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Electrolysis and energy

  1. Mar 29, 2010 #1
    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
     
    Last edited: Mar 29, 2010
  2. jcsd
  3. Mar 29, 2010 #2
    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 .
     
  4. Mar 29, 2010 #3
    Very much doubt it... but if you got an answer...
     
  5. Mar 29, 2010 #4
    you are trying eltrolzoye water or hydrogen .
     
  6. Mar 29, 2010 #5
    It doesn't say.
     
  7. Mar 29, 2010 #6
    well this might be the wrong way to go about this but it take
    4.52eV electron volts , to break the H2 bond.
     
  8. Mar 29, 2010 #7
    O maybe the answer to the problem will help - 0.13MJ
     
  9. Apr 2, 2010 #8
    nothing?....damnnnnnn
     
  10. Apr 2, 2010 #9
    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 .
     
  11. Apr 4, 2010 #10
    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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