# On thermodynamics of spontaneity of the reduction-oxidation chemical reactions

According to the first and second law for a closed system containing different chemicals 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 $dG\bigr|_{p,T}\le 0$, that is, the condition $dG\bigr|_{p,T}\le 0$ coincides with the second law only if the only work done by the system is the pressure work and no other kind.

In addition, 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, friction work etc.

Therefore, it is clear that:

1. A system cannot at the same be derived by $dG\bigr|_{p,T}\le 0$ and does e.g. an electric work $\delta W_{Electric}$;

2. If the system does have a $\delta W_{Electric}$ then its maximum value would be equal to $dG\bigr|_{p,T}$ but if so, then the process is already assumed reversible and all the inequalities should be substituted by equalities.

However, according to this Mc Graw-Hill link during a spontaneous reduction-oxidation chemical reaction the 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: $$\Delta G = W_{max}\le 0$$ I cannot understand this and have a number of problems with this derivation:

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 the work has attained its maximum value, then the process must be assumed as reversible, but the inequality in the formula above holds for irreversible processes!

3. If there is an Electric work done by the system then decreasing Gibbs free energy is no longer necessary due to the second law?

4. When both the donor and acceptor of electron in the chemical reaction lay inside the system the electric work will nowhere enter the formulation as the electric work is an internal work that does not crosses the boundary of the system!? But the whole formulation is here to study $W_{electric}$ in the system as spontaneity of the reaction should be related to the electromotive force of the reaction which has a chance to appear in $W_{electric}$ only! So what should be taken as system here?

Hint. In the books on thermodynamics that I have seen that discuss the electric works they are usually dealing with the problem of Electrochemical cells. But electrochemical cells work in an outer electric circuit and so if one assume the cell as the closed system yet the electric work will enter the discussion. Only one book was talking about open-circuit Emf but again I have the problems listed above with that as well.

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

• Yikes there is a lot here. Can you cut this down a bit so that it's easier to get to the core of your question? Commented Jul 15, 2014 at 5:26
• @BrandonEnright, actually I rewrote the question and managed to omit the contents unrelated to my main question, but I see it is still a long question. Commented Jul 15, 2014 at 6:22
• I may be wrong but I think that you are mistaken the notion of "work" here. In thermodynamics, $\delta W$ has to do with changes in the volume of the system and that's why it has a specific role to play. Other types of work will have to deal with something else. From what I understand, something as the electrical work has no reason not be understood as a contribution to the chemical potential (of the electrons) which is contained in the $\Delta G$ of your reaction. Do I miss something? Commented Sep 30, 2014 at 10:27
• @gatsu, you can be right but here I'm afraid what you say cannot be correct, at least all the references dealing with this problem assume the work to be a flow of energy crossing the boundaries, so they can no longer be contained in $\Delta G$ of the system. Commented Oct 1, 2014 at 11:24
• @topology: maybe I don't get the problem, but if you want electrons to move spontaneously from one point to another, you normally need a chemical potential gradient (for the electrons), that's how batteries work isn't it? The total work per particle corresponds to the chemical potential difference between the two ends. If I am wrong, what is the set-up you have in mind? Commented Oct 1, 2014 at 16:11

The material contained in the given link suffers from mistakes such as:

1. For a process carried out at constant temperature and pressure, the Gibbs free
energy change is equal to the maximum amount of work (wmax) that can be done by the process.

2. $G° = –nF$.

Coming to answer your questions, the second equality in your equations: δQ−δW = δQ−(pdV + other forms of work output) = dU for an arbitrary process (1st law) δQrev−δWrev = δQrev−δWmax = (pdV + other forms of work output)max= dU for a reversible process (1st law)

δQrev = TdS 2nd law

Combination of 1st and 2nd laws gives for a reversible process the equality sign and the inequality sign for an irreversible process.

$$δQ−δW=dU=TdS−pdV+∑_i μ_i \mathrm{d}N_i$$

applies only to reversible processes, since you have substituted δQ = TdS, which applies only for reversible processes. Consequently, your next step must read as:

$$⇒δW−p\mathrm{d}V+∑_i μ_i \mathrm{d}N_i=δQ−T\mathrm{d}S=0$$ ∵2nd law

and not as:

$$⇒δW−p\mathrm{d}V+∑_i μ_i \mathrm{d}N_i=δQ−T\mathrm{d}S ≤ 0$$ ∵2nd law

This leads to the last step with equality sign

$$⇒δW−p\mathrm{d}V=−\mathrm{d}G\bigr|_{p,T}$$

This equation says, the decrease in free energy of a system (redox reaction) corresponds to the maximum work that the system can provide other than pressure-volume (pdV) work.

The statement in the link: For a process carried out at constant temperature and pressure, the Gibbs free energy change is equal to the maximum amount of work ($W_{max}$) that can be done by the process. $ΔG=W_{max}$, is not correct.

When both the donor and acceptor of electron in the chemical reaction lay inside the system the electric work will nowhere enter the formulation as the electric work is an internal work that does not crosses the boundary of the system!? But the whole formulation is here to study Welectric in the system as spontaneity of the reaction should be related to the electromotive force of the reaction which has a chance to appear in Welectric only! So what should be taken as system here?

You can very well have the electrons getting transfered from the donor to the acceptor within the system. In such a case, you will be studying a corrosion reaction and the free energy will be liberated as heat and gets tranfered to the heat reservoir that maintains the system at constant temperature.

But when you want to study the process under reversible conditions (open circuit emf you take leads outside the cell.

When you take the electric leads out and connect your load, the electrons don't get tranfered from the donor to the acceptor directly, but, liberated from the donor, they pass through the external circuit and then reach the acceptor - while the ion transfers within the cell do the job of transporting the electric charge, thereby maintaing the continuity of passage of electricity though the cell.

If you have more questions or if your questions are not answered completely, you may seek clarifications and I will be happy to answer to the best of my knowledge.

• Hi user55356, welcome on Physics.SE. I have edited your post for readability, feel free to roll back if I should have changed anything in the content. We can use MathJax, a variant of $\LaTeX$ here, a basic tutorial for that is here. Also, it is not customary - and in fact discouraged - to sign posts with your name (real or otherwise). Aside from that, have fun! Commented Jul 15, 2014 at 11:15
• Dear @user55356, thank you for answering but what is written in that link is also available in many books, the most comprehensive of which I saw was that of Winterbone's. The problem is that I don't understand them so asked my problems with it here, not to mean they are wrong. Commented Jul 15, 2014 at 12:13
• Aside that, the second inequality that you said is not correct unless the process is reversible is not really true. The only exception is when the process is such irreversible that the Classical Thermodynamics cannot be used for its study. Otherwise, if we can talk about $dU$ then we can also use the equality $dU=T dS - p dV + \sum \mu dN$. In using this equality I have not assumed the equality $\delta Q= T dS$ for you to conclude my formulation is limited to a reversible process, especifically mention that the first line of that equation corresponds to only the first law of thermodynamics. Commented Jul 15, 2014 at 12:17
• About your answering my 4th point, yes you are right. But let assume I am to study the spontaneity of the Redox reaction, usually we are told to compare the reduction potentials of the sub-reactions involved, however, I want to prove this and for this sake I should seek for $W_{electric}$, being related to electric potential difference of cathode and anode, if it is positive or negative. This has been done in the link provided and many books that I have seen but not even a single of them has explained my points more thoroughly. Commented Jul 15, 2014 at 12:29