Why 0.0592? Doesn't the coefficient change with temperature?

Yes — the coefficient in the log-base-10 form is exactly RT·ln 10 / F, which at 298.15 K evaluates to 0.05916 V (textbooks universally round to 0.0592 V). It scales linearly with T, so at 310.15 K (body temperature) the same coefficient is 0.06154 V, and at 273.15 K (0 °C) it drops to 0.05418 V. Use the 0.0592/n shortcut only at 25 °C; this calculator uses the general RT/nF form so it works at any temperature you enter.

When is Q written products-over-reactants versus reactants-over-products?

Q is always products divided by reactants, but the identities of 'products' and 'reactants' depend on which direction you wrote the half-reaction. Because standard reduction potentials are tabulated for reductions (ox + n e⁻ → red), the Nernst equation as usually written uses Q = [red]/[ox] for a single half-cell — reduced form in the numerator, oxidized form in the denominator. For a whole cell reaction, Q follows the balanced overall equation. Flip the reaction and Q inverts (ln Q flips sign), so E° flips sign too — the net E stays the same.

What if my E° is tabulated as an oxidation potential instead of a reduction?

Modern IUPAC convention is to tabulate reduction potentials only. If you have an oxidation potential (older tables, or Latimer diagrams for the reverse direction), negate it before entering: E°(reduction) = −E°(oxidation). Do the same for Q if you also flipped the half-reaction direction. The safest workflow is: rewrite everything as reductions first, then plug into the Nernst equation.

Why does the calculator reject Q = 0 or negative Q?

Because ln Q is undefined for Q ≤ 0 — mathematically the equation has no solution there. Physically, Q = 0 would mean either zero product concentration (only reactants present, reaction hasn't started) or zero reactant concentration (reaction complete), both extremes at which E diverges to ±∞ under the ideal-solution assumption. Real electrochemistry near these limits is dominated by other effects (kinetic overpotential, diffusion, junction potentials) that this simple equation doesn't capture.

Can I use activities instead of concentrations for Q?

Yes — the rigorous form of Q uses activities, and activities equal concentrations only in the ideal-dilute limit. For a general lab problem at bench ionic strength (10 mM to 1 M), concentration-based Q typically drifts by 5–15 % from the activity-based value. For introductory-course accuracy the concentration form is fine; for careful analytical work, compute activity coefficients (Debye–Hückel or extended forms) and use activities.

Nernst Equation Calculator — E, E°, n, Q Solver

The Nernst equation predicts how the potential of an electrochemical cell shifts when the concentrations move away from standard state. It is the single equation that connects thermodynamic driving force ( $\Delta G$ ), the reaction quotient $Q$ , and the measurable cell voltage $E$ — and it is the backbone of every reference-electrode calibration, every ion-selective electrode, and every potentiometric titration.

Nernst Equation Calculator

Solve E = E° − (RT/nF)·ln Q for any of E, E°, n, or Q. Temperature defaults to 298.15 K (25 °C); the 25 °C shortcut E = E° − (0.0592/n)·log₁₀ Q is a display toggle. Concentration mode builds Q from [ox] and [red] using the reduction convention.

Equations used

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

E = E^{\circ} - \frac{0.0592}{n} lo g_{10} Q (25^{\circ} C only)

R = 8.314 462 618 J mol⁻¹ K⁻¹ and F = 96 485.332 12 C mol⁻¹ (CODATA 2018, exact by SI). At T = 298.15 K, RT·ln 10 / F ≈ 0.05916 V, commonly rounded to 0.0592 V in textbooks (Harris QCA 10e §14; Atkins PChem 11e §6.6). Q is written in the reduction direction (products / reactants for the reduction half-reaction). Intermediate arithmetic uses SI; rounding is a display-only step.

The equation

The Nernst equation is $E = E^\circ - (RT/(nF))\,\ln Q$ , where

E is the observed cell (or half-cell) potential in volts
E° is the standard potential — the value tabulated for unit activities of all species
R is the molar gas constant, $R = 8.314\,462\,618$ J mol $^{-1}$ K $^{-1}$ (CODATA 2018)
T is the absolute temperature in kelvin
n is the number of moles of electrons transferred in the balanced half-reaction (positive integer)
F is the Faraday constant, $F = 96\,485.332\,12$ C mol $^{-1}$ (CODATA 2018)
Q is the dimensionless reaction quotient, evaluated with activities (or approximated with concentrations)

The 25 °C shortcut

At $T = 298.15$ K, the group $RT\ln 10 / F$ evaluates to 0.05916 V, universally rounded to 0.0592 V in undergraduate texts (Harris QCA 10e §14; Atkins PChem 11e §6.6). Switching from natural log to base-10 log gives the familiar shortcut

$E = E^\circ - (0.0592/n)\,\log_{10} Q$ (25 °C only).

