Nernst Equation Made Simple — Cell Potential at Any Concentration
Standard electrode potentials in your textbook are measured under one very specific condition: every dissolved species at 1 M, every gas at 1 bar, temperature 298 K. Real cells almost never sit at those conditions. The Nernst equation is the correction that takes you from the table value E° to the actual voltage E of the cell in front of you. This guide explains every symbol, shows where the famous 0.0592 comes from, and works three problems completely.
The equation
At 298 K this becomes: E = E° − (0.0592 / n) log Q
Both forms are the same equation. The second is just the first with the constants put in and natural log converted to base-10 log. Use the log form for exams; use the ln form whenever the temperature is not 298 K.
What each symbol means
| Symbol | Meaning | Unit |
|---|---|---|
| E | Cell (or electrode) potential at the actual conditions | volt, V |
| E° | Standard potential, from the reduction-potential table | volt, V |
| R | Gas constant, 8.314 | J mol⁻¹ K⁻¹ |
| T | Absolute temperature | kelvin, K |
| n | Electrons transferred in the balanced cell reaction | no unit |
| F | Faraday constant, 96 485 | C mol⁻¹ |
| Q | Reaction quotient = products ÷ reactants, each raised to its coefficient | no unit |
Where 0.0592 comes from
It is not a magic number. Convert ln to log by multiplying by ln 10 = 2.3026, then put in the values at T = 298.15 K:
= (2478.8 ÷ 96 485) × 2.3026 = 0.025693 × 2.3026 = 0.0592 V
Because it contains T, the constant changes with temperature. At human body temperature, 310 K, the same arithmetic gives (8.314 × 310 ÷ 96 485) × 2.3026 = 0.026712 × 2.3026 = 0.0615 V. Blindly using 0.0592 in a 310 K question is a common way to lose marks.
Writing Q correctly — the step most students rush
Q is written from the balanced cell reaction, in the same way as an equilibrium constant expression, with three rules:
- Pure solids and pure liquids do not appear (their activity is 1).
- Dissolved species enter as their molar concentration.
- Gases enter as their partial pressure in bar.
For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), Q = [Zn²⁺] ÷ [Cu²⁺]. The two metals are solids, so they are simply absent.
Worked example 1 — a single electrode
Find the potential of a copper electrode dipped in 1.0 × 10⁻³ M Cu²⁺ at 298 K. E°(Cu²⁺/Cu) = +0.34 V.
Half reaction: Cu²⁺ + 2e⁻ → Cu, so n = 2.
Q = 1 ÷ [Cu²⁺] = 1 ÷ 0.0010 = 1000, so log Q = 3.
E = 0.34 − (0.0592 ÷ 2) × 3 = 0.34 − 0.0296 × 3 = 0.34 − 0.0888
E = +0.251 V ≈ +0.25 V
Diluting the Cu²⁺ made the electrode a weaker oxidising agent, so the reduction potential dropped. That direction should always feel reasonable before you accept a number.
Worked example 2 — a concentration cell
Cu | Cu²⁺ (0.0010 M) ‖ Cu²⁺ (0.100 M) | Cu at 298 K. Find E.
Both electrodes are copper, so E° = 0.34 − 0.34 = 0 V. The entire voltage comes from the concentration difference.
The cell drives copper from the dilute side into solution and plates it on the concentrated side, so Q = [dilute] ÷ [concentrated] = 0.0010 ÷ 0.100 = 0.010, and log Q = −2. Here n = 2.
E = 0 − (0.0592 ÷ 2) × (−2) = 0 + 0.0592
E = +0.0592 V
A cell with two identical electrodes still produces a measurable voltage. This is exactly how a pH meter and an ion-selective electrode work.
Worked example 3 — the hydrogen electrode and pH
Show that the potential of a hydrogen electrode at 1 bar H₂ is E = −0.0592 × pH, and evaluate it at pH 4.
Half reaction: 2H⁺ + 2e⁻ → H₂, E° = 0 V by definition, n = 2.
Q = p(H₂) ÷ [H⁺]² = 1 ÷ [H⁺]²
log Q = −2 log [H⁺] = +2 pH
E = 0 − (0.0592 ÷ 2) × 2 pH = −0.0592 pH
At pH 4: E = −0.0592 × 4 = −0.2368 V ≈ −0.237 V
This one line is the entire theory of potentiometric pH measurement.
Common mistakes
- Wrong sign on the log term. The equation subtracts. If Q > 1, E must be smaller than E°; if Q < 1, E must be larger. Check that before moving on.
- Q upside down. Q is products over reactants, for the reaction as you wrote it. Reversing it flips the sign of the correction.
- Wrong n. n is the electrons in the balanced overall reaction, not the charge on one ion. For 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu, n = 6, not 3.
- Multiplying E° when you balance. If you double a half reaction to balance electrons, n doubles but E° does not change. Potential is an intensive property.
- Using 0.0592 at every temperature. It is a 298 K value only.
- Including solids in Q. Zn(s), Cu(s), AgCl(s) never appear.
Where it appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 12 | EMF of a cell at given concentrations; effect of dilution on E |
| JEE / NEET | Concentration cells, E° to K conversion, pH from cell EMF |
| IIT-JAM / CUET-PG | Ksp from cell potential, non-298 K substitution |
| GATE / CSIR-NET | Coupled ΔG = −nFE problems, potentiometric titration curves |
Check your working in seconds. Enter E°, n, the concentrations and the temperature, and the Nernst Equation calculator returns E along with the value of Q it used — so you can see immediately whether your Q was the problem.
Open the Nernst Equation Calculator →