# Water splitting ways [closed]

Do you know the frequency you can split hydrogen from the water bonds? I have doing at now but i need to absorb more than 3 liters gas hydrogen per minute. My device working only with one ampere by some 40..50...100..dc volts. It is not fuel cell, not dry fuel cell , no membranes no antenna... but, it's working as well many years at now. So my question is to make it better to produce 100 liters gas hydrogen per minute with the same results of power input.

## closed as unclear what you're asking by Ivan Neretin, M.A.R., orthocresol♦, Jannis Andreska, bonJan 13 '16 at 12:55

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• Eh, um, er, em, um, uh, I dunno what you're talking about. – M.A.R. Jan 13 '16 at 10:35

So there is a theoretical minimum energy required to split the water. To count the energy your device can use for the conversion, you need the ampere and volt numbers. You can count the energy with $E = U \cdot I \cdot t$, where we measure the E (energy) in Joule the U (voltage) in Volts the I (electric current) in Amperes, t (time) in seconds. With your parameters $E = 100V \cdot 1A \cdot 60s = 6000J = 6kJ$ energy can be used to convert hydrogen on the maximum settings.
To produce $100L=0,1m^3$ volume of hydrogen in atmospheric pressure you can count the minimum energy required the following way. According to the ideal gas law $p \cdot V = n \cdot R \cdot T$. So in $25°C=298.15K$ temperature at $1atm=10^5Pa$ pressure (standard conditions) the amount of the hydrogen will be $n = \frac{p \cdot V}{R \cdot T} = \frac{10^5Pa \cdot 0.1m^3}{8.314\frac{J}{mol \cdot K} \cdot 298.15K}=4.105 mol$. To produce that much gas you need at least $E = 4.034 mol \cdot 286 kJ/mol = 1153kJ$ energy, which is not possible with your device. The maximum it can do in this conditions is $\frac{6kJ}{1153kJ} \cdot {100L/min} = 0.52L/min$. On higher temperatures and smaller pressures you may produce somewhat more than this in volume (but not in mass).