Why is the 0.0592/n shortcut only valid at 25°C?

0.0592 is RT·ln(10)/F evaluated specifically at T = 298.15 K. Since R and F are fixed constants, that numeric factor changes at any other temperature, using it at, say, 37°C body temperature introduces a small but real error, so the full E = E° − (RT/nF) ln Q form should be used whenever T isn't 25°C.

Nernst Equation Calculator: Cell Potential, Free

E°, standard cell potential (V)

n, electrons transferred

Q, reaction quotient

Temperature (K)

What is n in the Nernst equation?

n is the number of moles of electrons transferred in the balanced redox reaction, not the number of atoms or ions. Balance the two half-reactions so the electrons cancel, and n is that shared number. In the Daniell cell, Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu, so n = 2. Because n divides the log term, doubling it halves how much the potential shifts away from E°.

The Goldman equation (membrane potential)

The Nernst equation gives the equilibrium potential for one ion. A real membrane is permeable to several at once, and the Goldman-Hodgkin-Katz equation weights each by its permeability:

Vm = (RT/F) · ln [ (P_K[K⁺]out + P_Na[Na⁺]out + P_Cl[Cl⁻]in ) ------------------------------------------ (P_K[K⁺]in + P_Na[Na⁺]in + P_Cl[Cl⁻]out) ] RT/F = 26.7 mV at 37°C (61.5 mV if you use log10 instead of ln)

Chloride is flipped (out on top, in on the bottom) because it is an anion, so its charge sign reverses the ratio.

Worked, with typical mammalian values and permeability ratios P_K:P_Na:P_Cl = 1 : 0.04 : 0.45:

numerator = 1(5) + 0.04(145) + 0.45(10) = 15.3 denominator = 1(140) + 0.04(15) + 0.45(110) = 190.1 Vm = 26.7 × ln(15.3 / 190.1) = -67 mV (a normal resting potential)

Compare the single-ion Nernst potentials from the same concentrations: E_K = 26.7 × ln(5/140) = -89 mV and E_Na = 26.7 × ln(145/15) = +61 mV. Resting membrane potential sits close to E_K precisely because the membrane is far more permeable to potassium, and it swings toward E_Na during an action potential when sodium channels open.

How to use the Nernst equation

The Nernst equation adjusts a cell's standard potential for real, non-standard concentrations: E = E° − (RT/nF) ln Q. Here E° is the standard cell potential, R = 8.314 J/mol·K, T is temperature in kelvin, n is the number of electrons transferred, F = 96485 C/mol is Faraday's constant, and Q is the reaction quotient. At 25 °C (298 K) the RT/F term collapses to a constant, giving the common shortcut E = E° − (0.0592/n) log₁₀ Q. A positive E means the reaction is spontaneous as written under those conditions; at equilibrium Q equals K and E = 0.

Constants: R = 8.314 J/mol·K, F = 96485 C/mol. The 0.0592 factor is RT·ln(10)/F evaluated at 298 K, so it is only exact at 25 °C.

Worked example 1: the Daniell cell (the default)

E° = 1.10 V, n = 2 (Zn/Cu²⁺, the classic Zn|Zn²⁺||Cu²⁺|Cu cell), Q = 0.001, at 298.15 K, the tool's default values.

RT/nF = (8.314 × 298.15) / (2 × 96485) = 1.285 × 10⁻² V ln Q = ln(0.001) = −6.9078 E = E° − (RT/nF) ln Q = 1.10 − (1.285×10⁻²)(−6.9078) = 1.10 + 0.0887 = 1.1887 V → E > 0: spontaneous as written 25°C shortcut check: E = 1.10 − (0.0592/2) log(0.001) = 1.10 − (0.0296)(−3) = 1.1888 V

The full RT/nF form and the 0.0592/n shortcut agree to within 0.0001 V, as expected at exactly 25°C. A low reaction quotient (fewer products relative to reactants than at standard conditions) pushes E above E°, making the reaction more favorable than standard.

Worked example 2: a concentration cell (E° = 0)

Two Cu²⁺/Cu half-cells, one at 1.0 M and one at 0.001 M, connected as a concentration cell. Since both sides run the same electrode reaction, E° = 0, any voltage comes purely from the concentration difference. n = 2 for Cu²⁺ + 2e⁻ → Cu, Q = [dilute]/[concentrated] = 0.001/1.0 = 0.001.

RT/nF = 1.285 × 10⁻² V (same as example 1, same n and T) ln Q = ln(0.001) = −6.9078 E = E° − (RT/nF) ln Q = 0 − (1.285×10⁻²)(−6.9078) = +0.0887 V

Even with E° = 0, the cell produces a real +0.0887 V, driven entirely by the concentration gradient. This is the same principle behind resting membrane potentials in biology: no net chemical reaction is required for a voltage to exist, just a concentration difference across a barrier that ions can cross.

FAQ

What does it mean physically when E = 0?
The cell is at equilibrium, Q equals K, with no more net driving force to push electrons through the circuit. A fully discharged battery reads 0 V for exactly this reason.

How can a concentration cell generate voltage if E° = 0?
Same electrode reaction on both sides means E° = 0 by definition, but different concentrations mean Q ≠ 1, the −(RT/nF) ln Q term alone produces a nonzero E, as shown in example 2 above.

Why is 0.0592/n only valid at 25°C?
0.0592 is RT·ln(10)/F evaluated specifically at 298.15 K. At any other temperature that numeric factor changes, so the full E = E° − (RT/nF) ln Q form should be used whenever T isn't 25°C.

What's the difference between E and E°?
E° is the standard cell potential at 1 M, 1 atm, 25°C. E is the actual potential under real, non-standard conditions, found by correcting E° with Q via the Nernst equation.

Practice problems

1. A cell has E° = 0.80 V, n = 1, Q = 0.001, at 25°C. Find E.

Show answer

RT/nF = (8.314 × 298.15) / (1 × 96485) = 2.569 × 10⁻² V ln Q = ln(0.001) = −6.9078 E = 0.80 − (2.569×10⁻²)(−6.9078) = 0.80 + 0.1775 = 0.9775 V

2. A cell has E° = 0.20 V, n = 2, Q = 1,000,000, at 25°C. Find E.

Show answer

RT/nF = (8.314 × 298.15) / (2 × 96485) = 1.285 × 10⁻² V ln Q = ln(1,000,000) = 13.8155 E = 0.20 − (1.285×10⁻²)(13.8155) = 0.20 − 0.1775 = 0.0225 V A very large Q (far more product than reactant relative to standard conditions) pulls E down close to zero. This cell is nearly at equilibrium.