The formula
- General form
E_cell = E°_cell - (RT / nF)·ln Q- At 25 °C, using base-10 logs
E_cell = E°_cell - (0.0592 V / n)·log Q- Free energy from cell potential
ΔG = -nFE_cell- At equilibrium, where Q reaches K
E_cell = 0 and E°_cell = (0.0592 V / n)·log K
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
E_cell | Cell potential under the actual conditions | V |
E°_cell | Standard cell potential, with all solutes at 1 M and all gases at 1 bar | V |
R | Molar gas constant | 8.314 J/(mol·K) |
T | Absolute temperature | K |
n | Moles of electrons transferred in the balanced cell reaction | mol e⁻ |
F | Faraday constant, the charge on one mole of electrons | 96,485 C/mol e⁻ |
Q | Reaction quotient, products over reactants at the current concentrations | dimensionless |
K | Equilibrium constant for the cell reaction | dimensionless |
ΔG | Free energy change for the cell reaction | J |
When it applies
- Concentrations, pressures or both differ from standard state. When every species is already at standard state, Q is 1, the log term vanishes and E_cell equals E°_cell.
- The 0.0592 V shortcut is the RT/F group evaluated at 298 K, so it holds at 25 °C and nowhere else. Away from that, use RT/nF with T in kelvin.
- Q is built from the balanced cell reaction, leaving out pure solids and pure liquids, exactly as an equilibrium constant expression would.
- n counts electrons in the balanced overall reaction, not the electrons in one half-reaction as written in a table.
Worked example
Problem. A zinc-copper cell runs the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E°_cell = 1.10 V. The cell holds 1.0 M Zn²⁺ and 0.010 M Cu²⁺ at 25 °C. What is its potential?
- Count the electrons transferred. Zinc goes from 0 to +2 and copper from +2 to 0, so n = 2.
- Write the reaction quotient, leaving out the two solids: Q = [Zn²⁺] / [Cu²⁺] = 1.0 / 0.010 = 100.
- The temperature is 25 °C, so use the shortcut form: E_cell = 1.10 V - (0.0592 V / 2)·log(100).
- log(100) = 2, so the correction is (0.0296 V)(2) = 0.0592 V.
- E_cell = 1.10 - 0.0592 = 1.0408 V, which is 1.04 V to three significant figures. Product outweighing reactant pulls the potential below standard, as expected.
Answer. 1.04 V, about 0.06 V below the standard potential because the product ion is 100 times more concentrated than the reactant ion.
Common mistakes
- Mixing the two forms: using 0.0592 with a natural log, or RT/nF with a base-10 log. The 0.0592 V constant belongs with log, RT/nF belongs with ln.
- Inverting Q. Products go on top, so a build-up of product lowers the potential rather than raising it.
- Taking n from a single half-reaction. Both half-reactions have to be scaled to the same electron count first, and that shared number is n.
- Applying the 25 °C shortcut at another temperature. At 350 K the RT/F group is no longer 0.0592 V and the answer drifts.
Related formulas
- Hess's law:
ΔH_reaction = ΣΔH_steps - Gibbs free energy equation:
G = H - TS