The Nernst Equation

A copper coin dropped into dilute copper sulfate generates a different voltage than the same coin in concentrated solution. Standard electrode potentials alone cannot explain this. They describe cells only at one fixed, idealized composition. The Nernst equation fixes this gap. It tells you exactly how far a cell's potential drifts from its standard value as concentrations, pressures, or pH change. This single relation underlies pH meters, ion-selective electrodes, batteries under load, and the resting potential of every living cell. This article builds the Nernst equation from the Gibbs energy of reaction. You will see the derivation in full. Then you will apply it to two worked problems. Finally you will see where it breaks down and what it leads to next.

Key takeaways

The Nernst equation links cell EMF to the reaction quotient $Q$ : it corrects $E^\ominus$ for real, non-standard conditions.
It follows directly from $\Delta G = \Delta G^\ominus + RT\ln Q$ combined with $\Delta G = -nFE$ .
At 298.15 K, the equation reduces to a compact form using 0.0592 V (base-10 log) or $RT/F$ (natural log).
Concentration cells, with identical electrodes but different ion activities, produce a nonzero EMF from this equation alone.
Activities, not concentrations, are the rigorous variables; at low ionic strength the two nearly coincide.
Key idea: cell potential is a thermodynamic quantity that tracks $\Delta G$ of the actual, running reaction, not some fixed reference state.

The core insight

The Nernst equation is nothing more than $\Delta G = -nFE$ rewritten with the full, composition-dependent Gibbs energy expression $\Delta G = \Delta G^\ominus + RT\ln Q$ , so that the measured cell voltage automatically reports how far the reaction quotient $Q$ has pushed the system away from its standard state.

The physical picture

A galvanic cell converts chemical driving force into electrical work. The size of that driving force depends on how far the reaction is from equilibrium. At equilibrium the reaction quotient $Q$ equals $K$ , and no more work can be extracted: $E = 0$ . Far from equilibrium, $Q$ differs sharply from 1, and the cell delivers a larger voltage. Standard electrode potentials, $E^\ominus$ , are tabulated at the special point where every species sits at unit activity (1 mol/L solutes, 1 bar gases, pure solids and liquids). Real cells almost never sit exactly there. A concentration cell makes this vivid: two identical silver electrodes, one in 1.0 M AgNO $_3$ and one in 0.010 M AgNO $_3$ , have identical standard potentials for their half-reactions. Yet connecting them through a salt bridge produces a measurable voltage, driven purely by the concentration difference, with $E^\ominus_{\text{cell}} = 0$ exactly. The Nernst equation is the bridge between the standard number in a table and the actual voltage on a voltmeter. It says the deviation from standard behavior is logarithmic in $Q$ , scaled by $RT/nF$ — a direct fingerprint of Gibbs energy's own logarithmic dependence on concentration.

Figure

Cell voltage falls logarithmically as the reaction quotient rises toward K

Figure

Concentration cell: same electrode material, different activity, nonzero EMF

The core relations

\Delta G = \Delta G^\ominus + RT\ln Q

where $\Delta G^\ominus$ is the standard reaction Gibbs energy and $Q$ is the reaction quotient built from activities.

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

where $n$ is moles of electrons transferred and $F = 96{,}485\ \text{C mol}^{-1}$ .

E = E^\ominus - \frac{0.05916\ \text{V}}{n}\log_{10} Q \quad (T = 298.15\ \text{K})

where the 0.05916 V constant absorbs $RT\ln(10)/F$ at 25 °C.

E_{\text{cell}} = 0\ \Longleftrightarrow\ Q = K

where the cell has reached chemical equilibrium and can do no further electrical work.

Derivation AFrom Gibbs energy to the Nernst equation

Goal. Show that cell EMF depends on composition through the reaction quotient $Q$ , and derive the explicit logarithmic form.

Step 1. Start from the definition of the reaction Gibbs energy in terms of chemical potentials of products and reactants,

\Delta G = \sum_j \nu_j \mu_j

where $\nu_j$ are stoichiometric numbers (negative for reactants) and $\mu_j$ are chemical potentials of each species in its actual state.

Step 2. Each chemical potential departs from its standard value logarithmically in activity $a_j$ ,

\mu_j = \mu_j^\ominus + RT\ln a_j

where $\mu_j^\ominus$ is the chemical potential at unit activity.

