When do I use log vs ln in the Nernst equation?

Use the natural log in E = E° − (RT/nF) ln Q. At 25 °C, this simplifies to E = E° − (0.05916/n) log Q using base-10 logarithms.

How do I construct the reaction quotient Q?

Q is the product of activities of products raised to their stoichiometric coefficients divided by those of reactants. For dilute solutions, activities are approximated by molar concentrations. Pure solids/liquids and solvent water are omitted.

How does temperature affect the Nernst equation?

Temperature changes the factor RT/F and thus the slope. At 25 °C, RT/F ≈ 0.025693 V, giving the familiar 0.05916/n factor when expressed with base-10 logarithms.

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Nernst Equation Calculator

Compute the electrode (or cell) potential using the Nernst equation:
E = E° − (RT/nF) ln Q (general) or at 25 °C: E = E° − (0.05916/n) log Q. Build Q from concentrations or enter it directly. Includes clear steps.

Background

The Nernst equation connects the standard potential E° to conditions away from standard state via the reaction quotient Q and temperature T. Here, n is the number of electrons transferred and F is Faraday’s constant. At 25 °C, the factor RT/F becomes 0.025693 V, giving 0.05916 V when converting natural log to log base 10.

Choose mode and enter values:

Mode:

Quick (25 °C, use E = E° − (0.05916/n) log Q)

General (any T, use E = E° − (RT/nF) ln Q)

Standard potential, E° (V):

Electrons transferred, n:

How would you like to provide Q?

Enter Q directly

Build Q from concentrations

Reaction quotient, Q (unitless):

Enter Q as (activities) products^coeff / reactants^coeff. For dilute solutions, activities ≈ concentrations.

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Result:

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How to use this calculator

Quick (25 °C): Enter E°, n, and Q. We use E = E° − (0.05916/n) log Q.
General (any T): Enter E°, n, Q (or build Q), and T (K or °C). We use E = E° − (RT/nF) ln Q with your R and F (defaults provided).
Building Q: Add species as products or reactants with their concentrations and stoichiometric coefficients. Omit pure solids/liquids and solvent water.

Assumes dilute solutions (activities ≈ concentrations). Sign conventions follow the overall balanced reaction as written.

Example Problems & Step-by-Step Solutions

Example 1 (Zn/Cu at 25 °C)

Overall: Cu²⁺ + Zn → Cu + Zn²⁺ (n=2). E°=1.10 V. If [Zn²⁺]=0.010 M, [Cu²⁺]=1.0 M then Q=[Zn²⁺]/[Cu²⁺]=0.010. E = 1.10 − (0.05916/2)·log(0.010) ≈ 1.159 V.

Example 2 (Fe³⁺/Fe²⁺ at 25 °C)

Fe³⁺ + e⁻ ⇌ Fe²⁺ (n=1). E°=0.77 V. With [Fe²⁺]=0.10 M and [Fe³⁺]=1.0 M → Q=[Fe²⁺]/[Fe³⁺]=0.10. E = 0.77 − 0.05916·log(0.10) ≈ 0.829 V.

Example 3 (General T)

At 310 K, n=2, Q=0.010, E°=1.10 V: E = 1.10 − (RT/nF)lnQ with R=8.314, F=96485 → E ≈ 1.10 − (8.314·310/(2·96485))·ln(0.010) ≈ 1.164 V.

Frequently Asked Questions

Q: When do I use log vs ln?

Use ln in the general form E = E° − (RT/nF) ln Q. At 25 °C, converting to base-10 gives E = E° − (0.05916/n) log Q.

Q: What goes into Q?

Activities of products raised to their coefficients over reactants raised to theirs. For dilute solutions, activity ≈ molarity; omit pure solids/liquids and water.

Q: Does temperature matter a lot?

Yes. The factor (RT/F) scales with T, so higher T slightly changes the Nernst slope compared to 25 °C.