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Current Efficiency (Faradaic Efficiency) and Faraday’s Law Calculations for Electrochemical Reactors
This article explains how to calculate electrochemical reactor current efficiency (Faradaic efficiency) using Faraday’s law, with unit-consistent workflows, practical examples, and reporting checklists that can be applied directly to electrolysis, electrosynthesis, plating, and flow electrochemical reactors.
1. Why current efficiency matters in electrochemical reactor performance.
In an electrochemical reactor, electrical charge is the “reactant” that drives a target redox reaction at an electrode surface.
Current efficiency, often reported as Faradaic efficiency or coulombic efficiency, quantifies what fraction of the total charge actually produces the intended product instead of being consumed by side reactions.
Faraday’s law provides the theoretical link between charge passed and the maximum possible amount of product, so current efficiency becomes a rigorous selectivity metric that is comparable across reactor types and scales.
Note : Faradaic efficiency is not the same as energy efficiency because energy efficiency also depends on cell voltage, overpotentials, and ohmic losses, even when current efficiency is near 100%.
2. Core definitions and symbols used in Faraday law calculations.
2.1 Faraday’s law in reactor-friendly form.
For a constant current experiment, the total charge passed is calculated as Q = I × t, where Q is charge in coulombs, I is current in amperes, and t is time in seconds.
For a variable current profile, the total charge is the time integral Q = ∫ I(t) dt, evaluated over the electrolysis period.
The theoretical moles of electrons transferred are ne = Q / F, where F is the Faraday constant, approximately 96485 C/mol e−.
Key constants and unit relations. F = 96485 C/mol e−. 1 A = 1 C/s. Q (C) = I (A) × t (s).2.2 Electron stoichiometry for a product.
Let z be the number of electrons required to form one mole of a target product according to the balanced half-reaction and overall cell reaction accounting.
The theoretical moles of target product are N_theory = Q / (z × F) if 100% of charge goes to that product.
If the measured moles of product are N_meas, then current efficiency for that product is determined by comparing N_meas to N_theory.
2.3 Current efficiency and Faradaic efficiency calculation.
Faradaic efficiency for a single target product is FE(%) = (z × F × N_meas / Q_total) × 100.
For multiple products, the partial Faradaic efficiency for product i is FE_i(%) = (z_i × F × N_i / Q_total) × 100, and the sum of FE_i across quantified products provides a charge closure check.
When a mass measurement is used, N_meas is obtained from measured mass divided by molar mass, and the same definition applies.
Faradaic efficiency definitions. Q_total = total charge passed (C). z = electrons per mole product (mol e− / mol product). N_meas = measured moles of product (mol). FE(%) = (z × F × N_meas / Q_total) × 100.3. Step-by-step workflow to calculate product rate and current efficiency.
Step 1. Write the balanced half-reaction and determine z.
Identify the target product and write the half-reaction at the producing electrode, then determine how many electrons are required per mole of product.
For example, Cu2+ + 2 e− → Cu(s) gives z = 2 for copper deposition, and 2 H+ + 2 e− → H2(g) gives z = 2 for hydrogen.
Step 2. Compute Q_total from current and time.
Use Q_total = I × t for constant current, and ensure time is converted to seconds.
Step 3. Convert Q_total to theoretical product amount.
Compute N_theory = Q_total / (z × F), then convert moles to mass using molar mass or to gas volume using an equation of state and measured conditions.
Step 4. Measure product and convert to moles consistently.
For liquids or solids, use calibrated analytical results or gravimetry to obtain mass, then divide by molar mass to obtain moles.
For gases, use volumetric flow and composition, or calibrated gas analysis, then convert volume to moles using the measured temperature and pressure.
Step 5. Calculate Faradaic efficiency and perform charge closure checks.
Calculate FE(%) = (z × F × N_meas / Q_total) × 100 and repeat for each product if multiple products are present.
Compare the sum of partial FE values to 100% as a diagnostic of missing products, measurement bias, or unquantified parasitic currents.
Note : Always report whether Q_total includes startup transients, polarity reversals, and pauses, because inconsistent charge accounting is a common reason for non-reproducible current efficiency.
