About the Nernst Equation Calculator
This Nernst equation calculator corrects a standard cell potential E° for real conditions. Enter E°, the number of electrons transferred n, the temperature and either the reaction quotient Q directly or the product and reactant concentration terms, and it returns the actual potential E along with the Gibbs free energy change and the equilibrium constant.
It is aimed at chemistry students solving electrochemistry problems, at anyone building concentration cells or Daniell cells in the lab, and at biology students estimating membrane-style potentials from concentration ratios. The chart shows how E falls as Q rises — by 0.05916/n volts for every tenfold increase at 25 °C.
The calculator uses activities approximated by molar concentrations (or partial pressures in atm for gases), which is accurate for dilute solutions. Pure solids and liquids such as metal electrodes and water are left out of Q. At equilibrium E = 0 and Q = K, which is how the equilibrium constant is derived from E°.
With the default inputs, the cell potential e is 1.1296 V. Change any value above to recalculate instantly.
How to use the nernst equation calculator
- 1Enter the standard cell potential E° (cathode E° − anode E°).
- 2Enter n, the number of electrons transferred in the balanced equation.
- 3Set the temperature (25 °C for most textbook problems).
- 4Enter Q directly, or the product and reactant concentration terms.
- 5Read E, then check ΔG and K for the same reaction.
Formula and method
The Nernst equation links the potential of an electrochemical cell to the composition of the reaction mixture. It follows from ΔG = ΔG° + RT ln Q together with ΔG = −nFE: dividing by −nF gives E = E° − (RT/nF) ln Q. R is 8.314462618 J/(mol·K), F is 96,485.33212 C/mol, T is in kelvin and n is the moles of electrons transferred in the balanced reaction.
Q is the reaction quotient written like an equilibrium expression (products over reactants, each raised to its coefficient), leaving out pure solids and liquids. At equilibrium E = 0 and Q = K, so K = exp(nFE°/RT). The free energy change for the current conditions is ΔG = −nFE.
- E
- Cell potential under the given conditions (V)
- E°
- Standard cell potential (V)
- R
- Gas constant, 8.314462618 J/(mol·K)
- T
- Absolute temperature (K)
- n
- Electrons transferred per reaction
- F
- Faraday constant, 96,485.33212 C/mol
- Q
- Reaction quotient
Worked examples
Daniell cell with [Zn²⁺] = 0.10 M, [Cu²⁺] = 1.0 M
For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = 0.10 ÷ 1.0 = 0.10 and n = 2. E = 1.10 − (0.05916 ÷ 2) × log(0.10) = 1.10 + 0.0296 = 1.1296 V, and ΔG = −2 × 96,485 × 1.1296 ≈ −218.0 kJ/mol.
Silver half-reaction with Q = 1000
With n = 1 the slope is 0.05916 V per decade. Q = 1000 is three decades, so E = 0.80 − 3 × 0.05916 = 0.6225 V.
Daniell cell at 37 °C with Q = 0.005
At 310.15 K the slope rises to 2.303 × 8.314 × 310.15 ÷ (2 × 96,485) = 0.0308 V per decade. With Q = 0.01 ÷ 2 = 0.005, E = 1.10 − 0.0308 × log(0.005) ≈ 1.171 V.
Frequently asked questions
What is the Nernst equation used for?+
It calculates the potential of an electrochemical cell or half-cell when concentrations, pressures or temperature differ from standard conditions (1 M, 1 atm/bar, 25 °C). It is used for batteries, pH electrodes, corrosion and membrane potentials.
Where does 0.0592 come from in the Nernst equation?+
At 298.15 K, RT/F × ln 10 = 8.314 × 298.15 ÷ 96,485 × 2.3026 = 0.05916 V. That is why the 25 °C form is written E = E° − (0.0592/n) log Q.
How do I find n?+
Balance the overall redox reaction and count the electrons that cancel between the oxidation and reduction half-reactions. For Zn + Cu²⁺ → Zn²⁺ + Cu, two electrons move, so n = 2.
What happens to E at equilibrium?+
At equilibrium the cell can do no more work, so E = 0 and Q equals the equilibrium constant K. Rearranging gives K = exp(nFE°/RT), or log K = nE°/0.05916 at 25 °C.
Do solids and water appear in Q?+
No. Pure solids (such as metal electrodes) and pure liquids (such as water as solvent) have an activity of 1 and are left out of the reaction quotient. Gases enter as partial pressures.