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.
0 of 3 steps revealed.
Electrochemistry & Applications of ThermodynamicsElectrochemistryContent level: Core 27 min
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
Lesson
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.
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. Gases enter as partial pressures in atmospheres. Getting Q upside down is the single most common error in this topic.
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.
Discharge and death
During discharge, reactants are consumed and products build up, so Q rises continuously. E falls, slowly at first and then sharply as Q approaches K. At Q = K, E = 0 and no further work is available. Note again that E° never changes; only E does.
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.
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.
0 of 3 steps revealed.
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.
0 of 3 steps revealed.
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.
0 of 3 steps revealed.
Common mistakes
Inverting Q.
Why it's wrong: Q is products over reactants; inverting it flips the sign of the entire correction.
Check instead: Write the overall reaction first, then read Q from it.
Including solid electrodes in Q.
Why it's wrong: Pure solids and liquids have unit activity and never appear.
Check instead: List only aqueous species and gases.
Forgetting to divide 0.0592 by n.
Why it's wrong: The correction term scales inversely with the electrons transferred.
Check instead: Identify n from the balanced overall equation first.
Using the 0.0592 form at temperatures other than 25 °C.
Why it's wrong: That constant already has 298 K built in.
Check instead: Use the RT/nF form when the temperature differs.
Saying E° changes as the cell discharges.
Why it's wrong: E° is a fixed standard-state constant; discharge changes concentrations and therefore E.
Check instead: Track Q, not E°.
After the lesson · Question bank
Independent practice
Work through new problems on this topic on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
No questions are available for this topic yet. Explore the unit question bank.
Your next step
Lesson: Not marked completed
Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.
Continue through the unit’s lessons and available practice.
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
Sources and further reading
- Chemistry 2e, Section 17.4: Potential, Free Energy, and EquilibriumOpenStax · Rice University · Chapter 17.4View source
Access for free at openstax.org License