Clarifications on thermodynamic cell efficiency

Question

The thermodynamic efficiency of any cell (especially Fuel cells) is given as $$\frac{\Delta G}{\Delta H} \times 100$$

I understood this partly, that since $\Delta G$ is the useful work obtained in the ideal case, the efficiency expression must have it in the numerator. But I fail to see the intuition (or maths) behind the $\Delta H$ term in the denominator. I have two questions regarding this:

1. On applying this equation to ordinary (not fuel cells), for a general reaction $$\ce{A^+ {(aq)} + B {(s)} -> A {(s)} + B^+ {(aq)}}$$ the $\Delta G$ term is given as $-nF E_{\text{cell}}$ where $E_{\text{cell}}$ depends on the concentration of the two ionic species as given by the Nernst equation $\left(E=E^\circ-\frac{RT}{nF}\ln\frac{[\ce B^+]}{[\ce A^+]}\right)$. On the other hand, is the $\Delta H$ term independent of the concentration? and if not, how would it (and consequently efficiency) depend on the ionic concentrations?

2. For fuel cells, suppose the cell reaction is such that the $\Delta S$ is positive and since $\Delta G=\Delta H -T\Delta S$, therefore $|\Delta G|>|\Delta H|$ since the second term in the equation for $\Delta G$ becomes negative due to positive entropy change. Would this predict an efficiency greater than 100%?

Answer 1

1. ${\Delta G}$ and ${\Delta H}$ are normally expressed on a per mole basis. Your assertion that ${\Delta G}$ is related to concentration is not precise. It is related to the ratio of concentrations, and as such the incongruity with ${\Delta H}$ is not troubling.

2. If you found a fuel cell with a gain in entropy, according to your equation you would have more than 100% efficiency. I seriously doubt such a situation exists.