Nernst equation calculator

What is the Nernst equation?

E = E° − (RT ÷ nF) ln Q, where E° is the standard cell potential, n the moles of electrons transferred, Q the reaction quotient, R the gas constant and F the Faraday constant (96,485 C/mol).

What is the shortcut at 25 °C?

E = E° − (0.0592 V ÷ n) log Q. The calculator uses the full equation, so it works at any temperature; at 25 °C the factor is 0.05916 V.

Example

Co + Fe²⁺ → Co²⁺ + Fe with E° = −0.17 V, n = 2, [Fe²⁺] = 1.94 M and [Co²⁺] = 0.15 M: Q = 0.15 ÷ 1.94 = 0.077, so E = −0.17 − (0.0592 ÷ 2) log 0.077 = −0.14 V — not spontaneous.

How do I find Q?

Products over reactants, each concentration raised to its coefficient; pure solids and liquids are left out. In the example Q = [Co²⁺] ÷ [Fe²⁺].

What does E tell me?

A positive cell potential means the reaction runs spontaneously as written; negative means the reverse does. E = 0 at equilibrium.

How do I get E° in the first place?

From a table of standard reduction potentials: E°cell is the potential of the half-reaction that is reduced (the cathode) minus that of the half-reaction that is oxidized (the anode). In OpenStax Example 17.6, silver and iron give 0.7996 V − (−0.447 V) = 1.247 V. A half-reaction’s potential is used as listed even when the equation is doubled (OpenStax uses 0.7996 V for 2Ag⁺ + 2e⁻ → 2Ag, section 17.3); only n changes.

The equation and its terms

Under non-standard conditions the cell potential is E = E° − (RT ÷ nF) ln Q (OpenStax Chemistry 2e, section 17.4). Here n is the number of moles of electrons transferred in the balanced reaction, F is Faraday’s constant (96,485 C per mole of electrons), R is 8.314 J/(mol·K), T is in kelvin, and Q is the reaction quotient computed from the current concentrations. At 25 °C, (RT ÷ F) × ln 10 = 0.0592 V, which gives the familiar form E = E° − (0.0592 V ÷ n) log Q. This calculator keeps the full form, so you enter the temperature in °C and it converts. When Q = 1 the logarithm is zero and E = E°; each tenfold increase in Q lowers the potential by 0.0296 V for a two-electron reaction.

Worked example

A concentration cell has the same half-reaction on both sides and differs only in concentration, so E° = 0 and the whole potential comes from the Q term. OpenStax Example 17.8 pairs zinc half-cells at 0.10 M and 0.50 M, with n = 2. The spontaneous change moves zinc ions toward the dilute side, so Q = 0.10 ÷ 0.50 = 0.20, and E = 0 − (0.0592 ÷ 2) log 0.20 = 0.021 V. Enter E° = 0, n = 2, Q = 0.2 and 25 °C to see it. As the two sides even out, Q approaches 1 and E falls to zero, which is equilibrium. For the cobalt–iron example above, the potentials −0.28 V and −0.447 V give E° = −0.447 − (−0.28) = −0.167 V, which the book rounds to −0.17 V; with the unrounded value the calculator returns −0.134 V instead of −0.14 V.

Common mistakes

Inverting Q is the usual one: it is products over reactants, so for the example it is [Co²⁺] ÷ [Fe²⁺]; flipping it gives −0.20 V instead of −0.14 V. Next is the wrong n, which must come from the balanced equation (Al³⁺ to Al is 3 electrons per ion, but a reaction written for two aluminum atoms transfers 6). Third, putting pure solids and liquids into Q; they are left out. Fourth, entering the temperature in kelvin in a field that expects °C.

Sources: OpenStax Chemistry 2e, section 17.4 “Potential, Free Energy, and Equilibrium” (Nernst equation, concentration cells, Examples 17.6 to 17.8) and section 17.3 “Electrode and Cell Potentials” (E°cell = E°cathode − E°anode). Related: Gibbs free energy calculator, equilibrium constant calculator, pH calculator and Ohm’s law calculator.