Nernst Equation: Galvanic Cell EMF Under Non-Standard Conditions (Worked Examples)
By ChemistryIQ Team · April 27, 2026
Direct Answer: What the Nernst Equation Does
The Nernst equation calculates cell potential at non-standard concentrations: E = E° − (RT/nF) ln(Q), or at 25°C using common log: E = E° − (0.0592/n) log(Q). E° is the standard cell potential (1 M, 1 atm, 25°C); Q is the reaction quotient ([products]/[reactants]); n is the number of electrons transferred; F is the Faraday constant. When Q < 1 (more reactants than products), E > E° — the cell pushes harder. When Q > 1 (more products), E < E°. At equilibrium, Q = K and E = 0 — that's how galvanic cells 'die' when their cells run out of driving force. The equation also gives you the relationship E°cell = (0.0592/n) log K, which lets you compute equilibrium constants directly from standard reduction potentials.
The Equation: Two Common Forms
Original Nernst form (kelvin temperature explicit): E = E° − (RT/nF) ln(Q)
Where R = 8.314 J/(mol·K), T is temperature in K, n is moles of electrons, F = 96,485 C/mol.
Simplified form at 25°C (298 K) using log10: E = E° − (0.0592/n) log(Q)
This is the form used on most chemistry exams. Always check whether your problem is at 25°C — if not, use the temperature-explicit form.
Reaction quotient Q: same form as equilibrium expression. For aA + bB → cC + dD, Q = [C]^c[D]^d / [A]^a[B]^b. Solids, pure liquids, and water as solvent are excluded.
For the cell reaction Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s), Q = [Zn^2+]/[Cu^2+].
Worked Example 1: Standard Cell at Non-Standard Concentrations
Consider the Daniell cell: Zn(s) | Zn^2+(aq, 0.10 M) || Cu^2+(aq, 1.5 M) | Cu(s).
E° for this cell is +1.10 V (Cu^2+/Cu is +0.34 V; Zn^2+/Zn is -0.76 V; cell potential = +0.34 − (−0.76) = +1.10 V).
Electrons transferred: n = 2 (Zn → Zn^2+ loses 2; Cu^2+ → Cu gains 2).
Reaction quotient: Q = [Zn^2+]/[Cu^2+] = 0.10 / 1.5 = 0.067.
Apply Nernst: E = 1.10 − (0.0592/2) log(0.067) = 1.10 − (0.0296)(−1.18) = 1.10 + 0.035 = 1.135 V.
So at these non-standard concentrations, the cell pushes 35 mV harder than standard. The reason: low [Zn^2+] (product) and high [Cu^2+] (reactant) both favor the forward reaction by Le Chatelier's principle, increasing the driving force.
Worked Example 2: Concentration Cell (Same Half-Reaction Both Sides)
A concentration cell uses the same half-reaction on both sides — only the concentrations differ. The cell drives the system toward equal concentrations.
Example: Cu(s) | Cu^2+(aq, 0.001 M) || Cu^2+(aq, 0.10 M) | Cu(s).
E° = 0 V (same half-reaction on both sides — the standard potentials cancel).
Q = [Cu^2+, anode] / [Cu^2+, cathode] = 0.001 / 0.10 = 0.010.
E = 0 − (0.0592/2) log(0.010) = −(0.0296)(−2.0) = +0.059 V.
The cell produces about 60 mV — small, but nonzero. The Cu electrode in the dilute solution dissolves (anode) and the Cu electrode in the concentrated solution receives plating (cathode). The cell continues until both sides reach the same concentration, at which point E = 0.
This is the principle behind pH electrodes (H+ concentration cells), Pourbaix diagrams, and many biochemical electrode-based assays.
Worked Example 3: pH Dependence of Cell Potential
Many cell reactions involve H+ ions, making the EMF pH-dependent. Consider the half-reaction:
MnO4^- + 8H+ + 5e- → Mn^2+ + 4H2O, E° = +1.51 V
The Nernst expression: E = 1.51 − (0.0592/5) log([Mn^2+] / [MnO4^-][H+]^8)
At pH 0 ([H+] = 1 M, [Mn^2+] = 1 M, [MnO4^-] = 1 M): E = 1.51 V (standard).
At pH 4 ([H+] = 10^-4 M, others 1 M): Q = 1 / (1 × (10^-4)^8) = 10^32 E = 1.51 − (0.01184) log(10^32) = 1.51 − 0.379 = 1.131 V.
The potential drops by ~380 mV when pH rises from 0 to 4. The 8 H+ in the half-reaction means the potential is extremely pH-sensitive — every unit of pH change shifts the potential by (0.0592 × 8 / 5) = 95 mV. This is why permanganate is a much weaker oxidizer in basic solution.
The general rule: each H+ in the half-reaction contributes 59.2 mV per pH unit per electron transferred. For an aH+ + ne- → ... half-reaction at 25°C, dE/dpH = -0.0592a/n V/pH unit.
