half-reactions so the electrons cancel.

2. Get E°cell. Look up standard reduction potentials. The half-reaction with the more positive E° runs as written (cathode, reduction); the other flips and runs in reverse (anode, oxidation). E°cell = E°cathode − E°anode. Don’t flip the sign of E°anode and add — use the formula directly.

3. Identify n. It’s the number of electrons in the balanced overall reaction. If you scaled the reaction, n scales with it.

4. Write Q. Same form as the K expression: products over reactants, raised to stoichiometric powers. Pure solids and liquids have activity 1 and don’t appear. Concentrations in mol/L, gas pressures in atm.

5. Plug in.

Worked examples

1. Zinc-copper cell, non-standard. Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s). [Cu²⁺] = 0.010 M, [Zn²⁺] = 1.0 M.

E°(Cu²⁺/Cu) = +0.340, E°(Zn²⁺/Zn) = −0.763. E°cell = 0.340 − (−0.763) = 1.103 V. n = 2.

Q = [Zn²⁺]/[Cu²⁺] = 1.0/0.010 = 100. (Zn and Cu metals: activity 1, omitted.)

E = 1.103 − (0.05916/2) · log(100) = 1.103 − 0.0296 · 2 = 1.044 V.

The voltage drops from standard because product is enriched and reactant is depleted — Le Chatelier in voltage units.

2. Concentration cell. Two Ag electrodes, one in 0.10 M AgNO₃, one in 0.0010 M AgNO₃. Same metal, so E° = 0. Yet the cell drives current, because the system can lower its free energy by transferring Ag⁺ from concentrated to dilute side.

The dilute side is the cathode (Ag⁺ being reduced is favored where Ag⁺ is scarce — counterintuitive but follows from the math). Q = [Ag⁺]dilute/[Ag⁺]concentrated = 0.0010/0.10 = 0.010. n = 1.

E = 0 − 0.05916 · log(0.010) = −0.05916 · (−2) = 0.118 V.

Concentration cells are how pH meters work — a glass electrode is essentially a concentration cell sensitive to [H⁺].

3. Solve for unknown concentration. Zn-Cu cell measures 1.04 V at 25 °C with [Zn²⁺] = 1.00 M. Find [Cu²⁺].

1.04 = 1.103 − 0.02958 · log(1.00/[Cu²⁺]) → log(1/[Cu²⁺]) = 2.13 → [Cu²⁺] = 10⁻²·¹³ = 0.0074 M.

This is the principle behind ion-selective electrodes: measure E, back-calculate the unknown ion concentration.

4. Non-standard temperature. Zn-Cu cell at 37 °C (310.15 K) with [Cu²⁺] = 0.50 M, [Zn²⁺] = 0.10 M.

You can’t use the 0.05916 shortcut — that constant is hardcoded for 25 °C. Use the full equation. RT/nF = (8.314 × 310.15)/(2 × 96485) = 0.01336. ln(0.10/0.50) = ln(0.20) = −1.609.

E = 1.103 − 0.01336 · (−1.609) = 1.103 + 0.0215 = 1.125 V.

Q < 1 means more reactants than products, so E rises above E° — the reaction has more “room” to go.

E°, K, and ΔG are the same statement in different units

The Nernst equation collapses at equilibrium: E = 0 (no driving force) and Q = K. Plug those in and rearrange:

E° = (RT/nF) · ln(K) → at 25 °C: E° = (0.05916/n) · log(K)

This is one of the most useful relationships in physical chemistry. Measure a cell’s standard potential, get its equilibrium constant. Conversely:

• ΔG° = −nFE° (standard free energy from standard cell potential)
• ΔG = −nFE (free energy at any composition)

Voltage, equilibrium constant, and free energy are three faces of the same thermodynamic die.

Traps to watch for

Sign of E°anode. Standard tables list reduction potentials. E°cell = E°cathode − E°anode using the table values directly. Don’t manually flip the sign of the anode entry and then add — that’s the same operation done twice and you’ll get the wrong answer.

Inconsistent n and Q. If you scale the balanced equation by 2, n doubles and Q gets squared, but E° stays the same (it’s an intensive property). The ratio (1/n) · log(Q) is invariant under scaling, so the predicted E doesn’t change — but only if you scale n and Q together.

Solids and liquids in Q. Cu(s) and Zn(s) don’t appear. For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺]/[Cu²⁺]. Including the metals as if they had concentrations is a common slip.

Mixing ln and log. The full equation has ln. The 0.05916 shortcut bakes in the conversion factor 2.303 to use log₁₀. Pick one and stay there.

Forgetting Kelvin. The full Nernst equation needs T in K. Plugging in 25 instead of 298.15 turns a fine answer into nonsense by a factor of ~12.

Use the Electrochemistry Calculator to solve Nernst problems and convert between E°, K, and ΔG° automatically.