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How does mobile ions in a solution(electrolyte) conduct electricity?

by nyrychvantel
Tags: charge, conduct, electric, ion, mobile
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nyrychvantel
#1
Sep15-08, 02:12 AM
P: 13
I often read about how electrolyte conduct electricity, the websites I searched often explain only in terms of mobile ions are able to move freely in the solution, therefore they could carry charges throughout the solution bla bla bla. However, none of them explain how does that help conduct electricity.

Also, I read that electrons are not responsible for the conductivity of eletric charges in a solution, but the ions are, and involve some redox reactions. Again, they do not explain in great detail.

I am hoping some knowledgable guys here could explain in a more detailed scenario of how actually does ion capable of conducting electricity.

Thank you!
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Borek
#2
Sep15-08, 03:02 AM
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To transfer electricity through solution you have to move electrons through the phase boundary (that's where the redox systems come in handy) and then transport charge through the solution (that's where moving ions play its role).

Hard to help further if what you basically state is "I know it all, but I don't get it all".

Do you know how is electricity transferred in the wire? Electrons are freely moving through the metal. How does it differ from ions freely moving in the solution?
abcd8989
#3
Apr11-10, 11:27 PM
P: 43
I also shares exactly the same misunderstand with you. Can any guys explain in further detail?

Borek
#4
Apr12-10, 02:43 AM
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How does mobile ions in a solution(electrolyte) conduct electricity?

Yes, but not before you explain in details what you don't understand. Problem is, there are infinite ways to misunderstand the physical reality, but there is only one correct way this process goes. We have to know where the confusion is before trying to help.
abcd8989
#5
Apr12-10, 07:47 AM
P: 43
Quote Quote by Borek View Post
Yes, but not before you explain in details what you don't understand. Problem is, there are infinite ways to misunderstand the physical reality, but there is only one correct way this process goes. We have to know where the confusion is before trying to help.
Assume now we use conc. NaCl(aq) as an electrolyte , and a metal couple, Mg and Cu, is dipped into it to form a cell (a simple cell without salt bridge). So , Mg gives out e- and flows to Cu. It is the H+ in the electrolyte solution which is going to be discharged, but not Na+ , as Na+ is very stable and not react easily. But then, why do we need Na+? If Na+ more or less remains unreacted, not receiving elections, how does Na+ actually help the conduction of electricity, apart from its moving freely in the solution? Does moving freely in the solution is a factor for the conduction of electricity? If deionized water is used instead of brine, can electricity be produced? The amount of H+ in DI water is similar to that of brine, right? Or that the presence of sodium and chloride ions lead to an increase in the amount of H+? Though there is no Cl- ions in DI water, I think the OH- can be discharged instead, right? So, it turns to be pure water can also conduct electricity. I would like to know which of my concepts are wrong, leading to this absurd conclusion...thanks!
Borek
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Apr12-10, 10:47 AM
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To make things clear - you mean Mg and Cu electrodes connected outside of the solution?

Na+ (and Cl-) are there just to speed up the process. If they are not there, everything will happen the same way, just much more slowly.

Note that H+ that gets reduced doesn't have to go through the whole solution. As long as the reaction is slow enough there is enough H+ in the thin layer close to the copper electrode for the reaction to proceed. But we need a closed circuit and something (ions) must move through the bulk of the current to flow and for the reaction to proceed. When there are many ions, this process is much easier and much faster.
abcd8989
#7
Apr12-10, 08:02 PM
P: 43
Quote Quote by Borek View Post
To make things clear - you mean Mg and Cu electrodes connected outside of the solution?

Na+ (and Cl-) are there just to speed up the process. If they are not there, everything will happen the same way, just much more slowly.

Note that H+ that gets reduced doesn't have to go through the whole solution. As long as the reaction is slow enough there is enough H+ in the thin layer close to the copper electrode for the reaction to proceed. But we need a closed circuit and something (ions) must move through the bulk of the current to flow and for the reaction to proceed. When there are many ions, this process is much easier and much faster.

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As you mentioned, "When there are many ions, this process is much easier and much faster,"
do you mean that the movement of Na+ (and Cl-) can speed up the redox reaction? But I just wonder why even most Na+ actually are involved not in the redox reaction (due to its chemical stability), they can still help to speed up the reaction(the reduction of H+)? What is the principle behind?
Borek
#8
Apr13-10, 02:41 AM
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They facilitate the speed at which charge in transferred in the bulk of the solution, nothing else. They will never reach electrode surface and they will never react, however, they slightly shift - and that means current flows.
abcd8989
#9
Apr13-10, 03:03 AM
P: 43
Quote Quote by Borek View Post
They facilitate the speed at which charge in transferred in the bulk of the solution, nothing else. They will never reach electrode surface and they will never react, however, they slightly shift - and that means current flows.

