Chemistry

Nernst Equation

Correct a cell potential for nonstandard composition.

E = E^{\circ} - \frac{R T}{n F} ln Q

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Source & review

Difficulty

Intermediate

Read time

4 min

Prerequisites

Algebra, logarithms, and molar quantities

Review statusNeeds review

Source

OpenStax, Chemistry 2e, §17.4

Adapted from Teng-Jui Lin’s CHEM 155 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

FormuLab adapted notation and chemical symbols, added variable meanings, units, applicability conditions, and an original worked example. Corrected the equation name and retained the natural-log form with explicit activity-based Q.

Definition

Understanding the core concept

The Nernst equation connects cell potential to the activity-based reaction quotient. It uses natural logarithms in this form and predicts zero equilibrium cell potential when Q equals the equilibrium constant.

Variables & Units

Understanding each component

Symbol	Meaning	Units
$E$	Reversible cell potential at the current composition	$V$
$E^{\circ}$	Standard cell potential for the reaction as written	$V$
$R$	Molar gas constant	$J mo l^{- 1} K^{- 1}$
$T$	Absolute temperature	$K$
$n$	Number of electrons transferred per reaction as written	$1$
$F$	Faraday constant: charge per mole of electrons	$C mo l^{- 1}$
$Q$	Dimensionless reaction quotient built from activities	$1$

Real-World Applications

Where this formula is used in practice

Electrochemical cells

Estimate composition-dependent reversible voltage.

Concentration effects

Compare cell potentials as ion activities change.

Worked Example

Step-by-step calculation with real numbers

Problem

A two-electron cell has standard potential 1.10 V at 298.15 K. Find its reversible potential when Q = 10.0.

Given

E^{\circ} = 1.10 V

n = 2

Q = 10.0

T = 298.15 K

R = 8.314462618 J mo l^{- 1} K^{- 1}

F = 96485.33212 C mo l^{- 1}

Solution

E = E^{\circ} - \frac{R T}{n F} ln Q

E = [1.10 - \frac{( 8.314462618 ) ( 298.15 )}{( 2 ) ( 96485.33212 )} ln (10.0)] V

E \approx 1.07 V

Final Answer

E \approx 1.07 V

Formula Guide

Info

DifficultyIntermediate

Read Time4 min

PrerequisitesAlgebra, logarithms, and molar quantities

Tools

Source & review

StatusNeeds review

Last reviewedPending

Source

OpenStax, Chemistry 2e, §17.4

Adapted from Teng-Jui Lin’s CHEM 155 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

Assumptions

• Use a balanced redox reaction, positive electron-transfer number, and positive dimensionless Q.
• Use thermodynamic reversible potentials at the stated temperature; the formula excludes operating losses and overpotentials.

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