[Big-Daddy]

My book has this statement:

There are two diferent types of electrochemical experiment:
(i) Those in which an electrode dips into a solution so that a potential is established on it in accordance with the predictions of the Nernst equation. This is used to measure thermodynamic data (equilibrium constants, reaction free energies, enthalpies and entropies,and so on). This is the field of equilibrium electrochemistry. As the name implies, no sustained currents flow during such experiments.
(ii) Those in which a potential is applied between the electrode and the solution, thus causing the concentrations of the species in the cell to adjust themselves so as to conform to the Nernst equation. In order for this to happen current has to flow and electrolysis takes place.

In both cases, I think, an equilibrium involving electrons is set up when the electrode is placed in solution, e.g. Fe³⁺ (aq) + e^- (metal)  Fe²⁺ (aq). In case (ii), with a current, the system pushes itself towards the equilibrium ratios of concentration from the initial concentrations of the two iron species.

In case (i), however, I have two questions: firstly, how is the potential established, since the species' concentrations are not initially at equilibrium (as in case (ii), they get closer to equilibrium with time)? And secondly, for case (ii), do the current and potential need to be externally applied?

[Corribus]

In case one the potential is created by the simple fact that there is a nonequilibrium situation.  The potential difference between the electrode and the solution is directly related to the driving force for obtaining equilibrium (ΔG).  Once equilibrium has been reached (ΔG = 0), there is no longer an electronic potential difference.

[Big-Daddy]

Hmm ok now I'm a bit confused. Let me give it a shot and you can tell me what's wrong or right:

Case 1) Concentration of the species in the equilibrium (e.g. Fe³⁺ (aq), Fe²⁺ (aq)) do not match the equilibrium constant, as the equilibrium doesn't begin until the electrode is placed in solution. As soon as the electrode is, the concentrations begin to go towards equilibrium.

Shouldn't this be the realm of kinetics, then, which is specified in Case 2 rather than Case 1? And how exactly does having non-0 ΔG produce a potential difference?

Case 2 also begins with non-equilibrium concentrations of the species. Now a current is used to drive the reaction towards the same equilibrium constant and concentrations as in Case 1. So if this happens in Case 1, why is a current even necessary? And what, in terms of calculations, is the difference between passing a current (Case 2) and letting the equilibrium establish itself (Case 1)?

[Corribus]

Shouldn't this be the realm of kinetics, then, which is specified in Case 2 rather than Case 1? And how exactly does having non-0 ΔG produce a potential difference?

They are both kinetics... and both thermodynamics.

The difference between the two scenarios is what the equilibrium point is.

In scenario one, you are dealing essentially with a simple redox reaction.  If you put a stick of zinc in a dilute acid, there will be a reaction because there is an electrochemical driving force for the conversion of Zn⁰ to Zn²⁺ and H⁺ to H₂ gas.  We can represent this driving force by a Gibbs energy change, which will have an associated equilibrium constant.  The reaction will proceed until an equilibrium is reached and the driving force is zero due to a balance in the concentrations of reactants and products.  I think you know all of this because it is simple equilibrium chemistry - the only difference is that the reaction is a redox reaction, and thus the thermodynamical quantities can be expressed in terms of simple redox potentials.  (And in a way, it's really just a superficial difference - ALL reactions are redistributions of electrical charge, more or less; the only difference here is that we can make the approximation of "whole units of charge" moving from "predominantly one nucleus to another", which allows us the convenient use of redox potentials to express pretty much the same thermodynamical equations in a different form.)

In scenario two, all that changes is that there is an external potential energy applied.  What this really translates into is an additional force that resists or reinforces the predominant (natural, you might say) driving force for the reaction.  Essentially what it does is change the equilibrium point to favor more the products or the reactants.  Remember: ΔG is a potential energy concept, which means it is therefore a force concept.  The Gibbs energy tells you how a reaction will behave because all systems try to minimize potential energy.  By adding an external potential energy, you are therefore adding an additional force, which will shift the reaction equilibrium accordingly - and always away from the "natural" equilibrium point.

This is essentially how you charge a battery.  You apply a potential difference across a battery cell, which shifts the equilibrium because you are changing the reaction's driving force by adding a new force to the equation.  But what you are really doing is storing (potential) energy in the cell, because when the external potential difference is removed, now you are suddenly at a nonequilibrium position, and there is a natural driving force to return to the original equilibrium point.  So you remove the charging source (the external potential) and the electrochemical cell wants to return back to its starting point, and it'll do this by letting electrons flow back the way they came.  This electron flow is harnessed to do work elsewhere. 

