Nernst Equation

E = E° − (0.0592 V/n) log Q at 25 °C: corrects the standard cell potential for the actual, non-standard concentrations.

E = E^\circ - \frac{RT}{nF}\ln Q

E = E(standard) - (RT/nF) x ln Q ; at 25 C: E = E(standard) - (0.0592 V / n) x log Q

What it means

Standard potentials assume 1 M everything, which real cells rarely have. The Nernst equation slides the voltage up or down with the reaction quotient: build up products (large Q) and the voltage sags below E°; keep reactants rich (small Q) and it rises above E°. It is the same Qc you meet in equilibrium chemistry, just applied to a cell — and it explains why a battery's voltage fades as it discharges.

Variables

Symbol	Name	Unit	Meaning
$E$	Cell potential	V	The actual cell potential under the given non-standard conditions.
$E°$	Standard cell potential	V	The cell potential when all species are at standard-state conditions.
$R$	Gas constant	J/(mol·K)	The ideal gas constant, 8.314 J/(mol·K).
$T$	Temperature	K	Absolute temperature of the cell.
$n$	Electrons transferred	mol e⁻	Moles of electrons transferred per mole of reaction as balanced.
$F$	Faraday constant	C/mol	The charge carried by one mole of electrons, 96485 C/mol.
$Q$	Reaction quotient	dimensionless	The reaction quotient of the cell reaction at the current concentrations.

Assumptions

Temperature T is in kelvin in the RT/nF form.
The 0.0592 V shortcut is valid only at 25 °C (298 K).
Q takes the same products-over-reactants form as Kc for the cell reaction.
n is the moles of electrons transferred in the balanced cell reaction.

Derivation

Start from ΔG = ΔG° + RT ln Q and substitute ΔG = −nFE and ΔG° = −nFE°: −nFE = −nFE° + RT ln Q. Dividing everything by −nF gives E = E° − (RT/nF) ln Q. At 298 K, converting the natural log to base-10 introduces the factor 2.303, and 2.303RT/F evaluates to 0.0592 V.

Worked example

Problem

A Zn–Cu cell (E° = 1.10 V, n = 2) operates with [Zn²⁺] = 1.00 M and [Cu²⁺] = 0.0100 M at 25 °C. Calculate the actual cell potential E.

Given:

E° = 1.10 V for the Zn–Cu cell
n = 2 electrons transferred
Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu
[Zn²⁺] = 1.00 M, [Cu²⁺] = 0.0100 M

Write the reaction quotient: Q = [Zn²⁺] / [Cu²⁺] (solids omitted).
Substitute: Q = 1.00 / 0.0100 = 100.
Take the base-10 log: log Q = log(100) = 2.00.
Compute the prefactor: 0.0592 V / 2 = 0.0296 V.
E = 1.10 − (0.0296)(2.00) = 1.10 − 0.0592 = 1.041 V, rounded to 1.04 V.

E = 1.04 V

With Q = 100, the cell is product-heavy relative to standard conditions, so the measured 1.04 V sits just below E° = 1.10 V — as the cell discharges further, E keeps falling toward zero.