Entropy Changes in Electrolytic/Galvanic Cell?

[Dario56]

It's like any other reversible process. There is a slight pressure difference imposed across the membrane.

Yes, differential one.

 

[Dario56]

It's like any other reversible process. There is a slight pressure difference imposed across the membrane.

Although we did discuss the concept of equilibrium box, it didn't really answer my question. How can we connect what we learned to my question?

 

[Chestermiller]

Although we did discuss the concept of equilibrium box, it didn't really answer my question.

In Section 2.4 of Denbigh's book The Principles of Chemical Equilibrium, he discusses processes in closed systems operated such that (a) the only heat transferred to the system is from a reservoir which remains at the constant temperature T of the system (equal to the initial temperature of the system) and (b) the system is in contact with a constant pressure surroundings p (equal to the initial pressure of the system). This includes closed system in which chemical reactions are occurring within the system, including electrolytic reactions.

For such a closed system, the first law of thermodynamics tells us that $$\Delta U=q-w$$and the 2nd law of thermodynamics tells us that $$q=T\Delta S-\sigma T$$where q is the heat transferred from the surroundings to the system at temperature T, w is the work done by the system on the surroundings, and $\sigma$ is the entropy generated within the system due to irreversibility. If we combine these two equations, we obtain: $$\Delta U=T\Delta S-w-\sigma T$$or, equivalently, $$\Delta G=\Delta H-T\Delta S=-(w-p\Delta V)-\sigma T$$

OK so far?

 

[Dario56]

In Section 2.4 of Denbigh's book The Principles of Chemical Equilibrium, he discusses processes in closed systems operated such that (a) the only heat transferred to the system is from a reservoir which remains at the constant temperature T of the system (equal to the initial temperature of the system) and (b) the system is in contact with a constant pressure surroundings p (equal to the initial pressure of the system). This includes closed system in which chemical reactions are occurring within the system, including electrolytic reactions.

For such a closed system, the first law of thermodynamics tells us that $$\Delta U=q-w$$and the 2nd law of thermodynamics tells us that $$q=T\Delta S-\sigma T$$where q is the heat transferred from the surroundings to the system at temperature T, w is the work done by the system on the surroundings, and $\sigma$ is the entropy generated within the system due to irreversibility. If we combine these two equations, we obtain: $$\Delta U=T\Delta S-w-\sigma T$$or, equivalently, $$\Delta G=\Delta H-T\Delta S=-(w-p\Delta V)-\sigma T$$

OK so far?

I have this book, so I can check this chapter and we can discuss later.

 

[Dario56]

In Section 2.4 of Denbigh's book The Principles of Chemical Equilibrium, he discusses processes in closed systems operated such that (a) the only heat transferred to the system is from a reservoir which remains at the constant temperature T of the system (equal to the initial temperature of the system) and (b) the system is in contact with a constant pressure surroundings p (equal to the initial pressure of the system). This includes closed system in which chemical reactions are occurring within the system, including electrolytic reactions.

For such a closed system, the first law of thermodynamics tells us that $$\Delta U=q-w$$and the 2nd law of thermodynamics tells us that $$q=T\Delta S-\sigma T$$where q is the heat transferred from the surroundings to the system at temperature T, w is the work done by the system on the surroundings, and $\sigma$ is the entropy generated within the system due to irreversibility. If we combine these two equations, we obtain: $$\Delta U=T\Delta S-w-\sigma T$$or, equivalently, $$\Delta G=\Delta H-T\Delta S=-(w-p\Delta V)-\sigma T$$

OK so far?

Yes. I do understand what is being said.

 

[Chestermiller]

OK, The next step is to express the total work w as: $$w=P\Delta V+w'$$ where w' is work over and above pressure-volume work done by the system, which, in our case means electrical work done by the galvanic cell. So we are left with $$w'=-(\Delta H-T\Delta S)-\sigma T$$What this equation tells us is the, for all processes at constant external pressure on our system, the amount of galvanic work is equal to the decrease in Gibbs free energy of the system between the two specified end states minus the dissipated energy resulting from irreversible entropy generation during the process.

