Electrolysis cell

Calculate selected electrode products from current, time, product efficiencies and gas conditions.

About this tool

Calculate selected electrode products

Choose a reaction route, constant current I and duration. “Efficiencies and gas conditions” contains the charge share assigned to each chosen product and the temperature T and pressure p of the dry gas. Calculate shows the products at the end of the duration. Enter an elapsed time in seconds or use the slider to inspect earlier amounts; 0 s is the beginning. Changing a model input clears the previous result. Choosing an example returns the time to the end.

Faraday’s law

For elapsed time t, Q = It and n_e = Q/F. At each electrode the selected charge is ηQ and the product amount is n = ηQ/(zF), where η is entered as a percentage. Each H₂, Cl₂ or Cu corresponds to a transfer of z = 2 electrons; each O₂ to z = 4. The constants used are F = 96485.33212331 C/mol and R = 8.31446261815324 kPa·L/(mol·K). They follow from the SI-defined elementary charge, Avogadro constant and Boltzmann constant.

Both electrodes carry the same current. Their charge assignments are not separate amounts to add together. At each electrode, (1 − η)Q is unassigned to the selected product; the tool does not turn it into a named side product. η is a supplied product current efficiency, not a prediction of selectivity or energy efficiency.

Chosen half-reactions

Acidified water: cathode 2H⁺ + 2e⁻ → H₂; anode 2H₂O → O₂ + 4H⁺ + 4e⁻. Selected chloride route: cathode 2H₂O + 2e⁻ → H₂ + 2OH⁻; anode 2Cl⁻ → Cl₂ + 2e⁻. Copper deposition: cathode Cu²⁺ + 2e⁻ → Cu; the inert anode uses the water-oxidation reaction and produces O₂. A dissolving copper anode is not modeled.

The chloride choice specifies a reaction route; it does not predict pure chlorine from an arbitrary aqueous chloride solution. Chloride oxidation and water oxidation can compete. Electrode material, composition and operating conditions affect the actual products, and those effects are outside this calculation.

Gas volume, copper mass and example

Dry ideal-gas volume is V = nRT/p in liters. Copper mass is m = nM_Cu in grams, with M_Cu ≈ 63.546 g/mol as a standard approximation subject to natural isotopic variability. Copper has no gas-volume output. The plot compares both products in mmol on one common amount axis. Constant rates make the plotted curves straight lines.

The One faraday example sets I = 2 A and duration F/2 s, giving Q = F and n_e = 1 mol. At 100% product efficiencies, the chosen water reactions produce 0.5 mol H₂ and 0.25 mol O₂. Halving only the cathode efficiency halves its product amount while leaving the anode amount unchanged. Zero current or zero duration gives zero charge and products.

Scope

Input ranges are I = 0…100 A, duration 0…86400 s, each efficiency 0…100%, gas temperature 273.15…373.15 K and pressure 50…200 kPa. These are product and display limits. T and p specify the collected dry gas, not a calculated electrolyte temperature or apparatus geometry. Gas dissolution, water vapor, activities, finite reagent supplies, kinetics and voltage or current prediction are excluded.

Sources