Worked example 1
For Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), E° = 1.10 V. Calculate E when [Zn²⁺] = 1.00 M and [Cu²⁺] = 0.010 M at 25 °C.
Try it first: Predict the direction of the shift before calculating.
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ElectrochemistryCourse level: College Extension
The Nernst Equation and Nonstandard Conditions
What you'll be able to do: Use the Nernst equation to calculate a cell potential at nonstandard concentrations and predict how concentration changes shift the voltage.
Best after: Cell Potential, Free Energy, and the Equilibrium Constant
Introduction
Real cells almost never sit at 1 M. The Nernst equation is what connects the tidy standard potential to the messy actual one, and it is essentially Le Chatelier written in volts.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Write the reaction quotient for a cell reaction
- Apply the Nernst equation at 25 °C
- Predict whether a concentration change raises or lowers cell potential
- Explain what happens to E as a cell discharges
Where the equation comes from
Thermodynamics says ΔG = ΔG° + RT ln Q. Substituting ΔG = -nFE and ΔG° = -nFE° and dividing through by -nF converts every term from energy into voltage, giving the Nernst equation directly.
E = E° - (RT / nF) ln Q
The 25 degree form
With R = 8.314, T = 298 K, F = 96485, and a conversion from natural log to base 10, the coefficient RT/F times 2.303 becomes 0.0592 V. This is the version to use whenever a problem specifies room temperature.
Use E = E° − [RT/(nF)]ln Q for a general temperature. At 298.15 K, converting ln Q to 2.303 log₁₀Q gives E ≈ E° − (0.05916 V/n)log₁₀Q. The 0.0592 coefficient therefore belongs with base-10 log, while RT/(nF) belongs with natural log. R = 8.314 J/(mol·K), F = 96485 C/mol and T in K produce volts because J/C = V.
E = E° - (0.0592 / n) log Q
Writing Q correctly
Q uses the balanced overall cell reaction with product activities over reactant activities. Pure solids and pure liquids, including the electrode metals and water as solvent, are omitted. Gas activities are approximately P/P° for ideal gases; use the same standard-pressure convention as E°, normally 1 bar. Getting Q upside down is the single most common error in this topic.
For Zn + Cu²⁺ → Zn²⁺ + Cu, Q ≈ [Zn²⁺]/[Cu²⁺] and n = 2. At [Zn²⁺] = 1.00 M and [Cu²⁺] = 0.0100 M, Q = 100 and E = 1.10 − (0.0592/2)(2) ≈ 1.04 V. The pure metals have unit activity, so doubling a metal electrode's mass does not directly change Q. A ligand binding Cu²⁺ changes free Cu²⁺ activity, which can change E even if total dissolved copper stays the same.
for Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺] / [Cu²⁺]
Le Chatelier in volts
Increasing a reactant concentration lowers Q, makes the log term more negative, and raises E: the cell is pushed harder toward products. Increasing a product concentration does the opposite. If your Nernst answer disagrees with this qualitative check, you have a sign or setup error.
More reactant means more voltage; more product means less.
Follow actual voltage as composition changes
At fixed T and a fixed standard-state convention, E° stays constant while E depends on Q. In a simple closed cell reacting forward, Q rises toward K and the reversible E tends toward zero. The logarithmic equation alone does not imply one universal “slow then sharp” discharge curve; plateaus, phases, reactant inventories and transport can change its shape. Practical batteries can cease delivering useful current from depletion or internal losses before full thermodynamic equilibrium.
At Q = K, substituting E° = RT ln K/(nF) into the Nernst equation gives E = 0. At Q = 1, the correction is zero and E = E°, which generally is not equilibrium. These two conditions should not be confused.
at equilibrium: Q = K and E = 0
Measuring with the Nernst equation
Because E depends on log of concentration, a cell can serve as a sensor. For a one-electron process the potential changes 0.0592 V for every tenfold change in concentration, which is exactly how a pH electrode reports 59 mV per pH unit at room temperature.
