The electromotive force (EMF) of an electrochemical cell is the voltage it produces when no current flows. It comes from the difference in how strongly the two electrodes pull electrons: the half-reaction with the higher (more positive) reduction potential is reduced at the cathode, and the one with the lower potential runs backwards as an oxidation at the anode.
Standard reduction potentials are measured against the standard hydrogen electrode (0 V by definition) with every species at unit activity – roughly 1 mol/L for dissolved ions and 1 bar for gases – at 25 °C. Because both values are already reduction potentials, the cell EMF is simply the cathode value minus the anode value; you never flip a sign by hand. A positive E°cell means the reaction runs spontaneously as written and the cell can do electrical work, while a negative value tells you the reverse reaction is the spontaneous one.
The same number also fixes the thermodynamics: the standard Gibbs energy change is −nFE° and the equilibrium constant follows from ln K = nFE°/RT. A Daniell cell (zinc anode, copper cathode) gives about 1.10 V, which corresponds to K ≈ 10³⁷ – the reaction goes essentially to completion.
How to use the Cell EMF Calculator
- Cathode half-reaction (reduction)
- Cathode E° (custom)
- Electrons in cathode half-reaction
- Anode half-reaction (written as a reduction)
- Anode E° (custom)
- Electrons in anode half-reaction
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose the half-reaction at the cathode (reduction) and at the anode (oxidation) from the table, or choose “Custom” and type E° and the electrons in that half-reaction. The calculator subtracts the anode potential from the cathode potential, balances the electrons with the least common multiple of the two n values, and derives ΔG° and K at 25 °C.
Worked example
Daniell cell: copper cathode, Cu²⁺ + 2e⁻ → Cu, E° = +0.3419 V; zinc anode, Zn²⁺ + 2e⁻ → Zn, E° = −0.7618 V.
E°cell = 0.3419 − (−0.7618) = 1.1037 V. Both half-reactions involve 2 electrons, so n = 2 and ΔG° = −2 × 96 485 × 1.1037 = −212.98 kJ/mol. log₁₀K = 2 × 1.1037 / 0.05916 = 37.31, so K ≈ 2.0 × 10³⁷ – the zinc is oxidised essentially completely.
Silver–zinc (button cell chemistry): 0.7996 − (−0.7618) = 1.5614 V; the electrons balance with n = lcm(1, 2) = 2.
Standard reduction potentials used (25 °C)
| Half-reaction (reduction) | E° (V) | n |
|---|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.866 | 2 |
| H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O | +1.776 | 2 |
| PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O | +1.6913 | 2 |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.507 | 5 |
| Au³⁺ + 3e⁻ → Au | +1.498 | 3 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.3583 | 2 |
| O₂ + 4H⁺ + 4e⁻ → 2H₂O | +1.229 | 4 |
| Br₂(l) + 2e⁻ → 2Br⁻ | +1.066 | 2 |
| NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O | +0.957 | 3 |
| Ag⁺ + e⁻ → Ag | +0.7996 | 1 |
| Fe³⁺ + e⁻ → Fe²⁺ | +0.771 | 1 |
| O₂ + 2H⁺ + 2e⁻ → H₂O₂ | +0.695 | 2 |
| I₂ + 2e⁻ → 2I⁻ | +0.5355 | 2 |
| Cu⁺ + e⁻ → Cu | +0.521 | 1 |
| O₂ + 2H₂O + 4e⁻ → 4OH⁻ | +0.401 | 4 |
| Cu²⁺ + 2e⁻ → Cu | +0.3419 | 2 |
| Hg₂Cl₂ + 2e⁻ → 2Hg + 2Cl⁻ (calomel) | +0.2681 | 2 |
| AgCl + e⁻ → Ag + Cl⁻ | +0.2223 | 1 |
| Cu²⁺ + e⁻ → Cu⁺ | +0.153 | 1 |
| 2H⁺ + 2e⁻ → H₂ (SHE) | 0.00 | 2 |
| Pb²⁺ + 2e⁻ → Pb | −0.1262 | 2 |
| Sn²⁺ + 2e⁻ → Sn | −0.1375 | 2 |
| Ni²⁺ + 2e⁻ → Ni | −0.257 | 2 |
| Co²⁺ + 2e⁻ → Co | −0.28 | 2 |
| PbSO₄ + 2e⁻ → Pb + SO₄²⁻ | −0.3588 | 2 |
| Cd²⁺ + 2e⁻ → Cd | −0.403 | 2 |
| Fe²⁺ + 2e⁻ → Fe | −0.447 | 2 |
| Cr³⁺ + 3e⁻ → Cr | −0.744 | 3 |
| Zn²⁺ + 2e⁻ → Zn | −0.7618 | 2 |
| 2H₂O + 2e⁻ → H₂ + 2OH⁻ | −0.8277 | 2 |
| Al³⁺ + 3e⁻ → Al | −1.662 | 3 |
| Mg²⁺ + 2e⁻ → Mg | −2.372 | 2 |
| Na⁺ + e⁻ → Na | −2.71 | 1 |
| Ca²⁺ + 2e⁻ → Ca | −2.868 | 2 |
| K⁺ + e⁻ → K | −2.931 | 1 |
| Li⁺ + e⁻ → Li | −3.0401 | 1 |
Tips
- Standard potentials are reduction potentials at 25 °C, 1 M and 1 bar.
