The Nernst Equation

Drop a strip of zinc into a beaker of 0.001 M Cu²⁺ solution instead of the standard 1 M, and the cell voltage shifts measurably away from the textbook 1.10 V — even though nothing about the chemistry has changed. The electrons still flow from zinc to copper, the half-reactions are identical, and the Standard Electrode Potentials tables still apply, yet the number on the voltmeter depends on concentration. This is the everyday reality the Nernst equation exists to describe: cell potential is not a fixed constant of a reaction but a function of composition, temperature, and pressure, related to the standard potential by a logarithmic correction term. The equation matters because almost no real electrochemical system sits at standard state. Batteries discharge and their reactant concentrations fall continuously; blood pH sensors, ion-selective electrodes, and glucose meters all report concentration by measuring a voltage and inverting the Nernst relation; corrosion potentials in soils and pipelines depend on local ion activities and oxygen partial pressure; and every biological membrane potential — the resting potential of a neuron, the proton-motive force across a mitochondrial membrane — is a Nernst potential for one or more ions. The equation is the bridge between the thermodynamic quantity ΔG, established via Gibbs Energy, Cell Potential, and Spontaneity, and the directly measurable, concentration-sensitive quantity E. The big picture is simple once stated: E and ΔG are proportional (ΔG = −nFE), and ΔG itself depends logarithmically on reaction quotient Q through the standard isotherm ΔG = ΔG° + RT ln Q. Substituting one into the other and dividing through by −nF produces the Nernst equation directly — no new physics is required, only algebra applied to two relations already in hand. The equation therefore inherits everything about spontaneity, equilibrium, and the sign conventions of cell potentials, but re-expresses them as a function of instantaneous composition rather than a fixed reference state. This article rebuilds the equation from the reaction isotherm, walks through the reduction to the familiar 0.0592 V/n form at 25 °C, and works three problems of increasing difficulty: a concentration cell, a battery under partial discharge, and a gas-electrode potential requiring careful handling of activities and partial pressures. It closes with the equation's role in pH Dependence of Redox Potentials and in constructing Pourbaix Diagrams, both of which are just the Nernst equation with H⁺ written explicitly into Q.

Key takeaways

The Nernst equation converts the standard cell potential E° into the actual potential E at any composition, temperature, and pressure.
It is derived from two known relations: ΔG = −nFE and ΔG = ΔG° + RT ln Q — no independent postulate needed.
The working form is E = E° − (RT/nF) ln Q, which becomes E = E° − (0.0592 V/n) log Q at 25 °C.
Q uses activities: species in their standard states (pure solids/liquids, solvent) drop out; gases enter as P/P°, solutes as concentration/c°.
At equilibrium E = 0 and Q = K, giving ΔG° = −RT ln K = −nFE°, linking three quantities in one triangle.
Concentration cells (E° = 0) exist purely because of the ln Q term — proof that the equation adds real physics beyond E°.
Key idea: cell potential is a logarithmic thermometer for reaction quotient — it reports how far a redox system sits from equilibrium, not just which way it runs.

The core insight

Cell potential E is not an intrinsic property of a reaction; it is the "distance" (in RT/nF units, via ln Q) between the current composition and the equilibrium composition, anchored by the standard potential E° that fixes that distance's zero point at unit activities.

