Negative ions affecting galvanic cell voltage

[Biker]

I saw this more related to physics than chemistry.

So I have been trying to model how the voltage is generated in galvanic. I couldn't find any online and I thought it would be a good exercise.

I modeled ions as electric charges with different magnitude of charge. Each producing a different electric field depending on the nature of its components. For example Cu+2 has a strong electric field than Zn+2 which would explain the voltage.

Now I started to think about these electrical charges affect each other and said that there must be effect on each other because on such small scales the ions can do that. This would explain the higher concentration we have the higher the voltage is. All this interference and ions moving around, Somehow they behave like a terminal so that the voltage is constant.

Then I faced a problem, It seems that in questions they don't care about the effect of negative charges and their electric field. As though it wasn't even there. My question why is that? When they give us that the "Standard" voltage of Cu-Zn cell is 1.1 volts does that mean this is purely because of the interaction between positive ions?

Perhaps someone can set me straight and show the mistakes in my model.

 

[Borek]

It is more about what happens on the electrode where the ions interact with the surface than about interactions between ions in the solution.

 

[Dale]

For example Cu+2 has a strong electric field than Zn+2

Hmm, that doesn't seem right. They both have the same charge so by Gauss' law they should have the same field

 

[Biker]

It is more about what happens on the electrode where the ions interact with the surface than about interactions between ions in the solution.

Hmm, that doesn't seem right. They both have the same charge so by Gauss' law they should have the same field

Oh so it is not related to what I said?

Borek, Could you please explain your sentence thoroughly? I can't see how interactions with the electrode surface rather than differences (Which is wrong as dale pointed out). So If I use Graphite as an electrode and Cu+2 1 Molar solution, Would that change the voltage? Because it interacts differently with the electrode?
If that is true then it settles that my interpretation was wrong.

 

[Biker]

Hmm, that doesn't seem right. They both have the same charge so by Gauss' law they should have the same field

I am kind of familiar with Gauss' law only the basics though. But what about charge distribution? Why do we assume that the charge is in the center of the ion and it is symmetrical? If it is not can't be there differences in voltage between two ions?

 

[Borek]

When you put the metal electrode into the solution some of the atoms on the surface will dissolve becoming ions. The electrode will become electron rich and it gives it a bit of a potential. When you put the electrode into a solution containing metal ions, depending on their concentration there will be either some reduction or dissolution, charging the electrode (and - again - giving it a potential).

This is a bit more complicated when you use an inert electrode, as this simple explanation doesn't work then.

 

[Biker]

When you put the metal electrode into the solution some of the atoms on the surface will dissolve becoming ions. The electrode will become electron rich and it gives it a bit of a potential. When you put the electrode into a solution containing metal ions, depending on their concentration there will be either some reduction or dissolution, charging the electrode (and - again - giving it a potential).

This is a bit more complicated when you use an inert electrode, as this simple explanation doesn't work then.

Thank you for your answer, So by how much does the inert electrode affect? and back to the original question the answer would be because the ions dissolving from the metal electrode is what causing the voltage not the ions in the solution.

 

[Borek]

Hmm, that doesn't seem right. They both have the same charge so by Gauss' law they should have the same field

We can get closer to the smaller ion than to the large ion, so the maximum field is different. But I don't think it matters here.

 

[Biker]

We can get closer to the smaller ion than to the large ion, so the maximum field is different. But I don't think it matters here.

Moreever, The distribution of the charge is different. So there should be voltage between them. That is what I based my thinking on and the greater the concentration the greater this voltage is by somehow ions affecting each other.

But yea apparently that is wrong, Anyway. One more question, It wouldn't matter if I for example use CuSO4 or Cu(NO3)2. Depending on the idea of interaction with electrode that wouldn't affect.. Not entirely comfortable with the idea but have to go with it.

Is there is any experimental chart I could look at for this?

 

[Borek]

Thank you for your answer, So by how much does the inert electrode affect? and back to the original question the answer would be because the ions dissolving from the metal electrode is what causing the voltage not the ions in the solution.

TBH I can't think of a simple explanation of how the inert electrode works.

There is a nice analogy though, one that stems from the method used sometimes in the redox potential calculation. It happens that they can be easier if we assume presence of hydrated electrons. Voltage is then directly proportional to the log log of their concentration. Not that difficult imagining that such electrons do interact with the electrode surface charging it.

Trick is, while hydrated electrons are commonly present in some solvents, like ammonia, they are rather unheard of in water solutions (or at least difficult to observe, which suggests their concentration to be much lower than required for this model to work).

 

[Borek]

The distribution of the charge is different.

But it is symmetrical, so when you look from the outside it doesn't matter, that's how the Gauss law works here.

