Nernst Equation Calculator

Work out the reduction potential of a cell from its standard potential and activities.

Exact SI gas and Faraday constants, with temperatures in Celsius, Fahrenheit or kelvins.

Updated October 15, 2026

CreatorFrank Zhao

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Introduction

A standard reduction potential describes exactly one situation: every dissolved species at unit activity, every gas at one bar, and the cell at 25 °C. Real cells almost never sit there. This page takes the tabulated value as its anchor and answers the question you actually have — with my activities and my temperature, what potential does this couple really show?

Two numbers do the moving. The ratio of the two activities sets how far the reaction is from its standard state, and the temperature sets how much each step of that ratio is worth in volts. A couple whose reduced form is scarce is easier to reduce, so its potential rises above the tabulated value; the same couple sitting in a concentrated product is harder to reduce, and its potential falls. The logarithm is what keeps the shift gentle: a ten-fold change in the ratio is worth tens of millivolts at room temperature, not volts.

The potential is the output of this page. The five measurements — standard potential, temperature, electrons transferred and the two activities — stay yours, and the result box is read-only by design; the last question below rearranges the same equation if you ever need to go the other way. For coursework, that makes this a check on a hand calculation; for lab and battery work, it answers why a cell reads what it reads, and how far the reading will drift as the reaction proceeds.

If what you have is two half-cell values rather than one, the cell EMF calculator combines them into the standard potential for the whole cell, which is the number that belongs in the first box here.

Any temperature above 0 K

Entered in Celsius, Fahrenheit or kelvin

Five measurements in, one potential out

The result box is the only output

Exact gas and Faraday constants

Every digit is used, not a rounded 8.31 and 96 500

Quick start

Type the standard reduction potential

Take the standard potential from a table for the half-reaction you are working with, or for the whole cell if you already have that. The sign belongs to the reduction as written: the copper couple $\mathrm{Cu^{2+}/Cu}$ is tabulated at +0.34 V and the zinc couple $\mathrm{Zn^{2+}/Zn}$ at −0.76 V. Anything from a few volts below zero to a few volts above is a normal entry.

Set the temperature of the cell

Use the temperature the cell is actually at, and switch the box to °F or K if that is how you measured it — the equation needs kelvin, and the conversion is done for you. Absolute zero and anything below it is refused, because the thermal term has no meaning there.

Enter the electrons transferred

This is the electron count of the reaction exactly as written: two for the copper couple, one for a silver couple. It divides the thermal term, so a wrong count scales your whole correction. A fractional entry is flagged as implausible but still computed.

Enter both activities

The first box is the numerator of the quotient inside the logarithm and the second box the denominator — for a reduction written as $\mathrm{Ox} + z\,\mathrm{e^-} \rightarrow \mathrm{Red}$ that is the reduced form over the oxidized form. Both must be greater than zero, and pure solids and liquids go in as 1.

Read the reduction potential

The result box fills in once all five measurements are present. A positive value means the reduction you wrote is spontaneous against the reference; the size of the number is how strongly.

A worked case, step by step

Take a cell with a standard potential of 2.25 V at 25 °C, two electrons transferred, and an activity ratio of 0.020 over 0.200. The ratio is 0.1, below the standard state, so the correction term is positive and the answer comes out above 2.25 V:

\frac{R\,T}{z\,F}

=

\frac{8.31446261815324 \times 298.15}{2 \times 96485.3321233100184}

=

0.0128463\ \mathrm{V}

E

=

2.25 - 0.0128463 \times \ln(0.1)

=

2.25 + 0.0295797

=

2.27958\ \mathrm{V}

The page returns 2.27958 V, which is 29.6 mV above the tabulated 2.25 V. Notice the size of that step: ten-fold dilution of the reduced side is worth about 30 mV, because the thermal term at 25 °C is only 12.8 mV and the logarithm of ten is 2.303. Work the same numbers with a ratio of 10 instead of 0.1 and the answer lands 29.6 mV below 2.25 V.

The equation and its constants

The Nernst equation corrects a standard potential for the two conditions that are never standard in practice — the activities and the temperature:

E = E_0 - \frac{R\,T}{z\,F}\,\ln\frac{a_{\mathrm{red}}}{a_{\mathrm{ox}}}

E

The reduction potential the couple shows at your activities and temperature, in volts — the result box.

E_0

The standard reduction potential from the table: unit activity, one bar for gases, 25 °C.

R

Molar gas constant, 8.31446261815324 J K⁻¹ mol⁻¹.

T

Absolute temperature in kelvin; the page converts your scale before using it.

z

Moles of electrons transferred in the reaction as written.

F

Faraday constant, 96485.3321233100184 C mol⁻¹.

a_{\mathrm{red}}

Activity of the form on the reduced side of the half-reaction — the numerator.

a_{\mathrm{ox}}

Activity of the form on the oxidized side — the denominator.