Worked Example 4: Computing K from E°cell
At equilibrium, E = 0 and Q = K. Plugging into Nernst:
0 = E° − (0.0592/n) log K E° = (0.0592/n) log K log K = nE°/0.0592 K = 10^(nE°/0.0592)
For the Daniell cell: E°cell = +1.10 V, n = 2. log K = (2)(1.10)/0.0592 = 37.2 K = 10^37.2 ≈ 1.6 × 10^37
Massive K — the reaction goes essentially to completion. This makes physical sense: a cell potential of +1.10 V means a strongly downhill reaction in the forward direction.
For a near-zero cell potential reaction (E°cell = +0.05 V, n = 2): log K = (2)(0.05)/0.0592 = 1.69 K ≈ 49
Still favors products (K > 1) but only modestly.
If E°cell is negative, K < 1 and the reaction favors reactants in standard conditions — but the reaction can still go forward under non-standard conditions if Q is small enough.
Worked Example 5: Predicting Whether a Reaction Will Run
For Cu(s) + 2Ag+(aq, 0.0010 M) → Cu^2+(aq, 0.50 M) + 2Ag(s):
E° for this cell: Ag+/Ag = +0.80 V, Cu^2+/Cu = +0.34 V. E°cell = +0.80 − 0.34 = +0.46 V (Ag+ is reduced, Cu is oxidized).
n = 2 (Cu loses 2 electrons; 2 Ag+ each gain 1).
Q = [Cu^2+] / [Ag+]^2 = 0.50 / (0.0010)^2 = 5 × 10^5
E = 0.46 − (0.0592/2) log(5 × 10^5) = 0.46 − (0.0296)(5.70) = 0.46 − 0.169 = +0.291 V
E > 0, so the reaction is spontaneous as written under these conditions. Cu metal will dissolve and Ag will plate out.
If we changed the conditions so [Ag+] = 10^-9 M and [Cu^2+] = 1 M: Q = 1 / (10^-9)^2 = 10^18 E = 0.46 − (0.0296)(18) = 0.46 − 0.533 = −0.073 V
E < 0, so the reaction would now run BACKWARDS — Ag would dissolve and Cu would plate. Concentrations matter.
How ChemistryIQ Helps With Nernst Problems
Snap a photo of the cell setup or problem and ChemistryIQ identifies the cell reaction, looks up the standard reduction potentials, computes E°cell, identifies n, builds the reaction quotient Q, applies the Nernst equation, and produces the cell EMF at the specified conditions. For pH-dependent half-reactions, it computes the pH-shifted potential. For K-from-E° problems, it produces the equilibrium constant. Useful for ACS Gen Chem, MCAT, and analytical chemistry preparation.
Common Mistakes
First: forgetting that water and pure liquids/solids are excluded from Q. For Mn^2+ + 4H2O coming out of permanganate reduction, water doesn't appear in Q.
Second: getting the Q expression backwards. Q = [products]/[reactants], same form as K. Inverting it flips the sign of the log term.
Third: using R/nF without converting temperature to kelvin. The 0.0592 simplification ONLY applies at 25°C. For 50°C, you must use the full form: E = E° − (8.314)(323)/((n)(96485)) ln(Q).
Fourth: mixing up natural log and log10. RT/F at 25°C = 0.0257 V (using ln); using log10 introduces an extra factor of ln(10) ≈ 2.303, giving 0.0592 V. Always check which log you're using.
Fifth: forgetting that n is the electrons in the BALANCED cell reaction, not in either half-reaction alone. For Zn + Cu^2+ → Zn^2+ + Cu, n = 2. For 2Ag+ + Cu → 2Ag + Cu^2+, n is also 2 (Cu loses 2 electrons; 2 Ag+ each gain 1, totaling 2 electrons transferred per cycle of the balanced reaction).
FAQs
Common questions about nernst equation
Use Nernst whenever any concentration is not 1 M, any partial pressure is not 1 atm, or temperature is not 25°C. The standard E° applies only under standard conditions. In practice — especially in biology and electrochemistry labs — concentrations almost always deviate from 1 M, so Nernst is the working equation. The exam shortcut: if the problem gives you specific concentrations or asks about pH dependence, it expects Nernst.
At equilibrium, the forward and reverse rates are equal — there's no net driving force in either direction. The cell can do no work, so its potential is zero. Mathematically, at equilibrium Q = K, and Nernst becomes E = E° − (0.0592/n) log K. Combined with the relationship E° = (0.0592/n) log K, the right side equals zero. This is also why batteries 'die' — they reach internal equilibrium and no longer push current.
The cathode is the half-reaction with the higher (more positive) reduction potential — that's where reduction happens. The anode is the half-reaction with the lower reduction potential, but written as oxidation (so flip the sign). E°cell = E°cathode − E°anode (using both as reduction potentials). If E°cell is positive, the reaction as you wrote it is spontaneous; if negative, it runs the other way.
Yes. Snap a photo of the cell diagram or problem and ChemistryIQ identifies the half-reactions, computes E°cell from a built-in reduction potential table, builds the reaction quotient, applies the Nernst equation under the specified conditions, and produces a worked solution including unit checks. This content is for educational purposes only and does not constitute professional advice.