Cell EMF Calculator

Find a galvanic cell EMF from its two electrode potentials.

mV, V, kV and MV supported, with reverse solving for either electrode potential.

Updated October 14, 2026

CreatorFrank Zhao

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Introduction

The electromotive force of a cell is the voltage between its two electrodes with nothing drawing current — the difference between one electrode's potential and the other's. Fill in any two of the three boxes on this page and the third is worked out for you, in whichever direction you left open.

Both electrode boxes take standard reduction potentials: the potential of a half-reaction measured against the standard hydrogen electrode, which is fixed at $0\ \mathrm{V}$ by definition. Zinc's $\mathrm{Zn^{2+}/Zn}$ couple is listed at $-0.76\ \mathrm{V}$ , copper's $\mathrm{Cu^{2+}/Cu}$ couple at $+0.34\ \mathrm{V}$ , so the zinc/copper Daniell cell comes out at $1.1\ \mathrm{V}$ .

That covers three jobs that come up again and again: checking a worked exercise where two half-cells are given, predicting the voltage of a cell built from two known couples, and recovering a missing electrode potential from a cell voltage that was measured or looked up. Whoever is doing it — a student, a lab technician, or someone comparing battery chemistries — the question is the same: which electrode is the anode, and how big is the gap between them?

Any of the three

Pin two values and the third is solved

mV, V, kV, MV

Every box reads in its own voltage unit

−3.04 V to +2.87 V

Range covered by the standard tables

One thing this page deliberately does not do: it works with the tabulated standard values, not with a cell sitting on a bench. Why those two numbers differ is the subject of the last section.

Quick start

Write both electrodes as reduction potentials

Take both values from the same table, in reduction form. If a source prints a half-reaction as an oxidation — the reverse direction, with the electrons on the right — flip the sign before you type it in. The zinc couple is $\mathrm{Zn^{2+}} + 2e^- \rightarrow \mathrm{Zn}$ , $-0.76\ \mathrm{V}$ in reduction form.

Put the lower potential in Anode, the higher one in Cathode

Oxidation happens at the anode, and the anode is the electrode that gives up electrons most easily — the one with the lower potential. Against copper at $+0.34\ \mathrm{V}$ , zinc at $-0.76\ \mathrm{V}$ is the anode and copper is the cathode. If the two boxes end up the other way round, the page still answers, but it marks the EMF in red and tells you why.

Read EMF, or leave a different box empty

With both electrode potentials in, EMF is the one that gets worked out. To go the other way, clear the box you want answered and fill in the cell voltage plus the other electrode: the cleared box becomes the result. You never need to rearrange anything by hand.

Check the answer against the working

For the zinc/copper cell the page returns $1.1\ \mathrm{V}$ , which is exactly what the subtraction gives:

E_{\mathrm{cell}}

=

E_{\mathrm{cathode}} - E_{\mathrm{anode}}

=

0.34 - (-0.76)

=

1.1\ \mathrm{V}

The same $1.1\ \mathrm{V}$ is $1{,}100\ \mathrm{mV}$ . If a datasheet quotes millivolts, switch that box's unit menu instead of converting by hand — the answer does not move, only the reading does.

Read the red flags before you trust the number

Three things are flagged rather than hidden. A red EMF means the subtraction came out negative, so the electrode boxes are swapped. A red electrode potential means a single half-cell was quoted beyond the range the tables cover. A red cell voltage above 6 V means no pair of half-reactions can deliver it. In every case the number stays on screen next to the explanation, so you can see how far off it is.

How the answer is worked out

There is only one statement behind all three boxes: the anode's potential plus the cell's electromotive force equals the cathode's potential. Written the usual way round, that is a single subtraction.