- n is the number of electrons transferred in the balanced reaction.
- Faraday constant: 96 485 C per mole of electrons.
- Everything is calculated in your browser.
Common mistakes to avoid
- Flipping the sign of a potential and also subtracting it (counting the reversal twice).
- Multiplying a standard potential by the coefficient of the half-reaction – potentials are intensive.
- Using log base e where the Nernst equation uses log₁₀ (or the other way round).
Inputs explained
| Field | What to enter |
|---|---|
| Cathode half-reaction (reduction) | Choose one: F₂ + 2e⁻ → 2F⁻ (E° = +2.866 V), H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (E° = +1.776 V), PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O (E° = +1.6913 V), MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.507 V) … |
| Cathode E° (custom) | In V |
| Electrons in cathode half-reaction | A number |
| Anode half-reaction (written as a reduction) | Choose one: F₂ + 2e⁻ → 2F⁻ (E° = +2.866 V), H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (E° = +1.776 V), PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O (E° = +1.6913 V), MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.507 V) … |
| Anode E° (custom) | In V |
| Electrons in anode half-reaction | A number |
What each result means
- Cell EMF E°cell = [Cathode standard potential E°cathode] – [Anode standard potential E°anode]
- Standard Gibbs energy change ΔG° = -[Electrons transferred n (balanced)]×96485.33212×[Cell EMF E°cell]÷1000
- log₁₀ of equilibrium constant K = [Electrons transferred n (balanced)]×[Cell EMF E°cell]×96485.33212÷(8.314462618×298.15×ln(10))
- Equilibrium constant K = 10^[log₁₀ of equilibrium constant K]
Formula
E°cell = E°cathode − E°anode · ΔG° = −n·F·E°cell · log₁₀K = n·E°cell / 0.05916 V (25 °C) · n = lcm(n_cathode, n_anode)Worked example
Inputs — Cathode half-reaction (reduction): Cu²⁺ + 2e⁻ → Cu (E° = +0.3419 V); Anode half-reaction (written as a reduction): Zn²⁺ + 2e⁻ → Zn (E° = −0.7618 V).
Results — Cathode standard potential E°cathode = 0.3419 V; Anode standard potential E°anode = -0.7618 V; Cell EMF E°cell = 1.1037 V; Electrons transferred n (balanced) = 2; Standard Gibbs energy change ΔG° = -212.98 kJ/mol; log₁₀ of equilibrium constant K = 37.313; Equilibrium constant K = 2.0549e37; Direction = Spontaneous as written (galvanic cell).
Key terms
- Electromotive force (EMF)
- The open-circuit voltage of a cell – the energy delivered per unit charge when no current flows.
- Standard reduction potential E°
- Potential of a half-reaction written as a reduction, relative to the standard hydrogen electrode, at unit activities and 25 °C.
- Cathode
- Electrode where reduction takes place.
- Anode
- Electrode where oxidation takes place.
- Faraday constant F
- Charge of one mole of electrons, 96 485.332 C/mol.
Frequently asked questions
How do you calculate cell EMF?
Subtract the standard reduction potential of the anode from that of the cathode: E°cell = E°cathode − E°anode. For a Daniell cell, 0.3419 − (−0.7618) = 1.1037 V.
Which electrode is the cathode?
The one with the more positive reduction potential. Reduction happens at the cathode, oxidation at the anode, in both galvanic and electrolytic cells.
Do I multiply E° by the coefficients in the balanced equation?
No. Potentials are intensive – they do not change when you double a half-reaction. Only n, ΔG° and K depend on the number of electrons.
What does a negative EMF mean?
The reaction as written is not spontaneous; the reverse reaction is. In a real cell you would simply have the electrodes the wrong way round.
What is the EMF of a lead–acid cell?
PbO₂/PbSO₄ (+1.6913 V) as cathode and PbSO₄/Pb (−0.3588 V) as anode give 2.05 V per cell, so a 6-cell car battery gives about 12.3 V.
Why is the measured voltage lower than the EMF?
Under load, internal resistance and electrode overpotentials take part of the voltage, and non-standard concentrations change the potentials (use the Nernst equation for that).
What is the highest possible standard cell voltage?
Fluorine (+2.866 V) against lithium (−3.0401 V) gives about 5.9 V – which is why lithium is used in high-voltage batteries.
What formula does the Cell EMF Calculator use?
E°cell = E°cathode − E°anode · ΔG° = −n·F·E°cell · log₁₀K = n·E°cell / 0.05916 V (25 °C) · n = lcm(n_cathode, n_anode).