# What can we say about the Gibbs free energy if it decreases or not?

[Sorry if the question looks trivial but I have thought about it for days and red a number of books all with no success.]

In almost every thermodynamics book that has something to say about chemical reactions there is a proof for the fact that at constant pressure and temperature the process is spontaneous in the direction in which the Gibbs free energy decreases. In some of the books it is also remarked that this conclusion is for when there is no "additional work" (beside the work of pressure in expanding the system), if there is an additional work then except for $dG=0$ the reversible process will yield into that additional work gaining its maximum value:

In this regard I have a question: When we have some additional works there in the problem (like electric work in an electrochemical cell), so that the second inequality holds and the maximum work is assigned to $dG$ as $dG=\delta W_{Add,Max}$, why still we feel free to use the first inequality and claim $dG$ should be negative?

In the case of electrochemical cells we have: $\delta W_{E,Max}=-n\,F\,E_{cell}\,d\zeta$ wherein $\zeta$ measure the advancement of the reaction. So we would obtain in a reversible (i.e. infinitely slow) reaction $\Delta_rG=\frac{dG_{p,t}}{d\zeta}=-n\,F\,E_{cell}$. But all the books then state that since the reaction is spontaneous in the direction in which $\Delta_rG\le 0$ so we would obtain $E_{cell}\ge 0$, while we know the inequality $\Delta_rG\le0$ is obtained from $dG\le0$ which doesn't hold here.

My question in other wording reads like this. Does the electrochemical cell problem involve an electric work or not? If "yes" then from where do we know the Gibbs free energy should decrease? And if "no" then from where do we obtain the equality $\Delta_rG=-n\,F\,E_{cell}$?

I think the confusion may be because you are conflating two ideas about Gibbs free energy.

The first is that Gibbs energy is a thermodynamic potential - this means that on a given surface in phase-space (the NPT surface), it is the "driving" force for, well, everything. Including chemical reactions.

The second idea is that the net change in Gibbs energy when going from reactants to products indicates the direction that the reaction will (eventually) go, and that its magnitude tells you the maximum amount of work that could be done by that reaction.

The second idea follows from the first - this becomes easy to understand if you just picture the thermodynamic potential as a gravitational potential, since we all have a good intuitive understanding of gravitational potentials.

Imagine a hill representing the thermodynamic potential (in this case Gibbs energy): The rock represents a reaction coordinate - say, the concentration of a reactant. What will happen to the rock? It will go downhill. What will its maximum energy be when it reaches the bottom? Whatever the difference in potential energies is. What is the minimum energy it would need to get back the top? Again, whatever the net difference is in potential energies.

When you ask about $dG$ being negative, you are asking about the differential change - in other words, what is the slope? Obviously, if the reaction proceeds, the slope must be negative - "things roll downhill"

When you ask about $\Delta G$ being negative, you are asking whether the net change is positive or negative. The system can never do more work than the net change in $\Delta G$, because that is the maximum change in thermodynamic potential energy.

For your problem, the reaction involves moving an electron against a potential - the cell potential. The reaction can provide $\Delta G$ amount of energy with which to move that electron.

• Dear @thomij, thank you for your great explanation, only one point remains in the dark side for me, that the reaction itself involves the transfer of electron so moving the electron is not a work done by the reaction on the system's surrounding, am I wrong? – topology Aug 23 '14 at 10:28
• The electrons are moving against an external potential, which is the cell voltage. If all of the reactant molecules were in the same side of the cell, there would be no external potential and the total $\Delta G$ would be released as other forms of energy (heat mostly). Since in a Galvanic cell the reactants are separate, but electrons can flow through a circuit, then the voltage of the circuit is the external potential that the reaction is doing work against. – thomij Aug 23 '14 at 23:01