According to this Mc Graw-Hill link during a spontaneous redox reaction Gibbs Free Enthalpy must decrease and at the same time the change in the Gibbs Free Enthalpy is the maximum electric work that the reaction can do: $$\Delta G = W_{max}=-n F E_{cell}$$ in which $F$ stands for the Faraday constant, the amount of charge carried by one mole of electrons, and $n$ represents the number of moles of electrons to be transferred from the reducing agent to the oxidizing agent according to the balanced equation, and $E_{cell}$ is the electromotive force of the cell, which equals $V_{cathod} - V_{anode}$, or alternatively, $E_{oxidizer} - E_{reducer}$ wherein $E$ here stands for the reduction potential. As $\Delta G$ must be nagative in the spontanous direction so $E_{cell}$ must be positive.

However, according to the first and second law we have

\begin{align} &\delta Q - \delta W = dU = T dS - p dV +\sum_i \mu_i d N_i\\ &\Rightarrow\;\delta W - p dV + \sum_i \mu_i dN_i = \delta Q - T d S \le 0\qquad\because\text{2nd law}\\ &\Rightarrow\;\delta W - p dV + dG\bigr|_{p,T} \le 0\\ &\Rightarrow\;\delta W - p dV \le -dG\bigr|_{p,T}\\ \end{align}

If $\delta W = p dV$ then $dG\le 0$, that is, the condition $dG\le 0$ coincide with the 2nd law only id the only work done by the system is the pressure work.

Also if $\delta W = p dV + \text{other works}$, then $\text{other works}\le -dG\bigr|_{p,T}$. This means the change in the "minus Gibbs function" is the maximum work attainable from the system beside the pressure work. This extra work can be positive or negative, in the form of electric work or friction work or etc.

Now here I have some problems with the content of the link provided at the beginning of my question, as follows:

1. First of all, the maximum work that a system can do on its surrounding equals minus the change of the Gibbs Free Enthalpy, so the equality $\Delta G = W_{max}$ doesn't hold !?

2. If $\Delta G = W_{max}$ is valid, that is, I mean if the work attained is the maximum possible, then the process must be assumed as reversible , but if the process is reversible thenwe will need $\Delta G = 0$ and not $\Delta G < 0$, so that the process should be at equilibrium and spontaneous in no direction !?

3. That the Gibbs Free Energy must decrease in a closed system in the constant temperature and pressure is true only when the only type of work which is done is in the form of pressure work in expansion of the system's volume, not if other works like friction or electric works are also available, so that again using at the same time $\Delta G = W_{max}$ and $\Delta G \le 0$ is misleading !?

4. The electric work is an internal work for a system containing the Redox reaction and shouldn't enter the above formulation at all, as by work in the 1st law formulated above we mean only those work-based energy transport that occur at the boundaries of the system. So the electric force can never enter this formulation unless we change the control system under study, isn't it right ?

5. And the main question. if I was wrong in analyzing the content of the provided link (which is very probable) where is my mistakes and if I was correct then how can I obtain the same result (which is certainly correct), that "in a spontaneous redox reaction we should have $E_{oxidizer} - E_{reducer}\ge 0$"?

Sorry if the question was long, and thank you for answering

[I have asked the same question at the Chemistry.SE here but have not been convinced with the answer given there.]