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Kinetics of Electrolysis

  1. May 30, 2003 #1
    [SOLVED] Kinetics of Electrolysis

    Can you help me with this problem?

    Assumptions:
    1- Electrolysis of seawater at 1,000 ft.
    2- Gases are generated having buoyancy, probably chlorine, hydrogen, oxygen
    3- Gases are captured as they rise and mechanically converted to electrical energy to power electrolysis

    Question:

    Does the potential for work provided by the kinetic energy from the liberated gas exceed the energy required to sustain the electrolysis at 1,000... or greater ?

    Example: Assume you have a large boulder on a mountain side ready to fall with a small push. The "potential" release of energy is greater than the energy required to start the boulder rolling. I am thinking of the bubble as the boulder and the depth of the ocean as the mountain with electrolysis giving the "push"

    As the bubble is "pushed up the mountain" it's velocity will increase, mass will remain basically unchanged or slightly diminished. The amount of energy required to produce the initial "push" is fixed. But the height of the mountain and the "potential" for work can be changed by making the mountain taller or the reaction deeper without the need to increase the push.
     
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  3. May 30, 2003 #2

    chroot

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    The answer is no.

    - Warren
     
  4. May 30, 2003 #3
    Then the next question, of course, is WHY?
     
  5. May 30, 2003 #4

    chroot

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    It is not a hard calculation.

    - Warren
     
  6. May 30, 2003 #5
    It is if you are not a Mathematician/Physicist. I think it involves: Kinetics Formula, Chemistry of Electrolysis and Boyle's Law as well as Thermodynamics. I am neither Mathmetician or Physicist. Just curious.

    Can you tell me which law it would violate? Not conservation of energy since that would not include the kinetic potential of the rising gas.
     
  7. May 30, 2003 #6

    enigma

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    The _absolute_ largest energy you'd be able to pull would be

    ∫ (net bouyancy) ds

    which is not that much.

    Plus, you'd never even come close to pulling 100% out of it.
     
  8. Jun 1, 2003 #7
    I thought it would be more like:

    w = 1/2 * m * v2 - 1/2 * m * v02
     
  9. Jun 1, 2003 #8
    What about the increase in bouyancy that results from the change in depth? Would not the change be an increase in boyancy of about 14.7 pounds per square inch for avery 33 feet the bubble ascends? Can't this increase in "displacement" be translated into "work" see previous formula. [?]
     
  10. Jun 1, 2003 #9
    Can anyone answer this?

    Say you create, by electrolysis, a gas bubble having 1 pound of "pull" at a depth of 1,000 ft. The "pull" is measured by a spring scale (assume it is unaffected by any force other than the bubble for the time being). What would you expect the "pull" in pounds to be if the bubble ascends to a depth of ten feet?
     
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