Clarifications on thermodynamic cell efficiency 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:


On applying this equation to ordinary (not fuel cells), for a general reaction $$\ce{A^+ {(aq)} + B {(s)} -&gt; A {(s)} + B^+ {(aq)}}$$ 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?
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 $|\Delta G|&gt;|\Delta H|$ 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%?
${\Delta G}$ 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.
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.