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For electrochemical energy devices such as batteries or fuel cells working reversibly, efficiency is defined as: $$ \eta = \frac {\Delta_rG^⦵} {\Delta_rH^⦵} $$ Since cell operates reversibly Gibbs energy change of the system is equal to electrical work done by the cell, so we can instead just use standard Gibbs energy of reaction as energy output. What I am not clear about is why is enthalpy of reaction used as energy input? Since electrochemical energy devices transform chemical energy of electroactive materials to electrical energy, energy input should be chemical energy, but I am not sure why enthalpy of reaction represents chemical energy? Enthalpy of reaction is given by this equation: $$ \Delta_rH^⦵ = \Delta_rG^⦵ + T\Delta_rS^⦵$$

Since cell operates reversibly, $\Delta_rG^⦵$ represents reversible electrical work (maximum non-PV work done by the cell) and $T\Delta_rS^⦵$ represents heat exchanged with surroundings to keep cell at constant T in reversible operation since change in Gibbs energy as measure of non-PV work is only applicable for closed systems at constant p and T.

Why is chemical energy sum of these two quantities physically speaking since I do know how to derive this equation from 1st and 2nd law of thermodynamics, but I am not sure in its physical interpretation in context of electrochemical cells?

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  • $\begingroup$ I'm a bit unclear about what you are asking, is it why is $G=H-TS$ or why the efficiency is defined as it is ? $\endgroup$
    – porphyrin
    May 26 at 10:57
  • $\begingroup$ Why is efficiency defined as it is. $\endgroup$ May 26 at 11:00
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The efficiency is $\eta=\displaystyle \frac{w}{\Delta H}$ where $w$ is the useful work output, and $\Delta H$ the reaction enthalpy. This can be written as $\displaystyle \eta=\frac{(\Delta H-Q_{in})}{\Delta H}$ and $Q_{in}$ is the heat added where $Q_{in}=T\Delta S$ when operated reversibly giving the efficiency equation you quote. In effect the efficiency accounts for heat losses or gains relative to the reaction energy available. Gibbs was first to realise that because heat from the surroundings can be added to or removed in a reaction the work done can be greater (or less than) than $\Delta H$. In your definition the efficiency can therefore be $\gt 100$%.

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  • $\begingroup$ Yes, this is equation I've written since deltaG = deltaH - TdeltaS. Only, it didn't really answer my question why is enthalpy of reaction taken as energy input? It seems that chemical energy equals enthalpy of reaction since chemical energy is converted to electrical energy. What you said about efficiency being able to be greater than 100% is something I suspected, but I wasn't sure if that is true. Now you said it in your answer. $\endgroup$ May 26 at 13:02
  • $\begingroup$ What I am also not sure is that if entropy change of reaction is positive than heat comes from surroundings to system and in that case energy output or change in Gibbs energy of the system is less than in case if no heat is exchanged which means that efficiency is lower than in other case. I am not sure why this is? Why does adding heat to the system reduce its efficiency? $\endgroup$ May 26 at 13:16
  • $\begingroup$ The enthalpy would be the maximum amount of energy that could be recovered to do work in a chemical reaction if it were not for the fact that energy can be taken from or given to the surroundings. In other words some amount of heat, not from the chemical reaction, can be used to do work, ( or removed from doing work) and this amount under reversible conditions is $T\Delta S$. $\endgroup$
    – porphyrin
    May 26 at 14:38
  • $\begingroup$ the efficiency is $\eta = 1-T\Delta S/\Delta H$ so if $T\Delta S$ is positive the efficiency is reduced for a given $\Delta H$ $\endgroup$
    – porphyrin
    May 26 at 14:41
  • $\begingroup$ Yes, but why is it that if heat is given to the system A.K.A entropy of reaction is positive that efficiency is reduced? I can see that from equation as you stated, but I am not sure I understand physical significance/picture of it. $\endgroup$ May 26 at 15:08

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