Electrochemistry, Nernst Equation & Faraday's Law Calculator
Calculate electrolysis mass, current, time, valence z, and STP gas volume via Faraday's Law (Q = It, m = MIt/(zF)), solve non-standard cell potentials and concentration cells with the Nernst equation (E = E° − (RT/nF)ln Q), and determine standard cell EMF E°cell, Gibbs free energy ΔG° = −nFE°, and equilibrium constant K with a 1-click toggle between textbook F = 96,500 C/mol and exact CODATA F = 96,485.33212 C/mol.
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Key Formula & Conditions
Learning modelSolves Faraday's law of electrolysis (Q = It, m = MIt/(zF), STP gas volume, 4-way reverse solver for m, I, t, z), the Nernst equation (E = E° − (RT/nF)ln Q, concentration cells), and standard cell thermodynamics (E°cell, ΔG° = −nFE°, equilibrium constant K) with 1-click toggle between textbook F = 96,500 C/mol and CODATA F = 96,485.33212 C/mol.
Electrolysis Target & Half-Reaction Conditions
Reference Table: Standard Reduction Potentials E° (25°C) & Equivalent Masses per Mole of Electrons
| Half-Reaction (Written as Reduction) | E° (V vs SHE) | Electrons z | Molar Mass M | Mass / Volume per 1 mol e⁻ |
|---|---|---|---|---|
| Li⁺ + e⁻ ⇄ Li | −3.040 V | 1 | 6.94 g/mol | 6.94 g / mol e⁻ |
| Al³⁺ + 3e⁻ ⇄ Al | −1.660 V | 3 | 26.98 g/mol | 8.99 g / mol e⁻ |
| Zn²⁺ + 2e⁻ ⇄ Zn | −0.760 V | 2 | 65.38 g/mol | 32.69 g / mol e⁻ |
| 2H⁺ + 2e⁻ ⇄ H₂ (g) | 0.000 V (SHE) | 2 | 2.016 g/mol | 11.2 L (STP) / mol e⁻ |
| Cu²⁺ + 2e⁻ ⇄ Cu | +0.340 V | 2 | 63.55 g/mol | 31.78 g / mol e⁻ |
| Ag⁺ + e⁻ ⇄ Ag | +0.800 V | 1 | 107.87 g/mol | 107.87 g / mol e⁻ |
| O₂ (g) + 4H⁺ + 4e⁻ ⇄ 2H₂O | +1.229 V | 4 | 32.00 g/mol | 5.60 L (STP) / mol e⁻ |
Done
Electrochemistry, Nernst Equation & Faraday's Law Calculator
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Worked Examples & Verified Values
CuSO₄ Electrolysis (5.00 A, 1930 s)
Daniell Cell ([Zn²⁺]=0.01M, [Cu²⁺]=1.0M)
Daniell Cell Thermo (E°=+1.100 V, n=2)
Common Pitfalls & Calculation Rules
- •Multiplying current I (in A) directly by time in minutes or hours without converting to seconds. Because 1 C = 1 A·s, time t must be in seconds.
- •Forgetting electron stoichiometry z in half-reactions. Passing 0.10 mol of electrons deposits 0.10 mol of Ag (z = 1), 0.050 mol of Cu (z = 2), and evolves only 0.025 mol of O₂ gas at the anode (z = 4).
- •Multiplying standard reduction potential E° by stoichiometric coefficients when balancing half-reactions. Electrode potential E° (V = J/C) is an intensive property; scaling a half-reaction changes n in ΔG° = −nFE°, never E° itself.
Verification scope and citation information
Use worked examples within their stated inputs, units and assumptions. Tests cover specific cases at a specific time; they do not certify every input, expert review or standards compliance. Check the references and tool-specific verification notes, and enter your own access date when citing.
@misc{tools_hub_electrochemistry_calculator_2026,
author = {{Tools Hub}},
title = {Electrochemistry, Nernst Equation & Faraday's Law Calculator},
year = {2026},
url = {https://the-tools-hub.net/en/tools/electrochemistry-calculator/},
note = {Accessed: YYYY-MM-DD}
}How to use
- 1
Select Faraday constant mode & calculation tab
Toggle between High-School Textbook (F = 96,500 C/mol) and Exact SI/CODATA (F = 96,485.33212 C/mol), then choose Faraday's Law of Electrolysis, Nernst Equation, or Standard Cell EMF & ΔG°.
