The Nernst Equation, Explained
After reading this you can take a standard cell potential and correct it for real concentrations, temperatures and pressures using the Nernst equation, and you can read the sign and size of the result to say which way a cell will actually run.
A galvanic cell built from tables has a single number attached to it, the standard potential E^\circ. That number assumes every dissolved species sits at 1 mol/L and every gas at 1 bar. Real cells almost never do. A fresh battery, a half-drained battery and a concentration cell with the same electrodes on both sides all have different potentials, and the difference is exactly what the Nernst equation predicts.
Here is the hook. Take a copper-zinc cell with E^\circ = 1.10 V. Drain it until the product ions build up so the reaction quotient Q climbs to 10. The potential drops by 0.0296 V to about 1.07 V. Push Q all the way to 10^{37}, the equilibrium value, and the potential falls to zero. That is a dead battery, described by a single logarithm.
What the equation says
The Nernst equation connects the actual cell potential E to the standard potential and the reaction quotient.
The symbols:
- E
- Cell potential under the actual conditions, in volts.
- E°
- Standard cell potential, the value at 1 mol/L and 1 bar, in volts.
- R
- Gas constant,
8.314J/(mol·K). - T
- Absolute temperature in kelvin.
- n
- Moles of electrons transferred in the balanced cell reaction.
- F
- Faraday constant,
96485C/mol, the charge on one mole of electrons. - Q
- Reaction quotient, products over reactants each raised to its stoichiometric coefficient, using current (not equilibrium) activities.
The grouping RT/nF carries units of joules per coulomb, which is volts, so the subtraction is dimensionally clean. The logarithm is dimensionless because Q is a ratio of activities.
Where the RT/nF factor comes from
The equation is not a separate law. It falls out of thermodynamics. The free energy change of a reaction away from standard conditions is
Electrical work links free energy to potential through \Delta G = -nFE and \Delta G^\circ = -nFE^\circ. Substitute both:
Divide through by -nF and you have the Nernst equation. Nothing new was assumed. The n in the denominator is the same electron count that appears in the work term, which is why a two-electron reaction responds half as strongly to Q as a one-electron reaction.
The 0.0592 shortcut at 25 °C
Most textbook problems sit at 298.15 K. Collapse the constants at that temperature and convert the natural log to base 10:
So at 25 °C the equation becomes the form most people memorize:
Read that as a rule of thumb: every factor of 10 in Q shifts the potential by 0.0592/n volts. For n = 1 that is 59.2 mV per decade. For n = 2 it is 29.6 mV per decade. This is the slope you see in pH electrodes and ion-selective sensors.
The 0.0592 value is only correct at 298.15 K. At body temperature, 310.15 K, the factor rises to 0.0615 V. If your temperature field holds anything other than 298.15, the calculator uses the full RT/nF form and the shortcut no longer matches.
Worked example reproducing the demo
Copper-zinc cell with Q = 0.1
Use the demo defaults: E^\circ = 1.10 V, n = 2, Q = 0.1, T = 298.15 K.
- Compute the prefactor: RT/nF = (8.314 \times 298.15)/(2 \times 96485) = 0.01285 V.
- Take the natural log of Q: \ln 0.1 = -2.3026.
- Multiply: 0.01285 \times (-2.3026) = -0.02959 V.
- Subtract from E°: E = 1.10 - (-0.02959) = 1.1296 V.
The potential is 1.130 V, slightly above standard. That sign makes sense. A Q below 1 means products are scarce relative to reactants, so the forward reaction is more favorable than at standard conditions and the cell pushes harder.
Check with the shortcut, since we are at 25 °C: E = 1.10 - (0.0592/2)\log_{10}(0.1) = 1.10 - (0.0296)(-1) = 1.1296 V. Same answer.
How potential tracks the reaction quotient
Because the correction is logarithmic, the potential falls in a straight line when you plot it against \log_{10} Q. The slope is exactly -0.0592/n. For the demo cell that slope is -0.0296 V per decade.
Notice the vertical marker at \log_{10} Q = 0. There Q = 1, the log is zero, the correction vanishes, and E = E^\circ exactly. That is the second footnote on the tool, and it is the anchor point of the whole line.
When to use it, and when not
Reach for the Nernst equation whenever conditions differ from standard: dilute solutions, gas partial pressures away from 1 bar, temperatures other than 25 °C, or a discharging battery whose ion concentrations have drifted. It is the tool behind pH meters, reference electrodes and any equilibrium constant you extract from a cell measurement.
