Nernst Equation

9.10 Nernst Equation

Dial ion concentrations and watch a dot slide along a live E vs log Q chart. Voltmeter color shifts as E crosses zero — showing exactly when the cell dies.

E = E° − (0.0592/n)logQQ & K4 CellsΔG Connection

Topic 9.10

Cell Potential Under Nonstandard Conditions

Explain the relationship between deviations from standard cell conditions and changes in the cell potential.

In a real system under nonstandard conditions, the cell potential varies with the concentrations of the active species. The cell potential is a driving force toward equilibrium: the farther the reaction is from equilibrium, the greater the magnitude of the cell potential.

EK 9.10.A.2 issues a warning: equilibrium arguments such as Le Châtelier's principle do not apply to electrochemical systems, because the systems are not at equilibrium. Reason with Q and K, or with the Nernst equation — not with "the shift".

EK 9.10.A.3 gives the full logic:

E° corresponds to standard conditions, which means Q = 1.
As the system approaches equilibrium, the magnitude of the cell potential decreases, reaching zero at equilibrium (Q = K). A dead battery is a cell that has reached equilibrium.
Deviations that take the cell further from equilibrium than Q = 1 (i.e. more reactant, Q < 1) increase |E| relative to E°.
Deviations that take the cell closer to equilibrium than Q = 1 (i.e. more product, Q > 1) decrease |E| relative to E°.

Concentration cells are the extreme case: both half-cells contain the same species, so E° = 0. The cell still runs, driven purely by the concentration difference. The direction of electron flow is determined by considering which direction moves the system toward equilibrium — the dilute half-cell is the anode, since oxidation there increases its concentration.

EK 9.10.A.4 is explicit about how this is assessed: algorithmic calculations using the Nernst equation are insufficient to demonstrate understanding. Students should qualitatively understand the effects of concentration on cell potential and use conceptual reasoning, including qualitative use of:

E = E° − (RT/nF) ln Q

Read the equation structurally: larger Q makes the subtracted term larger, so E falls below E°. Smaller Q makes it negative, so E rises above E°.

Key points

E° corresponds to Q = 1; the farther from equilibrium, the larger |E|.
Increasing reactant concentration (Q < 1) raises E; increasing product concentration (Q > 1) lowers E.
E = 0 exactly at equilibrium — that is a dead battery.
In a concentration cell E° = 0 and the dilute half-cell is the anode.

Equations

$E_{ce l l} = E_{ce l l}^{\circ} - \frac{R T}{n F} ln Q$
on the exam sheetPrinted on the sheet, but the AP Exam assesses qualitative reasoning with it.
- $Q$ reaction quotient
- $n$ moles of electrons transferred
- $F$ 96 485 C mol⁻¹

Common mistakes

Applying Le Châtelier’s principle to an electrochemical cell. The CED explicitly rules this out — the system is not at equilibrium.
Thinking a dead battery has run out of atoms. It has reached equilibrium: Q = K and E = 0.
Assuming E° = 0 means no cell potential. A concentration cell has E° = 0 but a nonzero E.
Grinding through Nernst arithmetic when the question wants conceptual reasoning.

Worked example

For the cell Zn(s) | Zn²⁺ ‖ Cu²⁺ | Cu(s), E°_cell = +1.10 V. Predict whether E_cell is greater than, less than, or equal to 1.10 V when (a) [Cu²⁺] is increased to 2.0 M with [Zn²⁺] at 1.0 M, (b) [Zn²⁺] is increased to 2.0 M with [Cu²⁺] at 1.0 M, and (c) the cell has been running for a long time.

The reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), so Q = [Zn²⁺]/[Cu²⁺]. (Solids are excluded.)

(a) Increasing [Cu²⁺] to 2.0 M.
Q = 1.0/2.0 = 0.50, which is less than 1.

In E = E° − (RT/nF) ln Q, a Q below 1 makes ln Q negative, so the subtracted term is negative and E > 1.10 V.

Conceptually: raising the reactant concentration moves the system further from equilibrium, and cell potential measures distance from equilibrium.

(b) Increasing [Zn²⁺] to 2.0 M.
Q = 2.0/1.0 = 2.0, which is greater than 1.

ln Q is now positive, so a positive quantity is subtracted and E < 1.10 V. More product means closer to equilibrium and a smaller driving force.

(c) After running a long time. As the cell operates, Zn is oxidized (raising [Zn²⁺]) and Cu²⁺ is reduced (lowering [Cu²⁺]), so Q climbs steadily. E therefore falls continuously and reaches zero when Q = K — the cell is at equilibrium and the battery is dead. Note that plenty of zinc and copper remain; what has been exhausted is the driving force, not the material.

Full notes for topic 9.10 →