Electrochemical cell

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cepheid
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I have a problem that in my physics textbook, a source of EMF is just shown as a box having + and - terminals. The inner workings of that box need not be considered. They claim that although there is obviously an electrostatic force Fe due to this separation of charge, it is countered by a force F specific to the source that maintains the potential difference between the terminals, even in the open circuit case. Without F, charge would simply move between the terminals (accelerated by Fe), and the potential difference would be eliminated. My question is, isn't this claim that the potential difference between the terminals is maintained in an open circuit situation false in the case of an electrochemical cell?

Here's why I think so. In an open circuit situation (battery is not in use and not connected to anything), the redox reaction can't occur in the first place because the anode is not physically in contact with the oxidising agent (the cathode) so what the hell could possibly be oxidising it? Hence, neither the + nor the - terminal is actually charged, so the potential difference is zero. If I'm wrong, can you please explain why? It doesn't make sense to me that we can have a sustained redox reaction occurring in a battery that's just sitting there.
 

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GCT
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It doesn't make sense to me that we can have a sustained redox reaction occurring in a battery that's just sitting there.
Redox reaction does not occur when the system is closed. There's still a difference in electric potential, which can be partly due, in some cases, simply...to the differences in oxidation states. It's somewhat similar to the situation as if you were to have two separated conductors/insulators with different charge states, a potential difference exists between the two, you connect them with a conducting wire, and electron transfer will occur. Note that redox reaction is essentially a transfer of electrons; an chemical agent is oxidized (loses electrons), while the other is reduced (gains electrons).
 
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cepheid
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Thanks for the reply.

I'm not sure whether this means my reasoning is flawed, correct, or just not applicable. Could you clarify what you meant by, "when the system is closed?"

Much appreciated...
 
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cepheid
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I'm also still not clear on why there would be a potential difference between the anode and cathode if no reaction is occurring. It's been a while since I ran across the term "oxidation states", so if someone could clear up that, and what I asked about just above, that would be great.
 
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GCT
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Your initial reasoning is somewhat erroneous. Oxidation states take into account, the formal charge in addition to each element's electronegativity (I'll post the exact definition later, if I have the time). Formal charge pertains to whether the atom is electron deficient...basically whether it possess a overall positive charge/negative charge.

The redox reactions do not create the potential. The electric potential is there, potential for the system to do work by electrical means, note electric potential, and it is in electrostatic equilibrium. Potentials are set up by charges, charges create electrical potential, in the case of an electrochemical cell, the anode and the cathode may have differences in the electrical potential. Electrical potential simply describes the electric field state.

For instance, take away the anode/cathode, which ever you prefer; each still possess a potential (under most conditions) independent of the cathode. It is when you connect this two cells, which have differences in potential, with a conducting wire where you'll actually have a redox reaction occur, which is simply a transfer of electrons, until the two reach equal voltage states (delta V =0).
 
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cepheid
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GeneralChemTutor said:
The redox reactions do not create the potential. The electric potential is there, potential for the system to do work by electrical means, note electric potential, and it is in electrostatic equilibrium. Potentials are set up by charges, charges create electrical potential, in the case of an electrochemical cell, the anode and the cathode may have differences in the electrical potential. Electrical potential simply describes the electric field state.
If a redox reaction is not occurring, as in the case of the closed system, then there is no charge, and hence no electric field. What does that leave us with? If, as you claim, a difference in potential can be achieved by a difference in oxidation states, and the formal charge is zero (both electrodes are electrically neutral), then the only thing that is left that could contribute to a difference in oxidation states is a difference in electronegativities (ie the fact that copper "wants" to lose its electrons more so than zinc, for example). But as to how this simple tendency to lose electrons on the part of the anode can actually lead to a difference in potential before it actually does lose those electrons, leading to a charge imbalance, is a mystery to me. Nevertheless, the potential is somehow there. It makes no sense to me, without a net charge. (I highlighted this because it is the crux of my confusion). I remember vaguely in grade 12 chem having a table that listed these "potentials due to electronegativity" quantitatively, in volts, with the symbol [itex] \mathcal{E} [/itex], so I guess I can't refute it. Each was measured relative to that of hydrogen, and the potential for the copper / zinc cell could be computed by subtracting the [itex] \mathcal{E}'s [/itex] for Cu and Zn, or something like that. I'm sorry for my persistence, I'm not trying to refute you, only to understand how this potential difference could arise. Thanks for your patience man.
 
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cepheid said:
Here's why I think so. In an open circuit situation (battery is not in use and not connected to anything), the redox reaction can't occur in the first place because the anode is not physically in contact with the oxidising agent (the cathode) so what the hell could possibly be oxidising it? Hence, neither the + nor the - terminal is actually charged, so the potential difference is zero. If I'm wrong, can you please explain why? It doesn't make sense to me that we can have a sustained redox reaction occurring in a battery that's just sitting there.
If you put piece of copper into the water it will dissolve - only a little bit, but some of the copper will change to ionic form and the electrode itself will get charged with with excess electrons. Second electrode will do the same - it will be just charged differently. You have two charged electrodes and that's the source of potential.

Now, both redox reactions are sustained - but it is a dynamical situation. On each electrode there are two reactions takin place - dissolution and deposition. Their speeds depend on the ions concentration and the potential of the electrode (remember, it is charged with excess electrons). Electrode is in the state of equilibrium so the speed of dissolution and deposition is the same - looking at the whole situation you see a 'sustained' reaction - but it is not 'freezed'.
 
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GCT
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If a redox reaction is not occurring, as in the case of the closed system, then there is no charge, and hence no electric field.
Nevertheless, the potential is somehow there. It makes no sense to me, without a net charge.
why of course there is.......observe the half reactions, in most cases you're dealing with solvated ions. Simply put a voltage is set up by a charge system, you can't argue with that. So we're dealing with charges here one way or another, I think that in order for you to understand this better, you'll need to refresh or study more in depth the concept of voltage (I'm actually on the chapter myself, physics). Unfortunately the concept of voltage or rather electrical potential is not so specific, in most cases it seems that a mathematical analysis is required even to determine an absence/existence of a voltage state or difference.

Some redox reactions pertain to ionization energies and electron affinities, where eletron affinity = -second ionization energy (I'll have to read up a bit on this later). If you can imagine the case where a metal of a particular ionization energies gets sufficiently close to an nonmetal with a particular electron affinity, a redox reaction will occur. Why? Well, it's a bit complicating overall, one will need to consider the whole picture (e.g. Mg2+ ions form due to the overall stability of an ionic compound). We can make a guess that the transfer of electrons (where one becomes a cation and the other an anion) occurs in general due to differences in charge states and distribution (I'm sure there are others ways for a redox reaction to occur). The tendency of an electron to relocate, despite the pull of the cationic component, indicates a difference in electrical potential. The atoms themselves establish the electrical potentials.

Although redox reactions can occur through other means, for instance through a conducting solution, another way to occur despite significant separation is by connecting separated, often identical solutions with a conducting wire (basically the same situation as if there were to be a redox reaction in a particular solution).

Redox reactions = transfer of electrons, transfer of electrons does not occur without an initial potential difference, electrons cannot simply translocate through air (with particular conditions they can).

I'll try finding a nice link detailing redox reactions.
 

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