Where this lives today

Every glass pH electrode is a live demonstration of the Nernst equation: the glass membrane develops a potential that follows $E = \text{const} - 0.05916\,\text{pH}$ at 25 °C, and pH meters are simply voltmeters with the inverse relation pre-programmed in — this is the direct analytical descendant of Example 3 above, and the starting point for the full treatment in pH Dependence of Redox Potentials. Ion-selective electrodes for K⁺, Na⁺, Ca²⁺, F⁻, and NH₄⁺ used in clinical blood-gas analyzers, water-quality testing, and industrial process control all report concentration by measuring a Nernstian voltage and solving for activity. In biology, the Nernst potential for a single permeant ion is the starting building block of the Goldman-Hodgkin-Katz equation used to compute resting membrane potentials of neurons; the ≈ −70 mV resting potential of a typical neuron reflects Nernst potentials for K⁺, Na⁺, and Cl⁻ weighted by their membrane permeabilities. Mitochondrial and chloroplast bioenergetics quantify the proton-motive force driving ATP synthase using an explicitly Nernstian term for the transmembrane pH gradient. In materials and energy engineering, battery state-of-charge estimation algorithms in electric vehicles and grid storage use Nernst-equation-based models (layered with kinetic and resistive corrections) to infer remaining capacity from open-circuit voltage, exactly the logic of the lead-acid example above. Corrosion engineers use the equation to predict the potential of metal surfaces in soils or seawater as a function of dissolved oxygen partial pressure and pH, feeding directly into the construction of Pourbaix Diagrams that map regions of corrosion, passivation, and immunity for a metal across pH and potential space. Fuel cells operating at elevated temperature must use the temperature-corrected form of the constant (not 0.0592 V), since solid-oxide fuel cells run near 800–1000 °C, where the RT/F prefactor is nearly three times its room-temperature value.

Ask the tutor: How does the Nernst equation combine with acid–base equilibria to generate the sloped boundary lines on a Pourbaix diagram?

The physical picture

Start with the Daniell cell: Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s). At standard conditions — both ions at 1 M activity, 298 K, 1 bar — the cell delivers E° = +1.10 V, obtained from tabulated standard reduction potentials as E°cell = E°cathode − E°anode = (+0.34 V) − (−0.76 V). But run the same cell with 2.0 M Zn²⁺ and 0.010 M Cu²⁺: now the reaction quotient for

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

is Q = [Zn²⁺]/[Cu²⁺] = 200, far above 1. A large Q means products are "over-supplied" relative to standard state, so the forward reaction has less driving force left — the voltage must drop below 1.10 V. The Nernst equation quantifies exactly how much. The mechanism is thermodynamic, not electrical: nothing new happens at the electrode surfaces. The electrons still transfer with the same intrinsic energy per mole of reaction; what changes is the concentration-dependent part of the Gibbs energy, exactly the RT ln Q correction familiar from equilibrium theory. Because ΔG and E are linearly related through ΔG = −nFE at every composition (not just at standard state), the same ln Q correction that shifts ΔG must shift E, scaled by the constant −1/(nF). A particularly clean illustration is the concentration cell: two identical electrodes, say Ag | Ag⁺, one in 1.0 M Ag⁺ and one in 0.0010 M Ag⁺, joined by a salt bridge. Since both half-reactions are chemically identical, E° = 0 exactly — tables would predict zero voltage. Yet a real voltmeter reads a nonzero value, because Q ≠ 1: silver ion is being diluted from the concentrated side to the dilute side, and this dilution alone is thermodynamically downhill, generating a measurable EMF with no net "chemistry" beyond concentration equalization. This is the strongest evidence that the ln Q term in the Nernst equation is a genuine, independent physical effect, not just a bookkeeping label on E°.

Figure

Daniell cell wired to show the Nernst shift from standard to actual conditions

Figure

Concentration cell: identical chemistry, nonzero voltage from Q alone

The core ideas

The Nernst equation in its general form:

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

where: $E$ is the actual cell potential (V); $E^{\circ}$ is the standard cell potential (V, all activities = 1); $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ ; $T$ is absolute temperature (K); $n$ is the number of moles of electrons transferred in the balanced overall reaction; $F = 96{,}485\ \text{C mol}^{-1}$ is the Faraday constant; $Q$ is the reaction quotient built from activities of products over reactants, each raised to its stoichiometric coefficient. Reading: as Q rises above 1 (products favored/accumulated), the correction term is subtracted, lowering E; as Q falls below 1 (reactants still abundant), ln Q is negative, so E rises above E°. At Q = 1 the correction vanishes and E = E° exactly, regardless of T — this is why E° is defined at unit activity, not at any particular temperature. At 298.15 K, converting to base-10 logarithms and inserting constants:

E = E^{\circ} - \frac{0.05916\ \text{V}}{n}\log_{10} Q

where: the numeral 0.05916 V is $\dfrac{RT\ln 10}{F}$ evaluated at $T = 298.15$ K; all other symbols as above. Reading: this is the form used in essentially all room-temperature laboratory calculations; the constant scales with T, so it is not 0.0592 at body temperature (310 K, constant ≈ 0.0615 V) or in a high-temperature fuel cell. At equilibrium, no net current flows and E = 0, while Q takes its equilibrium value K:

