The Nernst equation
The Nernst equation adjusts cell potential for actual concentrations: E = E° − (0.0592/n)·log Q at 25 °C. Includes a fully verified Daniell cell calculation.
Quick answer
The Nernst equation gives the cell potential under nonstandard conditions: E = E° − (RT/nF)·ln Q, which at 25 °C becomes E = E° − (0.0592 V/n)·log Q.
Example: For Zn|Zn2+(0.10 M)||Cu2+(1.0 M)|Cu: Q = 0.10/1.0 = 0.10, log Q = −1, n = 2, so E = 1.10 − (0.0592/2)(−1) = 1.10 + 0.0296 = 1.1296 ≈ 1.13 V.
What you will learn
- State the Nernst equation in both the general and 25 °C forms
- Explain what Q, n, R, T, and F represent with correct units
- Compute a cell potential at nonstandard concentrations step by step
- Show that E = E° when Q = 1 and E → 0 as Q → K
- Use the sign of the log term to predict whether E rises above or falls below E°
From standard conditions to real conditions
Standard potentials assume every solute is at 1 M — a condition real batteries almost never meet. The Nernst equation corrects for reality by folding the reaction quotient Q into the potential:
E = E° − (RT/nF)·ln Q
Here E° is the standard cell potential, and Q is the reaction quotient written from the balanced equation (products over reactants, each raised to its stoichiometric coefficient). When concentrations are standard, Q = 1, ln Q = 0, and E = E° exactly.
The equation also tells the life story of a battery: as the cell discharges, products accumulate and reactants deplete, Q grows, and E shrinks. When Q reaches the equilibrium constant K, E hits zero — the battery is dead.
The 25 °C shortcut
At 25 °C (298.15 K), the constants combine into a convenient number. Evaluating RT/F with R = 8.314 J/(mol·K) and F = 96485 C/mol gives 0.025693 V; converting the natural log to base-10 log multiplies by 2.303, yielding 0.05916 V, universally rounded to 0.0592 V:
E = E° − (0.0592 V / n)·log Q (valid at 25 °C only)
The sign logic is worth memorizing: if Q < 1 (reactants favored relative to standard), log Q is negative, the subtracted term becomes positive, and E > E°. Diluting a product or concentrating a reactant raises the voltage. If Q > 1, the voltage drops below E°.
Solids, pure liquids, and solvents are omitted from Q — only aqueous ions and gases with their concentrations or partial pressures appear.
Worked example: Daniell cell at nonstandard concentrations
Cell: Zn(s)|Zn2+(0.10 M)||Cu2+(1.0 M)|Cu at 25 °C.
Step 1 — standard potential. E°cell = 0.34 − (−0.76) = 1.10 V.
Step 2 — write Q. Overall reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Solids are omitted, so Q = [Zn2+]/[Cu2+] = 0.10/1.0 = 0.10.
Step 3 — count electrons. Each half-reaction transfers 2 electrons, so n = 2.
Step 4 — substitute. log Q = log(0.10) = −1.
E = 1.10 − (0.0592/2)·(−1) = 1.10 + 0.0296 = 1.1296 ≈ 1.13 V.
Step 5 — sense-check. The product ion Zn2+ is dilute (0.10 M) relative to standard, so Q < 1 and the voltage should exceed E° = 1.10 V. It does: 1.13 V > 1.10 V. The arithmetic and the chemistry agree.
From cell potential to equilibrium constants
A frequent exam variation asks for the equilibrium constant from a standard potential. At equilibrium E = 0 and Q = K, so:
0 = E° − (0.0592/n)·log K, which rearranges to log K = nE°/0.0592.
For the Daniell cell: log K = (2)(1.10)/0.0592 = 37.16, so K ≈ 1.4 × 10^37. The astronomically large K confirms what the +1.10 V already said: the reaction goes essentially to completion.
The same rearrangement works in reverse: measure a cell's voltage at known concentrations, solve for Q, and you can determine an unknown ion concentration — the principle behind ion-selective electrodes, including the glass pH electrode.
Variables and units
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| gas constant | J/(mol·K) | Ideal gas constant | |
| temperature | K | Absolute temperature | |
| moles of electrons | mol e− | Moles of electrons transferred in the balanced equation | |
| Faraday constant | C/mol | Faraday's constant, charge per mole of electrons |
Common mistakes
It applies to every electrochemical cell at nonstandard conditions. Concentration cells are just the special case where E° = 0.
Q uses current concentrations; K is Q at equilibrium. As a cell runs, Q approaches K and E approaches zero — that is why dead batteries read 0 V.
n is the number of electrons transferred in the balanced overall equation after the half-reactions are combined and electrons cancel — for the Daniell cell, n = 2, not 4.
Real-world applications
Predicting real battery voltage as reactants are consumed
Computing cell potential at nonstandard concentrations
Finding ion concentrations from a measured cell voltage
Understanding pH electrodes, which are Nernstian devices
Explaining why a cell's voltage fades as it discharges
Frequently asked questions
Does 0.0592 V work at any temperature?
No. The derivation assumes temperature in kelvin, and RT/F changes with T. At other temperatures use the general form E = E° − (RT/nF)·ln Q with T in kelvin.
Do I multiply by 2.303 when using 0.0592?
No — 0.0592/n already includes the conversion. The 0.0592 V form pairs with log base 10; the RT/nF form pairs with natural log. Mixing them is a common error.
If Q = 1, does the Nernst equation just give E°?
Yes, exactly. At Q = 1 the log term is zero and E = E°. Standard potentials are simply the Nernst equation evaluated at unit activities.
Why does a battery's voltage drop as it runs down?
As the cell discharges, products build up and reactants deplete, so Q grows toward K and E falls toward zero. A dead battery is a cell at equilibrium.