Electrochemistry Calculator
Nernst equation cell potentials and Faraday's law deposition masses.
Nernst: E = E° − (R·T/(n·F))·ln Q, T in kelvin, R = 8.314 J/(mol·K), F = 96 485 C/mol; at 25 °C use E ≈ E° − (0.05916/n)·log₁₀Q. Faraday: Q = I·t, n(e⁻) = Q/F, m = M·Q/(n·F).
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What it does
Electrochemistry connects thermodynamics to measurable voltages and currents. The Nernst equation tells you how far a cell's potential departs from its standard value once concentrations are no longer unit activity, which is the normal situation in a real cell. Faraday's law goes the other way, turning a current and a duration into the amount of substance that reacted at an electrode, which is the basis of electroplating, coulometry and battery capacity estimates.
When to use it
Use the Nernst mode to predict the voltage of a concentration cell, to check whether a redox reaction is still spontaneous after the reactants have been partly consumed, or to see how a pH-dependent half-cell shifts the cell potential. Use the Faraday mode to estimate plating thickness from current and time, to plan a deposition experiment, or to convert a measured charge into moles and mass for a coulometry calculation.
FAQ
- What does the Nernst equation correct for?
- The standard potential E° applies only when every species is at unit activity. The Nernst equation subtracts (R·T/(n·F))·ln Q to account for actual concentrations or partial pressures, where n is the number of electrons transferred and Q is the reaction quotient with products over reactants. At 25 °C the factor R·T/F·ln10 equals 0.05916 V, which is where the familiar 0.05916/n·log Q form comes from.
- How is E°cell built from two half-cells?
- E°cell = E°cathode − E°anode, using both values as reduction potentials. The cathode is where reduction happens and the anode where oxidation happens, so the anode potential is subtracted rather than added. The tool uses this route whenever the standard cell potential field is left empty and both half-cell fields are filled.
- How much metal does electrolysis deposit?
- Faraday's law gives m = M·I·t/(n·F). The charge passed is Q = I·t in coulombs, one mole of electrons carries 96 485 C, and n electrons are needed per metal ion, so the moles deposited are Q/(n·F) and the mass follows from the molar mass. The calculation assumes 100 % current efficiency and no competing reactions such as hydrogen evolution.
- Why do I need to enter the number of electrons?
- Both equations are per mole of reaction, and n sets how many electrons that mole moves. For Ag⁺ + e⁻ → Ag, n = 1; for Cu²⁺ + 2 e⁻ → Cu, n = 2; for Al³⁺ + 3 e⁻ → Al, n = 3. Using the wrong n scales the Nernst correction and the Faraday mass by the same factor, which is the most common mistake in these problems.