Cu(s) vs Cu2+([H2O]6) | Who Can Explain?

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In summary: First of all, I feel I have posted a legitimate mainstream physics question fully compliant with PF terms & conditions, clinically devoid of any voodoo.I have initially categorized my question as "electrodynamics" related and posted it in "classical physics" because I'm interested in learning about the charge dynamics under the given conditions. My post was subsequently moved to "chemical engineering" by the administrators.Your question "Does current flow through an open circuit?" is what in my native language is called a 'charged question' no pun intended i.e. the premise for the question is not clear, but it is definitely there, bordering to entrapment. I'm not inferring that you are doing
  • #1
vis viva
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What happens in the boundary area between Cu(s) and Cu2+(a)? How does the various particles respond to each others presence?
Please behold my newest graphical masterpiece here below. The winner of this expensive artwork will be the first that can help me understand how the various particles respond to each others presence?

copper-what-happens.png

The Cu2+ is supposed to depict CuSO4 pentahydrate in water, so I guess the correct term would be [Cu(H2O)6]2+ if we are going to nitpick.
 
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  • #2
Nothing, as there's only one electrode. Get yourself another electrode, then things start to happen.
 
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  • #3
mjc123 said:
Nothing, as there's only one electrode. Get yourself another electrode, then things start to happen.

Really? Color me surprised and hit me in the face with a blunt instrument of your personal preference.

I would have thought at least one of two things would happen once the power is turned on:

1. The positive charges in the electrode will repel the positive charges in the aquatic solution
2. A neutrally charged boundary area will form at the electrode.
- or -
3. The passing negative charges (e.g. electrons) will attract the positive charges in the aquatic solution to the electrode.

And nothing of this would/could happen you say?
 
Last edited:
  • #4
@mjc123 's point is that all of the voltage is dropped across the resistor (unless for some reason the resistor has a higher resistance than the air around the beaker).
 
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  • #5
TeethWhitener said:
@mjc123 's point is that all of the voltage is dropped across the resistor (unless for some reason the resistor has a higher resistance than the air around the beaker).

I appreciate that. After further consideration I think I have provided too few details in my drawing, I should have inserted a switch between the power+ and the electrode/resistor net. This way it would have been clear that the circuit has two distinct states ON and OFF.

I sincerely apologize for my grave omission.
 
  • #6
vis viva said:
I appreciate that. After further consideration I think I have provided too few details in my drawing, I should have inserted a switch between the power+ and the electrode/resistor net. This way it would have been clear that the circuit has two distinct states ON and OFF.

I sincerely apologize for my grave omission.
Sarcasm is even more hilarious when it’s unintentionally clueless.

Imagine your resistor was gone, so that you just have a broken wire in its place. What would your circuit look like then? Keeping in mind that “circuit” comes from the same root as “circle.”
 
  • #7
The negative electrode don’t get no respect, I tell ya...
1584055884634.jpeg
 
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  • #8
TeethWhitener said:
Sarcasm is even more hilarious when it’s unintentionally clueless.

I always try to be curb my natural occurring sarcasm tendencies, because I feel that sarcasm is the lowest form of human communication. However I don't see where I have used sarcasm in this post, but I can understand that you were amused anyway, so all is not that bad.

TeethWhitener said:
Imagine your resistor was gone, so that you just have a broken wire in its place. What would your circuit look like then? Keeping in mind that “circuit” comes from the same root as “circle.”

To address your question first: If the resistor was gone, there would be no "circuit".

Now please tell me what you believe my lesson should be from the above , I'd rather not be making assumptions when I can simply ask you directly.
 
  • #9
chemisttree said:
The negative electrode don’t get no respect, I tell ya...

A facebook type meme.. OK, even though I always thought of PF more like this:
HatTip.jpg
 
  • #10
vis viva said:
To address your question first: If the resistor was gone, there would be no "circuit".
Yes. Does current flow through an open circuit?
 
  • #11
TeethWhitener said:
Yes. Does current flow through an open circuit?

First of all, I feel I have posted a legitimate mainstream physics question fully compliant with PF terms & conditions, clinically devoid of any voodoo.

I have initially categorized my question as "electrodynamics" related and posted it in "classical physics" because I'm interested in learning about the charge dynamics under the given conditions. My post was subsequently moved to "chemical engineering" by the administrators.

I appreciate that if you look at my question now as a "chemical engineering" question through acid-splatter-proof-safety-googles-of-the-highest-pay-grade, then you could be forgiven for assuming that my question is related to electrolysis, which it is absolutely not.

