Why delta G = -nFE and not -F * integral(dn*E)?

$\Delta G$ equals the maximum work that a system can do and in redox reactions it's the work done by electrons. According to physics, $W = E \cdot q$ ; but that's true only if $E$ is constant, however during the reaction, $E$ decreases until it's zero, so mustn't it be like the integral $\int dn \cdot E$ ? Why do we instead say that $\Delta G = -F \int dn \cdot E$ ?

• Because $\Delta G$, or more properly speaking $\Delta_\mathrm{r}G$, is not referring to the Gibbs free energy change of the system as $E$ slowly decreases to $0$. It's referring to the slope of $G$ when plotted against $\xi$, the so-called "extent of reaction". In this context, when you transfer $\mathrm{d}n$ moles of electrons, the Gibbs free energy change is $\mathrm{d}G = -FE\,\mathrm{d}n$. The quantity $\Delta_\mathrm{r}G$ is not related to $\int \mathrm{d}G$, but is instead related to $\mathrm{d}G/\mathrm{d}n$ (that's also why it has units of kJ/mol and not kJ). – orthocresol Apr 12 '17 at 17:10
• Since it is a gradient, $\Delta_\mathrm{r}G$ is defined only at a single state of a system, and therefore a single instantaneous value of $E$. The explanation of the meaning of $\Delta_\mathrm{r}G$ should be in most physical chemistry textbooks, and there is also a good (but fairly involved) article on it: J. Chem. Educ. 2014, 91, 386 – orthocresol Apr 12 '17 at 17:13
• @orthocresol , sorry it's off-topic, but can I somehow chat with you privately ? – Saba Tavdgiridze Apr 12 '17 at 18:07
• The Anslyn / Dougherty book also covers this concept quite well. – Zhe Apr 12 '17 at 20:29