Enthalpy of Reaction and Efficiency of Electrochemical Energy Devices

Question

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?

Answer 1

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 $\gt 100$%.