These charge/discharge cycles cannot be repeated indefinitely, because there are usually other irreversible, minor processes that happen during each cycle (typically the quality of the electrode declines over time, because the metal doesn't re-deposit perfectly - this results in imperfect condition, the generation of excess heat, corrosion of other parts of the battery, etc.).

[Big-Daddy]

In scenario one, you are dealing essentially with a simple redox reaction.  If you put a stick of zinc in a dilute acid, there will be a reaction because there is an electrochemical driving force for the conversion of Zn⁰ to Zn²⁺ and H⁺ to H₂ gas.  We can represent this driving force by a Gibbs energy change, which will have an associated equilibrium constant.  The reaction will proceed until an equilibrium is reached and the driving force is zero due to a balance in the concentrations of reactants and products.  I think you know all of this because it is simple equilibrium chemistry - the only difference is that the reaction is a redox reaction, and thus the thermodynamical quantities can be expressed in terms of simple redox potentials.  (And in a way, it's really just a superficial difference - ALL reactions are redistributions of electrical charge, more or less; the only difference here is that we can make the approximation of "whole units of charge" moving from "predominantly one nucleus to another", which allows us the convenient use of redox potentials to express pretty much the same thermodynamical equations in a different form.)

I see, so in Case 1, the presence of the electrode makes new reactions (involving electrons from the electrode) possible and thus a Gibbs' energy change drives forward the new equilibria towards the constants, creating a potential. In the case of zinc and dilute acid, we would have two equilibria, Zn  Zn²⁺ + 2e^- and 2H⁺ + 2e^-  H₂, both of them driven by a Gibbs' energy change that becomes possible when the electrode is placed in solution. We can do the kinetics calculations if we want, but what my books referred to in the OP was the fact that we usually use this to establish a potential difference between the electrode and solution in order to measure something thermodynamic (e.g. an equilibrium constant), rather than to calculate concentration-with-time values, which is the business of Case 2.

In scenario two, all that changes is that there is an external potential energy applied.  What this really translates into is an additional force that resists or reinforces the predominant (natural, you might say) driving force for the reaction.  Essentially what it does is change the equilibrium point to favor more the products or the reactants.  Remember: ΔG is a potential energy concept, which means it is therefore a force concept.  The Gibbs energy tells you how a reaction will behave because all systems try to minimize potential energy.  By adding an external potential energy, you are therefore adding an additional force, which will shift the reaction equilibrium accordingly - and always away from the "natural" equilibrium point.

OK, so let's say we've got the same 2 equilibria in solution: Zn  Zn²⁺ + 2e^- and 2H⁺ + 2e^-  H₂. Does my application of an external pd change the ΔG values for these reactions? If not, how does the pd then manage to affect these equilibria?

[Corribus]

@BD

In the case of zinc and dilute acid, we would have two equilibria, Zn  Zn²⁺ + 2e^- and 2H⁺ + 2e^-  H₂, both of them driven by a Gibbs' energy change that becomes possible when the electrode is placed in solution.

Well, you really only have one equibrium because those are half reactions.  (Neither one of those exists independently, really, because electrons typically don't just freely float around in solution.  The two reactions are concerted, in other words, and it makes no practical sense to separate them.

We can do the kinetics calculations if we want, but what my books referred to in the OP was the fact that we usually use this to establish a potential difference between the electrode and solution in order to measure something thermodynamic (e.g. an equilibrium constant), rather than to calculate concentration-with-time values, which is the business of Case 2.

"Establishing a potential difference between the electrode and solution" is identical to saying "Calculating ΔG for the reaction".  So yes, you are right to a point, although there is no reason you cannot determine kinetics of the reaction if you want.

OK, so let's say we've got the same 2 equilibria in solution: Zn  Zn²⁺ + 2e^- and 2H⁺ + 2e^-  H₂. Does my application of an external pd change the ΔG values for these reactions? If not, how does the pd then manage to affect these equilibria?