E_{\mathrm{cell}} = E_{\mathrm{cathode}} - E_{\mathrm{anode}}

Because the same statement can be rearranged, the page can also answer from the other side. Each empty box has its own form, and the result is the same arithmetic read in a different order:

E_{\mathrm{cell}} = E_{\mathrm{cathode}} - E_{\mathrm{anode}}

\qquad

E_{\mathrm{cathode}} = E_{\mathrm{anode}} + E_{\mathrm{cell}}

\qquad

E_{\mathrm{anode}} = E_{\mathrm{cathode}} - E_{\mathrm{cell}}

E_{\mathrm{cell}}

Electromotive force of the cell, always written positive for the spontaneous direction

E_{\mathrm{cathode}}

Standard reduction potential of the electrode where reduction happens, the higher of the two

E_{\mathrm{anode}}

Standard reduction potential of the electrode where oxidation happens, the lower of the two

f_{\mathrm{unit}}

Factor that converts the unit you selected into volts before the subtraction runs

Units never change the answer

Each box carries its own unit menu, and the four factors are exact powers of ten, so a potential typed in millivolts and the same potential typed in volts produce identical results. Converting happens on the display side only:

E_{\mathrm{V}} = E_{\mathrm{entered}} \times f_{\mathrm{unit}}

Menu entry	Multiply the entered value by	So 0.76 V is shown as
millivolts (mV)	0.001	760 mV
volts (V)	1	0.76 V
kilovolts (kV)	1 000	0.00076 kV
megavolts (MV)	1 000 000	0.00000076 MV

The two references both potentials have to share

Every standard potential is quoted against the same zero: the standard hydrogen electrode, defined as $0\ \mathrm{V}$ at 25 °C, unit activity and 1 bar. The values come in one table for exactly that reason — the whole tabulated range fits between lithium at $-3.04\ \mathrm{V}$ and fluorine at $+2.87\ \mathrm{V}$ . A textbook table such as OpenStax Chemistry 2e, section 17.3 lists them the same way, so any pair you take from one column belongs together.

Two conventions that decide the sign

The cathode is the higher-potential electrode by definition, not the one with the bigger number in some other sense. Take both values as reduction potentials from one table and the subtraction sorts itself out. Mix a reduction potential from one source with an oxidation potential from another and the sign of the answer flips.

What the page accepts, and where it pushes back

Zero is a real answer. Two identical electrodes give $E_{\mathrm{cell}} = 0$ — no driving force — and the page reports it without complaint.
A negative cell voltage is not a number to keep. If you type one, or if the subtraction produces one, the box turns red and says that the cathode has to sit above the anode. Swap the two potentials and the same magnitude comes back positive.
Reality ceilings are warnings, not errors. A single electrode potential beyond $\pm 4\ \mathrm{V}$ or a cell voltage above $6\ \mathrm{V}$ is kept on screen and marked, because both are far outside anything the half-cell tables contain. That usually means a unit was misread rather than a discovery was made.

Worked examples

Three cases, all checked against the arithmetic above: a forward solve from two halves, a reverse solve from a measured voltage, and the same difference read in millivolts.

A magnesium/copper cell

Situation: You want the voltage of a cell built from a magnesium electrode and a copper electrode before ordering anything.

Entered

Anode−2.36 V

Cathode+0.34 V

EMFleave empty

The page returns

E_{\mathrm{cell}}

=

0.34 - (-2.36)

=

2.7\ \mathrm{V}

A single box with a tenth of a volt either way, well inside the range the tables cover: the EMF box comes back as $2.7\ \mathrm{V}$ with no flag.

Recovering a missing electrode potential

Situation: An open-circuit reading of 0.46 V is measured for a cell whose positive electrode is silver, and you want to know which other couple was on the other side.

Entered

Cathode+0.80 V (Ag⁺/Ag)

EMF0.46 V

Anodeleave empty

The page returns

E_{\mathrm{anode}}

=

0.80 - 0.46

=

0.34\ \mathrm{V}

The anode finished at $+0.34\ \mathrm{V}$ , which is the $\mathrm{Cu^{2+}/Cu}$ couple — so the cell under test was a silver/copper pair, and the silver electrode was the cathode because it sits higher.