0 = E^{\circ} - \frac{RT}{nF}\ln K \quad\Longrightarrow\quad \ln K = \frac{nFE^{\circ}}{RT}

where: $K$ is the thermodynamic equilibrium constant for the overall redox reaction, and all other symbols are as above. Reading: this closes the triangle ΔG° ↔ E° ↔ K — any one determines the other two — and is the bridge used constantly in Pourbaix Diagrams to convert tabulated potentials into equilibrium activities of H⁺ and dissolved species.

Derivation AFrom the reaction isotherm to the Nernst equation

Goal. Show that E = E° − (RT/nF) ln Q follows directly from the van't Hoff reaction isotherm and the definition of cell potential, with no additional assumption.

Step 1 — the reaction isotherm. For any reaction at fixed T and P, the Gibbs energy change under actual (non-standard) activities is related to the standard Gibbs energy change by

\Delta G = \Delta G^{\circ} + RT\ln Q

This is a general thermodynamic identity — it holds for any reaction, redox or not — derived from the chemical potentials $\mu_i = \mu_i^{\circ} + RT\ln a_i$ of each species, summed with stoichiometric coefficients. It carries no electrochemical content yet.

Step 2 — the electrical work identity. For a reversible electrochemical cell, the maximum non-expansion work extractable equals the Gibbs energy change, and that work equals the charge moved times the potential:

\Delta G = -nFE \qquad \text{and} \qquad \Delta G^{\circ} = -nFE^{\circ}

Here $n$ is the number of electrons in the balanced overall equation, $F$ converts moles of electrons to coulombs, and the minus sign encodes the convention that spontaneous cells (ΔG < 0) produce positive E.

Step 3 — substitute. Insert both Step 2 expressions into the isotherm of Step 1:

-nFE = -nFE^{\circ} + RT\ln Q

Step 4 — isolate E. Divide every term by $-nF$ :

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

This is the Nernst equation. Notice that no new empirical constant was introduced: R, T, F, and n were already present in the isotherm and the work identity. The equation is a re-expression, not a new law — which is exactly why it must reduce, at Q = 1, to E = E° with zero correction, and why setting E = 0 must reproduce the standard ΔG°–K relation from Gibbs Energy, Cell Potential, and Spontaneity.

Consistency check (limiting case). As $Q \to 1$ (all species at standard-state activity, the defining condition for E°), $\ln Q \to 0$ and $E \to E^{\circ}$ exactly, independent of $T$ — confirming the algebra introduces no spurious temperature dependence at the reference point.

Go deeper · Derivation B · 2 minThe half-cell (Nernst) potential and why solids/solvent drop out of Q

Derivation BThe half-cell (Nernst) potential and why solids/solvent drop out of Q

Goal. Derive the single-electrode form of the Nernst equation and show precisely why pure solids, pure liquids, and solvent do not appear in Q, using the general half-reaction M $^{n+}$ + n e⁻ → M(s) as the working example, then generalize.

Step 1 — write the half-reaction isotherm. Treat the reduction M $^{n+}$ (aq) + n e⁻ → M(s) as its own reaction with its own ΔG. The chemical potential of each species enters as $\mu_i^\circ + RT\ln a_i$ . For the metal ion, $a_{\text{M}^{n+}} = [\text{M}^{n+}]/c^{\circ}$ (dimensionless, referenced to $c^{\circ}=1$ M); for the solid metal, because pure condensed phases are taken at their own standard state (the pure substance itself), $a_{\text{M(s)}} \equiv 1$ identically, not approximately — it is a definition of the activity scale for pure solids/liquids, not an approximation that becomes exact only in some limit.

Step 2 — build Q for the half-reaction. For reduction (electrons as reactants, written on the left),

Q_{\text{half}} = \frac{a_{\text{M(s)}}}{a_{\text{M}^{n+}}} = \frac{1}{[\text{M}^{n+}]/c^{\circ}}

so the solid contributes a factor of exactly 1 and disappears from the expression — not because its "concentration is large" but because its activity is defined as unity in its standard state.