It wouldn't matter if I for example use CuSO4 or Cu(NO3)2. Depending on the idea of interaction with electrode that wouldn't affect

The only thing that matters is the concentration of the Cu²⁺, counterion doesn't matter (unless it somehow reacts with the ion, complexing it or something).

 

[Biker]

But it is symmetrical, so when you look from the outside it doesn't matter, that's how the Gauss law works here.The only thing that matters is the concentration of the Cu²⁺, counterion doesn't matter (unless it somehow reacts with the ion, complexing it or something).

Isn't that when you look at it for a kind of far distance? But if you get down to nano scales then you should take care of distribution. Wont it greatly affect the field?
Is there is a experimental chart I could look at? Just to convince myself that it doesn't. Without doubt I take your word. Just for more convincing argument so that I don't ask about it again

 

[Biker]

One last question, Why when we have a question about electrolysis with inert electrode we uses the reduction table to calculate the minimum voltage needed. Even though isn't that table made for ions with their metal electrode?

 

[Borek]

Isn't that when you look at it for a kind of far distance? But if you get down to nano scales then you should take care of distribution. Wont it greatly affect the field?

This is tricky. When it comes to an isolated ion distribution of charge is symmetrical. When it becomes surrounded by water molecules it is not, as these molecules create an external field, but then we can't get close enough to observe these differences.

Is there is a experimental chart I could look at?

Sorry, you have lost me. Experimental chart of what?

Why when we have a question about electrolysis with inert electrode we uses the reduction table to calculate the minimum voltage needed. Even though isn't that table made for ions with their metal electrode?

Just reverse the thinking we started with when we put an electrode into the solution. When you put an inert electrode in the solution some of the ions present get reduced depositing first atoms of the metal on the surface, from this moment on the situation is identical.

I know it is a bit handwavy, I have no doubt there do exist better explanations/models. This approach works for me.

 

[Biker]

This is tricky. When it comes to an isolated ion distribution of charge is symmetrical. When it becomes surrounded by water molecules it is not, as these molecules create an external field, but then we can't get close enough to observe these differences.
Sorry, you have lost me. Experimental chart of what?
Just reverse the thinking we started with when we put an electrode into the solution. When you put an inert electrode in the solution some of the ions present get reduced depositing first atoms of the metal on the surface, from this moment on the situation is identical.

I know it is a bit handwavy, I have no doubt there do exist better explanations/models. This approach works for me.

1) But electrons get very close. I am not supporting of course the model I created but it seems though on those tiny scales that electrons reach, those differences matter

2) Experimental chart of difference solutions but the same positive ion and using them to create cells. To just see the effect.

3) I guess that works with me too. The problem that bothers me is that we create a lot of advanced stuff using batteries: mobile phones, laptops...etc but we don't get taught a good model about how this works and I have to be satisfied with The word interaction with electrode makes voltage and concentration affect this interaction by pushing the dissolution of the atoms of electrodes left or right

 

[Borek]

But electrons get very close. I am not supporting of course the model I created but it seems though on those tiny scales that electrons reach, those differences matter

On the atom scale single electron as a point charge doesn't matter, what matters is a charge distribution.

Experimental chart of difference solutions but the same positive ion and using them to create cells. To just see the effect.

I take you mean "different solutions". Well, no need for the experimental data if you can calculate potentials using Nernst equation - and it will be exactly the same number.

I guess that works with me too. The problem that bothers me is that we create a lot of advanced stuff using batteries: mobile phones, laptops...etc but we don't get taught a good model about how this works and I have to be satisfied with The word interaction with electrode makes voltage and concentration affect this interaction by pushing the dissolution of the atoms of electrodes left or right

I think we have several quite good models, you are just at the bottom of the cliff and you still don't see how it all works.

Note that the real model must take into account zillion of details - ions are almost never isolated, they are solvated (or hydrated in water), and the model has to account for solvation thermodynamics and kinetics. Then there is a double layer on the electrode surface, which changes how the electrode behaves. Then there are transport phenomena, which define how the battery behaves under load. Then there complexation and ion pair creation issues, which influence every other part of the battery inner workings. And that's still not all. Initially we ignore most of these effects, sadly, when we want to get into an even simplified model they start to become important and the simple model doesn't want to work without taking more advanced stuff into account.

 

[Biker]

Well, I checked Nernest equation. It obviously going to give me the same result. It doesn't have dependence on Negative ion concentration.

Anyway, Yes Indeed. I am still in high school but thought I could give it try. I searched a lot of websites for a good model but I couldn't find and by what you said there are still a lot to be learned.

I guess I have to stick to this explanation even though I don't understand the things behind it. Thank you Borek.