# Faradaic efficiency calculator for gas and liquid products

> Faradaic efficiency from GC ppm and gas flow rate, or from moles of product and charge passed. H₂, O₂, CO, formate, CH₄, NH₃ or any product. Free.

URL: https://www.labcanvas.com/tools/faradaic-efficiency-calculator

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.

## Quick answer

- **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:** FE = z·n·F ÷ Q, F = 96,485 C/mol; gas by GC: FE = z·F·x·v·p ÷ (R·T·I).
- **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 %.

Free, in the browser, no account needed. Updated 2026-11-08.

## Features

- GC mode: several gas products at once from ppm or % in the outlet stream, flow rate, pressure, temperature and current
- Moles mode: any product (gas or liquid) from its amount and the charge, or the current × time
- Electron counts for H₂, O₂, CO, HCOOH/formate, CH₄ and NH₃ (from nitrate), each cited, or type any \(z\)
- Faradaic efficiency per product and the total, with the share of charge not accounted for
- Production rate in µmol/h for each gas product
- Results as CSV; nothing is uploaded

## How to use it

1. **Choose the method**: Gas products by GC (on-line measurement in the outlet stream) or from moles and charge (for liquids or collected gas).
2. **Enter the cell conditions**: GC: total current, outlet gas flow, pressure and temperature. Moles: the charge, or a constant current and the time.
3. **Add the products**: Pick each product (its electron count fills in) and enter its GC concentration or amount; use “Other” to type any z.
4. **Read the efficiencies**: Read the Faradaic efficiency of each product, the total, and how much of the charge is not in the products entered.
5. **Export**: Copy the results or download them as CSV.

## 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 |

## 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](https://doi.org/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](https://doi.org/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](https://doi.org/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](https://doi.org/10.3390/nano12132209)); two for CO and for formic acid from CO₂ ([doi:10.1021/acs.accounts.4c00283](https://doi.org/10.1021/acs.accounts.4c00283)); eight for CH₄ from CO₂ ([doi:10.1002/anie.202508497](https://doi.org/10.1002/anie.202508497)); eight for NH₃ from nitrate ([doi:10.3390/molecules31132307](https://doi.org/10.3390/molecules31132307)). For other products type \(z\) yourself.

For the reaction figures to go with the numbers, see the [electrocatalysis templates](https://www.labcanvas.com/templates/electrocatalysis).

## 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 \(\mathrm{FE} = 2 \times 120\times10^{-6} \times 96\,485 / 36\) \(= \mathbf{64.3\,\%}\).

## 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.