Unit 9 · Thermodynamics and Electrochemistry

AP Chemistry Nernst Equation

How concentrations change a cell's voltage, how to predict the direction and size of the change, and how it connects to free energy, in four worked examples.

The standard cell potential E° applies only when every solution is 1.0 M, and a real cell rarely stays there. The Nernst equation gives the potential at any concentrations, and the reasoning it supports is what the exam leans on: whether E rises or falls as Q changes, and what E becomes at equilibrium. The Unit 9 review states that the qualitative reasoning matters more than plugging numbers into the equation; this page covers both, so you can handle a question that does ask for a value. E° itself comes from the method in galvanic cells and cell potential.

The Nernst Equation and What Q Does to Cell Potential

• Nernst equation: E = E° − (RT/nF) ln Q, where Q has the products over the reactants for the cell reaction as written, with solids and pure liquids left out.
• At 25°C in base 10: RT/F = 0.02569 V and multiplying by 2.303 gives 0.0592 V, so E = E° − (0.0592/n) log Q.
• Q < 1: ln Q is negative, so E > E°. Q = 1: E = E°. Q > 1: E < E°.
• Q = K: E = 0, and the cell is at equilibrium, which is what a dead battery is.

As a cell runs, reactants are used up and products build, so Q grows and E falls steadily until Q reaches K. The graph shows that trend for the zinc-copper cell. Because E depends on the logarithm of Q, a tenfold change in concentration shifts E by only about 0.03 V when n = 2, so cell voltages stay close to E° until Q becomes extreme.

Worked Example 1: Predicting and Calculating E When Q Is Greater Than 1

A zinc-copper cell, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), has E° = +1.10 V. The solutions are 1.0 M Zn²⁺ and 0.010 M Cu²⁺. Is E larger or smaller than 1.10 V, and what is it?

Q: [Zn²⁺]/[Cu²⁺] = 1.0 ÷ 0.010 = 100. Solid zinc and copper are left out.
Prediction: Q > 1, so E is smaller than E°; the mixture has relatively more product than at standard conditions.
Calculation: n = 2, so E = 1.10 − (0.0592/2) log 100 = 1.10 − (0.0296)(2) = 1.04 V.

Writing the prediction before the arithmetic is a good habit: it catches a sign error, since an answer above 1.10 V would contradict Q > 1. The equation sheet gives the form with ln, so at 25°C convert as shown above or use RT/nF directly.

Worked Example 2: Predicting and Calculating E When Q Is Less Than 1

The same cell now has [Zn²⁺] = 0.0010 M and [Cu²⁺] = 2.0 M. Predict whether E is above or below 1.10 V, then calculate it.

Q: 0.0010 ÷ 2.0 = 5.0 × 10⁻⁴, so log Q = −3.30.
Prediction: Q < 1, so E is larger than E°; there is plenty of reactant and little product, so the reaction is pushed harder.
Calculation: E = 1.10 − (0.0296)(−3.30) = 1.10 + 0.098 = 1.20 V.

The change from standard conditions is only 0.10 V even though the concentrations differ from 1.0 M by factors of about 1000 and 2, which shows how gentle the logarithm is. The rule that Le Chatelier language does not apply to a cell that is running is explained in the Unit 9 review; reason with Q and E instead.

Worked Example 3: Finding the Voltage of a Concentration Cell

A cell has two copper electrodes, one in 0.010 M Cu²⁺ and the other in 1.0 M Cu²⁺. Find E, say which side is the anode, and say what happens as the cell runs.

E°: both half-cells are the same, so E° = 0.
Direction: the cell moves toward equal concentrations. In the concentrated half-cell Cu²⁺ is reduced (cathode), and in the dilute half-cell Cu is oxidized (anode) to raise its concentration.
Q: for the net change from the concentrated to the dilute solution, Q = 0.010/1.0 = 0.010.
E: E = 0 − (0.0592/2) log 0.010 = 0 + (0.0296)(2) = +0.059 V.
As it runs: the concentrations approach each other, Q approaches 1, and E falls to 0.

A voltage appears even though the two sides are made of the same material, because the difference in concentration alone is enough to drive the reaction. The standard-state E° method gives nothing for such a cell, so the Nernst equation is the only way to find its voltage. The electrode labels follow the same rules as in galvanic cells.

Worked Example 4: Connecting the Nernst Equation to Free Energy and K

For the zinc-copper cell, find ΔG° and K from E° = 1.10 V, then find ΔG for the conditions of Example 1 (E = 1.04 V).

ΔG°: −nFE° = −(2)(96,485)(1.10) = −212 kJ/mol.
K: at equilibrium E = 0, so 0 = E° − (0.0592/2) log K, giving log K = (2)(1.10)/0.0592 = 37.2 and K ≈ 1.5 × 10³⁷, which is why the cell runs almost to completion.
ΔG at E = 1.04 V: −(2)(96,485)(1.04) = −201 kJ/mol.

That last value agrees with ΔG = ΔG° + RT ln Q = −212 + (0.0083)(298)(ln 100) ≈ −201 kJ/mol, so the Nernst equation is the free-energy relationship divided by −nF. The same link between ΔG° and K is used in the Gibbs free energy page.

Five Nernst Equation Practice Questions With Answers

1. For the zinc-copper cell, Q is less than 1. Is E greater than, equal to, or less than 1.10 V?
   Show answer

   Greater than 1.10 V. For Q < 1, ln Q is negative, so subtracting it raises E above E°.
2. Find E for the zinc-copper cell with [Zn²⁺] = 0.10 M and [Cu²⁺] = 1.0 M.
   Show answer

   Q = 0.10, log Q = −1, so E = 1.10 − (0.0296)(−1) = 1.13 V.
3. What is E when a cell reaches equilibrium, and what does that mean for the cell?
   Show answer

   E = 0, because Q = K. No further work can be done, which is what a dead battery means.
4. A concentration cell uses silver electrodes in 0.10 M and 1.0 M Ag⁺. Find E, using n = 1.
   Show answer

   E° = 0 and Q = 0.10, so E = 0 − (0.0592/1) log 0.10 = +0.059 V.
5. In a concentration cell, which half-cell is the anode?
   Show answer

   The more dilute one. Metal is oxidized there to increase the ion concentration, while ions are reduced in the more concentrated half-cell.

Common Nernst Equation Mistakes

• Using E° when the concentrations are not 1.0 M. Use E, and find it from Q.
• Inverting Q. Q is products over reactants for the reaction as written, so reversing the reaction inverts Q and changes the sign of E°.
• Using the wrong n. It is the moles of electrons transferred in the balanced cell reaction as written.
• Mixing log and ln. The constant 0.0592 goes with log base 10 at 25°C; with ln Q the factor is RT/nF.
• Putting solids in Q. Solid electrodes are left out; only aqueous species and gases appear.
• Treating a cell that is running as a system at equilibrium. A working cell is not at equilibrium, so reason with Q and E, not Le Chatelier's principle.

Qualitative questions on Q and E are common on both sections, and the arithmetic is short once the prediction is clear. To see how a stronger Unit 9 would shift your overall estimate, the AP Chem Score Calculator turns section scores into a 1–5 range.

Related Resources

• AP Chem Score Calculator
• Unit 9 Review: Thermodynamics and Electrochemistry
• AP Chemistry Galvanic Cells and Cell Potential
• AP Chemistry Gibbs Free Energy
• AP Chemistry Electrolysis
• AP Chemistry Equilibrium Constant Kc and Kp
• AP Chemistry Reference Sheet
• AP Chemistry Study Guide

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