$\Delta G$ 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 $\int \mathrm dn\cdot E$? Why do we instead say that $\Delta G=-F \int \mathrm dn\cdot E$?

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    $\begingroup$ 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). $\endgroup$ Apr 12, 2017 at 17:10
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    $\begingroup$ 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 $\endgroup$ Apr 12, 2017 at 17:13
  • $\begingroup$ The Anslyn / Dougherty book also covers this concept quite well. $\endgroup$
    – Zhe
    Apr 12, 2017 at 20:29
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    $\begingroup$ n is not real number with meaning of molar amount but integer parameter with meaning of number of exchanged electrons per 1 ion or molecule. Therefore it is not differentiable. $\endgroup$
    – Poutnik
    Dec 20, 2023 at 5:47

1 Answer 1


We can find too many descriptions of how to use the formula ΔG = - zFΔE, but I have never seen how to derive it. However, I believe that this equation is probably not that difficult to derive. Perhaps the following logic can be used to derive it;

Let the W be "the energy obtained as a result of the transformation of 1 mol of a certain reactant into a product in an isothermal and isobaric process".

And, we assume that all this energy was used to accelerate z mol of electrons. Suppose that this effect is represented by the potential difference ΔE,

W = - zFΔE

Conversely, if ΔE is given, and W is the "energy of z mol electrons moving to the location of potential difference ΔE," then

W = - zFΔE

Here, ΔG = G(product) - G(reactant)

and theΔG means "maximum energy obtained in the process of changing 1 mol of a certain reactant to a product in an isothermal and isobaric process"

So, assuming that the maximum energy can be extracted in this isothermal isobaric process and that it is all used to accelerate z mol of electrons, ΔG causes the same effect as the following ΔE occurs between the anode and cathode

ΔG = -zFΔE

Consider the situation where a beanbulb is connected to a battery. An isothermal isobaric process is taking place in the dry cell, This isothermal isobaric process produces ΔG, but what is flowing through the beanbulb is electrons, not ions.

So, when considering the electromotive force of a battery, it would be natural to consider the flow of electrons. In the 18th and 19th centuries, ignorant mankind decided that current flows from the anode to the cathode. But actually, electrons flow from the cathode to the anode (The charge of the particles mediating the current was positive, not negative). Therefore, when thinking about electrons, the ± signs of potential always tends to cause such confusion.

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    $\begingroup$ The equation is simply the definition of electrical work: charge times voltage. In this case, written for a mole of electrons (Faraday constant). $\endgroup$ Nov 12, 2023 at 2:10
  • $\begingroup$ @Matt Hanson You are probably right in explaining the simplest ΔG=-nFE. However, the subject of this question is not the derivation of this equation itself, but why the sign is "negative". So the simplest answer to this question would be to explain the process of derivation with attention to the sign. $\endgroup$ Nov 20, 2023 at 3:24

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