View Full Version : Electrolysis and energy
Patrickas
Mar29-10, 02:03 AM
How much electric energy needs to be used, to get 2.5l of hydrogen, when T=298K, pressure 10^5Pa? Electrolysis happens at 5V, efficiency is 75%.
I am quite lost here. Not sure were to start. I tried finding the needed current by getting hydrogen to kg and then by comparing to 0.0104*10^-6kg/C (electrolysis). I get 21.6*10^3C.
Energy - E=21600*5*0.75=60.75*10^3J
Of course wrong.
Second try:
n(H)=PV/RT i get that hydrogen has 0.1mole=0.1g=100C to electrolyse it.
Then E=100*5*0.75=375J - WRONG
maybe look at the Bond dissociation energy , or the coulomb energy ,
E= q1*q2/(4pi*e*r)
q1 and q2 are the charges , e=permitivity constant r is the radius between the charges .
Patrickas
Mar29-10, 02:43 AM
maybe look at the Bond dissociation energy , or the coulomb energy ,
E= q1*q2/(4pi*e*r)
q1 and q2 are the charges , e=permitivity constant r is the radius between the charges .
Very much doubt it... but if you got an answer...
you are trying eltrolzoye water or hydrogen .
Patrickas
Mar29-10, 02:59 AM
you are trying eltrolzoye water or hydrogen .
It doesn't say.
well this might be the wrong way to go about this but it take
4.52eV electron volts , to break the H2 bond.
Patrickas
Mar29-10, 03:27 AM
O maybe the answer to the problem will help - 0.13MJ
Patrickas
Apr2-10, 10:14 AM
nothing?....damnnnnnn
ya man i would like to help but I’m at a loss , I don’t know why other people haven’t jumped in yet .
Patrickas
Apr4-10, 08:32 AM
If anyone is interested how to solve it read along. :smile:
"You have 0.1 mole of hydrogen gas produced (from pV=nRT)
So far so good
Multiply by Avogadro's number to find number of hydrogen molecules
Each molecule needs 2 hydrogen atoms
Each atom had a charge = 1.6 x 10^-19 C
Multiply number of atoms by this to get total charge
Energy transfer = voltage times charge, E=VQ
With only 75% efficiency the energy actually needed is this value divided by 0.75
Plug in the numbers and you get the quoted answer."
This is a quote from the solver.
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