In Gibbs Free Energy, why multiply -T?

[Bystander]

Where should I use ΔG and where should I use ΔG0?

Use "ΔG" when looking at a system for which reactants and products are NOT in their standard states. Use ΔG⁰ and the activity relation when you are trying to find ΔG.

ΔG = ΔH - TΔS Is this right?

This is the general expression for change in Gibb's free energy. It's always valid.

ΔG =nFE or ΔG0=nFE?

ΔG = - nFE gives you the emf of an actual cell, and ΔG⁰ = - nFE⁰ gives the emf of a cell in which reactants and products are in their standard states.

 

[gracy]

Thanks for your all answers.Please bear with me.How can we apply gibbs free energy concept in electrochemical cells?I mean what are reactants and products in electrochemical cells?We will write over all reaction equation (combining oxidation and reduction half cell equation )and can figure out reactants and product of the reaction.Then we can apply all gibbs energy concepts,right?

 

[gracy]

Two different kinds of metal can be placed in a conducting solution to form galvanic cell,right?So one electrode of Na and one of chlorine can not make galvanic cell as chlorine is not a metal,right?

 

[Chestermiller]

Gracy: I think it would better for you to take a step backwards temporarily and first get a good solid understanding of how all this works for ideal gases that are undergoing chemical reaction. After you master that, you can safely move on to more complicated systems with confidence. Are you familiar with the Van't Hoff Equilibrium Box?

Chet

 

[gracy]

Are you familiar with the Van't Hoff Equilibrium Box?

Van't Hoff Equilibrium Box is a kind of vessel which has constant volume and the various substance which take part in any reaction are in equilibrium.The wall of this vessel is permeable to only some substances.This box is only applicable for homogeneous systems.

 

[Chestermiller]

Van't Hoff Equilibrium Box is a kind of vessel which has constant volume and the various substance which take part in any reaction are in equilibrium.The wall of this vessel is permeable to only some substances.This box is only applicable for homogeneous systems.

This is kind of correct, but let's make it more precise. Is there only one semipermeable membrane? What about the temperature?

We've been talking about ΔG and ΔG⁰. If you had a Van't Hoff Equilibrium Box available to you, how would use use it to experimentally measure ΔG⁰ for a reaction involving ideal gases? (Use of supplementary equipment, such as cylinders with pistons, is allowed.)

Chet

 

[gracy]

Is there only one semipermeable membrane?

I think it depends on number of walls of van't hoff box.

What about the temperature?

At constant temperature.I guess.

 

[gracy]

We've been talking about ΔG and ΔG0. If you had a Van't Hoff Equilibrium Box available to you, how would use use it to experimentally measure ΔG0 for a reaction involving ideal gases? (Use of supplementary equipment, such as cylinders with pistons, is allowed.)

I don't know about this.Please help me.

 

[Chestermiller]

I think it depends on number of walls of van't hoff box.

No. The is a separate semipermeable membrane for each reactant and each product of the reaction.

 

[gracy]

No. The is a separate semipermeable membrane for each reactant and each product of the reaction.

So,how many?

 

[Chestermiller]

So,how many?

As I said, one membrane for each reactant and one membrane for each product. Does every reaction have the same number of products and reactants?

Chet

 

[Chestermiller]

I don't know about this.Please help me.

It would really help with your understanding if you could describe how a Van't Hoff equilibrium box could, at least conceptually, be used to determine ΔG⁰ for a reaction. Rather than my trying to go into the details of this, I'm hoping you will consider reading up about this in your Thermodynamics book or online. I will be glad to answer any questions. FYI, ΔG⁰ refers to the difference in Gibbs free energy between the following two thermodynamic equilibrium states:

1. Pure reactants (in separate containers), each at 1 atm pressure and temperature T
2. Pure products (in separate containers), each at 1 atm pressure and temperature T

Chet

 

[Ygggdrasil]

So ,you mean at equilibrium ΔG =0 but ΔGo is not zero instead ΔGo = -RT ln(K).Right?
At time 1:13 is he wrong?

He's being very loose with the terminology and mixing up the standard change in Gibbs free energy and the real change in Gibbs free energy.

