Your reactor might consist of "a cell", but, more than likely, it consists of multiple cells. A "cell" consists of two electronically conducting phases (at the plates) connected by an ionically conducting phase (through the aqueous electrolyte between them). In a single "cell", as an electrical current passes, it must change from electronic current at the surface of the positive face of one plate to ionic current through the electrolyte and back to electronic current at the negative face of the other plate. These changes of conduction mode, in each and every single "cell", are always accompanied by oxidation/reduction reactions. A single "cell" can consist of either two bipolar plates (improperly refered to as "neutral plates") or two monopolar plates (plates connected, externally, to positive and/or negative terminals) or a combination of a monopolar plate and a bipolar plate. It makes no difference which is which. Furthermore, this holds true for nested tubes or any other types of electrode configurations you can dream up.
Next,
Faraday Efficiency is a term that crops up in the HHO community over and over. There have been claims made that some of the best and the brightest in the HHO community have calculated it every way from Sunday and they swear that, as it turns out, 100% Faraday Efficiency is equal to somewhere in the neighborhood of 7.5 MMW (Milliliters per Minute per Watt). That is PURE UNADULTERATED HOGWASH! It is simply NOT true and the reason why should be blatantly obvious to you once you have read and understand this article, but if it still isn't clear to you, in part two of this article, I will use the data from an actual YouTube video, that you can watch, to explain exactly why it's not even anywhere close to being true, but, first, you need to understand how Faraday Efficiency (Refered to as Current Efficiency in Electrochemistry reference manuals and materials) is calculated.
For the purposes of this article, I am going to round off some numbers, but I would like to encourage you to do the math yourself to double check any and all of my numbers.
.627 Liters per hour per amp is representative of 100% Faraday Efficiency at 32 oF and 1 atm pressure. The ratio of .627 liters per hour per amp per cell is the same thing as .627 divided by 60 minutes to get .01045 liters per minute per amp per cell and then, because there are 1000 milliliters in a liter, multiplied by 1000 to get 10.45 milliliters per minute per amp per cell.
So, let's take a look at another way to get there and try to construct a proof along the way.
We'll use Faraday's Laws and perform the calculation for the half reactions to find the theoretical volumes of Hydrogen and Oxygen produced per minute per amp per cell during the electrolysis of water, but there are a few preliminary considerations to get out of the way first.
The volume of Hydrogen, Oxygen, Air, HHO or any gas for that matter, per mole is a given value. At standard pressure ( 1 atmosphere) and temperature (273.15 oK or 298 oK depending on who you ask), the volume of Hydrogen per mole is 22.414 Liters or 22,414 Milliliters at 273.15 oK which, by the way, is the Ideal Gas Constant (0.0820574587) multiplied by that particular "Standard Temperature"(represented in degrees Kelvin). Also, this is the point in the calculations where, by compensating for the volume per mole, temperature corrections are made.
For example, some people will use what is commonly referred to as "room temperature" (25 oCentigrade (synonymous with Celsius) = 77 oFahrenheit = 298oKelvin) as the Standard Temperature to make these calculations which makes the volume of gas per mole = Ideal Gas Constant (0.0820574587) multiplied by room temperature in Kelvins (298oK) = 24.4531226926 Liters or 24,453 Milliliters per mole. In order to make this more clear, I will carry out the example for both Standard Temperatures throughout these calculations, but we could insert any temperature at all, as long as it is represented in degrees Kelvin, and it would work exactly the same.
Anyway, here are a couple more things to understand first. Electrical Charge in Coulombs (C) = t (time) multiplied by I (current)...(60 seconds x 1 Amp) = 60 Coulombs Also, 1 mole of Hydrogen yields 2 moles of electrons. The electrical charge of one mole of electrons (Faraday's Constant) is given as 96,485 Coulombs (1 Faraday). Since there are two moles of electrons in one mole of Hydrogen, the electrical charge delivered by one mole of Hydrogen = 2 x 96,485 Coulombs or 192,970 Coulombs.
Hydrogen volume per minute per amp per cell = Electrical charge in Coulombs (60 Coulombs) / (divided by) Electrical charge delivered by one mole of Hydrogen (192,970 Coulombs) multiplied by the Ideal Gas Constant (0.0820574587) multiplied by the temperature (represented in degrees Kelvin) which, fo example, is equal to 22,414 milliliters (at 273.15oK or 0oC or 32oF) or 24,453 milliliters (at 298oK or 25oC or 77oF).
Now, when you perform those half reaction calculations for Hydrogen at a temperature of 273.15oK (0 oC or 32 oF), it turns out like this:
Hydroxy contains 100% more or twice as much Hydrogen than Oxygen, so 50% or half of 6.97 milliliters (3.485 Milliliters) should equal the volume of Oxygen produced per minute per amp per cell and when we add the two together, that brings us up to 10.45 milliliters per minute per amp per cell.
Okay, now, let's go ahead and carry out our proof and verify that again by performing the calculations for the other half reaction for Oxygen and adding it to our results for Hydrogen.
Instead of 2 moles of electrons like we had for Hydrogen, we have 4 moles of electrons for Oxygen, so 4 x 96,485 C = 385,940 C/mole.
Now, again, when you perform those half reaction calculations for Oxygen at a temperature of 273.15 oK (0 oC or 32 oF), it turns out like this:
To correct for pressure, you just divide that final number by the atmospheric pressure represented in units of atm (atmospheres). Most local weather stations report atmospheric pressure in millibars or hectopascals (both the same thing). You can convert barometric pressure to atmospheres by multiplying the value given in millibars or hectopascals by .0009869233
That's the nub of it, but keep in mind that although I rounded off many of the figures for the purpose of slightly improving readablility of this article, for the sake of precision and accuracy, my calculator does not round off anything during the calculations. It's only the final output that gets rounded off!