Step 3. Substitute Step 2 into Step 1 and collect the standard terms separately from the activity terms,

\Delta G = \sum_j \nu_j \mu_j^\ominus + RT\sum_j \nu_j \ln a_j

The first sum is by definition $\Delta G^\ominus$ ; the second sum, using $\sum \nu_j \ln a_j = \ln \prod a_j^{\nu_j}$ , is $RT\ln Q$ , giving

\Delta G = \Delta G^\ominus + RT\ln Q

Step 4. Bring in the electrical work relation. The maximum non-expansion work available from the cell reaction equals the Gibbs energy change, and this work is done by moving $n$ moles of electrons through potential difference $E$ ,

\Delta G = -nFE

and correspondingly at standard conditions,

\Delta G^\ominus = -nFE^\ominus

Step 5. Substitute both electrical relations from Step 4 into the Gibbs energy expression from Step 3,

-nFE = -nFE^\ominus + RT\ln Q

Step 6. Divide through by $-nF$ to isolate $E$ ,

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

Step 7. Convert to base-10 logarithms for practical calculation, using $\ln Q = \ln(10)\log_{10}Q$ , and insert $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T = 298.15\ \text{K}$ , $F = 96{,}485\ \text{C mol}^{-1}$ ,

\frac{RT\ln(10)}{F} = \frac{(8.314)(298.15)(2.303)}{96{,}485} = 0.05916\ \text{V}

This yields the boxed working form:

\boxed{E = E^\ominus - \dfrac{0.05916\ \text{V}}{n}\log_{10} Q}

Worked examples

Example 1 — Foundational: a Daniell cell away from standard conditions. Consider the classic cell

\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}

with $n = 2$ and $E^\ominus_{\text{cell}} = E^\ominus_{\text{Cu}^{2+}/\text{Cu}} - E^\ominus_{\text{Zn}^{2+}/\text{Zn}} = 0.34\ \text{V} - (-0.76\ \text{V}) = 1.10\ \text{V}$ . Suppose $[\text{Cu}^{2+}] = 0.0100\ \text{M}$ and $[\text{Zn}^{2+}] = 1.00\ \text{M}$ , both dilute enough that activity coefficients are taken as 1, and both solids have activity 1. The reaction quotient, with solids omitted, is

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1.00}{0.0100} = 100

Apply the Nernst equation at 298.15 K:

E = 1.10\ \text{V} - \frac{0.05916\ \text{V}}{2}\log_{10}(100)

E = 1.10\ \text{V} - (0.02958\ \text{V})(2)

E = 1.10\ \text{V} - 0.0592\ \text{V} = 1.041\ \text{V}

Sanity check. Raising product-side $\text{Zn}^{2+}$ and lowering reactant-side $\text{Cu}^{2+}$ pushes $Q$ above 1, so the driving force should drop below $E^\ominus$ . It does: 1.041 V is slightly below 1.10 V, by a modest 0.059 V — consistent with the mild, logarithmic sensitivity of $E$ to concentration.

Example 2 — Applied: a concentration cell used to back out an unknown Ag⁺ concentration. Two silver electrodes are placed in AgNO $_3$ solutions of different concentration, connected by a salt bridge:

\text{Ag(s)} \mid \text{Ag}^+(0.0100\ \text{M}) \parallel \text{Ag}^+(x\ \text{M}) \mid \text{Ag(s)}

Both half-reactions are $\text{Ag}^+ + e^- \rightarrow \text{Ag(s)}$ with identical $E^\ominus$ , so $E^\ominus_{\text{cell}} = 0$ exactly — the electrode materials are the same. The measured cell voltage is $+0.0680\ \text{V}$ , with the right-hand electrode (concentration $x$ ) wired as the cathode (positive terminal). Here $n = 1$ (one electron per Ag⁺ reduced), and the overall cell reaction, cathode minus anode, is

\text{Ag}^+(x\ \text{M}) \rightarrow \text{Ag}^+(0.0100\ \text{M})

Q = \frac{[\text{Ag}^+]_{\text{anode side, produced}}}{[\text{Ag}^+]_{\text{cathode side, consumed}}} = \frac{0.0100}{x}

Apply the Nernst equation:

0.0680 = 0 - \frac{0.05916}{1}\log_{10}\!\left(\frac{0.0100}{x}\right)

Divide both sides by $-0.05916$ :

\log_{10}\!\left(\frac{0.0100}{x}\right) = -1.149

Exponentiate:

\frac{0.0100}{x} = 10^{-1.149} = 0.0710

Solve for $x$ :

x = \frac{0.0100}{0.0710} = 0.141\ \text{M}

Sanity check. The cathode side is more concentrated, since reduction is thermodynamically favored where activity is higher, which drives current in the observed direction. A concentration ratio of about 14 to 1 producing 0.068 V is reasonable: each factor of 10 in ratio contributes 0.0592 V for $n=1$ , and $\log_{10}(14.1) \approx 1.15$ , matching $1.15 \times 0.0592\ \text{V} = 0.068\ \text{V}$ — the numbers close the loop.