It doesn't change the inherent ΔG for the reaction, but the external potential is essentially added to the reaction ΔG to determine the new equilibrium point that is reached under the externally applied potential.  Think of it this way.  Let's say you have an electrochemical cell.  Current flows from one end to the other because there is a natural drive to obtain equilibrium.  Once equilibrium is reached, there is no more spontaneous current flow because the ΔG = 0.  In order to "recharge" the battery, you have to make current flow in the opposite direction.  This takes work because you are moving from an equilibrium point to a non-equilibrium point.  This work is the external potential you apply.  Effectively what it does is create a new, temporary equilibrium that makes it spontaneous for the electrons to flow back the other way.  One this new equilibrium is established, the external potential is removed, and therefore it is spontaneous once again for the electrons to flow in the original direction - provided there is contact between the two cells.  (The battery stores the energy until it is needed, usually by putting it in a flashlight or something, which completes a circuit, which allows the electrons to flow.)

[Big-Daddy]

Well, you really only have one equibrium because those are half reactions.  (Neither one of those exists independently, really, because electrons typically don't just freely float around in solution.  The two reactions are concerted, in other words, and it makes no practical sense to separate them.

Hmm. In my book (the same as the quote in the OP comes from) it says that a potential difference is given by Φ_M-Φ_S=constant-RT/F*log_e([Fe²⁺]/[Fe³⁺]) when an electrode (M) is placed in the solution, as the reaction Fe³⁺ (aq) + e^- (metal)  Fe²⁺ (aq) begins to proceed towards equilibrium. That would suggest that this equilibrium need not to be coupled with any other which is also ocurring?

"Establishing a potential difference between the electrode and solution" is identical to saying "Calculating DG for the reaction".  So yes, you are right to a point, although there is no reason you cannot determine kinetics of the reaction if you want.

It doesn't change the inherent DG for the reaction, but the external potential is essentially added to the reaction DG to determine the new equilibrium point that is reached under the externally applied potential.  Think of it this way.  Let's say you have an electrochemical cell.  Current flows from one end to the other because there is a natural drive to obtain equilibrium.  Once equilibrium is reached, there is no more spontaneous current flow because the ΔG = 0.  In order to "recharge" the battery, you have to make current flow in the opposite direction.  This takes work because you are moving from an equilibrium point to a non-equilibrium point.  This work is the external potential you apply.  Effectively what it does is create a new, temporary equilibrium that makes it spontaneous for the electrons to flow back the other way.  One this new equilibrium is established, the external potential is removed, and therefore it is spontaneous once again for the electrons to flow in the original direction - provided there is contact between the two cells.  (The battery stores the energy until it is needed, usually by putting it in a flashlight or something, which completes a circuit, which allows the electrons to flow.)

So the effective ΔG of the reaction in the solution changes depending on your externally applied pd. This in turn changes the equilibrium constant (ΔG° is constant of course).

So if effective ΔG=inherent ΔG+f(V), where f(V) is a function of the externally applied pd, then because f(V) is never permanently 0, effective ΔG is never permanently 0 and so the reaction keeps being driven on until there are no reactants left to convert to products. Is that correct?

[Corribus]

Hmm. In my book (the same as the quote in the OP comes from) it says that a potential difference is given by Φ_M-Φ_S=constant-RT/F*log_e([Fe²⁺]/[Fe³⁺]) when an electrode (M) is placed in the solution, as the reaction Fe³⁺ (aq) + e^- (metal)  Fe²⁺ (aq) begins to proceed towards equilibrium. That would suggest that this equilibrium need not to be coupled with any other which is also ocurring?

Well the electrons must go somewhere.  They do not just float about in solution.

So the effective ΔG of the reaction in the solution changes depending on your externally applied pd. This in turn changes the equilibrium constant (ΔG° is constant of course).

So if effective ΔG=inherent ΔG+f(V), where f(V) is a function of the externally applied pd, then because f(V) is never permanently 0, effective ΔG is never permanently 0 and so the reaction keeps being driven on until there are no reactants left to convert to products. Is that correct?

ΔG for the reaction and the external potential are opposing forces.  Let's assume you hold the external potential at a constant level.  The magnitude of ΔG is proportional, in a way, to the relative concentrations of products and reactants (actually, to the logarithm of Q, the reaction quotient with ΔG° as a reference value).  Once the external potential is applied, the magnitude of external potential initially exceeds that of ΔG and electrons will flow backwards in such as way to produce reactants (actually, a lower value of Q).  This will begin to decrease the value of ΔG.  Eventually you create so many reactants that the driving force to go in the "natural" direction becomes larger and larger, and eventually will match the reverse driving force created by the external potential.  This is a pseudo-equilibrium point because all the rates (the forward rate caused by ΔG [the "natural" thermodynamic rate] and the backward rate caused by the external potential are balanced, so the net effect is no change in either reactants or products.  What this means is that the reverse reaction keeps being driven until there are no reactants left to convert to products ONLY in the case that the pseudo-equilibrium constant species that this happens (that is, if Q can never equal the new pseudo-equilibrium constant).   I suppose it might be more accurate to say that the external potential is an amendment of ΔG° - a reaction reference point - not ΔG as I suggested before, because it actually creates a new equilibrium constant that ΔG needs to be compared to in order to determine the reaction direction.  I apologize that my post before was misleading because of that small but important detail.

