The Nernst Equation

Why a battery's voltage sags as it drains — and how to calculate the exact voltage at any concentration.

Intro Gen ChemUni Year 1
⏱️ About 18 min

A fresh AA battery reads a little above its rated voltage; a nearly-dead one reads lower, even though it's the same chemistry inside. Standard cell potential can't explain that — it assumes perfect 1 M conditions that never last. The Nernst equation is the correction that connects voltage to the actual concentrations in the cell.

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The big idea: The real cell potential depends on concentrations through the reaction quotient Q. The Nernst equation adjusts E° up or down: as reactants are consumed and products build up, Q rises and the voltage falls — until, at equilibrium, E hits zero and the battery is dead.
🎯 By the end, you'll be able to
  • Write the Nernst equation in its 25°C form and identify every term
  • Compute the reaction quotient Q for a cell reaction
  • Calculate cell potential E at non-standard concentrations
  • Explain why E falls as a cell discharges and reaches 0 at equilibrium
📎 Helpful to know first

Standard conditions rarely last

The standard cell potential E° is measured with every dissolved species at 1 M and gases at 1 bar. The instant a cell starts working, those concentrations drift: reactants deplete, products accumulate. The measured voltage drifts too. To track it, we need concentrations in the equation — and that's exactly what the reaction quotient Q brings in.

\[ E = E^{\circ} - \frac{0.0592}{n}\,\log Q \quad (\text{at } 25^{\circ}\text{C}) \]
The Nernst equation at 25°C. E° is the standard potential, n is the moles of electrons transferred, and Q is the reaction quotient.
🔑 What each symbol means
  • E — the actual cell potential right now (volts).
  • E° — the standard cell potential (from E°cathode − E°anode).
  • n — moles of electrons transferred in the balanced reaction.
  • Q — the reaction quotient: products over reactants, same form as K, using current concentrations.
The 0.0592 already bundles the gas constant, temperature (298 K), and Faraday's constant for you — but only at 25°C.
✨ Reading the direction of the shift
If Q < 1 (more reactants than products), log Q is negative, so E is above E° — a fresh cell pushes a little harder. As the cell runs, products build up, Q rises, log Q climbs, and E drops. When the reaction reaches equilibrium, Q = K and E = 0: no more push, the battery is dead.
📝 Worked example: For the Zn/Cu cell, Zn + Cu²⁺ → Zn²⁺ + Cu, E° = 1.10 V and n = 2. Find E when [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M.
  1. Write Q: solids don't appear, so Q = [Zn²⁺] / [Cu²⁺] = 1.0 / 0.010 = 100.
  2. log Q = log(100) = 2.
  3. Nernst: E = 1.10 − (0.0592 / 2)(2) = 1.10 − (0.0296)(2).
  4. = 1.10 − 0.0592 = 1.0408 V.
✓ E ≈ 1.04 V — slightly below E° = 1.10 V, because product Zn²⁺ is high and reactant Cu²⁺ is low (Q > 1).
⚠️ Two easy slips
First: n is the electrons in the balanced reaction, not a guess — for Zn/Cu it's 2. Second: it's log (base 10) in the 0.0592 form, not natural log. (The ln version uses RT/nF ≈ 0.0257/n instead.) Mixing these up is the most common Nernst error.
✏️ Practice: A Zn/Cu cell has E° = 1.10 V and n = 2. Now [Zn²⁺] = 0.10 M and [Cu²⁺] = 1.0 M, so Q = [Zn²⁺]/[Cu²⁺] = 0.10. Find E at 25°C, in volts. (log 0.10 = −1.)
V
Solution
  1. Q = [Zn²⁺]/[Cu²⁺] = 0.10 / 1.0 = 0.10, so log Q = −1.
  2. E = E° − (0.0592/n)·log Q = 1.10 − (0.0592/2)(−1).
  3. = 1.10 − (0.0296)(−1) = 1.10 + 0.0296.
  4. = 1.13 V. Q < 1, so E rose above E° — plenty of reactant, little product.

Check your understanding

1. In the Nernst equation E = E° − (0.0592/n)·log Q, what is n?
n is the number of moles of electrons transferred in the balanced cell reaction (2 for the Zn/Cu cell).
2. As a galvanic cell discharges toward equilibrium, its cell potential E:
Products build up, Q rises, and E falls. At equilibrium Q = K and E = 0 — the cell can do no more work.
3. If Q < 1 for a cell reaction at 25°C, then compared with E° the actual E is:
Q < 1 makes log Q negative, and subtracting a negative raises E above E° — a fresh cell rich in reactants pushes a bit harder than standard.
🔎 Related lessons
Previous: Galvanic Cells & Cell Potential Next: Electrolysis & Faraday's Laws
✅ Key takeaways
  • Real cell potential depends on concentration; the Nernst equation supplies the correction.
  • At 25°C: E = E° − (0.0592/n)·log Q, where n = moles of electrons and Q = products/reactants.
  • Q < 1 raises E above E°; Q > 1 lowers it below E°.
  • As a cell discharges, Q climbs and E falls; at equilibrium Q = K and E = 0 (dead battery).
  • Use log (base 10) with 0.0592, and take n straight from the balanced reaction.
➡️ So far the electrons have flowed on their own, driven by a positive E°cell. But you can also run the whole thing in reverse — pour electrical energy IN to force a non-spontaneous reaction. That's electrolysis, and it comes with its own quantitative laws.
Want to test yourself on this? Try the Chemistry practice test →