Both constants are exact numbers

Since the 2019 revision of the SI, the gas constant and the Faraday constant are not measured quantities any more: each is an exact product of defining constants. The page evaluates them at their full value rather than substituting a rounded 8.31 and 96 500, so the answer carries no avoidable rounding of its own:

R = k\,N_{\mathrm{A}}

=

8.31446261815324\ \mathrm{J\,K^{-1}\,mol^{-1}}

F = N_{\mathrm{A}}\,e

=

96485.3321233100184\ \mathrm{C\,mol^{-1}}

NIST lists both as exact constants — the molar gas constant and the Faraday constant — and the uncertainty column reads "exact" for each.

What one decade of activity ratio is worth

Collapse the thermal term into a single slope and the equation becomes easy to hold in your head: dividing or multiplying the ratio by ten changes the potential by a fixed number of millivolts, whatever the starting point.

\frac{R\,T}{F}\ln 10

=

0.0591593\ \mathrm{V}

\approx

0.05916\ \mathrm{V}

\text{per decade}

Divide by the number of electrons for other reactions: 0.0296 V per decade when two electrons move, 0.0197 V when three do. The slope is proportional to temperature rather than fixed, which is what makes this equation sensitive to how warm the cell is.

0 °C

0.0541988 V per decade

25 °C

0.0591593 V per decade

37 °C

0.0615404 V per decade

A tenth of a volt per decade is the reason ion-selective and pH electrodes are quoted as 59.16 mV per decade at room temperature and about 61.5 mV at body temperature: the electrode slope is this same thermal term.

Activities, and when a concentration will do

An activity is a dimensionless, effective concentration, and in dilute solution it is close enough to the molarity that the two can be used interchangeably. That is the assumption most hand calculations make. It stops holding as the solution gets stronger or the ions carry more charge, because the ions start to shield one another and the activity drops below the molarity.

When you need the real number, the activity is the activity coefficient times the concentration, $a = \gamma\,c$ , and the activity coefficient calculator gives you that coefficient from the ionic strength and the ion charge. Feeding an activity rather than a molarity is the single easiest way to make this page agree with a careful measurement.

What goes in the two boxes

A half-reduction: for $\mathrm{Ox} + z\,\mathrm{e^-} \rightarrow \mathrm{Red}$ the first box takes the reduced form and the second the oxidized form, so the quotient is $a_{\mathrm{red}}/a_{\mathrm{ox}}$ .
A whole cell: the same two boxes take the product side over the reactant side of the overall reaction — the reaction quotient Q. For the zinc–copper cell that is $a_{\mathrm{Zn^{2+}}}/a_{\mathrm{Cu^{2+}}}$ , and the two solids never appear.
Pure solids and liquids: enter 1. A metal electrode and liquid water are in their standard state by definition, so they drop out of the logarithm.
Gases: use the fugacity, which for the pressures a lab cell works at is the partial pressure in bar.

The page solves one way, on purpose

The potential is the output box and it cannot be typed into: the five measurements drive it, and the page will not silently decide which of your inputs to overwrite. If you have a measured potential and want an activity back, the rearrangement is short — it is the first question below.

Worked examples

A Daniell cell with a dilute product ion

Situation: A zinc–copper cell is set up with 0.0010 mol/L zinc sulfate and 1.0 mol/L copper sulfate, both at 25 °C. The standard potential of the cell is 1.10 V, the sum of +0.34 V for the copper couple and the sign-flipped −0.76 V for the zinc couple, with two electrons transferred.

What goes in

Standard potential

E_0 = 1.10\ \mathrm{V}

Temperature25 °C

Electrons transferred

z = 2

Reduced-form box (numerator)

a_{\mathrm{Zn^{2+}}} = 0.0010

Oxidized-form box (denominator)

a_{\mathrm{Cu^{2+}}} = 1.0

Result

Q = \frac{0.0010}{1.0} = 1.0 \times 10^{-3}

E

=

1.10 - 0.0128463 \times \ln(1.0 \times 10^{-3})

=

1.10 + 0.0887

=

1.18874\ \mathrm{V}

In practice: The cell starts 89 mV above its standard voltage because the product ion is scarce, and the reading falls as the cell runs: with the zinc side at 0.10 mol/L the same inputs return 1.12958 V, so two decades of product cost about 59 mV at this electron count.

A concentration cell, where the standard potential is zero

Situation: Two copper electrodes sit in copper sulfate solutions of 0.0010 mol/L and 1.0 mol/L, at 25 °C. Because both electrodes are the same couple, the standard potential of the transition from the concentrated solution to the dilute one is exactly 0 V — every volt of the reading comes from the ratio.

What goes in

Standard potential

E_0 = 0\ \mathrm{V}

Temperature25 °C

Electrons transferred

z = 2

Reduced-form box (numerator)

a_{\mathrm{Cu^{2+}}} = 0.0010

Oxidized-form box (denominator)

a_{\mathrm{Cu^{2+}}} = 1.0

Result

E = -\frac{R\,T}{2F}\,\ln\frac{0.0010}{1.0}

E = 0.0128463 \times 2.302585 \times 3 = 0.088739\ \mathrm{V}

In practice: The whole output is a measurement of the ratio: the dilute side is three decades below the concentrated one, and each further ten-fold step adds another 0.0296 V.

