Decomposition potential and Gibbs Free Energy

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SUMMARY

The decomposition potential is consistently higher than the theoretically determined potential according to thermodynamic principles, represented by the equation E=η+Eeq, where E is the decomposition potential, η is the overpotential, and Eeq is the equilibrium potential. The relationship ΔG=-nFE indicates that a higher decomposition potential results in a more negative Gibbs Free Energy, which contradicts the intuitive understanding that higher voltage requires more energy. This discrepancy arises because thermodynamic values are based on infinitely slow processes through equilibrium positions, which do not reflect real system behaviors, similar to the Carnot cycle's theoretical efficiency limits.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically Gibbs Free Energy and its equations.
  • Familiarity with electrochemical concepts, including decomposition potential and overpotential.
  • Knowledge of the Carnot cycle and its implications for efficiency in thermodynamic systems.
  • Basic grasp of electrochemistry, particularly the relationship between voltage and energy in electrochemical cells.
NEXT STEPS
  • Research the implications of overpotential in electrochemical reactions.
  • Study the derivation and applications of the Gibbs Free Energy equation in various chemical processes.
  • Explore the limitations of the Carnot cycle and its relevance to real-world thermodynamic systems.
  • Investigate advanced electrochemical modeling techniques to better understand decomposition potentials.
USEFUL FOR

Students and professionals in chemistry and electrochemistry, particularly those focusing on thermodynamics, energy systems, and electrochemical processes.

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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