[vmelkon]

I calculated the rate of production of H2 and I'm not sure if it is correct.
With 30 A, it should produce 104 mL/min of H2 at STP.
If I use Faraday's Law, I get 205 mL/s of H2 at STP.
That is a big difference. Is it correct?
(I'm assuming 1 mol of H2 occupies 22 L)

In my first calculation, I assumed that each electron in the current would get captured by a H+ and end up as H, which eventually ends up as H2(g).

There is this website that talks about 40 A and getting 2.0 to 4.5 L of (H2+O2) per minute, which means 1.33 L to 3.0 L of H2/min
http://www.aardvark.co.nz/hho_fraud.shtml

and that seems rather high.

[Borek]

30A yields around 200 mL per minute, assuming 100% efficiency.

But if you have 12 volts, you can run several cells in row, so you will get several times more gas.

[vmelkon]

I just calculated again

1 A = 6.241 * 10^18 electrons/s
30 A = 1.8723 * 10^19 electrons/s
which means 1.8723 * 10^19 H atoms/s
which means 9.3615 * 10^18 H2/s
which means 1.55429 *10^-5 H2 mol/s (if 6.023 * 10^23 = 1 mol)
which means 0.00034194421 H2 L/s (if 1 mol occupies 22 L)
which means 0.34194421 H2 mL/s
which mean 20.5166 H2 mL/min

Which is no where near Faraday's Law equation. Faraday's Law gives me 205 mL/min. I made a mistake in the first post when I wrote  mL/sec

I am only considering 1 electrolysis cell. Never mind the case of having it in series because that would x2, x3, and whatever the results.

[sjb]

I just calculated again

1 A = 6.241 * 10^18 electrons/s
30 A = 1.8723 * 10^19 electrons/s

Check this step...

[vmelkon]

Looks like the 19 in 30 A = 1.8723 * 10^19 electrons/s needs to be 20

1 A = 6.241 * 10^18 electrons/s
30 A = 1.8723 * 10^20 electrons/s
which means 1.8723 * 10^20 H atoms/s
which means 9.3615 * 10^19 H2/s
which means 1.55429 *10^-4 H2 mol/s (if 6.023 * 10^23 = 1 mol)
which means 0.0034194421 H2 L/s (if 1 mol occupies 22 L)
which means 3.4194421 H2 mL/s
which mean 205.166 H2 mL/min

Thanks, now both results are the same.