The coefficient scales linearly with T, so the shortcut is only valid near room temperature. This calculator does the arithmetic in the general RT/nF form so it works at any T you enter; the checkbox above swaps the display to the 0.0592/n form when you’re at 25 °C.

Sign conventions

Two rules keep sign errors out of Nernst problems:

Every half-reaction is written as a reduction. Standard reduction potentials (SRP) are tabulated only for the reduction direction. If a problem gives an oxidation potential, negate it before use.
Q is products over reactants for the reaction as written. For a half-reaction Ox + n e⁻ → Red, that means Q_half = [Red] / [Ox]. Flip the half-reaction to oxidation and Q inverts — but E° flips sign at the same time, so the net E is unchanged.

For a whole cell, subtract the anode reduction potential from the cathode reduction potential: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ . Then write $Q_{\text{cell}}$ from the overall balanced reaction.

Worked example — Daniell cell (25 °C)

The classic Zn/Cu Daniell cell:

Zn(s) | Zn²⁺(0.10 M) ‖ Cu²⁺(1.0 M) | Cu(s)

with cathode Cu²⁺ + 2 e⁻ → Cu (E° = +0.34 V) and anode Zn²⁺ + 2 e⁻ → Zn (E° = −0.76 V), so E°_cell = 0.34 − (−0.76) = 1.10 V. The overall reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) has Q = [Zn²⁺]/[Cu²⁺] = 0.10/1.0 = 0.10 and n = 2.

Daniell cell at 25 °C

E°: 1.10 V
n: 2
Q: [Zn²⁺]/[Cu²⁺] = 0.10
T: 298.15 K

Formula
$E = E^{\circ} - \frac{0.0592}{n} lo g_{10} Q$
Substitute
$E = 1.10 - \frac{0.0592}{2} lo g_{10} (0.10)$
Note
Q < 1 pushes E above E° — the reaction is farther from equilibrium than at standard state. Source: Harris QCA 10e Example 14-1, Atkins PChem 11e §6.6.
Result
$E = 1.10 + 0.02960 = 1.1296 V$

Temperature effects

The Nernst coefficient $RT/(nF)$ scales linearly with T. At 273.15 K (0 °C) the base-10 coefficient is 0.05418 V; at 298.15 K it is 0.05916 V; at 310.15 K (body temperature — the working point for physiological pH sensors and mitochondrial membrane potentials) it is 0.06154 V. A pH electrode calibrated at room temperature and used in a 37 °C incubator will read about 4 % low on the mV-per-decade slope unless the meter’s temperature compensation is active.

Common pitfalls

Using 0.0592/n at anything other than 25 °C. Use the general RT/nF form or scale 0.0592 by T/298.15.
Reversing Q. Q for a single half-reaction is [reduced]/[oxidized] under the reduction convention. It is easy to write it upside down; the sign of the answer will be wrong by the sign of ln Q.
Forgetting that solids and pure liquids have activity 1. Metals in a Daniell cell drop out of Q.
Mixing pKₐ tables with reduction potentials. They live at different temperatures; SRP tables are 25 °C and infinite dilution, activity-based.
Using Q ≤ 0 as an edge case. The equation is undefined there — this calculator rejects it explicitly rather than returning NaN.

For the acid–base couple that behaves electrochemically like a Nernstian sensor, see the pH calculator. To prepare the electrolyte solutions that go into Q, use the molarity calculator and dilution calculator. Buffer preparation for reference-electrode fills is covered by the buffer calculator.

Practice problems

Attempt each on paper, then expand the worked solution to check your arithmetic and sig-figs.

Problem 1 Easy A Daniell cell (Zn | Zn²⁺(0.10 M) || Cu²⁺(1.0 M) | Cu) at 25 °C has E° = 1.10 V and n = 2. What is E?

Answer: E ≈ 1.130 V

Solve for E — Daniell cell

E°: 1.10 V
n: 2
Q: [Zn²⁺]/[Cu²⁺] = 0.10 / 1.0 = 0.10
T: 298.15 K

Formula
$E = E^{\circ} - \frac{0.0592}{n} lo g_{10} Q$
Substitute
$E = 1.10 - \frac{0.0592}{2} lo g_{10} (0.10)$
Note
Because [Zn²⁺] is a tenth of [Cu²⁺], Q < 1, and −log₁₀ Q is positive — E climbs above E°. The general RT/nF form gives the same answer to four decimals at 25 °C.
Result
$E = 1.10 + 0.0296 = 1.1296 \approx 1.130 V$

Problem 2 Intermediate For Fe³⁺/Fe²⁺ (E° = +0.771 V, n = 1) with [Fe³⁺]/[Fe²⁺] = 100 at 25 °C, what is E?

