Understanding Free Energy in Fuel Cell Reactions

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SUMMARY

This discussion focuses on the calculation of electrical work and waste heat in fuel cell reactions, specifically using methane as fuel. The reaction under consideration is (CH4) + 2(O2) --> 2(H2O) + (CO2). It is established that the Gibbs free energy change (ΔG) represents the maximum electrical work obtainable from the system, with ΔG = W(electrical). Additionally, the relationship between waste heat and enthalpy is clarified, revealing that waste heat (Q) is calculated as Q = ΔG - ΔH, rather than ΔH = ΔU + PΔV.

PREREQUISITES
  • Understanding of Gibbs free energy (ΔG) and its significance in thermodynamics.
  • Familiarity with fuel cell reactions and their chemical equations.
  • Knowledge of the first law of thermodynamics, including concepts of enthalpy (ΔH) and internal energy (ΔU).
  • Basic principles of non-mechanical work in thermodynamic systems.
NEXT STEPS
  • Study the derivation and implications of Gibbs free energy in electrochemical cells.
  • Learn about the thermodynamic relationships between ΔG, ΔH, and ΔU in detail.
  • Explore the calculations of electrical work and waste heat in various fuel cell types.
  • Investigate the role of entropy in fuel cell efficiency and performance.
USEFUL FOR

Students and professionals in chemical engineering, electrochemistry, and energy systems, particularly those focusing on fuel cell technology and thermodynamic efficiency.

lustrog
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I'm given a fuel cell that uses methane as fuel:

(CH4) + 2(O2) --> 2(H2O) + (CO2)

And I'm asked to find (a) the electrical work I can get out for each mole of methane, and (b) the amount of waste heat produced.

I understand that ΔG is defined as the amount of useful non-mechanical work that can be obtained from the system. So my first thought was that ΔG = Q + W(electrical). However, I'm told it's actually the case that ΔG = W(electrical) only. So why exactly is that? Isn't Q a form of non-mechanical work? Secondly, when I'm asked to compute the amount of waste heat, I go ahead and accept the confusing notion that ΔG = W(electrical) and attempt to compute the waste heat simply as ΔH = ΔU + PΔV = ΔU - W = Q. However, I'm told the waste heat is actually Q = ΔG - ΔH. Can anyone explain these two areas of confusion for me?
 
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Just read the sticky. Perhaps this should have been in the HW section. But it is a conceptual question. I just pulled a relevant textbook question to demonstrate my areas of confusion about free energy.
 

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