Gas yield and Faraday efficiency

How much gas can this current possibly produce, and how much of that did the cell actually make?

The formula

H = \frac{I \cdot t}{2 F} \cdot V_{m} G = 1.5 \cdot H

LaTeX

H = \frac{I \cdot t}{2F} \cdot V_m \qquad G = 1.5 \cdot H \qquad e = \frac{V_{measured}}{G}

Work it out

I Cell current A

Average current actually passing through the electrolyte — not the supply current, which may include the driver’s own draw.

t Run duration s

How long that current flowed.

T Gas temperature °C

Temperature of the collected gas. A volume without one is not a quantity of anything.

V Measured gas volume

Millilitres actually collected, if you have the figure. Leave at zero to see the ceiling alone.

Multiply the current by the time to get the charge passed, in coulombs. This is the whole of the input to Faraday’s law — voltage, frequency and waveform do not appear.
Divide by 2F, where F is 96,485 coulombs per mole of electrons. The 2 is because reducing two protons to one hydrogen molecule takes two electrons. The result is moles of hydrogen.
Multiply by the molar volume of a gas at the collection temperature — 22.414 litres per mole at 0 °C, rising by about 1/273 of that per degree.
For the total of the mixed gases, multiply by 1.5: splitting water gives two volumes of hydrogen for every one of oxygen, so the total is one and a half times the hydrogen.
If you measured a volume, divide it by this ceiling. That fraction is the Faraday efficiency, and it is the honest figure of merit for a cell.

Every electron that crossed the cell reduced a proton. Real cells lose some to heating the electrolyte, to corroding the electrodes and to side reactions, which is why measured efficiency is below 1 and never above it.
The current used is the current through the electrolyte. Supply current includes whatever the driver itself consumes, and using it inflates the denominator — which understates efficiency rather than overstating it.
The gas is dry and is hydrogen and oxygen. Gas collected over water carries water vapour, which at 20 °C is about 2 % of the volume and rises steeply with temperature. Uncorrected, that vapour is counted as product.
The gas is at atmospheric pressure and the stated temperature. Collected under any head of water it is compressed, and the volume read is smaller than the volume produced.
An efficiency above 1 is a measurement problem, not a result. The usual causes are supply current mistaken for cell current, vapour counted as gas, a pulsed current measured as its peak rather than its mean, or a leak admitting air.