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Faradaic efficiency calculator for gas and liquid products

Q: How do I calculate Faradaic efficiency from GC data?

For a gas product in the outlet stream, FE = z · F · x · v · p / (R · T · I): x is its volume fraction (ppm × 10⁻⁶), v the outlet flow rate in m³/s, p the pressure, T the gas temperature and I the total current.

Faradaic efficiency is the fraction of the charge passed that ends up in a product: FE = z·n·F / Q, with z the electrons per molecule, n the moles of product, F = 96 485 C/mol and Q the charge. For a gas product measured on-line by GC, the moles come from its volume fraction x in the outlet flow v, so FE = z·F·x·v·p / (R·T·I), with I the total current.

What it does: Calculates the Faradaic efficiency of each product from GC ppm and gas flow rate, or from moles of product and charge passed, for H₂, O₂, CO, formate, CH₄, NH₃ or any z.
Formula: $\begin{aligned} &\mathrm{FE} = z \cdot n \cdot F \div Q, \quad F = 96{,}485\ \mathrm{C/mol} \\ &\text{gas by GC: } \\ &\quad \mathrm{FE} = z \cdot F \cdot x \cdot v \cdot p \div (R \cdot T \cdot I) \end{aligned}$
Example: 10 mA for 1 hour passes 36 C; finding 120 µmol of H₂ gives FE = 2 × 120 × 10⁻⁶ × 96,485 ÷ 36 = 64.3 %.

Last updated 8 November 2026

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At a glance

Faradaic efficiency calculator at a glance
Formula	$\mathrm{FE} = z \cdot n \cdot F / Q$ ; gas by GC: $z \cdot F \cdot x \cdot v \cdot p$ ${}/ (R \cdot T \cdot I)$
Constants	$F = 96\,485\ \mathrm{C/mol}$ , $R = 8.314\ \mathrm{J/(mol{\cdot}K)}$
Products	H₂, O₂, CO, HCOOH, CH₄, NH₃ (electron counts cited) or any $z$
Input	GC ppm or %, flow rate, current; or moles and charge
Output	FE per product, total, production rates; CSV
Runs in	Your web browser; nothing is uploaded
Price	Free, no account needed

What it does

How to use the faradaic efficiency calculator

Choose the methodGas products by GC (on-line measurement in the outlet stream) or from moles and charge (for liquids or collected gas).
Enter the cell conditionsGC: total current, outlet gas flow, pressure and temperature. Moles: the charge, or a constant current and the time.
Add the productsPick each product (its electron count fills in) and enter its GC concentration or amount; use “Other” to type any z.
Read the efficienciesRead the Faradaic efficiency of each product, the total, and how much of the charge is not in the products entered.
ExportCopy the results or download them as CSV.

How it works

Faraday’s law links charge and product: making one molecule takes $z$ electrons, so $n$ moles take $z \cdot n \cdot F$ coulombs. The Faradaic efficiency is that charge as a fraction of the charge $Q$ actually passed (Small 2026, doi:10.1002/smll.74606):

$\mathrm{FE} = \frac{z \cdot n \cdot F}{Q}$

with $F = 96\,485\ \mathrm{C\,mol^{-1}}$ and $Q$ from the integrated chronoamperogram, or current × time at constant current. It is the ratio of the product found to the product expected at 100 % Faradaic efficiency (doi:10.1039/d1ra08530a).

Gas products by GC. For a product measured on-line in the outlet gas (ACS Appl Energy Mater 2025, doi:10.1021/acsaem.5c01865):

$\mathrm{FE} = \frac{z \cdot F \cdot x \cdot v \cdot p}{R \cdot T \cdot I}$

where $x$ is the volume fraction of the product ( $\mathrm{ppm} \times 10^{-6}$ ), $v$ the outlet gas flow rate in m³/s, $p$ the pressure (101.325 kPa), $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ , $T$ the gas temperature in K, and $I$ the total current. The term $x \cdot p \cdot v / RT$ is the product’s molar flow by the ideal gas law. Use the flow leaving the cell at the T and p you enter.

Electrons per molecule. Two for H₂ and four for O₂ (the two-electron HER and four-electron OER, doi:10.3390/nano12132209); two for CO and for formic acid from CO₂ (doi:10.1021/acs.accounts.4c00283); eight for CH₄ from CO₂ (doi:10.1002/anie.202508497); eight for NH₃ from nitrate (doi:10.3390/molecules31132307). For other products type $z$ yourself.

For the reaction figures to go with the numbers, see the electrocatalysis templates.

Worked example

CO₂ reduction, GC. A cell runs at 100 mA with an outlet gas flow of 20 mL/min at 25 °C and 101.325 kPa. The GC finds 18 000 ppm CO, 3500 ppm H₂ and 400 ppm CH₄. For CO, the molar flow is $\begin{aligned} &\frac{0.018 \times 101\,325\ \mathrm{Pa} \times 3.333\times10^{-7}\ \mathrm{m^3/s}}{8.314 \times 298.15} \\ &= 2.45\times10^{-7}\ \mathrm{mol/s}\ \ (883\ \upmu\mathrm{mol/h}), \end{aligned}$ and $\begin{aligned} \mathrm{FE} &= \frac{2 \times 96\,485 \times 2.45\times10^{-7}}{0.100} \\ &= \mathbf{47.3\,\%}. \end{aligned}$ H₂ gives 9.2 % and CH₄ ( $z = 8$ ) 4.2 %, 60.7 % in total; the other 39.3 % of the charge went to products not measured here (liquids, for example) or was lost.

Water electrolysis, moles. 10 mA for one hour passes 36 C. Finding 120 µmol of H₂ gives $\mathrm{FE} = 2 \times 120\times10^{-6} \times 96\,485 / 36$ $= \mathbf{64.3\,\%}$ .

Frequently asked questions

How do you calculate Faradaic efficiency?

Divide the charge that went into the product by the charge passed: $\mathrm{FE} = z \cdot n \cdot F / Q$ , with $z$ the electrons per molecule, $n$ the moles of product, $F = 96\,485\ \mathrm{C/mol}$ and $Q$ the total charge in coulombs.

How do I calculate Faradaic efficiency from GC data?

For a gas product in the outlet stream, $\mathrm{FE} = z \cdot F \cdot x \cdot v \cdot p / (R \cdot T \cdot I)$ : $x$ is its volume fraction ( $\mathrm{ppm} \times 10^{-6}$ ), $v$ the outlet flow rate in m³/s, $p$ the pressure, $T$ the gas temperature and $I$ the total current.

How many electrons are needed for each product?

Two for H₂, four for O₂, two for CO and for formate from CO₂, eight for methane from CO₂ and eight for ammonia from nitrate. For other products enter $z$ yourself.

Why do my Faradaic efficiencies not add up to 100 %?

Some charge may go into products you did not measure (liquid products when only the gas is analysed, for example), into side reactions, or be lost. A total above 100 % usually means a wrong electron count, flow rate or unit.

Which flow rate should I use?

The gas flow leaving the cell, at the pressure and temperature you enter, because the GC concentration refers to that stream.

What charge should I use?

The total charge over the electrolysis, from the integrated chronoamperogram (absolute value), or current × time if the current was constant.

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