For a galvanic cell, it is possible to carry out a constant external pressure process reversibly by controlling the external voltage applied to the cell so that it nearly matches the cell potential. In such a case, the entropy generated will be zero ($\sigma = 0$), and we will have:
$$w'_{rev}=-\Delta G=-(\Delta H-T\Delta S)$$ and $$q_{rev}=T\Delta S$$ From this it follows that $-\Delta G$ can be interpreted as the reversible galvanic work and $T\Delta S$ can be interpreted as the reversible heat transfer.

 

[Dario56]

OK, The next step is to express the total work w as: $$w=P\Delta V+w'$$ where w' is work over and above pressure-volume work done by the system, which, in our case means electrical work done by the galvanic cell. So we are left with $$w'=-(\Delta H-T\Delta S)-\sigma T$$What this equation tells us is the, for all processes at constant external pressure on our system, the amount of galvanic work is equal to the decrease in Gibbs free energy of the system between the two specified end states minus the dissipated energy resulting from irreversible entropy generation during the process.

For a galvanic cell, it is possible to carry out a constant external pressure process reversibly by controlling the external voltage applied to the cell so that it nearly matches the cell potential. In such a case, the entropy generated will be zero ($\sigma = 0$), and we will have:
$$w'_{rev}=-\Delta G=-(\Delta H-T\Delta S)$$ and $$q_{rev}=T\Delta S$$ From this it follows that $-\Delta G$ can be interpreted as the reversible galvanic work and $T\Delta S$ can be interpreted as the reversible heat transfer.

Well, result here did lead us towards the equation I've written in my main post which doesn't really answer my question. What I am also not sure is if #T\Delta S# is heat exchanged which isn't clear to me in this context, what does enthalpy of reaction represent in context of galvanic/electrolytic cell since enthalpy of reaction usually represents heat of reaction at constant p and T needed to keep system at constant T. In this context it might have different meaning. I think it represents total energy released/abosrbed in a cell operating at constant p and T which is equal to sum of electrical energy released/absorbed and #T\Delta S# term I don't understand. What are your thoughts since I don't really get this well?

 

[Chestermiller]

My thoughts are that you are very close to having this worked out correctly.

The enthalpy change is equal to the heat only if the pressure is constant throughout the process and the only kind of work done is PV work. We saw that in the reversible path case of ideal gas reactions, the only kind of work is PV work, but the pressure is not constant during steps 1 and 3, in which the pure gases are being expanded and compressed. In those steps, there is heat exchanged, but the enthalpy changes are zero. For an irreversible version of the ideal gas reactions, steps 1 and 3 are not preformed, and the heat in this case turns out to equal to the standard enthalpy change.

In the case of the electrolytic reactions, the pressure is constant throughout the process, but PV work is not the only kind of work. This leads us to $$\Delta H=q-w'$$For the reversible path, we have the same $\Delta H$ being equal to the reversible heat, $T\Delta S$, minus the reversible work $w'_{rev}$: $$\Delta H=q_{rev}-w'_{rev}=T\Delta S-w'_{rev}$$. This is the maximum work that the system can do at constant T and P. In neither case is the enthalpy change equal to q unless the work in the irreversible case is zero (the cell is short circuited).

 

[Dario56]

My thoughts are that you are very close to having this worked out correctly.

The enthalpy change is equal to the heat only if the pressure is constant throughout the process and the only kind of work done is PV work. We saw that in the reversible path case of ideal gas reactions, the only kind of work is PV work, but the pressure is not constant during steps 1 and 3, in which the pure gases are being expanded and compressed. In those steps, there is heat exchanged, but the enthalpy changes are zero. For an irreversible version of the ideal gas reactions, steps 1 and 3 are not preformed, and the heat in this case turns out to equal to the standard enthalpy change.

In the case of the electrolytic reactions, the pressure is constant throughout the process, but PV work is not the only kind of work. This leads us to $$\Delta H=q-w'$$For the reversible path, we have the same $\Delta H$ being equal to the reversible heat, $T\Delta S$, minus the reversible work $w'_{rev}$: $$\Delta H=q_{rev}-w'_{rev}=T\Delta S-w'_{rev}$$. This is the maximum work that the system can do at constant T and P. In neither case is the enthalpy change equal to q unless the work in the irreversible case is zero (the cell is short circuited).