- 2
Enter electrolysis or cell parameters
Input current I, time t, electron stoichiometry z, and molar mass M (or pick CuSO₄, AgNO₃, Al₂O₃, or brine presets), or enter standard potentials E°, temperature T, and ion concentrations.
- 3
Inspect results, Nernst plot & step-by-step working
Review deposited mass (g, mg), STP gas volume (mL), non-standard cell voltage E with interactive E vs. log₁₀ Q chart, ΔG° and equilibrium constant K, and complete substitution steps.
Features
- One-click toggle between High-School Textbook (F = 96,500 C/mol) and Exact SI / CODATA (F = 96,485.33212 C/mol) Faraday constants
- Four-way Faraday's law solver for deposited mass m, required current I, electrolysis time t, and ionic valence z
- Simultaneous calculation of total charge Q (Coulombs & mAh), moles of electrons n(e⁻), and STP gas evolution volume (O₂, Cl₂, H₂ at 22.4 L/mol)
- Built-in electrolysis presets for CuSO₄ refining, AgNO₃ electroplating, Hall-Héroult Al smelting, chlor-alkali brine, and lead-acid battery discharge
- Bidirectional Nernst equation solver for cell potential E, standard potential E°, and reaction quotient Q (direct or concentration ratio)
- Real-time interactive SVG Nernst slope chart plotting E against log₁₀ Q with live operating point
- Thermodynamic bridge calculator for E°cell = E°cathode − E°anode, standard Gibbs energy ΔG° = −nFE°, and equilibrium constant K
- IUPAC cell line notation generator, step-by-step equation substitution viewer, and 1-click Markdown report export
Use cases
High-School & AP/IB Chemistry Exam Verification
Check electrolysis mass and gas volume problems using either F = 96,500 C/mol or 96,485.33 C/mol with complete step-by-step working.
Analytical & Physical Chemistry Lab Reports
Compute non-standard Nernst potentials, concentration cell voltages, Gibbs free energy ΔG°, and equilibrium constants K across any temperature.
Electroplating & Battery Capacity Engineering
Back-calculate required plating time or current incorporating Faradaic current efficiency η (%), and convert seamlessly between Coulombs (C) and mAh.
Details
Electrochemistry governs the interconversion between chemical free energy (Gibbs energy change ΔG) and electrical work (charge Q × potential difference E) in oxidation-reduction (redox) reactions. In electrolysis, an external power source drives a non-spontaneous reaction (ΔG > 0, E < 0); in a galvanic (voltaic) cell, a spontaneous redox reaction (ΔG < 0, E > 0) performs electrical work on an external circuit.
Formulated by Michael Faraday in 1833, Faraday's laws of electrolysis state that the amount of substance reacted at an electrode is directly proportional to the total electric charge passed, Q = I · t (current I in amperes × time t in seconds), and inversely proportional to the stoichiometric number of electrons z transferred per mole of substance. The charge carried by one mole of electrons is the Faraday constant F = e · NA = 96,485.33212 C/mol under the 2019 SI redefinition (while high-school chemistry curricula commonly use the rounded value F = 96,500 C/mol). Accounting for current efficiency η (%), the deposited mass is m = (M · I · t / (z · F)) · (η / 100).
Under non-standard concentrations or partial pressures, electrode and cell potentials follow the Nernst equation: E = E° − (RT / nF) ln Q = E° − (2.3026 RT / nF) log₁₀ Q. At 25.0°C (298.15 K), the thermal prefactor (RT ln 10) / F equals 0.05916 V (59.16 mV / decade). Consequently, a tenfold increase in the reaction quotient Q shifts the cell voltage by −59.16 mV for a one-electron transfer (n = 1) or −29.58 mV for a two-electron transfer (n = 2, such as the Daniell cell). In a concentration cell where both half-cells share identical electrodes (E°cell = 0 V), a concentration gradient alone generates a positive voltage E = −(0.05916 / n) log₁₀(C_anode / C_cathode).
Thermodynamically, electrical work relates directly to Gibbs free energy via ΔG = −nFE and ΔG° = −nFE°cell. At chemical equilibrium (E = 0 V and Q = K), the Nernst equation yields ΔG° = −RT ln K = −nFE°cell, allowing equilibrium constants K = 10^(nFE° / 2.3026RT) to be determined directly from potentiometric voltage measurements.