I should clarify too that I don't deal often with electrochemistry and here only apply my knowledge of general thermodynamic principles to this problem.  Like everything else electrochemistry is a fundamentally thermodynamical matter but it's easy to miss things when applying thermodynamical principles to areas you haven't thought about in a long time.  Therefore this post and the few before represent my evolving thinking on the matter.  I may yet not be wholly correct, so please keep asking questions if you perceive there is still a problem in your understanding.  It might be a problem in your understanding, it might be a problem in the way I'm expressing my understanding, or it could be I've missed something critical in a hasty analysis of the question.

[Big-Daddy]

[Well the electrons must go somewhere.  They do not just float about in solution.

From what I understand, the electrons are either in the metal electrode itself, or they join onto an Fe³⁺ ion floating around the solution to make an Fe²⁺. But my book makes no mention of any other equilibria that are going on at the same time. It seems that if there is just 1 electrode, then there is just 1 half-reaction. But yeah I see what you're saying, I want to know where the electrons are going or coming from ...

ΔG for the reaction and the external potential are opposing forces.  Let's assume you hold the external potential at a constant level.  The magnitude of ΔG is proportional, in a way, to the relative concentrations of products and reactants (actually, to the logarithm of Q, the reaction quotient with ΔG° as a reference value).  Once the external potential is applied, the magnitude of external potential initially exceeds that of ΔG and electrons will flow backwards in such as way to produce reactants (actually, a lower value of Q).  This will begin to decrease the value of ΔG.  Eventually you create so many reactants that the driving force to go in the "natural" direction becomes larger and larger, and eventually will match the reverse driving force created by the external potential.  This is a pseudo-equilibrium point because all the rates (the forward rate caused by ΔG [the "natural" thermodynamic rate] and the backward rate caused by the external potential are balanced, so the net effect is no change in either reactants or products. I suppose it might be more accurate to say that the external potential is an amendment of ΔG° - a reaction reference point - not ΔG as I suggested before, because it actually creates a new equilibrium constant that ΔG needs to be compared to in order to determine the reaction direction.  I apologize that my post before was misleading because of that small but important detail.

Ah that mostly makes sense, thanks. Except for this bit:

What this means is that the reverse reaction keeps being driven until there are no reactants left to convert to products ONLY in the case that the pseudo-equilibrium constant species that this happens (that is, if Q can never equal the new pseudo-equilibrium constant).

I'm getting the idea that the external potential and Gibbs' free energy push in opposite directions so that eventually the rates of reaction caused by each are equal and this is a pseudo-equilibrium point. But in what case is this point not reached? Q is just going to move until it reaches the (pseudo)-equilibrium point, no? I ask about this specific case because I thought the business of electrolysis was to keep going until there are no reactants left to convert to products - or is it just an extremely high equilibrium constant?

[Corribus]

Re: the first part.  You can't just have a half reaction.  Electrons are always going somewhere.  Otherwise there is no driving force for the reduction or reduction.  Something has to be doing the reducing/oxidizing.  It can even be a proton in solution.  I'm not sure where your iron example is coming from.  Maybe that would help.

Re: the second part.  I'm sorry that was confusing.  I only mean this: there is always an equilibrium.  You can't ever convert 100% of the reactants in a reaction, because there is always some probability of back reaction, no matter how large the driving force is.  Of course, it may be so large that the back reaction is effectively nonexistent (such that it's impossible to measure) in which we can speak as though the reaction yields complete conversion.  We can also get almost 100% yields by continually removing a product, thereby continually shifting the equilibrium.

In this specific case, even a very high external potential will still eventually yield an equilibrium, no matter how large you make it.  Using a large enough potential you can effectively drive all products back to reactants, but it's never "complete", so to speak.  The bit about Q never equaling K was a sloppy way of saying if one reactant is gone, then no equilibrium can be obtained.  It's like if you have a reaction A + B  C, with K = 5.  If the starting concentration of B and C are zero, there will never be reaction, and therefore no equilibrium.  Trivial result, of course, so I shouldn't have mentioned it since it was obviously confusing.