I will check this out in Denbigh in some more detail.

 

[Dario56]

I will check this out in Denbigh in some more detail.

My thoughts are that you are very close to having this worked out correctly.

The enthalpy change is equal to the heat only if the pressure is constant throughout the process and the only kind of work done is PV work. We saw that in the reversible path case of ideal gas reactions, the only kind of work is PV work, but the pressure is not constant during steps 1 and 3, in which the pure gases are being expanded and compressed. In those steps, there is heat exchanged, but the enthalpy changes are zero. For an irreversible version of the ideal gas reactions, steps 1 and 3 are not preformed, and the heat in this case turns out to equal to the standard enthalpy change.

In the case of the electrolytic reactions, the pressure is constant throughout the process, but PV work is not the only kind of work. This leads us to $$\Delta H=q-w'$$For the reversible path, we have the same $\Delta H$ being equal to the reversible heat, $T\Delta S$, minus the reversible work $w'_{rev}$: $$\Delta H=q_{rev}-w'_{rev}=T\Delta S-w'_{rev}$$. This is the maximum work that the system can do at constant T and P. In neither case is the enthalpy change equal to q unless the work in the irreversible case is zero (the cell is short circuited).

Hey again. I was busy with other stuff and didn't do much about this. Can you please point me to reference in Denbigh or some other textbooks for equations and statements you've written in your last answer. Thank you.

 

[Chestermiller]

Hey again. I was busy with other stuff and didn't do much about this. Can you please point me to reference in Denbigh or some other textbooks for equations and statements you've written in your last answer. Thank you.

I already referenced you to the relevant sections of Denbigh. Regarding the galvanic reaction case, I did not say a single thing that cannot be deduced from the Denbigh development.

At some point, your understanding has got to progress to the point where you can analyze things for yourself beyond that presented in books and references. That is what I have done for the ideal gas reaction case that I discussed above.

If you have doubts about something I said, please articulate them so that I can elaborate.

 

[Dario56]

I already referenced you to the relevant sections of Denbigh. Regarding the galvanic reaction case, I did not say a single thing that cannot be deduced from the Denbigh development.

At some point, your understanding has got to progress to the point where you can analyze things for yourself beyond that presented in books and references. That is what I have done for the ideal gas reaction case that I discussed above.

If you have doubts about something I said, please articulate them so that I can elaborate.

Yes. Well, our discussion did lead us towards the equation I've written in my main post. This equation is the last one you've written. So, if TdeltaS term represents reversible heat exchanged and standard deltaG represents maximum non - PV work, what does standard deltaH represent? Is it the total energy input or output for reaction in a cell at constant p and T? For electrolytic cell it would be total energy we need to put in for reaction to happen at constant p and T. What still isn't clear to me is that TdeltaS term as I can't connect heat transfer with standard entropy of reaction even though I do understand how this equation is derived fundamentally as it is derived in Denbigh. According to my understanding entropy of the system should change during a reaction regardless of heat transfer with surroundings or even if heat weren't exchanged.

 

[Chestermiller]

Yes. Well, our discussion did lead us towards the equation I've written in my main post. This equation is the last one you've written. So, if TdeltaS term represents reversible heat exchanged and standard deltaG represents maximum non - PV work, what does standard deltaH represent?

It represents the enthalpy of the products in the standard state minus enthalpyof the reactants in the standard state. So it is "material related," not "process related." In certain types of processes, it can also be interpreted in terms of the heat and work, and measured in that way. For example, in the case of a galvanic reaction run reversibly, it turns out to equal to the reversible heat minus the reversible work. In the case of a short-circuited galvanic process, it turns out to be equal to the heat.

What still isn't clear to me is that TdeltaS term as I can't connect heat transfer with standard entropy of reaction even though I do understand how this equation is derived fundamentally as it is derived in Denbigh. According to my understanding entropy of the system should change during a reaction regardless of heat transfer with surroundings.