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ur.. please let me ask in deep, "They facilitate the speed at which charge in transferred in the bulk of the solution, nothing else.", but how?

Well, I just found a segment in Wikipedia explaining this matter:
For example, in a solution of ordinary salt (sodium chloride, NaCl) in water, the cathode reaction will be

2H2O + 2e− → 2OH− + H2

and hydrogen gas will bubble up; the anode reaction is

2NaCl → 2 Na+ + Cl2

and chlorine gas will be liberated. The positively charged sodium ions Na+ will react towards the cathode neutralizing the negative charge of OH− there, and the negatively charged oxide ions O2− will react towards the anode neutralizing the positive charge of H+ there. Without the ions from the electrolyte, the charges around the electrode would slow down continued electron flow; diffusion of H+ and OH− through water to the other electrode takes longer than movement of the much more prevalent salt ions.

However, how do the Na+ "react towards" the cathode? There shouldn't be any chemical reactions, right?
Borek
#10
Apr13-10, 03:16 AM
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Remember - current is just a charge transfer. If you move an ion from one place to another, you have current flowing thrugh the solution. The more free flowing ions, the easier the charge transfer is. In pure water there are only minute amounts of ions (from water autodissociation), so such a water has very high resistivity. That means very small currents and very slow charge transfer. In solution of NaCl there are many ions, they can move freely, so in the bulk of the solution charge can be tranferred easily - and fast.

Rmemeber we are talking about charge transfer by ions that will never react on the electrode. All they do is they are transferring charge by moving. Ions that will react were already close to the electrode and they are from water autodissociation.
abcd8989
#11
Apr13-10, 03:27 AM
P: 43
Quote Quote by Borek View Post
Remember - current is just a charge transfer. If you move an ion from one place to another, you have current flowing thrugh the solution. The more free flowing ions, the easier the charge transfer is. In pure water there are only minute amounts of ions (from water autodissociation), so such a water has very high resistivity. That means very small currents and very slow charge transfer. In solution of NaCl there are many ions, they can move freely, so in the bulk of the solution charge can be tranferred easily - and fast.

Rmemeber we are talking about charge transfer by ions that will never react on the electrode. All they do is they are transferring charge by moving. Ions that will react were already close to the electrode and they are from water autodissociation.

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i get what you means, however, i'd like to know whether movement of the charged particles, i.e. Na+, can really increase the voltage of the cell. I think the voltage of the cell is limited by the amount of H+, since only H+ can be discharged, but most probably not Na+. The discharge process of H+ removes the electrons given by the metal, and remains the neutral charge of the solution. If there is less H+ to be discharged, the voltage should drop significantly, right? But how can Na+ affect the voltage then? I understand that there is current flowing when Na+ is moving, but do they really contribute to the voltage of the cell?
Borek
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Apr13-10, 03:41 AM
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Do you know what ohmic drop is?
abcd8989
#13
Apr13-10, 03:48 AM
P: 43
Quote Quote by Borek View Post
Do you know what ohmic drop is?

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sorry, i ve not learnt that before, is it related to my question?
Borek
#14
Apr13-10, 05:03 AM
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Yes, it is related. Do you know Kirchhoff's voltage law? It can be expressed as "sum of voltages from batteries and ohmic drops on resistances equals zero". Ohmic drop is iR - that is, current times resistance. The higher the resistivity and the current, the higher the ohmic drop is, hence the lower the potential in the rest of the loop (that's not exactly correct, but let's not be too nitpicky). That's exactly what happens here.
abcd8989
#15
Apr13-10, 05:32 AM
P: 43
Quote Quote by Borek View Post
Yes, it is related. Do you know Kirchhoff's voltage law? It can be expressed as "sum of voltages from batteries and ohmic drops on resistances equals zero". Ohmic drop is iR - that is, current times resistance. The higher the resistivity and the current, the higher the ohmic drop is, hence the lower the potential in the rest of the loop (that's not exactly correct, but let's not be too nitpicky). That's exactly what happens here.
do you mean the higher the resistivity of the solution, due to the lack of mobile ions, the lower the voltage that the cell can provide? I think I get it. However, here come another misunderstanding: How can the movement of the charged particles actually reduce the resistance of the solution? Yes, I do know that their movement, by definition, is current, but why shouldn't the resistance of the cell be related to the rate of the discharged of the H+ ions? If there is a lack of H+ ions, even there is a myraid of Na+, the resistance should still be extremely high, which, in other words, the movement of the Na+ / Cl- particles don't really contribute the the reduction of resistance. Shall I have any misconceptions regarding the above points? Please kindly retify them, thanks!
Borek
#16
Apr13-10, 06:22 AM
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Resistance of the cell is not related to the concentration of ions that react at electrode, but potential at which they react does. That's what Nernst equation describes.

