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Maximum Useful Work from a Fuel Cell

  1. Nov 25, 2007 #1
    What is the maximum useful work which can be obtained from a hydrogen-oxygen fuel cell that produces 1.96 kg of water at 25°C?



    How do we approach this problem?

    I guess -Ecell x charge = w?
     
  2. jcsd
  3. Nov 25, 2007 #2
    2H2(g) -----> 4H+ + 4e
    O2(g) + 4H+ + 4e -------> 2H2O(g)


    E cell = 1.2 V


    anyhelp for the rest
     
  4. Nov 25, 2007 #3

    ShawnD

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    It looks like you're on the right track. The reduction table I have says the voltage is 1.23 while yours says 1.2, so that's good.

    What you're supposed to do is figure out how many moles of water is 1.96kg

    From there you use the mole ratio between water and elections to see how many moles of electrons is that. The way you wrote it, it looks like 4 electrons makes 2 waters.

    Then how many coulombs of electrons is that. The conversion rate between moles and coulombs is called Faraday's Number, = 96,485.3383 coulomb/mole

    1 Volt = 1 Joule / 1 Coulomb
     
    Last edited: Nov 25, 2007
  5. Nov 25, 2007 #4

    arivero

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    what about the maximum power?
     
  6. Nov 25, 2007 #5

    ShawnD

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    Impossible to tell. You would need to know the internal resistance of the cell.
     
  7. Nov 25, 2007 #6
    can you check this:
    I need to find charge
    ..............(1 mol)
    1960 g x --------- x 2 moles e = 217.536 moles e
    ..............(18.02g)

    217.536 moles e x 96485 C/mole e = 20988960.96 C


    w = -1.23 x 20988960.96 =-25816422 J


    according to my assignment online that is wrong :( HELPP
     
    Last edited: Nov 25, 2007
  8. Nov 25, 2007 #7
    Hint: Note that work is conventionally defined as being done ON the system. Here, however, you are asked for the (useful) work that can be obtained FROM the system.
     
  9. Nov 26, 2007 #8

    chemisttree

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    Maximum useful work always makes me think about the definition of Gibbs energy.

    Do not approach this problem by counting charges. It will lead you nowhere.
     
    Last edited: Nov 26, 2007
  10. Nov 26, 2007 #9
    G = -nFEcell


    (2)(96485)(1.2) = work
    ?
     
  11. Nov 27, 2007 #10

    chemisttree

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    try ΔG = ΔH - TΔS for the reaction:

    [tex]H_2 + 1/2O_2 \rightarrow H_2O[/tex]
     
    Last edited: Nov 27, 2007
  12. Nov 27, 2007 #11
    so just do that for this question, i dont have to take into account that mass of 1.96 kg?
     
  13. Nov 27, 2007 #12

    chemisttree

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    You might be able to find molar gibbs energy in a table somewhere. From there convert the molar energy into the energy contained in 1.96 Kg.
     
  14. Nov 27, 2007 #13
    -237.192 kJ/mol is the delta G formation

    so ur saying just multiply that by the following

    1.96 kg = 1960 g x 1mol / 18.02 g x-237.192 kJ/mol = -25798 kJ

    ???????????/
     
  15. Nov 27, 2007 #14
    If i use

    G = -nFEcell
    (2)(96485)(1.23) = work
    then multiply by 1960/18.02 it gives me

    -25816 kJ

    so which should i put I only have 1 try left but those answers seem to be slightly different!!!!!!!!!!!!!!!

    -(2)(96485)(1.23)(1960)/(18.02) = -25816 kJ

    or

    1960 g x 1mol / 18.02 g x-237.192 kJ/mol = -25798 kJ

    also: im guessing i put positive of those values
     
  16. Nov 27, 2007 #15

    chemisttree

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    This is the answer you obtained in post #6, isn't it? Was it correct?

    These answers are very close to each other. Does your assignment require you to use significant figures?
    How many significant figures for 1.2 V?
    For 1.23 V?
    For 1.96 Kg?
     
  17. Nov 27, 2007 #16
    i never treid any of these answers unfortunately, I was doing it wrong the first few times. But it doesnt really require sig figs because I remember I put in for an answer 2104 mols for another question and it showed me that I was correct and the correct answer was 2104.23 mols

    but im still hesitant as to what I should put :(
     
  18. Nov 28, 2007 #17
    It would be positive. So, the answer is, in your case, +25816 kJ
     
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