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Decomposition potential and Gibbs Free Energy

  1. Jun 27, 2015 #1
    The decomposition potential is always higher than the theoretically determined potential by thermodynamics. E=η+Eeq where E is the decomposition potential, η is the overpotential and Eeq is the theoretically determined potential.

    And ΔG=-nFE and if we were to substitute the decomposition potential into this equation, the higher the decomposition potential the more negative the Gibbs Free Energy. This seems wrong because intuitively I feel like the higher the decomposition potential the more energy is required. Like in any electrical appliance the greater the voltage the more energy is needed. So what is wrong with my concept here?
     
  2. jcsd
  3. Jun 27, 2015 #2

    Borek

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    Staff: Mentor

    Overpotential covers loses, not the original process.
     
  4. Jul 2, 2015 #3
    So should the equation be E=Eeq-η instead? But still why would the decomposition potential for the process be higher than what was thermodynamically determined E? And how would I calculate the delta G this way?
     
  5. Jul 2, 2015 #4

    Borek

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    Staff: Mentor

    Because the thermodynamical values are for a process that goes infinitely slow through a series of equilibrium positions. No real system behaves this way.

    This is not different conceptually from the way Carnot cycle describes heat engine. It gives a theoretical estimation of the maximum efficiency, but no system ever will be able to reach it.
     
  6. Jul 2, 2015 #5
    Hmm but isn't it counter intuitive? Since in a normal cell the higher the E the more negative delta G is. But in this case it seems like the higher the E the more positive the delta G should be. Why is this so?
     
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