Step 3 — apply the Nernst equation to the half-reaction.

E_{\text{half}} = E^{\circ}_{\text{half}} - \frac{RT}{nF}\ln\!\left(\frac{1}{a_{\text{M}^{n+}}}\right) = E^{\circ}_{\text{half}} + \frac{RT}{nF}\ln a_{\text{M}^{n+}}

This shows explicitly that increasing metal-ion activity raises the reduction potential of that couple — more oxidized species available to be reduced makes reduction more favorable, exactly as chemical intuition demands.

Step 4 — assemble the full cell and generalize the solid/solvent rule. For any overall reaction, $Q$ is built the same way, term by term: solutes contribute $[\,i\,]/c^\circ$ , gases contribute $P_i/P^\circ$ (with $P^\circ = 1$ bar), and every pure solid, pure liquid, and — critically — the solvent when it is present in vast excess (its mole fraction barely changes, so its activity stays pinned near 1) are omitted, each contributing a multiplicative factor of exactly 1 to Q. This is why, for example, the acidified permanganate half-reaction

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)

has water omitted from Q (it is the solvent) but H⁺ retained explicitly with exponent 8 — the seed of pH Dependence of Redox Potentials, where that eighth-power H⁺ term makes MnO₄⁻ potentials swing dramatically with pH.

Consistency check (limiting case). Setting $n=1$ and a single univalent ion recovers exactly the textbook single-ion Nernst potential $E = E^{\circ} + (RT/F)\ln[\text{M}^+]$ , the form used directly for membrane-potential calculations of monovalent ions such as K⁺ and Na⁺.

Worked examples

Example 1 — Foundational: the Daniell cell away from standard state. Take Zn(s) | Zn²⁺(2.0 M) ‖ Cu²⁺(0.010 M) | Cu(s), T = 298 K. From tables, $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34$ V, $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76$ V, so

E^{\circ}_{\text{cell}} = 0.34 - (-0.76) = 1.10\ \text{V}

The overall reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with $n = 2$ (two electrons transferred per formula unit — read directly off either half-reaction, which each involve 2 e⁻). Both metals are pure solids (activity 1), so

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{2.0}{0.010} = 200

Apply the 25 °C form:

E = 1.10 - \frac{0.05916}{2}\log_{10}(200)

\log_{10}(200) = 2.301

E = 1.10 - \frac{0.05916}{2}(2.301) = 1.10 - 0.0681 = 1.032\ \text{V}

The voltage drops by about 68 mV from standard, because the actual composition (Zn²⁺-rich, Cu²⁺-depleted) is closer to the products-favored side than the standard 1 M/1 M reference. Limiting case check: if both concentrations were forced to 1 M, Q = 1, log Q = 0, and E → 1.10 V = E° exactly, confirming the correction term vanishes at the defining reference composition.

Example 2 — Applied: voltage of a partially discharged lead-acid battery cell. The lead-acid cell reaction is

\text{Pb(s)} + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)

with $E^\circ = 2.05$ V and $n = 2$ . Sulfuric acid participates explicitly as H⁺ and HSO₄⁻/SO₄²⁻ in the detailed mechanism, but a common simplified treatment writes the overall $Q$ using the activity of H₂SO₄ (through H⁺ and HSO₄⁻ combined) squared in the denominator, while PbSO₄(s), Pb(s), PbO₂(s), and H₂O(l) — all pure condensed phases or solvent — drop out:

Q = \frac{1}{a_{\text{H}^+}^{2}\,a_{\text{HSO}_4^-}^{2}}

Suppose discharge has lowered the effective sulfuric acid activity product so that $a_{\text{H}^+}^2 a_{\text{HSO}_4^-}^2 = 0.15$ (well below the standard-state value of 1, since acid is consumed as the battery discharges and PbSO₄ builds up on the plates). Then

Q = \frac{1}{0.15} = 6.67

E = 2.05 - \frac{0.05916}{2}\log_{10}(6.67) = 2.05 - \frac{0.05916}{2}(0.824)

E = 2.05 - 0.0244 = 2.026\ \text{V}

A drop of only about 24 mV per cell for a substantial degree of acid depletion — consistent with the everyday observation that lead-acid batteries hold a fairly flat voltage over much of their discharge curve, only falling sharply near full depletion where $a_{\text{H}^+}$ collapses and $Q$ grows very large. Limiting case check: as the battery approaches full discharge and acid activity products fall toward zero, $Q \to \infty$ , $\log Q \to +\infty$ , and E is driven toward zero (and eventually negative), correctly predicting the characteristic voltage "knee" at end of discharge.