Here's a key thing to remember, ΔG depends on concentration, and the sign of ΔG tells you whether the reaction will be spontaneous for a given concentrations of product and reactant. ΔGo is independent of concentration and it tell you about the position of equilibrium for a reaction. ΔGo > 0 means equilibrium favors the reactants while ΔGo < 0 means equilibrium favors the products. Even if ΔGo > 0, you can still have some conversion of reactant to product if the reaction lies too far on the reactant side of the equilibrium (i.e if the reaction quotient Q < Keq). Many unfavorable (ΔGo > 0) reactions in biology happen this way because the cell is quickly depleting the product, which keeps Q < Keq and keeps ΔG < 0 to allows these transformations to occur spontaneously.

Where should I use ΔG and where should I use ΔG0?I mean the formula
ΔG = ΔH - TΔS Is this right?or it is only applicable for standard gibbs free energy ..

This is technically correct, but here, you cannot use ΔSo for the calculation. ΔS will depend not only on the standard entropies of the reactants and products, but also how the reaction partitions the atoms in the reaction between reactants and products. ΔS = ΔSo + RT lnQ For example, if you consider the reaction A --> B, having a system consisting of all A molecules and no B molecules has a very low entropy. In the absence of any intrinsic differences in the enthalpies or entropies of molecules A and B, a situation with 50% A and 50% B would be most entropically favorable.

In general, however, I would mostly say to use ΔGo = ΔHo - TΔSo

 

[gracy]

I want to know about negative EMF.What does Negative emf indicates?What is concept behind Negative emf ?I am not talking about Negative induced emf,I am talking in context of electrochemical cell.
I am asking in the sense that
E°cell = E°red (cathode) – E°red (anode)
so can you explain negative emf in terms of his formula,

 

[Bystander]

Negative emf indicates that the reaction as written does not "go." Remember that "del G nought" is equal to the negative of "n F E nought."

 

[gracy]

But Negative EMF indicates non-spontaneity,and in galvanic cell reactions are always spontaneous that means negative EMF is not possible in galvanic cell ,right?

 

[gracy]

I found this line about electrolytic cell
The potential/voltage input + the cell potential must be > 0 for the reactions to occur.
What is potential/voltage input ,and the cell potential ?How these two differ?

 

[Bystander]

in galvanic cell reactions are always spontaneous

That appears to be the way "galvanic" is interpreted/defined/used. I don't remember the distinction being quite so absolute fifty years ago, but life is lots easier if we bow to current convention and use.

negative EMF is not possible

As a consequence of the current definition, that is correct.

 

[gracy]

Is there any cell potential in electrolytic cell?I am not talking about voltage given by external source,i am referring to it's own potential i.e because of it's electrodes.

 

[Bystander]

The potential/voltage input + the cell potential must be > 0 for the reactions to occur.
What is potential/voltage input ,and the cell potential ?How these two differ?

"Potential/voltage input" is some voltage source external to the electrolytic cell of interest, like a battery charger.

" ... potential/voltage input + the cell potential must be > 0 ..." has to be a misprint or typographical error. If it reads "... potential/voltage input minus the cell potential must be > 0 ..." it makes a little more sense to me.

 

[gracy]

If it reads "... potential/voltage input minus the cell potential must be > 0 ..." it makes a little more sense to me.

How?I mean why it makes more sense?Can you please elaborate?

 

[Bystander]

Is there any cell potential in electrolytic cell?I am not talking about voltage given by external source,i am referring to it's own potential i.e because of it's electrodes.

Yes. You can do electrolysis of water by running a current through it at a voltage sufficient to decompose it (~ 1 volt at standard conditions if memory serves), or you can allow the oxygen and hydrogen to react at the two electrodes to form water and generate electricity in a fuel cell.

 

[gracy]

But in all my textbook examples of electrolytic cell ,both electrodes i.e cathode as well as anode are of same element,so how can there be cell potential?As in one of my threads 'why cathode is positively charged' we have discussed that potential depends on element/material of electrodes.

 

[Bystander]

How?I mean why it makes more sense?Can you please elaborate?

Consider a "dead" automotive battery. To charge it, a power/voltage source is connected positive terminal to positive terminal of the battery, and negative terminal to negative terminal of the battery. The difference between the voltage of the source and the voltage in the battery is what drives the charging process. Connecting in series, as is implied by "... input + the cell potential must be greater than zero ..." is what starts car fires, battery explosions, and all sorts of other mayhem by short-circuiting two power sources through each other. Looking at the difference (parallel connection) we have one source (that with the larger magnitude) charging the other.

electrodes i.e cathode as well as anode are of same element,so how can there be cell potential?

