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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: $$\Delta G = W_\text{max}=-n F E_\text{cell}$$ 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 $E_\text{cell}$ is the electromotive force of the cell, which equals $V_\text{cathod} - V_\text{anode}$, or alternatively, $E_\text{oxidizer} - E_\text{reducer}$ wherein $E$ here stands for the reduction potential. As $\Delta G$ must be nagative in the spontanous direction so $E_\text{cell}$ must be positive.

However, according to the first and second law we have

\begin{align} &\delta Q-\delta W=\mathrm dU=T\mathrm dS-p\mathrm dV+\sum_i\mu_i\mathrm dN_i\\ &\Rightarrow\;\delta W-p\mathrm dV+\sum_i\mu_i\mathrm dN_i=\delta Q-T\mathrm dS\le0\qquad\because\text{2nd law}\\ &\Rightarrow\;\delta W-p\mathrm dV+\mathrm dG\bigr|_{p,T}\le0\\ &\Rightarrow\;\delta W-p\mathrm dV\le-\mathrm dG\bigr|_{p,T}\\ \end{align}

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

Also if $\delta W=p\mathrm dV+\text{other works}$, then $\text{other works}\le-\mathrm dG\bigr|_{p,T}$. 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 $\Delta G=W_\text{max}$ doesn't hold!?

  2. If $\Delta G=W_\text{max}$ 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 $\Delta G=W_\text{max}$ and $\Delta G\le0$ 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 $E_\text{oxidizer}-E_\text{reducer}\ge0$"?

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Question 1:

You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done by the system on the surroundings is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy change of the system is negative (energy is lost). Chemistry is concerned ultimately only with the system (usually the chemical reaction under investigation), and not the surroundings (apparatus, equipment, experimenter, laboratory, and the rest of the universe).

Thus, for pressure-volume work:

$$W=-p\Delta V$$ $$\delta W=-pdV$$

We need this negative sign. The system does pressure-volume work on the surroundings when the system expands (think a piston in an automotive engine) so $dV>0$. Without the negative sign, we would assume that the system gains energy and the surroundings lose energy when the system expands.

Thus, in the chemistry formulation of the first law (assuming only pressure-volume work) is:

$$dU=\delta Q + \delta W =TdS - pdV$$

In physics, some consideration is given to the surroundings (and in some cases the surroundings may be more interesting than the system, such as in the proper definition of entropy). Especially in the early days of physics, work leaving the system was useful in that it could be utilized. Thus, in physics $W=p\Delta V$ means work is being done by the system on the surroundings. Thus, the first law as formulated in physics is:

$$dU=\delta Q - \delta W = TdS -pdV$$

Note however, regardless of the sign convention in the definition of $W$ as a function of $p$ and $\Delta V$, we still end up with $$dU=TdS-pdV$$

The sign of $W_{\text{max}}$ in chemistry needs to be negative to follow the same conditions of chemistry. Thus, the equation in your chemistry textbook is correct. An equivalent definition in physics would be:

$$W_{\text{max}}=nFE_{\text{cell}}$$ $$\Delta G_\text{cell}=-W_\text{max}=-nFE_\text{cell}$$

In chemistry the negative sign goes in the definition of work. In physics the negative sign goes in front of the work term. Don't mix and match conventions, and you will get consistent answers.

Question 2:

$\Delta G = 0$ is the definition of equilibrium not the conditions for reversibility. A reaction can be reversible (and most electrochemical reactions are) and still be far from equilibrium at the initial conditions.

Question 3:

Temperature is not required to be constant here. Also, note that the term $W_\text{max}$ refers to the maximum possible electrochemical work. Usually less electrochemical work is done because heat is also generated (which changes the temperature). Also, most electrochemical systems are not closed. Even a battery is not a closed system (by the strict physics definition). It's hard to maintain constant pressure in a closed system.

Question 4:

Technically the formulation of the first law that you used is explicitly for pressure-volume work done by closed gaseous systems. As most redox reactions occur in solution in open systems, that formulation is invalid for a different reason.

Question 5:

Given that your initial equation is correct, then for a spontaneous reaction:

$$\Delta G < 0$$ $$\therefore -nFE_\text{cell} < 0$$ $$nFE_\text{cell} > 0$$ $$E_\text{cell} > 0$$ $$E_\text{red} - E_\text{ox} > 0$$

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  • $\begingroup$ Thanks for mentioning the sign convention in chemistry about work, but that really solves no problem, since work is substituted by an expression and the minus would go inside that expression. / About the reversibility and equilibrium mention why $\Delta G=0$ is used to define the equilibrium. According to the second law we should always have $dG\le 0$ so that if $G$ reaches a local minimum that would imply an equilibrium. But a local minimum is reached whenever $dG=0$, which also occurs when the equality in the $dG\le 0$ holds. (to be continued) $\endgroup$ – topology Jun 29 '14 at 20:27
  • $\begingroup$ That is, a reversible process is an idealistic process always being at equilibrium, a quasi equilibrium process would be still irreversible though is very close to being reversible. Therefore, if one assumes the problem as reversible then the process is always at a local equilibrium and $dG=0$ and not $dg<0$. Maybe here we should run a little bit out of precise speaking and say the process is quasi equilibrium and thus $dG<0$ although $dG\approx 0$, say, a very very slow process. That the formulation works for points far from equilibrium is another point which is interesting at its own place. $\endgroup$ – topology Jun 29 '14 at 20:30
  • $\begingroup$ Yet I'm not convinced with the fourth part of your answer, which is the key part for me to understand where I am wrong. Actually I don't understand why you assume the system as open, in a Redox reaction there exists no source of electrons other than the reactants, and if we consider the reactants inside the system, as they have been considered by writing the term $\sum_i \mu_i dN_i$ in the internal energy change of the system, then the whole electric current and so the whole electrochemical cell is inside the system with no electric current flowing outside the cell. please explain more, thanks $\endgroup$ – topology Jun 29 '14 at 20:37
  • $\begingroup$ I'm using the open/closed definition of open = system can exchange matter and energy with the surroundings and closed = system can exchange energy with the surroundings but not matter. $\endgroup$ – Ben Norris Jun 30 '14 at 1:08
  • $\begingroup$ Yes, and you assume a Redox reaction involves matter transport to the environment but to me if we are considering both the reactants and the products inside the system then the system would be closed, not open, and we are indeed assuming such when $dU$ is expanded by $T dS - p dV + \sum_i \mu_i dN_i$ with $i$ counting all the reactants and products involved in the reaction. You don't agree? $\endgroup$ – topology Jun 30 '14 at 15:28

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