To some extent you can argue that's the same - usually the lower the concentration, the higher the potential required for the reaction (ignoring potential signs/reaction type for the clarity of the argument), which can be explained as higher resistance of the "reaction" requiring higher potential for the same current to flow (ohmic drop again). But that's mainly a matter of semantics.
abcd8989
#17
Apr13-10, 09:44 PM
P: 43
Thank you for your explanation regarding resistance of cell. Notwithstanding, I have just found an extract from the Wikipedia regarding how NaCl increase the conductivity of the solution, which I still don't quite understand (mainly the bold statements as shown). Would you please also elaborate a bit?

Quote from:http://en.wikipedia.org/wiki/Electrolyte

Electrochemistry
Main article: electrolysis

When electrodes are placed in an electrolyte and a voltage is applied, the electrolyte will conduct electricity. Lone electrons normally cannot pass through the electrolyte; instead, a chemical reaction occurs at the cathode consuming electrons from the anode, and another reaction occurs at the anode producing electrons to be taken up by the cathode. As a result, a negative charge cloud develops in the electrolyte around the cathode, and a positive charge develops around the anode. The ions in the electrolyte move to neutralize these charges so that the reactions can continue and the electrons can keep flowing.

For example, in a solution of ordinary salt (sodium chloride, NaCl) in water, the cathode reaction will be

2H2O + 2e− → 2OH− + H2

and hydrogen gas will bubble up; the anode reaction is

2NaCl → 2 Na+ + Cl2

and chlorine gas will be liberated. The positively charged sodium ions Na+ will react towards the cathode neutralizing the negative charge of OH− there, and the negatively charged oxide ions O2− will react towards the anode neutralizing the positive charge of H+ there. Without the ions from the electrolyte, the charges around the electrode would slow down continued electron flow; diffusion of H+ and OH− through water to the other electrode takes longer than movement of the much more prevalent salt ions.
In other systems, the electrode reactions can involve the metals of the electrodes as well as the ions of the electrolyte.

It mentions that "The positively charged sodium ions Na+ will react towards the cathode neutralizing the negative charge of OH− there, and the negatively charged oxide ions O2− will react towards the anode neutralizing the positive charge of H+ there.". Concerning the first statement, I think there is not much OH- ions around the anode. It is because the dissociation of water molecule is slow in nature. There ought to be only little OH- ions around the cathode. So, relatively speaking, only little Na+ ions should be enough for "neutralizing the charge of OH-" (though I don't know how the neutralization process is). Hence, I deduce that the excess Na+ ions are of no use. However, the fact is that the more concentrated the NaCl(aq) is, the larger the voltage produced, provided that all other variables, such as temperature, metal couple used, distance of electrodes, are kept constant. It seems that the excess Na+ ions still have other functions apart from "neutralizing the charge of OH-", but what are they?
Borek
#18
Apr14-10, 02:34 AM
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Higher voltage can be explained by the ohmic drop, period.

I am not sure what wiki article author meant. My bet is that he is referring to the electroneutrality, and that "react" doesn't mean "chemical reaction". Reduction of water at the anode means buildup of OH- ions close to the surface, that means solution is no longer neutral and becomes charged, which can disrupt ion transport. But it also means negatively charged solution attracts cations, which will travel to neutralize the charge. The higher the concentration of NaCl, the faster and easier the process, that's all.

Thing that is confusing here is the fact that salt that helps in conducting electricity is not reacting at the electrodes. However, that also means we are only describing temporary situation, that is, if we would like to do long electrolysis, with large charge passing through the solution, our description will fail and we will observe different phenomena occuring. Not entirely different, but sooner or later reactions of our "inert" electrolyte or "inert" electrodes are inevitable. Could be you are trying to extend the model and that's source of your confusion.

Note: from what I remember dissociation of water is not so slow, if there are not many OH+ ions thats rather because neutralization reaction is incredibly fast.

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