Getting the inputs right

Measure the cell, not the room

The slope is proportional to temperature, so a cell warm from its own reaction is not at the temperature written on the bench. The Daniell example above returns 1.18874 V at 25 °C and 1.19231 V at 37 °C with every other number unchanged — a 3.6 mV difference from temperature alone.

Balance first, then scale freely

The electron count is the one in the reaction exactly as written. Doubling every coefficient doubles the count and squares the ratio, and the potential does not move: $z = 4$ with a ratio of $1.0 \times 10^{-6}$ returns the same 1.18874 V as $z = 2$ with $1.0 \times 10^{-3}$ . A mismatch between the count and the ratio is a real error; scaling the whole reaction is not.

One number per box — pre-combine the coefficients

Each activity box takes a single number, so a coefficient in the balanced equation has to be worked in by hand: for $\mathrm{2H^+ + 2e^- \rightarrow H_2}$ the denominator is the square of the hydrogen ion activity. At pH 3 that square is $10^{-6}$ , and with the hydrogen partial pressure at 1 bar the page returns −0.177478 V.

Keep the direction consistent

Reversing a reaction flips both the sign of $E_0$ and the ratio. Do one or the other, not both. If your measured potential is negative and you expected positive, check which couple you wrote as the reduction before you touch either number — a double flip is the easiest way to hide a real sign error.

Solids, liquids and the reference

A pure metal electrode or the solvent water is 1 and contributes nothing to the logarithm, while a dissolved ion or a gas needs a real number. And remember what the tabulated $E_0$ is measured against — see the question on reference electrodes below if your reading comes from a silver chloride or calomel probe.

Check one number you already know

The standard hydrogen electrode is the sanity check built into this page: with both activities at 1 it returns exactly 0 V at any temperature, because the ratio is one and the logarithm vanishes. If your setup gives 0 V there, the inputs are wired the right way round.

Questions

Can I solve for one of the activities instead of the potential?

Not on this page — the potential is the output and the measurements are the inputs. The rearrangement takes one line, though: the ratio you need is the exponential of the potential difference scaled by the thermal term.

\frac{a_{\mathrm{red}}}{a_{\mathrm{ox}}}

=

e^{\,(E_0 - E)\,z\,F/(R\,T)}

Put the potential you measured in place of E, and divide or multiply by whichever activity you already know. One warning worth repeating: this assumes the cell is at equilibrium and that no other couple is setting the reading.

Why does the page sometimes warn about an input but still show a number?

There are two different kinds of complaint, and they behave differently on purpose.

A hard refusal — an activity at or below zero, or a temperature at or below absolute zero. There is no value for the logarithm or the thermal term, so the entry keeps your draft on screen with a red error and does not enter the calculation; the result stays at its previous value or blank.
A plausibility flag — a fractional electron count, or a potential more than 6 V away from zero. The arithmetic still runs and the number stays visible, drawn in red, because a finite answer you can inspect beats a blank box. Treat the flag as "check this input", not as a verdict on your chemistry.

What else can I work out from a cell potential?

Two things, both one multiplication away. The electrical work the reaction can do is the free energy change,

\Delta G = -z\,F\,E

and the standard potential is also the bridge to the equilibrium constant of the reaction, $\ln K = z\,F\,E_0/(R\,T)$ . If it is the energy you want rather than the voltage, the Gibbs free energy calculator takes it from there, and the equilibrium constant calculator converts in the other direction.

Does it matter which reference electrode I measured against?

It does. Tabulated standard potentials are quoted against the standard hydrogen electrode, which is assigned exactly 0 V by convention, so a half-cell potential is always a difference rather than an absolute quantity. A reading taken against a silver/silver chloride or calomel reference sits on a different zero.

Before comparing a measured value with a table, add your reference's own potential against the hydrogen electrode — the value printed on your probe's certificate, or the tabulated value for that reference electrode at your working temperature. Only then is the number on the same scale as the standard potential you typed in.

How far to trust it

This is the equilibrium potential, not the voltage you will measure. The equation is thermodynamics: it says nothing about how fast the electrode reactions run. Activation overpotential, concentration overpotential, liquid-junction potentials and the cell's own internal resistance all pull a real reading away from the calculated value, usually downwards.
Molarity is a stand-in for activity. In dilute solution the two agree closely enough, which is why the substitution is so common. In concentrated solutions, in solutions with multiply charged ions, and in anything with a high ionic strength the activity falls below the molarity, and the correction stops being small.
A single potential is only meaningful against its reference. The hydrogen electrode convention is what makes a table of standard potentials comparable at all. Mixing readings from different reference electrodes without converting them puts your numbers on two different scales.
The flags are plausibility checks, not validation. A fractional electron count and a potential more than 6 V from zero are almost always typing slips, and the page says so in red while keeping the number. They cannot tell you whether the standard potential you looked up belongs to the couple you meant, or whether your balanced equation is right.
Use it to plan and to check, not to certify. For battery design, corrosion assessment, industrial electrolysis or any decision where being tens of millivolts out carries a real cost, compare the result with measured data and have a qualified electrochemist review the conclusion.

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