Electrochemistry Calculator

Cell potential and free energy in one step

An electrochemical cell turns a spontaneous redox reaction into electrical work. Its standard cell potential, E°cell, tells you the voltage the cell would deliver when every species is at standard conditions (1 M solutions, 1 bar gases, usually 25 °C). A positive value means the reaction runs spontaneously as written; a negative value means it needs an external voltage, as in electrolysis or battery charging.

This calculator takes the standard reduction potentials of the cathode and the anode, which you can look up in any table of standard electrode potentials, and returns E°cell together with the corresponding Gibbs free energy change. It is useful for checking homework, deciding which half-cell acts as the cathode in a galvanic cell, and quickly judging whether a proposed redox pairing is favorable.

The formulas behind the result

Both half-cell values must be reduction potentials. The anode is the electrode where oxidation happens, but you still enter its reduction potential from the table; the subtraction takes care of the reversal. The calculator then converts voltage to energy with the Faraday constant:

E°cell = E°cathode − E°anode, in volts.
ΔG° = −n F E°cell, where F = 96,485 C/mol of electrons is the Faraday constant.
n is the number of electrons transferred in the balanced reaction. This calculator uses n = 1, so its ΔG is per mole of electrons; multiply by your reaction's n for the full-reaction value.
The result is displayed in kJ/mol after dividing joules by 1000.

Worked example

Consider a zinc–copper (Daniell) cell. Copper is reduced at the cathode with E° = +0.34 V, and zinc is oxidized at the anode with a reduction potential of −0.76 V.

E°cell = 0.34 − (−0.76) = 1.10 V. Then ΔG = −(1)(96,485)(1.10) = −106,133.5 J, which the calculator shows as −106.13 kJ/mol, with E°cell displayed as 1.100 V.

The reaction Zn + Cu²⁺ → Zn²⁺ + Cu transfers n = 2 electrons, so the full-reaction free energy is twice the displayed value, about −212.3 kJ per mole of reaction. The negative sign confirms the reaction is spontaneous.

Common mistakes and how to interpret the result

Flipping the sign of the anode value. Enter the anode's reduction potential exactly as tabulated. The subtraction already accounts for oxidation, so negating it yourself gives the wrong answer.
Multiplying half-cell potentials by stoichiometric coefficients. Potentials are intensive quantities and do not scale when you balance the equation. Only ΔG scales with n.
Forgetting n. The kJ/mol figure here assumes one electron. For a two-electron reaction, double it before comparing with thermodynamic tables.
Applying standard values to non-standard conditions. Concentration, pressure and temperature shift the real voltage; use the Nernst equation when conditions differ from standard.

Frequently Asked Questions

Which electrode should I enter as the cathode?

The cathode is where reduction occurs. In a galvanic cell it is the half-reaction with the more positive standard reduction potential. If you enter the pair the other way round, the result simply comes out negative, indicating the reaction is not spontaneous in that direction.

Why is the free energy negative when the voltage is positive?

Because ΔG = −nFE°. A positive cell potential means the system can do electrical work on its surroundings, which corresponds to a decrease in free energy and a spontaneous reaction.

Does this calculator account for the number of electrons?

No. It computes ΔG for one mole of electrons (n = 1). Multiply the displayed kJ/mol by the number of electrons in your balanced equation to get the free energy for the whole reaction.

When is the standard cell potential not enough?

When concentrations are not 1 M, gases are not at 1 bar, or the temperature is not 25 °C. Then the actual potential differs and you need the Nernst equation or a full equilibrium treatment.