Your question "Does current flow through an open circuit?" is what in my native language is called a 'charged question' no pun intended i.e. the premise for the question is not clear, but it is definitely there, bordering to entrapment. I'm not inferring that you are doing so deliberately.

I would like to know the English term for a 'charged question' as this is not the first time I encounter one, nor will it be the last, so I would appreciate to learn the English equivalent if it exists because it would make me appear more eloquent and knowledgeable and elevate my social status tremendously.

Your premise is, and sorry for assuming your assumption, but otherwise this will become very lengthy very quickly, feel free to correct me: Any flow of charge requires a closed circuit.

The latter statement is not technically correct, it is however correct that: A CONTINUOUS flow of charge requires a CLOSED CIRCUIT.

That is the reason why I apologized for omitting the switch, because without it, it was not initially clear that my question was related to a definitive two state system with a transition state in between.

If I redraw the schematic based on "if the resistor was gone" it would look like this:
dipole-vessel.png

And that is nothing but a dipole antenna with one pole dipped into a vat of Cu2+

A power supply is supplying power by charge separation i.e. a power supply is nothing more than a charge pump that pumps negative charges from the positive pole to the negative pole.

When you turn on the power supply above, it will pump negative charges from the positive pole to the negative pole rendering the positive pole positively charged and the negative pole negatively charged.

Therefore the positive pole, which is positively charged, is surrounded by positive Cu2+ charges, and therefore my original question still stands unanswered: "What happens in the boundary area between Cu(s) and Cu2+(a)? How does the various particles respond to each others presence?"

If my above elaboration can not yield a intelligible answer, then apparently I do not know how to pose a simple physics question on PF that will yield an equally simple answer. And in that case I would request the assistance of a friendly mentor that can guide me prior to me posting anything else on PF.

Thank you for your kind attention.
 
  • #12
I'm trying to get you to answer your own question with the knowledge you already have. So I'll say this as nicely as I can: please lose the condescending attitude. If the next post is as ridiculous as the last two, someone else can help you, because I'm over it.

What I was trying to elicit was a simple "no." You can caveat it if you want, but the fact is that a DC open circuit will very quickly reach a steady state of zero current. This allows us to say something about the circuit's resistance (or conductance, if you like). Namely, that it is infinite. My hope was that continuing this line of inquiry would get you to see that replacing an infinite resistance resistor with a finite resistance resistor will cause all the voltage to drop across the finite resistance resistor.

vis viva said:
Your premise is, and sorry for assuming your assumption, but otherwise this will become very lengthy very quickly, feel free to correct me: Any flow of charge requires a closed circuit.

The latter statement is not technically correct, it is however correct that: A CONTINUOUS flow of charge requires a CLOSED CIRCUIT.
If you want to think about it this way, then the really accurate statement is that any charge flow requires a potential gradient. So, where is the potential gradient in your first diagram that would allow anything to happen at the copper electrode?
 
  • #13
@vis viva - your post is unhelpful. And it seems to me, you may be less knowledgeable than you believe.

The help you are getting is from a PhD chemist and other PhD folks like me. We are scientists and we are trying to help. Not criticize.
You have to realize that if you are an auto-didact you probably use terms and concepts in non-standard ways.

We have reasons, execellent reasons, why things are presented using defined terminology - so we are all on the same page. Please join us there.

Thanks for posting.
 
  • #14
We call questions of entrapment trick questions. The car driving over a hill problem that was given to Einstein supposedly was an example of a trick question with no solution.
 
  • #15
TeethWhitener said:
I'm trying to get you to answer your own question with the knowledge you already have.

Unfortunately that is not how it comes over, I would be more comfortable with questions rather than assumptions.

TeethWhitener said:
So I'll say this as nicely as I can: please lose the condescending attitude.

And I'll reciprocate your niceness; I'll drop whatever you want me to drop, but you will have to assist me, because I have no idea of where you feel I was out of line, some examples would be helpful. Are you sure you basically just can't palete the way I express myself, and regardless of what I say you will feel that I'm out of line? I'm asking a simple question here, I'm not deliberately trying to provoke you.

TeethWhitener said:
What I was trying to elicit was a simple "no." You can caveat it if you want, but the fact is that a DC open circuit will very quickly reach a steady state of zero current.

Why would I want to caveat that? Everything you say is correct under the lumped element model, but I don't want to use that model, I feel it is too inadequate for discussing fundamental particle dynamics.

TeethWhitener said:
If you want to think about it this way, then the really accurate statement is that any charge flow requires a potential gradient. So, where is the potential gradient in your first diagram that would allow anything to happen at the copper electrode?

