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Nernst Equation Calculator (Cell Potential)
In short: Nernst equation calculator: solve E, E°, Q or n at any temperature, and pick two half-cells from a table of standard reduction potentials to get E°cell, ΔG° and K. It is free, works in your web browser and needs no download.
Use the top calculator for E = E° − (RT/nF) ln Q with any one unknown. Use the cell builder to choose a cathode and an anode from 38 common couples. It finds E°cell, n, ΔG°, the equilibrium constant K and, if you give Q, the actual cell voltage.
E = E° − (RT/nF) ln Q. Fill in three of E, E°, n and Q and leave one blank.
Cell EMF from two half-cells
Nernst equation and related formulas
E = E° − (RT / nF) ln Q at 298.15 K: E = E° − (0.0592 / n) log₁₀ Q
E°cell = E°cathode − E°anode ΔG° = −nFE°cell ln K = nFE°cell / RT log K = nE° / 0.0592 (298 K)
R = 8.314 J mol⁻¹ K⁻¹, F = 96485 C mol⁻¹, T in kelvin and n is the number of electrons transferred in the balanced reaction. Q is written like an equilibrium expression, with products over reactants. Leave out pure solids and liquids. Gases enter as partial pressures in bar.
| Sign of E°cell | ΔG° | K | Under standard conditions |
|---|---|---|---|
| Positive | Negative | Greater than 1 | Spontaneous (galvanic cell) |
| Zero | Zero | 1 | At equilibrium |
| Negative | Positive | Less than 1 | Non-spontaneous (needs electrolysis) |
Concentration cells
Both electrodes are the same metal, so E° = 0. The voltage comes only from the concentration difference. For Cu | Cu²⁺ (0.010 M) || Cu²⁺ (1.0 M) | Cu, n = 2 and Q = 0.010/1.0, so E = 0 − (0.0592/2) log 0.010 = +0.0592 V. The dilute side is the anode.
Standard reduction potentials (25 °C)
| Half-reaction | E° (V) |
|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 |
| S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻ | +2.01 |
| H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O | +1.78 |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.51 |
| Au³⁺ + 3e⁻ → Au | +1.50 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 |
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | +1.33 |
| O₂ + 4H⁺ + 4e⁻ → 2H₂O | +1.23 |
| Br₂ + 2e⁻ → 2Br⁻ | +1.07 |
| NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O | +0.96 |
| Hg²⁺ + 2e⁻ → Hg | +0.85 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Fe³⁺ + e⁻ → Fe²⁺ | +0.77 |
| I₂ + 2e⁻ → 2I⁻ | +0.54 |
| Cu⁺ + e⁻ → Cu | +0.52 |
| O₂ + 2H₂O + 4e⁻ → 4OH⁻ | +0.40 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| Hg₂Cl₂ + 2e⁻ → 2Hg + 2Cl⁻ | +0.27 |
| AgCl + e⁻ → Ag + Cl⁻ | +0.22 |
| Sn⁴⁺ + 2e⁻ → Sn²⁺ | +0.15 |
| 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Pb²⁺ + 2e⁻ → Pb | −0.13 |
| Sn²⁺ + 2e⁻ → Sn | −0.14 |
| Ni²⁺ + 2e⁻ → Ni | −0.25 |
| Co²⁺ + 2e⁻ → Co | −0.28 |
| Cd²⁺ + 2e⁻ → Cd | −0.40 |
| Fe²⁺ + 2e⁻ → Fe | −0.44 |
| Cr³⁺ + 3e⁻ → Cr | −0.74 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| 2H₂O + 2e⁻ → H₂ + 2OH⁻ | −0.83 |
| Mn²⁺ + 2e⁻ → Mn | −1.18 |
| Al³⁺ + 3e⁻ → Al | −1.66 |
| Mg²⁺ + 2e⁻ → Mg | −2.37 |
| Na⁺ + e⁻ → Na | −2.71 |
| Ca²⁺ + 2e⁻ → Ca | −2.87 |
| Ba²⁺ + 2e⁻ → Ba | −2.91 |
| K⁺ + e⁻ → K | −2.93 |
| Li⁺ + e⁻ → Li | −3.04 |
Common textbook values versus the standard hydrogen electrode, rounded to 0.01 V. Sources differ by about 0.01 to 0.03 V for some couples (for example Cr₂O₇²⁻/Cr³⁺ is listed as 1.33 or 1.36 V, and Au³⁺/Au as 1.50 or 1.52 V).