This is similar to the answer I gave for the enthalpy change. The standard entropy change is equal to the entropy of the products in the standard state minus the entropy of the reactants in the standard state. So it is "material related," not "process related." In certain types of processes, it can also be interpreted in terms of the heat and work, and measured in that way. For example, for a reaction process carried out reversibly at constant temperature between the standard states, it is the heat transferred from the reservoir to the system in the process divided by the temperature. For a reaction process carried out irreversibly in contact with a constant temperature reservoir, it exceeds the heat transferred from the reservoir to the system in the process divided by the reservoir temperature.

My point is this: Stop thinking of these thermodynamic functions in terms of process. They are physical properties of the material that have very specific values for the state in which the material resides. They are material properties in very much the same sense as viscosity, thermal conductivity, or diffusivity. It is just that certain specific kinds of processes can be used to measure the changes in these properties. And, if these properties are already known as functions of state, they can be used to predict the heat or work for certain kinds of processes.

 

[Dario56]

It represents the enthalpy of the products in the standard state minus enthalpyof the reactants in the standard state. So it is "material related," not "process related." In certain types of processes, it can also be interpreted in terms of the heat and work, and measured in that way. For example, in the case of a galvanic reaction run reversibly, it turns out to equal to the reversible heat minus the reversible work. In the case of a short-circuited galvanic process, it turns out to be equal to the heat.This is similar to the answer I gave for the enthalpy change. The standard entropy change is equal to the entropy of the products in the standard state minus the entropy of the reactants in the standard state. So it is "material related," not "process related." In certain types of processes, it can also be interpreted in terms of the heat and work, and measured in that way. For example, for a reaction process carried out reversibly at constant temperature between the standard states, it is the heat transferred from the reservoir to the system in the process divided by the temperature. For a reaction process carried out irreversibly in contact with a constant temperature reservoir, it exceeds the heat transferred from the reservoir to the system in the process divided by the reservoir temperature.

My point is this: Stop thinking of these thermodynamic functions in terms of process. They are physical properties of the material that have very specific values for the state in which the material resides. They are material properties in very much the same sense as viscosity, thermal conductivity, or diffusivity. It is just that certain specific kinds of processes can be used to measure the changes in these properties. And, if these properties are already known as functions of state, they can be used to predict the heat or work for certain kinds of processes.

Yes, I actually agree with your point and I did before. Maybe you didn't get what I asked, but as we said TdeltaS represents reversible heat transferred and my question is what does entropy change of reaction (material related) has to do with heat transferred in reversible process (process related)? That is all I am asking and what I don't get. Basically where is the connection between material related change (entropy of reaction) with process related change (reversible heat transfer) since entropy change of system during reaction happens simply because products and reactants are different compounds with different structure and aggregate state. What does this fact have to do with heat exchanged in reversible process? I don't know if you understand what I am asking.

 

[Chestermiller]

Yes, I actually agree with your point and I did before. Maybe you didn't get what I asked, but as we said TdeltaS represents reversible heat transferred and my question is what does entropy change of reaction (material related) has to do with heat transferred in reversible process (process related)? That is all I am asking and what I don't get. Basically where is the connection between material related change (entropy of reaction) with process related change (reversible heat transfer) since entropy change of system during reaction happens simply because products and reactants are different compounds with different structure and aggregate state. What does this fact have to do with heat exchanged in reversible process? I don't know if you understand what I am asking.

Well, that all goes back to the issue of why the statistical mechanical representation of entropy and entropy change (natural log of the number of quantum mechanical states) turns out to be equal to the integral of $dq_{rev}/T$. This is covered in a typical statistical thermo course, at least for an ideal gas. Have you had a course in statistical thermo?

 

[Dario56]

Well, that all goes back to the issue of why the statistical mechanical representation of entropy and entropy change (natural log of the number of quantum mechanical states) turns out to be equal to the integral of $dq_{rev}/T$. This is covered in a typical statistical thermo course, at least for an ideal gas. Have you had a course in statistical thermo?