A fuel cell uses Pt for both electrodes with a supply of oxygen at one and a supply of hydrogen at the other. The Pt catalyzes dissociation of hydrogen and of oxygen, and of the electron transfers necessary for formation of water. Same electrode materials, but different processes occurring at each electrode.

Electrolytic refining of copper occurs by dissolving copper at the anode, immersed in a sulfuric acid-copper sulfate solution, transport of the copper ions through the solution to the cathode, and reduction of the copper ions at the copper cathode to yield pure copper. In principle, there is a zero potential between the electrodes; in practice, the composition of the impure copper is sufficiently different from that of pure copper that it is more easily oxidized, but that potential is more than offset by the resistance to transport of the copper ions between the electrodes.

Bottom line: cell potentials are meaningful only for ZERO current flow, and only under special circumstances of cell design. You don't have to learn all of this at one time. There will be upper level courses that cover it in more detail.

 

[gracy]

Consider a "dead" automotive battery

Is that galvanic cell?As far as I know all batteries are galvanic cell.Anyways ,I am asking about electrolytic cell in my 73rd post.

 

[fernandofortes]

I haven't heard that term used with Gibbs free energy as I really don't know much about this. But I've learned that when ΔG is positive, a favorable (spontaneous) reaction will occur and non-spontaneous reactions when its negative.

opposite you said: negative favorable or spontaneous; positive unfavorable or non-spontaneous. You could use also negative possible; positive impossible. Or negative allowed; positive prohibited

 

[fernandofortes]

Sorry I meant favorable reactions at negative Gibbs Energy. Also are you saying that reactions only occur at negative Gibbs energy?

Better say negative (or positive) DELTA Gibbs energy, since we deal with changes.

 

[fernandofortes]

A positive standard Gibbs free energy change does not mean that the reaction will not occur at all. It simply means that the conversion at which the reactants will be at equilibrium with the products will be small.

Chet

Or we could say that at standard P, T, and, most important, 1 M all participants, it will not occur at all, and worst, you will see the opposite: reactants increase above 1 M and products decrease bellow 1M.

 

[gracy]

Or we could say that at standard P, T, and, most important, 1 M all participants, it will not occur at all, and worst, you will see the opposite: reactants increase above 1 M and products decrease bellow 1M.

Is it right?

 

[fernandofortes]

Chet's just pointing out that Gibb's free energy has nothing to do with whether a reaction "goes" or does "not go." It limits the degree of "completion" of a reaction, allowing quantitative calculation of quantities/concentrations of products.

I think that since you write (substance A) arrow (substance B) you refer tot A as reactant and B as product, so A must decrease and B increase IF your reaction is possible. DeltaG negative will say that you correctly assumed reactants and products, and your reaction is allowed to "go".

 

[fernandofortes]

You seem to be ascribing some special significance to the situation where the equilibrium constant is unity. This value of K is no more special than any other value of K. Remember also that the standard free energy change refers to the change in free energy starting out with the pure reactants in stoichiometric proportions at 1 atm., and ending up with the pure products in corresponding proportions at 1 atm. In the equilibrium constant K, you are talking about the reactants and products not necessarily in stoichiometric proportions nor at 1 atm pressure, and the partial pressures of the reactants and products even don't have to be equal; in fact, even, if there are only single moles involved, the only requirement is that the product of the reactant partial pressures must match the product of the partial pressures of the products. If the final number of moles is different from the initial number of moles, not even this is required.

Chet

And remember that say "Why system has no available energy to do work if it's reactants and products have same concentration?" is wrong if you forget that you are talking about equilibrium constant equal 1. I think there is no example of reactants and products at same concentration and the system at equilibrium.

 

[AndrewBworth]

Thermodynamics only determines whether or not a reaction can occur and what the equilibrium will be at completion. The rate at which a reaction occurs is determined by kinetics. So oxygen and hydrogen don't spontaneously react due to reasons of kinetics -but the thermodynamics is correct in that it is a downhill reaction.