4. Tables you can reuse for fast and correct calculations.
4.1 Common electrochemical products and electron numbers.
| Target product. | Example half-reaction at producing electrode. | z (e− per mole product). | Typical measurement approach. |
|---|---|---|---|
| Hydrogen, H2. | 2 H+ + 2 e− → H2. | 2. | Gas flow + composition, or volumetric collection. |
| Oxygen, O2. | 2 H2O → O2 + 4 H+ + 4 e−. | 4. | Gas analysis and flow, or dissolved oxygen balance. |
| Chlorine, Cl2. | 2 Cl− → Cl2 + 2 e−. | 2. | Gas analysis or iodometric titration of absorbed Cl2. |
| Copper deposition, Cu(s). | Cu2+ + 2 e− → Cu(s). | 2. | Mass gain by gravimetry, thickness by metrology. |
| CO from CO2 reduction. | CO2 + 2 H+ + 2 e− → CO + H2O. | 2. | Gas chromatography with calibrated flow. |
| Formate from CO2 reduction. | CO2 + H+ + 2 e− → HCOO−. | 2. | Ion chromatography or NMR quantification. |
4.2 Unit conversions that prevent common errors.
| Item. | Correct conversion. | Practical reminder. |
|---|---|---|
| Hours to seconds. | t(s) = t(h) × 3600. | Do not mix hours in Q = I × t. |
| Charge from current. | Q(C) = I(A) × t(s). | 1 A already means 1 C/s. |
| Moles from mass. | N(mol) = m(g) / M(g/mol). | Keep grams and g/mol consistent. |
| Moles from gas volume. | N = P × V / (R × T). | Use measured P and T, and consistent units. |
5. Worked example A: copper electroplating with current efficiency.
Consider a plating reactor running at I = 10 A for t = 2.0 h depositing copper via Cu2+ + 2 e− → Cu(s), so z = 2.
First convert time to seconds, t = 2.0 × 3600 = 7200 s.
Compute total charge, Q_total = 10 × 7200 = 72000 C.
Theoretical moles of copper deposited at 100% Faradaic efficiency are N_theory = 72000 / (2 × 96485) = 0.373 mol, using F = 96485 C/mol e−.
With copper molar mass M = 63.546 g/mol, the theoretical mass is m_theory = 0.373 × 63.546 = 23.7 g.
If the measured mass gain is 21.8 g, then N_meas = 21.8 / 63.546 = 0.343 mol.
Faradaic efficiency is FE = (2 × 96485 × 0.343 / 72000) × 100 = 92.0%.
Example A summary. I = 10 A. t = 2.0 h = 7200 s. Q_total = 72000 C. z = 2 for Cu2+ + 2 e− → Cu. N_theory = Q_total / (zF) = 72000 / (2×96485) = 0.373 mol. m_theory = N_theory × 63.546 = 23.7 g. m_meas = 21.8 g → N_meas = 0.343 mol. FE = (zFN_meas/Q_total)×100 = 92.0%.Note : In plating, current efficiency can be reduced by hydrogen evolution and by dissolution or flaking of deposit, so a mass-based FE should be paired with visual inspection and thickness uniformity checks.
6. Worked example B: hydrogen generation and Faraday law gas calculations.
Consider a water electrolysis cathode producing hydrogen via 2 H+ + 2 e− → H2, so z = 2.
The reactor runs at I = 50 A for t = 30 min, so t = 1800 s, and Q_total = 50 × 1800 = 90000 C.
The theoretical moles of H2 are N_theory = 90000 / (2 × 96485) = 0.466 mol.
If hydrogen is collected and measured as 10.0 L at 1.00 atm and 298 K, then N_meas = P × V / (R × T) = (1.00 atm × 10.0 L) / (0.082057 L·atm·mol−1·K−1 × 298 K) = 0.409 mol.
The Faradaic efficiency is FE = (2 × 96485 × 0.409 / 90000) × 100 = 87.6%.
In this example, the missing charge can plausibly be attributed to parasitic reactions, gas crossover, dissolution, or collection losses, but the correct conclusion depends on measurements that close the mass balance.
Example B summary. I = 50 A. t = 30 min = 1800 s. Q_total = 90000 C. z = 2 for H2. N_theory = 90000 / (2×96485) = 0.466 mol. Measured: V = 10.0 L, P = 1.00 atm, T = 298 K. N_meas = PV/(RT) = 0.409 mol. FE = (2×96485×0.409/90000)×100 = 87.6%.Note : Gas-product Faradaic efficiency is highly sensitive to whether the reported gas volume is dry or wet and to whether pressure and temperature are measured at the sampling point, so these details must be reported explicitly.
7. How to handle multiple products and partial current efficiencies.
In electrosynthesis and CO2 electrolysis, multiple products can form simultaneously, so the applied current splits into partial currents associated with each product-forming pathway.
Compute each product’s partial Faradaic efficiency using its own z value and measured molar amount, then sum the partial efficiencies to assess closure.