Worked examples
Daniell cell
Zn | Zn²⁺ || Cu²⁺ | Cu. E° = 0.34 − (−0.76) = 1.10 V, n = 2.
ΔG° = −2 × 96485 × 1.10 = −212 kJ/mol. log K = 2 × 1.10 / 0.0592 = 37.2.
E° = 1.10 V, ΔG° = −212 kJ/mol, K ≈ 1.5 × 10³⁷
Non-standard Daniell cell
[Zn²⁺] = 0.10 M, [Cu²⁺] = 1.0 M. Q = [Zn²⁺]/[Cu²⁺] = 0.10.
E = 1.10 − (0.0592/2) log 0.10 = 1.10 + 0.0296
E = 1.13 V
NCERT-style Mg | Cu cell
Mg | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu. E° = 0.34 − (−2.37) = 2.71 V.
Q = 0.001/0.0001 = 10. E = 2.71 − (0.0592/2) log 10
E = 2.68 V
Hydrogen electrode at pH 10
2H⁺ + 2e⁻ → H₂, E° = 0, P(H₂) = 1 bar. Q = 1/[H⁺]² = 10²⁰.
E = 0 − (0.0592/2) × 20 = −0.0592 × pH
E = −0.592 V
Equilibrium constant (JEE style)
Cu + 2Ag⁺ → Cu²⁺ + 2Ag. E° = 0.80 − 0.34 = 0.46 V, n = 2.
log K = 2 × 0.46 / 0.0592 = 15.5
K ≈ 3.6 × 10¹⁵
Permanganate oxidises Fe²⁺
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. E° = 1.51 − 0.77.
n = 5 (the Fe half-reaction is multiplied by 5, but E° is not).
E° = +0.74 V, spontaneous
Units and common mistakes
- Never multiply E° by stoichiometric coefficients. E° is intensive. Multiplying a half-reaction by 5 leaves its E° unchanged.
- Subtract reduction potentials. E°cell = E°cathode − E°anode, using both values as reduction potentials from the table. Do not flip the sign of the anode as well.
- Get n from the balanced equation. For Al | Cu²⁺, n = 6 (2Al → 2Al³⁺ + 6e⁻).
- log or ln. 0.0592/n goes with log₁₀. RT/nF (0.02569/n V at 25 °C) goes with ln.
- 0.0592 V only at 25 °C. At other temperatures use 2.303RT/F. At 37 °C it is 0.0615 V.
- Q is products over reactants. Omit solids such as Zn(s) and Cu(s). For a cell, use the overall reaction, not a half-reaction.
- Kelvin and joules. T in K, ΔG = −nFE gives J/mol; divide by 1000 for kJ/mol.
Drawing an electrochemical cell? MolDraw has lab apparatus and structure tools for diagrams, reaction schemes and notes. It is free and online.
Open MolDraw editorFrequently asked questions
What is the Nernst equation?
E = E° − (RT/nF) ln Q. It gives the potential of an electrode or cell when concentrations are not 1 M. At 25 °C it becomes E = E° − (0.0592/n) log Q.
How do I calculate the standard cell potential?
Look up both standard reduction potentials and use E°cell = E°cathode − E°anode. For the Daniell cell, 0.34 − (−0.76) = 1.10 V. A positive value means the cell reaction is spontaneous.
Where does 0.0592 come from?
It is 2.303RT/F at 298.15 K: 2.303 × 8.314 × 298.15 / 96485 = 0.0592 V. It changes with temperature, so use the full RT/F form away from 25 °C.
How are E°, ΔG° and K related?
ΔG° = −nFE° and ΔG° = −RT ln K, so ln K = nFE°/RT. A positive E° gives a negative ΔG° and K greater than 1.
What is a concentration cell?
A cell with the same electrode and ion on both sides at different concentrations. E° = 0, so E = −(0.0592/n) log(dilute/concentrated). Electrons flow until the concentrations are equal.
What happens to E as the cell runs?
Products build up and reactants are used, so Q rises and E falls. When E reaches zero, Q equals K and the cell is flat (at equilibrium).
What is n in the Nernst equation?
The number of moles of electrons transferred per mole of reaction as written. It must match the balanced overall equation, for example n = 2 for Zn + Cu²⁺ and n = 5 for MnO₄⁻ oxidising Fe²⁺.
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