The same difference in millivolts

Situation: A sensor datasheet gives a half-cell reading of 760 mV against a reference at 0.00 V, and you want the cell voltage in volts.

Entered

Anode0.00 V

Cathode760 mV

EMFleave empty

The page returns

760\ \mathrm{mV} \times 10^{-3}

=

0.76\ \mathrm{V}

=

E_{\mathrm{cell}}

Reading the anode in volts and the cathode in millivolts is safe here: the factors are exact, so $0.76\ \mathrm{V}$ and $760\ \mathrm{mV}$ are the same number, and switching the EMF box to millivolts shows it as $760\ \mathrm{mV}$ .

Next step: once the voltage is settled, write the cell reaction itself. The chemical equation balancer takes both sides of the overall reaction and returns the smallest whole-number coefficients — useful for turning the two half-reactions into one equation with the electrons cancelled.

Questions

Why is the cell EMF flagged as negative?

Because the two electrodes are in the wrong boxes. A galvanic cell is written with the cathode on the high-potential side, so the difference $E_{\mathrm{cathode}} - E_{\mathrm{anode}}$ is positive for the direction the reaction runs on its own. A negative result is the same cell read backwards: swap the two electrode potentials and the identical magnitude appears with a plus sign.

It is also the honest answer if you really did mean the reverse direction — an external supply driving the reaction the other way is electrolysis, and it needs a voltage at least that large applied across the cell.

Why does a voltmeter read less than the EMF this page calculates?

The page returns the standard-state potential difference. A cell on a bench differs from that for at least four reasons:

Concentration. The standard value assumes unit activity, roughly $1\ \mathrm{mol\,L^{-1}}$ . At other activities the potential moves by the Nernst term $-\frac{RT}{nF}\ln Q$ , which can shift a cell by tens of millivolts.
Current drawn. EMF is the open-circuit value. As soon as the cell supplies current, internal resistance and electrode overpotential pull the terminal voltage below it.
Junctions and reference drift. The measurement includes liquid-junction potentials and any error in the reference electrode, neither of which appears in a tabulated difference.

A high-impedance meter at open circuit, with the reference electrode kept in good condition, is the closest a bench reading gets to the number this page returns.

Can I use it for several cells in series?

Yes, with one addition: cells in series add their voltages, so three zinc/copper cells give $3 \times 1.1 = 3.3\ \mathrm{V}$ . Enter that total in EMF, plus one electrode potential, and the page solves for the other — or work through the cells one at a time. What it will not do is treat a series stack as one cell: above $6\ \mathrm{V}$ the EMF box is flagged, because no single pair of half-reactions reaches that voltage.

What the result does not cover

Standard-state numbers, not bench measurements

Both electrode potentials and the cell EMF describe the standard state: 25 °C, unit activity and 1 bar, with every species in its tabulated form. A real cell is usually somewhere else, and the difference is not an error in the arithmetic.

Activity, not just concentration. The Nernst term in the FAQ above works with activities. If you need the activity coefficient that turns a measured molarity into one, the activity coefficient calculator returns it from the ion charge and the ionic strength of the solution.
Temperature. Tabulated potentials are quoted at 25 °C. Moving a cell tens of degrees shifts the equilibrium potentials and changes which reaction is even the favourable one.
Kinetics are not in the number. A positive EMF says the reaction is thermodynamically downhill; it says nothing about how fast it runs, whether an oxide layer blocks it, or whether a gas will evolve instead. Some half-cells with excellent potentials are practically unusable without a catalyst.
A voltage is not a capacity. The answer is a potential difference only. It tells you nothing about how much charge the cell can deliver, how it ages, or what current it can sustain — those come from the electrode masses, the electrolyte and the cell design.
Not a substitute for a measurement. Treat the result as the theoretical ceiling for the pair of couples you entered, and confirm anything that matters with a real cell and a proper reference electrode.

Rule of thumb: if the number is more than a volt away from what you expected, suspect the sign convention or the unit before you suspect the chemistry. The red flags on the page point at both.

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