100% Faraday Efficiency = 9.282 MMW!

[BoyntonStu]

COPIED FROM ANOTHER FORUM

Faraday Efficiency
Faraday electrolysis efficiency uses quantities of electrons (coulombs) and gas volume to determine the efficiency of an electrolytic cell.

According to Faraday law, 4 moles of electrons moving through a cell will create 2 moles of H2 gas, and 1 mole of O2 gas, at 100 percent Faraday Efficiency.

How to calculate 100 percent Faraday Efficiency:

First convert 4 moles of electrons into amps. (4 moles) * (avagadro's number) =
4 * 6.0221417e23 = 2.408856716e24 (electrons)

Then divide the total electrons by 1 Coulomb (quantity of electrons)

2.4088567e24 (electrons) / 6.24150947e18 (1 Coulomb) = 385941 Coulombs

Now calculate the Amp Hours:

Since (1 Amp) = (1 Coulomb * 1 Second)

(385941 C) / (3600 Seconds) = 107.205 Amp Hours (Ah)

Now figure out how much gas is in 2 moles of H2, and 1 mole of O2.
According to the ideal gas law, one mole of gas has a volume of 24.446 Liters at 25 C, 1 Atm. So:

(3 moles) * 24.446 L/mol = 73.338 Liters of HHO gas

So this is as exciting as Faraday Efficiency gets, it deals with Amps and Moles only (not energy or voltage).

So 107.205 Amps over one hour will generate 73.338 Liters of H2 O2 gas at 100 percent efficiency. That's it, 100 percent Faraday Efficiency.



It gets interesting when you throw the Gibbs Free Energy of water into the equation.
At 25 C the Gibbs Free Energy (see post below) 237.18 Joules are required to convert 1 mole of H2O into 1 mole of H2 gas and a 1/2 mole of O2 gas.


So from the above calculations we know: 107.205 Amps continuous for 1 Hour will create 3 moles of H2/O2 gas, which has a volume of 73.338 Liters.

If we multiply the Gibbs Free Energy (energy used to create 1.5 moles of gas) by 2, we should get the actual energy required to make the above quantity, 3 moles of gas.

237.18 kJ * 2 = 474.36 kJ

Convert 474.36 kJ to Watts:

474360 Joules / 3600 seconds = 131.7666 Watts

Now put everything together (amps + voltage)

Since: Watts = Amps * Volts

131.7666 Watts = 107.204 Amp * Volts

Volts = 1.229 V Which is considered the minimum voltage for electrolysis to occur at 25C, 101.325 kPa. Which is also 100 percent Faraday, and "Gibbs" Efficiency.


Calculate W/LPM and MMW:

73.338 L/Hr / 60 Minutes = 1.223 LPM
Requires 131.76 Watts.

1 LPM / 1.223 LPM = .81766
.81766 * 131.76 Watts = 107.73

107.73 Watts will generate 1 LPM at 100 percent efficiency, or 9.282 MMW, at 25C 101.325 kPa.

This information is a candidate for a sticky!

BoyntonStu

[Super Fuel FX]

Stu, you lost me at "Faraday"... kinda like Tim Taylor hearing funny things... Glad you're here to figure all this out. My head hurts now just "thinking about thinking about it"