Two limits deserve care. First, at equilibrium the net potential is zero, so setting E = 0 gives \ln K = nFE^\circ/RT, a clean route from cell voltage to the equilibrium constant K. Second, the equation uses activities, not concentrations. At the low millimolar range that distinction stays under a percent or two, but in concentrated brines the activity coefficient can pull the true potential tens of millivolts off the concentration-based estimate. Treat any Nernst prediction in a strong electrolyte as an approximation, not a measurement.
Pure solids and pure liquids have activity 1 and never appear in Q. A zinc electrode dissolving into Zn²⁺ contributes only the ion term, not the metal. Leaving solids out of Q is the single most common source of wrong answers.
Reading and interpreting the result
The sign of the correction tells you the direction. When Q \lt 1 the log is negative, the correction adds to E^\circ, and the cell is more energetic than standard. When Q \gt 1 the correction subtracts, and the cell is weaker. This table shows the demo cell across five decades.
| Q | log₁₀ Q | Correction (V) | E (V) |
|---|---|---|---|
| 0.001 | -3 | +0.0888 | 1.1888 |
| 0.1 | -1 | +0.0296 | 1.1296 |
| 1 | 0 | 0 | 1.1000 |
| 10 | 1 | -0.0296 | 1.0704 |
| 1000 | 3 | -0.0888 | 1.0112 |
The whole span from Q = 0.001 to Q = 1000, a factor of a million, moves the potential by only 0.178 V. That compression is the practical meaning of a logarithm: concentration changes have to be large before the voltage moves much. It is also why a battery holds a nearly flat voltage through most of its discharge, then drops sharply only as Q approaches K.
Common mistakes
Watch for these specific errors, each of which produces a plausible-looking but wrong number:
- Wrong n. Use the electrons in the balanced overall reaction, not per half-cell as written. For Cu²⁺ + Zn the answer is n = 2, not 1.
- Q inverted. Products go on top. Putting reactants on top flips the sign of the correction and moves the potential the wrong way.
- Log base confusion. The
0.0592shortcut uses base-10 log. The RT/nF form uses natural log. Mixing them scales the answer by 2.303. - Forgetting temperature. If the problem is at 37 °C or 60 °C, the prefactor changes and the 0.0592 rule is wrong by several percent.
- Including solids in Q. Solids and pure liquids drop out. So does the solvent in dilute aqueous work.
Related tools
Getting Q right depends on a correctly balanced reaction, so start with the Chemical Equation Balancer to fix the coefficients and the electron count. If your Q comes from measured concentrations, the Concentration Converter turns g/L or ppm into the molarity the equation expects, and the pH Calculator supplies hydrogen-ion activity for cells that involve H⁺. When you set E = 0 to find where the reaction rests, the Equilibrium (ICE Table) Solver handles the concentrations at that point. For the thermodynamic side, the free energy link ties back to concepts you can explore with the Ideal Gas Law Calculator when gas partial pressures enter Q.
Frequently asked questions
Why does E° stay fixed while E changes?
E° is a defined reference at 1 mol/L and 1 bar, a property of the electrode couple. E is the actual voltage under your conditions. The Nernst equation is the bridge: it corrects the fixed reference for the real reaction quotient.
What happens to E at equilibrium?
At equilibrium Q equals K and the cell can do no more work, so E = 0. Setting E = 0 in the equation gives E^\circ = (RT/nF)\ln K. For the demo cell, E° = 1.10 V and n = 2 give ln K = 85.7, so K is about 1.7 × 10³⁷. That enormous number is why the cell runs essentially to completion.
Can Q be greater than 1 and still give a positive E?
Yes, up to a point. The cell keeps a positive potential as long as Q stays below K. For the demo cell that ceiling is around 10³⁷, so any realistic Q leaves E positive. E only turns negative if Q exceeds K, which means you would have to force the reaction backward.
Does the equation work for concentration cells?
It works well. In a concentration cell E° = 0 because both electrodes are identical, so E = -(RT/nF)ln Q depends only on the ratio of concentrations. A tenfold concentration difference with n = 1 gives 0.0592 V, which is how these cells are measured.
Why is the voltage change so small for big concentration swings?
Because the dependence is logarithmic. A hundredfold change in Q is only two decades, worth 2 × 0.0592/n volts. For n = 2 that is 0.0592 V total. You need many orders of magnitude in Q to move the potential by a large amount.
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