A decade in Q shifts voltage by 0.0592/n V
At 25 °C, E=E°-(0.0592/n)log Q. A tenfold rise in Q lowers E by 0.0592/n V, while a tenfold fall raises it by the same amount. This mental check catches sign mistakes when concentrations favor reactants or products; pure solids remain absent from Q.
Scale Q and n together if the equation is scaled. Doubling the Zn/Cu equation gives Q′ = Q² and n′ = 4. Since log(Q²)/4 = log Q/2, E is unchanged. A tenfold increase in a concentration is not necessarily a tenfold increase in Q: a squared product term increases Q by a factor of 100, doubling its voltage correction. First build Q from the balanced reaction, then apply decade reasoning.
For n=2, each decade in Q changes E by about 0.0296 V.
Key ideas
Rule
Nernst equation
E = E° - (0.0592/n) log Q at 298 K.
Definition
Q
Reaction quotient for the overall cell reaction, excluding pure solids and liquids.
Rule
Q < 1
E is greater than E°.
Rule
Q > 1
E is less than E°.
Rule
Dead battery
Q = K, E = 0, ΔG = 0.
Equation
Nernst equation
E = E° - (R T / n F) ln Q
- Q = reaction quotient at the moment considered
- n = moles of electrons transferred
Equation
Room-temperature form
E = E° - (0.0592 / n) log Q
- 0.0592 = volts, valid at 298 K
Worked examples
Worked example 2
For the same zinc-copper cell (E° = 1.10 V, n = 2), what [Zn²⁺] gives E = 1.13 V when [Cu²⁺] = 1.00 M?
Try it first: E is above E°, so Q must be less than 1.
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Worked example 3
A cell has E° = 0.46 V with n = 2. At a moment during discharge Q = 1.0 × 10⁶. Find E.
Try it first: With Q far above 1, expect a noticeably reduced voltage.
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Worked example 4
For Zn + Cu²⁺ → Zn²⁺ + Cu, E°=1.10 V at 25 °C. Find E when [Zn²⁺]=0.010 M and [Cu²⁺]=1.00 M.
Try it first: Q=[Zn²⁺]/[Cu²⁺]=0.010 and n=2.
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Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Inverting Q. | Q is products over reactants; inverting it flips the sign of the entire correction. | Write the overall reaction first, then read Q from it. |
| Including solid electrodes in Q. | Pure solids and liquids have unit activity and never appear. | List only aqueous species and gases. |
| Forgetting to divide 0.0592 by n. | The correction term scales inversely with the electrons transferred. | Identify n from the balanced overall equation first. |
| Using the 0.0592 form at temperatures other than 25 °C. | That constant already has 298 K built in. | Use the RT/nF form when the temperature differs. |
| Saying E° changes as the cell discharges. | E° is a fixed standard-state constant; discharge changes concentrations and therefore E. | Track Q, not E°. |
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After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
The Nernst equation corrects a standard cell potential for actual concentrations and pressures: E = E° - (RT/nF) ln Q, which at 25 °C simplifies to E = E° - (0.0592/n) log Q. Q is the ordinary reaction quotient for the overall cell reaction, written with products over reactants and omitting pure solids and liquids. When Q is less than 1, meaning reactants are relatively concentrated, the log term is negative and E is larger than E°; when Q exceeds 1, E is smaller than E°. As a cell discharges, products accumulate and Q climbs toward K, so E falls steadily until it reaches zero at equilibrium, which is a dead battery. Setting E = 0 and Q = K recovers the relationship E° = (0.0592/n) log K from the previous lesson, showing that this is one equation family, not two. The Nernst equation also underpins measurement: a pH meter is a cell whose potential varies by about 0.0592 V per pH unit.
- E = E° - (0.0592/n) log Q at 25 °C
- Q follows the usual rules, with solids and pure liquids omitted
- Excess reactant raises E; excess product lowers it
- Discharge drives Q toward K and E toward zero
- E° is constant; only E responds to concentration
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Sources and further reading
- Chemistry 2e, Section 17.4: Potential, Free Energy, and EquilibriumOpenStax · Rice University · Chapter 17.4View source
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