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One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$ If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.

In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.

What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

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  • $\begingroup$ This wall of text is hard to read: please use paragraphs. $\endgroup$ – Ed V May 6 at 15:46
  • $\begingroup$ I have separated your question into paragraphs, I hope that's okay. If not feel free to edit it again. But please pay attention to the formatting in future, no one would bother to read a wall of text. $\endgroup$ – S R Maiti May 6 at 15:59
  • $\begingroup$ Thanks, man. I've completely forgotten about it. $\endgroup$ – Dario Mirić May 6 at 16:17
  • $\begingroup$ $\Delta G^0$ is the amount of electrical work that must be done to bring about the dissociation and $T\Delta S^0$ is the amount of heat that must be added to the system to hold it at constant temperature T. $\Delta H^0$ is the enthalpy of the products of the dissociation minus the enthalpy of the reactants. $\endgroup$ – Chet Miller May 6 at 20:39
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    $\begingroup$ It is also equal to the heat transferred from the surroundings to the system divided by the temperature for a process that takes the system reversibly from its initial state to its final state. $\endgroup$ – Chet Miller May 6 at 23:12
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[OP] One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$

This is just the definition of the Gibbs energy. By itself, it does not give you any insight into chemical reactions.

[OP] If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.

This is not the usual interpretation. Instead $ \Delta_rG_m^⦵ $ is the maximal work you can get out of the reaction (or, if the reaction moves away from equilibrium as written, the minimal work you have to put in to run it in that direction). $ \Delta_rH_m^⦵ $ is the sum of heat and non-PV work exchanged with the system. It could be negative or positive irrespective of whether this is an electrolytic or galvanic cell. Finally, the entropy of reaction is the same as for any reaction - roughly how much the dispersion of energy and particles changes within the system when the reaction proceeds.

[OP] In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.

You can't make any generalizations about the heat transfer when there is also work exchanged between system and surrounding. In many thermodynamic arguments, you talk about processes that happen in the absence of non-PV work, and in a reversible manner (i.e. the entropy of the universe is near-constant, which is the same as saying it is a near-equilibrium process).

[OP] What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.

Without a specific link, it is hard to understand what statement on the hyperphysics site you are referring to, and it what context it was made. In general, the entropy change of the system reflects changes in the system.

EDIT: The remainder was written after OP provided the link.

[Hyperphysics] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of $-\frac{\Delta H}{T}$. TΔS is not directly related to the change of entropy in the surrounding. The reason that $\Delta H$ is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though). The hyperphysics quote does not say that the entropy change of the system is associated with heat exchange. Their (implicit) argument is the second law:

$$T \Delta S_\mathrm{system} + T \Delta S_\mathrm{surrounding} > 0 $$

So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Typically, though, we don't want a super-slow reaction, so you would apply a higher voltage than necessary, and the heat transfer would go in the other direction (as you can tell by the heat typically dissipated from a fast battery charger).

Applying this to the first law, some of the energy input for the reaction can be in the form of heat transfered from the surrounding, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.

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  • $\begingroup$ Yes, thanks for your notes. My question remains what does entropy change of reaction have to do with heat exchanged with surroundings? Change in entropy due to reaction in a system doesn't happen because of heat transfer, it has to do with the fact that products and reactants have different entropies because of their structure and aggregate state regardless of heat transfer (which you agreed with me). If so, why is TdeltaS(reaction) interepreted as heat transfered since entropy change here has a different origin which doesn't come from heat transfer. $\endgroup$ – Dario Mirić May 7 at 9:02
  • $\begingroup$ You asked for a link to hyperphysics page: hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html Last two paragraphs in section electrolysis talk about what I mentioned/quoted. $\endgroup$ – Dario Mirić May 7 at 9:18
  • $\begingroup$ No non-PV work? They are talking about Gibbs energy of reaction being a electrical work which is a non-PV work. $\endgroup$ – Dario Mirić May 7 at 10:38
  • $\begingroup$ I am not sure I get what you are saying. I thought you said they didn't take into account electrical work. $\endgroup$ – Dario Mirić May 7 at 11:47
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    $\begingroup$ @DarioMirić If there is no reaction (to simplify things), you can't transfer heat from cold to hot place because the entropy gain in the hot place is smaller than the entropy loss in the cold place (because T is in the denominator of $\Delta S = \frac{q_\mathrm{rev}}{T}$). So what stops you is the second law. $\endgroup$ – Karsten Theis May 7 at 12:24
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The link between entropy and heat follows from the balance between entropy change of system and surroundings during a reversible process. During such a process the total change in entropy is zero, and the change in the entropy of the surroundings is given by the second law of thermodynamics as $\Delta S_\mathrm{surr}=-q_\mathrm{rev}/T$ (assuming constant T), from which it follows necessarily that $$\Delta S_\mathrm{system}=\frac{q_\mathrm{rev}}{T}$$ When there is non-expansion work under constant T and p, the enthalpy change can be equated with the heat plus the non-pV work. When in addition the process is reversible, the equation you posted follows by writing $\Delta G = w_\mathrm{rev,non-pV}$.

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