If the sum is far below 100%, the most common causes are unmeasured products, inaccurate flow calibration, condensation losses, or analytical response factors that are not validated.
If the sum exceeds 100%, the most common causes are inconsistent time windows between charge integration and sampling, double counting between liquid and gas phases, or concentration values reported outside their calibration range.
Multiple product accounting. For each product i: FE_i(%) = (z_i × F × N_i / Q_total) × 100. Charge closure check: FE_sum(%) = Σ FE_i(%).8. Separating current efficiency from reactor productivity and energy efficiency.
8.1 Productivity metrics that complement Faradaic efficiency.
Current efficiency describes selectivity with respect to charge, but reactor design also requires rate and throughput metrics.
Common metrics include production rate in mol/s or g/h, areal production rate per electrode area, and space-time yield per reactor volume.
These metrics often improve when current density increases, but Faradaic efficiency may decrease if side reactions scale more strongly at higher overpotential.
8.2 Voltage and energy efficiency concepts.
Energy per mole product depends on both current and cell voltage because electrical power is P = I × V_cell.
Even at high Faradaic efficiency, a large overpotential increases energy consumption per unit product.
A practical reporting approach is to present Faradaic efficiency alongside cell voltage at the same operating point, because both are needed to understand operating cost and thermal management load.
9. Practical causes of Faradaic losses in electrochemical reactors.
Faradaic losses occur when electrons flow to reactions that do not form the intended product or when products are lost before quantification.
Common causes include competitive hydrogen evolution, corrosion or dissolution of electrodes, redox shuttles in the electrolyte, product crossover through membranes, and homogeneous reactions that consume intermediates.
In flow electrochemical reactors, shunt currents and non-uniform current distribution can reduce apparent current efficiency when measurements are taken at a single outlet point that does not represent the whole reactor.
Note : If Faradaic efficiency changes strongly with current density, report the full FE versus current density trend rather than a single value, because the reactor operating window is often determined by the FE drop-off point.
10. Reporting checklist for publishable current efficiency data.
For electrochemical reactor results to be reproducible, the calculation basis must be fully specified.
| Reporting item. | What to provide. | Why it matters for current efficiency. |
|---|---|---|
| Charge accounting window. | Start and stop times, and whether transients are included. | FE depends directly on Q_total, so inconsistent windows bias results. |
| Current and mode. | Constant current, constant voltage, or waveform, with logged data. | Variable I requires integration, not I×t. |
| Electron stoichiometry. | Balanced half-reaction and z for each product. | Wrong z gives systematically wrong FE. |
| Analytical method. | Calibration range, detection limits, and uncertainty estimate. | N_meas is the second pillar of FE, and must be defensible. |
| Gas handling details. | Dry versus wet gas, pressure, temperature, and flow calibration. | Gas moles from volume are sensitive to conditions. |
| Product phase partitioning. | How products are split between gas and liquid and accounted for. | Double counting or missing phase causes FE sum errors. |
FAQ
What is the difference between current efficiency, Faradaic efficiency, and coulombic efficiency.
In most electrochemical reactor literature these terms are used interchangeably to mean the fraction of total charge that produces a specified product, computed as FE(%) = (z × F × N_meas / Q_total) × 100.
When authors distinguish them, the distinction is typically contextual rather than mathematical, so it is best practice to state the exact formula and all variables used.
How do I choose z for Faraday law calculations in electrosynthesis.
Choose z from the balanced electrode half-reaction that forms the target product, where z equals the number of electrons transferred per mole of product formed.
If the mechanism is complex, z is still determined from net stoichiometry between reactants and product under the defined electrochemical pathway, and it must be consistent with how N_meas is defined.
Why can the sum of partial Faradaic efficiencies be less than 100%.
A sum below 100% typically indicates unquantified products, side reactions such as hydrogen evolution, crossover losses, or analytical undercounting due to condensation, adsorption, or incomplete sampling.
It can also occur when Q_total is integrated over a longer period than the sampling window used to quantify products.
Can Faradaic efficiency exceed 100% in my calculations.
A value above 100% indicates an inconsistency in units, time windows, z values, calibration factors, or double counting between phases.
It should be treated as a diagnostic signal to audit the charge integration and product quantification workflow.
How should I report Faradaic efficiency for a continuous flow electrochemical reactor.
Report Q_total over the same steady-state window used for outlet sampling, and report flow rate, residence time, electrode area, current density, and the analytical method used to determine N_meas for each product.
When possible, include a closure table showing partial FE for all quantified products and the sum over the same time interval.