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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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
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.
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
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
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⁺]
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.
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
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.
E = E° - (0.0592/n) log Q at 298 K.
Reaction quotient for the overall cell reaction, excluding pure solids and liquids.
E is greater than E°.
E is less than E°.
Q = K, E = 0, ΔG = 0.
E = E° - (R T / n F) ln Q
E = E° - (0.0592 / n) log Q
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.
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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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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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.
Why it's wrong: Pure solids and liquids have unit activity and never appear.
Check instead: List only aqueous species and gases.
Why it's wrong: The correction term scales inversely with the electrons transferred.
Check instead: Identify n from the balanced overall equation first.
Why it's wrong: That constant already has 298 K built in.
Check instead: Use the RT/nF form when the temperature differs.
Why it's wrong: E° is a fixed standard-state constant; discharge changes concentrations and therefore E.
Check instead: Track Q, not E°.
No practice questions are available for this topic yet. You can still practice the whole unit.
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.
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