Example 3 — Exam/competition level: a hydrogen-gas half-cell potential with non-unit pressure and pH. Consider the standard hydrogen electrode operated at non-standard conditions: $P_{\text{H}_2} = 0.50$ bar, pH = 4.00, T = 298 K. The half-reaction (as reduction) is

2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g), \qquad E^\circ = 0.000\ \text{V (by definition)}

with $n = 2$ . Build Q for the reduction as written (product over reactant activities):

Q = \frac{P_{\text{H}_2}/P^\circ}{a_{\text{H}^+}^{2}}

From pH = 4.00, $a_{\text{H}^+} = 10^{-4.00} = 1.00\times10^{-4}$ , so $a_{\text{H}^+}^2 = 1.00\times10^{-8}$ . With $P_{\text{H}_2}/P^\circ = 0.50$ :

Q = \frac{0.50}{1.00\times10^{-8}} = 5.0\times10^{7}

\log_{10}Q = \log_{10}(5.0\times10^{7}) = 7.699

E = 0.000 - \frac{0.05916}{2}(7.699) = -0.2278\ \text{V}

So this hydrogen electrode reads about −0.228 V versus the standard hydrogen electrode, purely because it is more acidic-poor... wait — check the sign physically: lowering $a_{\text{H}^+}$ (less acidic, higher pH) makes the reduction 2H⁺ + 2e⁻ → H₂ less favorable (fewer H⁺ available to reduce), so E for this couple should indeed be pushed negative relative to 0.000 V — matching the calculation. This exact relation, $E_{\text{H}^+/\text{H}_2} = -0.05916\times\text{pH}$ V at 1 bar H₂ and 25 °C, is the single most-used identity in constructing the water-stability lines of Pourbaix Diagrams. Limiting case check: at pH = 0 with $P_{\text{H}_2} = 1$ bar exactly (true standard state), $Q = 1$ , and $E \to 0.000$ V = E°, as required.

Limiting cases, scope & common traps

Situation	What the Nernst equation predicts	Why
Q = 1 (all activities standard)	E = E° exactly, for any T	ln Q = 0 kills the correction term
Q → equilibrium value K	E → 0	no net driving force; matches ΔG = 0
Pure solids/liquids/solvent in Q	Contribute a factor of exactly 1	activity is defined as 1 in their own standard state, not merely "large amount"
Very dilute redox-active species	ln Q term dominates, E can deviate strongly from E°	log divergence as concentration → 0 (E → ±∞ formally; in practice other effects intervene first)
Same couple, two concentrations (concentration cell)	E° = 0, E ≠ 0	ln Q alone drives a real, measurable voltage
High temperature	The 0.0592 V/n constant itself scales with T	constant is (RT ln10)/F, not fixed at 0.0592
Multi-electron, multi-H⁺ half-reactions (e.g. MnO₄⁻/Mn²⁺)	E extremely sensitive to pH	H⁺ enters Q raised to a large exponent (8 here)

The chemist behind it

Origins · 1 minWalther Hermann Nernst and how this was discovered

Walther Hermann Nernst

1864–1941 · German physical chemist, Berlin/Göttingen, turn of the 20th century

Nernst built his reputation studying the thermodynamics of electrochemical cells and dilute solutions, work that culminated in the 1889 equation bearing his name, derived using the emerging tools of chemical thermodynamics from Gibbs and van't Hoff. He later became one of the founders of modern physical chemistry through the Nernst Heat Theorem (an early statement of the Third Law of Thermodynamics), for which he won the 1920 Nobel Prize in Chemistry. Nernst was a demanding, sharp-tongued experimentalist who also dabbled in unlikely inventions — an electrically heated glass-filament lamp (the "Nernst lamp") among them — before his later years were darkened by the rise of Nazism, under which his Jewish ancestry cost him his position and, tragically, several family members their lives.