Limiting cases & common traps

Situation	What happens	Why
$Q = 1$	$E = E^\ominus$	Logarithm term vanishes
$Q = K$	$E = 0$	Reaction at equilibrium, no net driving force
$Q \to 0$	$E \to +\infty$ (formally)	Real cells are limited by mass transport and side reactions before this is reached
Very dilute ionic solutions	Concentration ≈ activity	Activity coefficients approach 1 as ionic strength → 0
Concentrated or multiply-charged ions	Concentration ≠ activity	Debye–Hückel corrections needed; use $a_i = \gamma_i [i]/c^\ominus$
Gases in $Q$	Use partial pressure ratio $p_i/p^\ominus$	Standard state for gases is 1 bar, not 1 M

The chemist behind it

Walther Nernst derived this relation in 1889 while extending the young discipline of chemical thermodynamics into electrochemistry. He recognized that the electromotive force of a cell was not an independent electrical curiosity, but a direct, measurable window onto the Gibbs energy of the driving reaction — bridging Gibbs' abstract thermodynamic formalism with a benchtop voltmeter reading. His broader program, culminating in the third law of thermodynamics (the "Nernst heat theorem"), earned him the 1920 Nobel Prize in Chemistry.

Origins · 1 minWalther Hermann Nernst and how this was discovered

Walther Hermann Nernst

1864–1941 · German physical chemist who unified electrochemical potential with reaction Gibbs energy and later formulated the third law of thermodynamics

Ask the tutor: How do I extend the Nernst equation to extract $\Delta H^\ominus$ and $\Delta S^\ominus$ from the temperature dependence of measured cell potentials?

Why does the Nernst equation use activities instead of concentrations?

Chemical potential is rigorously a function of activity, $\mu = \mu^\ominus + RT\ln a$ , because real ions interact electrostatically and deviate from ideal dilute behavior. Concentration is only a convenient stand-in. At low ionic strength, activity coefficients $\gamma_i$ approach 1, so $a_i \approx [i]/c^\ominus$ and concentration works well. In concentrated or highly charged ionic solutions, Debye–Hückel or extended theories are needed to get accurate voltages, especially in precise potentiometry.

Does the Nernst equation apply to any redox reaction, or only to full cells?

It applies to a single half-reaction potential (using the SHE-referenced $E^\ominus$ ) and to a full cell potential, using the same logic in both cases. For a half-reaction it predicts how the electrode potential shifts with the activities of the species in that half-reaction alone. For a full cell, you use the overall $n$ and the overall reaction quotient built from every species in the net balanced equation.

How is the Nernst equation connected to pH measurement?

A pH glass electrode behaves like a hydrogen-ion-selective half-cell whose potential follows the Nernst equation with $n = 1$ . At 25 °C, each unit change in pH shifts the electrode potential by 0.05916 V, since $\log_{10}[\text{H}^+] = -\text{pH}$ . Calibrating the electrode against buffers of known pH converts this predictable voltage response directly into a pH readout, which is the working principle behind every commercial pH meter.

Why can real cells not reach the infinite voltage the equation suggests as Q approaches zero?

The equation is a thermodynamic limit assuming ideal, fully equilibrated activities down to vanishingly small values. Real systems hit practical floors first: solubility limits, trace impurities, junction potentials, and diffusion-limited mass transport all cap how low an active-species activity can actually go. Kinetic and transport barriers intervene long before the thermodynamic singularity is approached, so measured voltages always stay finite.

Is the Nernst equation valid only at equilibrium?

No — this is a common confusion. The Nernst equation describes a cell held at a fixed, measured composition, giving the reversible EMF for that instantaneous state, whether or not the reaction has reached equilibrium overall. Equilibrium is simply the special case where $Q = K$ and $E = 0$ . Away from equilibrium, $Q \neq K$ , and the equation correctly reports a nonzero, thermodynamically meaningful voltage.