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 →Electrochemistry is one of the highest-scoring chapters in Class 12 once the sign conventions click. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.