History

- 1889 — Nernst publishes the equation relating electrode potential to ion activity, built on van't Hoff's reaction-isotherm work.

- 1893 — Nernst's textbook Theoretische Chemie helps establish physical chemistry as a discipline, spreading the equation's use widely.

- 1906 — Nernst proposes the Heat Theorem, precursor to the Third Law of Thermodynamics.

- 1920 — Nernst awarded the Nobel Prize in Chemistry for thermochemistry work.

- 1930s — Ion-selective electrode theory and early pH glass electrodes formalize the equation's role in analytical chemistry.

Why does E° not depend on concentration, but E does?

E° is defined at a fixed reference composition — all activities equal to 1 (and gases at 1 bar) — so it is a single number characteristic of the reaction itself, tabulated once and used universally. E is the potential at whatever composition actually exists in the cell at the moment of measurement. The Nernst equation is precisely the bridge: E equals E° corrected by −(RT/nF) ln Q, where Q encodes the deviation of the real composition from the standard reference. As a reaction proceeds and concentrations change, Q changes continuously, so E changes continuously too, even though E° remains fixed.

Why do pure solids and liquids not appear in the reaction quotient Q?

Activity for a pure solid or pure liquid is defined as exactly 1 relative to its own standard state (the pure substance at 1 bar and the temperature of interest). Adding more or less of a pure solid does not change the chemical potential of the substance already present — there is no internal concentration gradient to speak of within a single pure phase. Because Q is built from activities, not amounts, and each of these activities equals 1 by definition, they contribute a multiplicative factor of 1 to Q and simply do not appear when Q is written out.

Does the 0.0592 V constant ever change?

Yes — the constant is (RT ln10)/F, and 0.05916 V is its value only at T = 298.15 K. At body temperature (310 K) it becomes about 0.0615 V; in a solid-oxide fuel cell operating near 1000 K it is roughly 0.198 V, more than triple the familiar room-temperature value. Any calculation performed at a temperature other than 25 °C must recompute this prefactor from R, T, and F directly rather than reusing 0.0592.

Can the Nernst equation predict a negative cell potential?

Yes, and this is chemically meaningful: a negative E signals that the reaction as written is non-spontaneous under the actual conditions, even if E° for that same reaction was positive. This happens when Q is large enough that the −(RT/nF) ln Q correction exceeds E° itself. Practically, this is what happens as a battery nears full discharge — Q grows very large as reactants are depleted, and E falls, eventually toward zero and, if pushed further, negative, at which point the reaction reverses direction spontaneously.

How is the Nernst equation used to build a Pourbaix diagram?

A Pourbaix diagram plots potential E against pH, showing which redox and acid–base species are thermodynamically stable in each region. Each line on the diagram is the Nernst equation for one specific half-reaction, rewritten with H⁺ appearing explicitly in Q (as in the MnO₄⁻/Mn²⁺ or H⁺/H₂ examples above) so that E becomes a linear function of pH with a slope proportional to (number of H⁺)/(number of electrons). Boundaries between stability regions are literally Nernst-equation lines; the whole diagram is a Nernst equation solved and plotted for every relevant couple simultaneously.

Why does a concentration cell eventually stop producing voltage?

As the cell operates, ions migrate from the concentrated half-cell to the dilute half-cell (or the redox couple reacts to transfer charge), driving Q toward 1. Since E = −(RT/nF) ln Q and E° = 0 for a concentration cell, E depends entirely on how far Q is from 1. As the two concentrations approach each other, ln Q → 0 and E → 0 — the cell has reached the equilibrium composition where no further net driving force exists, exactly analogous to any other redox cell approaching Q = K.

Is the Nernst equation exact, or an approximation?

Written in terms of true thermodynamic activities, the Nernst equation is exact — it follows rigorously from the combination of the reaction isotherm and the ΔG = −nFE identity, both themselves exact thermodynamic relations for a reversible cell. The approximations enter only when activities are replaced by molar concentrations (ignoring activity coefficients) or when a cell is not operating reversibly (significant current flow introduces kinetic overpotentials not captured by this purely thermodynamic equation).