Well, I didn't learn statistical thermo in detail, but I do know basics. I do know that entropy is macroscopic quantitiy of number of microstates system can exist at certain macrostate and I do know that entropy change of a system is determined by integral of reversible heat needed to bring system from state 1 to state 2. Bringing heat increases average kinetic energy and number of possible microstates of molecules. Molecules at high temperatures ocuppy many available kinetic energy states and on low temperatures they don't, so the same amount of heat increases entropy more on low temperatures than on high. We can use this fact to define temperature.

 

[Chestermiller]

Well, I didn't learn statistical thermo in detail, but I do know basics. I do know that entropy is macroscopic quantitiy of number of microstates system can exist at certain macrostate and I do know that entropy change of a system is determined by integral of reversible heat needed to bring system from state 1 to state 2. Bringing heat increases average kinetic energy and number of possible microstates of molecules. Molecules at high temperatures ocuppy many available kinetic energy states and on low temperatures they don't, so the same amount of heat increases entropy more on low temperatures than on high. We can use this fact to define temperature.

I have no idea. I'm a continuum thermo guy myself, and my background in statistical thermo is very weak.

 

[Dario56]

I have no idea. I'm a continuum thermo guy myself, and my background in statistical thermo is very weak.

Well, I don't know much about it either simply because engineers don't learn much about statistical thermo on their curriculum (I am on my last year before getting masters in ChemE). I do know some basics from physical chemistry, thermo class and something I learned by myself as I studied thermo and other fields in ChemE. But, yes, I am not sure how to connect entropy change of reaction (material related) with reversible heat exchanged (process related) since entropy change of system isn't due to heat transfer, but has to do with chemical reaction in the system.

 

[Chestermiller]

Well, I don't know much about it either simply because engineers don't learn much about statistical thermo on their curriculum (I am on my last year before getting masters in ChemE). I do know some basics from physical chemistry, thermo class and something I learned by myself as I studied thermo and other fields in ChemE. But, yes, I am not sure how to connect entropy change of reaction (material related) with reversible heat exchanged (process related) since entropy change of system isn't due to heat transfer, but has to do with chemical reaction in the system.

As I learned it as a ChE, the way to get the entropy change of a closed system is to evaluate dq/T for a reversible path. If you are interested in seeing how it comes in for chemical reactions, including entropy generation in irreversible paths, see Chapter 24 of BSL.

 

[Dario56]

As I learned it as a ChE, the way to get the entropy change of a closed system is to evaluate dq/T for a reversible path. If you are interested in seeing how it comes in for chemical reactions, including entropy generation in irreversible paths, see Chapter 24 of BSL.

Yes, these two are connected somehow. Entropy change because of reaction has to do something with reversible change between states 1 and 2 as you mentioned. What is BSL?

 

[Chestermiller]

Yes, these two are connected somehow. Entropy change because of reaction has to do something with reversible change between states 1 and 2 as you mentioned. What is BSL?

In ChE, is Bird, Stewart, and Lightfoot, Transport Phenomena not a major resource and text any more? I referred to it extensively during my career at DuPont.

 

[Dario56]

In ChE, is Bird, Stewart, and Lightfoot, Transport Phenomena not a major resource and text any more? I referred to it extensively during my career at DuPont.

It is. I didn't recognize the acronym. You worked at DuPont? I've heard of that company.

 

[Dario56]

In ChE, is Bird, Stewart, and Lightfoot, Transport Phenomena not a major resource and text any more? I referred to it extensively during my career at DuPont.

I have one more question which is how do we determine sign convention of additional work (electrical work in galvanic/electrolytic cell)? I guess we say that electrical work is positive if surroundings does work on system increasing Gibbs energy of system/reaction. So, electrical work is positive in electrolytic cell where Gibbs energy of reaction increases and negative in galvanic cell? So, electrical work has the same sign as change of G of reaction. Did I explain this correctly?

 

[Chestermiller]

I have one more question which is how do we determine sign convention of additional work (electrical work in galvanic/electrolytic cell)? I guess we say that electrical work is positive if surroundings does work on system increasing Gibbs energy of system/reaction. So, electrical work is positive in electrolytic cell where Gibbs energy of reaction increases and negative in galvanic cell? So, electrical work has the same sign as change of G of reaction. Did I explain this correctly?

It depends on the sign convention. In ChE, work done on the surroundings is usually taken as positive.