Answer: E ≈ 0.889 V

Solve for E — Fe³⁺/Fe²⁺

E°: +0.771 V
n: 1
Q: [Fe²⁺]/[Fe³⁺] = 1/100 = 0.01
T: 298.15 K

Formula
$E = E^{\circ} - \frac{0.0592}{n} lo g_{10} Q$
Substitute
$E = 0.771 - 0.0592 \cdot lo g_{10} (0.01)$
Note
Under reduction convention Q = [red]/[ox] = [Fe²⁺]/[Fe³⁺]. More Fe³⁺ than Fe²⁺ means Q is small, which makes the log term positive and pulls E above the standard value.
Result
$E = 0.771 + 0.1184 \approx 0.889 V$

Problem 3 Intermediate A cell with E° = +0.34 V and n = 2 measures E = +0.28 V at 25 °C. What is Q?

Answer: Q ≈ 108

Solve for Q

E: +0.28 V
E°: +0.34 V
n: 2
T: 298.15 K

Formula
$Q = exp (\frac{( E ^{\circ} - E ) n F}{R T})$
Substitute
$Q = exp (\frac{( 0.34 - 0.28 ) ( 2 ) ( 96485 )}{( 8.3145 ) ( 298.15 )})$
Note
E below E° means Q > 1: products dominate, driving the reaction closer to equilibrium. This is a common titration-cell setup — as more product forms during a titration, E drifts toward E°_eq.
Result
$Q = exp (4.674) \approx 107.1$

Frequently asked questions

Why 0.0592? Doesn't the coefficient change with temperature?: Yes — the coefficient in the log-base-10 form is exactly RT·ln 10 / F, which at 298.15 K evaluates to 0.05916 V (textbooks universally round to 0.0592 V). It scales linearly with T, so at 310.15 K (body temperature) the same coefficient is 0.06154 V, and at 273.15 K (0 °C) it drops to 0.05418 V. Use the 0.0592/n shortcut only at 25 °C; this calculator uses the general RT/nF form so it works at any temperature you enter.
When is Q written products-over-reactants versus reactants-over-products?: Q is always products divided by reactants, but the identities of 'products' and 'reactants' depend on which direction you wrote the half-reaction. Because standard reduction potentials are tabulated for reductions (ox + n e⁻ → red), the Nernst equation as usually written uses Q = [red]/[ox] for a single half-cell — reduced form in the numerator, oxidized form in the denominator. For a whole cell reaction, Q follows the balanced overall equation. Flip the reaction and Q inverts (ln Q flips sign), so E° flips sign too — the net E stays the same.
What if my E° is tabulated as an oxidation potential instead of a reduction?: Modern IUPAC convention is to tabulate reduction potentials only. If you have an oxidation potential (older tables, or Latimer diagrams for the reverse direction), negate it before entering: E°(reduction) = −E°(oxidation). Do the same for Q if you also flipped the half-reaction direction. The safest workflow is: rewrite everything as reductions first, then plug into the Nernst equation.
Why does the calculator reject Q = 0 or negative Q?: Because ln Q is undefined for Q ≤ 0 — mathematically the equation has no solution there. Physically, Q = 0 would mean either zero product concentration (only reactants present, reaction hasn't started) or zero reactant concentration (reaction complete), both extremes at which E diverges to ±∞ under the ideal-solution assumption. Real electrochemistry near these limits is dominated by other effects (kinetic overpotential, diffusion, junction potentials) that this simple equation doesn't capture.
Can I use activities instead of concentrations for Q?: Yes — the rigorous form of Q uses activities, and activities equal concentrations only in the ideal-dilute limit. For a general lab problem at bench ionic strength (10 mM to 1 M), concentration-based Q typically drifts by 5–15 % from the activity-based value. For introductory-course accuracy the concentration form is fine; for careful analytical work, compute activity coefficients (Debye–Hückel or extended forms) and use activities.

Sources

Harris, Quantitative Chemical Analysis, 10th ed. (2020), §14 — Fundamentals of Electrochemistry (Example 14-1)
Atkins & de Paula, Physical Chemistry, 11th ed. (2018), §6.6 — The response of cells to concentration
IUPAC Gold Book — Nernst equation
CODATA 2018 recommended value for the Faraday constant, via NIST
CODATA 2018 recommended value for the molar gas constant R, via NIST

Last updated: 2026-09-18.