The Nernst Equation & Nonstandard Cells

You’ll be able to

Explain qualitatively how concentration shifts a cell potential away from E°
Use the Nernst equation to calculate E under nonstandard conditions
Predict the direction a cell voltage moves as reactant and product concentrations change

Standard conditions are the exception, not the rule

E°cell assumes every solute is exactly 1 M and every gas 1 bar. Real cells rarely sit there, and as a cell discharges the concentrations continuously change. The Nernst equation corrects the standard potential for the actual conditions using the reaction quotient Q — the same Q from equilibrium, built from current (not standard) concentrations. As a cell runs, reactants fall and products rise, Q climbs, and the voltage sags toward zero; a dead battery is simply a cell that has reached equilibrium (Q = K, E = 0).

The Nernst equation (at 298 K)

E = E° − (0.0592 / n) log Q

General form: E = E° − (RT/nF) ln Q. At 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it.

Which way does the voltage move?

Reading the sign of the correction is often enough. More reactant (or less product) makes Q smaller, log Q negative, and −(0.0592/n) log Q positive, so E rises above E°. More product (or less reactant) makes Q larger and E falls below E°. This is Le Châtelier for cells: raising the concentration of a species that gets consumed gives the reaction more push, and more push means more voltage.

Worked example

A Daniell cell (Zn | Zn²⁺ || Cu²⁺ | Cu, E° = +1.10 V, n = 2) runs with [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M. Find E and compare it to E°.

1.Write Q for Zn + Cu²⁺ → Zn²⁺ + Cu. Solids are left out, so Q = [Zn²⁺] ÷ [Cu²⁺] = 1.0 ÷ 0.010 = 100.
2.Apply the Nernst equation at 298 K: E = E° − (0.0592/n) log Q = 1.10 − (0.0592/2) log(100).
3.log(100) = 2, and 0.0592/2 = 0.0296, so the correction is 0.0296 × 2 = 0.0592 V.
4.E = 1.10 − 0.0592 = 1.04 V.
5.Q > 1 (product Zn²⁺ far exceeds reactant Cu²⁺), so E came out below E° — exactly as the sign of the correction predicts.

Answer: E ≈ 1.04 V, slightly below the standard 1.10 V.

Checkpoint

For the Daniell cell (Zn + Cu²⁺ → Zn²⁺ + Cu), how does raising [Cu²⁺] above 1 M while keeping [Zn²⁺] at 1 M affect the cell potential?

Checkpoint

A cell has E° = +0.60 V, n = 2, and operates at 298 K with Q = 1.0 × 10⁴. Approximately what is E?

Tip

You rarely need to finish the arithmetic on the AP exam — often just the direction. Ask whether the change makes Q bigger or smaller: bigger Q pulls E down, smaller Q pushes E up. Get the sign right and most multiple-choice options fall away.

On the exam

A concentration cell is the purest test of the Nernst idea: identical electrodes, same E° of 0, so E depends entirely on the concentration difference through −(0.0592/n) log Q. Electrons flow to equalize the two sides, and the voltage dies the instant the concentrations match.

Answer the 2 checkpoints as you read.