Nernst Equation & Cell Potential Calculator
Calculate the actual cell potential at non-standard conditions using the Nernst equation. Find how concentration, temperature, and reaction quotient affect electrochemical cell voltage.
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Nernst Equation Guide
What's the key thing to understand about The Nernst Equation?
E = E° − (RT/nF) × ln(Q). At 25°C simplified: E = E° − (0.0257/n) × ln(Q) = E° − (0.05916/n) × log₁₀(Q). Where E = actual cell potential, E° = standard cell potential (all concentrations 1 mol/L, 25°C, 1 atm), R = 8.314 J/mol·K, T = temperature in Kelvin, n = moles of electrons transferred, F = Faraday constant (96,485 C/mol), Q = reaction quotient. The Nernst correction becomes zero when Q = 1 (standard conditions) or when ln(Q) = 0.
What's the key thing to understand about Reaction Quotient Q?
Q = [products]^stoich / [reactants]^stoich (same form as K but using actual concentrations, not equilibrium). For Zn/Cu cell: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Q = [Zn²⁺] / [Cu²⁺]. At equilibrium: Q = K, and E = 0 (dead battery). E > 0 means reaction is spontaneous (Q < K). E < 0 means reaction is non-spontaneous (Q > K, product-favoured conditions). Standard potential E° > 0 tells you the reaction is spontaneous under standard conditions.
What should I know about ΔG and Cell Potential?
The relationship between cell potential and Gibbs energy: ΔG = −nFE. At standard conditions: ΔG° = −nFE°. Relationship to K: ΔG° = −RT ln(K) = −nFE°. Therefore: ln(K) = nFE°/RT = nE°/0.0257 at 25°C. A zinc-copper cell (E° = 1.10V, n = 2): ln(K) = 2 × 1.10/0.0257 = 85.6. K = e^85.6 ≈ 10^37 — overwhelmingly product-favoured at equilibrium. ΔG° = −2 × 96485 × 1.10 = −212,267 J/mol = −212.3 kJ/mol. Highly spontaneous.
What's the key thing to understand about Concentration Cells?
A concentration cell has identical electrodes and electrolytes at different concentrations — the potential arises entirely from the concentration gradient. E = −(RT/nF) × ln(Q) = −(0.05916/n) × log([high concentration side]/[low concentration side]). Two copper electrodes: 0.1 mol/L Cu²⁺ vs 1.0 mol/L Cu²⁺. E = −(0.05916/2) × log(0.1/1.0) = −0.02958 × (−1) = +0.030 V. Biological example: nerve impulses rely on concentration cell potentials across cell membranes (Nernst potential for each ion dete