FAQ
What is the difference between F = 96,500 C/mol and F = 96,485.33212 C/mol?
The Faraday constant F is the magnitude of electric charge per mole of electrons (F = e · NA). Under the 2019 SI redefinition, its exact value is 96,485.33212 C/mol. High-school chemistry textbooks in Japan and introductory exams commonly round F to 3 significant figures (9.65 × 10⁴ = 96,500 C/mol) and design problem durations (such as 193 s or 1930 s) to divide cleanly by 965. This calculator lets you toggle between both values.
Why do we divide moles of electrons by z = 4 for oxygen (O₂) gas evolution at the anode?
During the electrolysis of aqueous sulfate or nitrate solutions with inert electrodes, water is oxidized at the anode according to 2H₂O → O₂ + 4H⁺ + 4e⁻. Producing 1 mole of O₂ gas requires transferring 4 moles of electrons (z = 4), so n(O₂) = n(e⁻) / 4. By contrast, chlorine evolution (2Cl⁻ → Cl₂ + 2e⁻) and hydrogen evolution (2H⁺ + 2e⁻ → H₂) require 2 moles of electrons per mole of gas (z = 2).
Why don't we multiply standard reduction potential E° when scaling a half-reaction?
While Gibbs free energy ΔG° (J/mol) is an extensive property that scales with stoichiometric coefficients, electrode potential E° = −ΔG° / (nF) (V = J/C) is an intensive property representing energy per unit charge. Doubling a half-reaction doubles both ΔG° and n, leaving their ratio E° unchanged. Thus, E°cell is always computed directly as E°cathode − E°anode.
How do electrode masses change during lead-acid battery discharge?
At the anode (Pb + SO₄²⁻ → PbSO₄ + 2e⁻), metallic Pb (207.2 g/mol) converts to insoluble PbSO₄ (303.3 g/mol), gaining 96.1 g per 2 moles of electrons (+48.0 g per mole of e⁻). At the cathode (PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O), PbO₂ (239.2 g/mol) converts to PbSO₄ (303.3 g/mol), gaining 64.1 g per 2 moles of electrons (+32.0 g per mole of e⁻).
How do you convert between Coulombs (C) and milliampere-hours (mAh)?
One ampere-hour (1 Ah) is 1 A × 3,600 s = 3,600 C. Therefore, 1 mAh = 0.001 A × 3,600 s = 3.6 C. Divide charge Q in Coulombs by 3.6 to obtain mAh, or multiply mAh by 3.6 to get Coulombs.
When should I use F = 96,500 C/mol versus 96,485.33212 C/mol?
Use F = 96,500 C/mol for high-school chemistry exams where problem times (like 193 s or 1930 s) are specifically chosen to divide cleanly by 965. Use the exact 2019 SI / CODATA constant F = 96,485.33212 C/mol for university physical/analytical chemistry and research.
Why does a concentration cell generate voltage even though E°cell = 0 V?
Because both half-cells use the same redox couple, E°cell = 0 V. However, the spontaneous tendency of concentrations to equalize (entropy of mixing, ΔG < 0) drives oxidation in the dilute half-cell (anode) and reduction in the concentrated half-cell (cathode), producing E = −(0.05916/n) log₁₀(C_dilute / C_conc) > 0 V.
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Verified: Checked November 10, 2026 via independent calculation and Playwright tests in both languages: CuSO₄ electrolysis (5.00 A, 1930 s, F=96,500) gives Q=9,650 C, n(e⁻)=0.1000 mol, 3.178 g Cu, and 560.0 mL STP O₂; brine electrolysis (10.0 A, 1.00 h, F=96,485.33) gives 14.92 g NaOH; Daniell cell Nernst ([Zn²⁺]=0.010 M, [Cu²⁺]=1.00 M, 25°C) gives E=+1.1592 V and ΔG=-223.68 kJ/mol; Ag⁺ concentration cell (0.0010 M / 0.100 M) gives E=+0.1183 V; standard Daniell thermodynamics (E°=+1.100 V, n=2) gives ΔG°=-212.27 kJ/mol and K=1.54×10³⁷. Applies to stated ideal models excluding overpotential and internal resistance.