Decomposition potential and Gibbs Free Energy

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Discussion Overview

The discussion revolves around the relationship between decomposition potential and Gibbs Free Energy, particularly in the context of thermodynamic principles. Participants explore the implications of overpotential and how it affects the calculation of Gibbs Free Energy in electrochemical processes.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the decomposition potential is always higher than the theoretically determined potential, leading to confusion regarding the relationship with Gibbs Free Energy.
  • Another participant suggests that overpotential accounts for losses in the system, rather than the original thermodynamic process.
  • A question is raised about whether the equation should be adjusted to E=Eeq-η, and how this would affect the calculation of ΔG.
  • It is proposed that thermodynamic values apply to processes that occur infinitely slowly, which does not reflect real system behavior.
  • A participant expresses confusion about the apparent contradiction between higher decomposition potential and its effect on ΔG, questioning why a higher E would lead to a more positive ΔG.

Areas of Agreement / Disagreement

Participants express differing views on the implications of decomposition potential and overpotential, with no consensus reached on the correct interpretation or calculation methods.

Contextual Notes

There are unresolved questions regarding the assumptions underlying the relationship between decomposition potential and Gibbs Free Energy, as well as the definitions of overpotential and equilibrium potential.

sgstudent
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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?
 
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Overpotential covers loses, not the original process.
 
Borek said:
Overpotential covers loses, not the original process.
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?
 
sgstudent said:
why would the decomposition potential for the process be higher than what was thermodynamically determined E?

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.
 
Borek said:
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.

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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