You have to know that I initially wanted to use the 2nd diagram, but I was concerned that my question would drown in ridicule because I would present a open circuit so I went with the first which I now regret. But now that the cat is out of the bag I would like to use the 2nd diagram. If you can allow me to do that then I can explain why I think something should be happening.
 
  • #16
jim mcnamara said:
@vis viva - your post is unhelpful.

How and where exactly?

jim mcnamara said:
The help you are getting is from a PhD chemist and other PhD folks like me. We are scientists and we are trying to help. Not criticize.

I always appreciate when people are trying to help and I don't mind criticism. What I mind is assumptions.

jim mcnamara said:
You have to realize that if you are an auto-didact you probably use terms and concepts in non-standard ways.

I understand and realize fully that I occasionally use concepts in non-standard ways which draws attention away from the essence of my question. This is always very frustrating for me, and I understand and realize fully that the ultimate responsibility for being understood is mine alone.

jim mcnamara said:
We have reasons, execellent reasons, why things are presented using defined terminology - so we are all on the same page. Please join us there.

I know you have reasons, excellent reasons, and I'm not questioning any of your reasons. I too want to use defined terminology, I just don't want to use the lumped element model for this post/question, I want to use fundamental electrodynamics/statics.

jim mcnamara said:
Thanks for posting.

You're welcome.
 
  • #17
jedishrfu said:
We call questions of entrapment trick questions. The car driving over a hill problem that was given to Einstein supposedly was an example of a trick question with no solution.

Yeah, but a 'charged question' isn't necessarily malicious and 'trick' implies malice. Are there perhaps other suitable terms in the English language?
 
  • #18
vis viva said:
You have to know that I initially wanted to use the 2nd diagram, but I was concerned that my question would drown in ridicule because I would present a open circuit so I went with the first which I now regret. But now that the cat is out of the bag I would like to use the 2nd diagram. If you can allow me to do that then I can explain why I think something should be happening.
The second diagram is basically a capacitor. Something interesting is happening there because there is a potential difference between the electrode/electrolyte and the opposite electrode. The accumulation of positive charge on the immersed electrode will attract a negative countercharge at the surface.

This doesn’t happen in the first diagram because all of the voltage is dropped across the resistor (I.e., there is no externally applied potential at the electrode).

However, at a molecular level, when a metal is dipped in an electrolyte, there’s almost always some inherent potential difference that causes accumulation of charges at the metal/electrolyte interface (regardless of whether an external potential is applied). Google “electrical double layer” for more information about this phenomenon.
 
  • #19
vis viva said:
However I don't see where I have used sarcasm in this post
This:
vis viva said:
I sincerely apologize for my grave omission.
(unless it's just a language translation issue)

Anyway, the bottom line here is that you keep asking about how your incomplete circuit will behave, and you keep getting informed that it will not act like anything, since the circuit is incomplete. I'll see if I can find a basic circuits page to help clear it up for you.
 
  • #21
berkeman said:
This:
However I don't see where I have used sarcasm in this post
(unless it's just a language translation issue)

I gave a reason for my sincere apology. And I wrote the word "sincere" which I understand means "Being without hypocrisy or pretense; true." in the English language. I endeavour to choose my words with care, I don't consume or abuse words or use them as random filler, I respect words.

berkeman said:
Anyway, the bottom line here is that you keep asking about how your incomplete circuit will behave, and you keep getting informed that it will not act like anything, since the circuit is incomplete. I'll see if I can find a basic circuits page to help clear it up for you.

That's not the bottom line.
 
  • #22
vis viva said:
That's not the bottom line.
Okay, then please help me out. What is the bottom line?
 
  • #23
TeethWhitener said:
The second diagram is basically a capacitor. Something interesting is happening there because there is a potential difference between the electrode/electrolyte and the opposite electrode. The accumulation of positive charge on the immersed electrode will attract a negative countercharge at the surface.

This doesn’t happen in the first diagram because all of the voltage is dropped across the resistor (I.e., there is no externally applied potential at the electrode).

However, at a molecular level, when a metal is dipped in an electrolyte, there’s almost always some inherent potential difference that causes accumulation of charges at the metal/electrolyte interface (regardless of whether an external potential is applied). Google “electrical double layer” for more information about this phenomenon.

First I would like to sincerely thank you for the above reply, that is exactly what I was wishing for, and it would be an excellent foundation for further discussion. There is so many things I wish to learn about this subject. And I'm sorry for the things we had to go through to come to this point and understanding.

However due to personal health concerns I need to withdraw myself from further discussions.
 
  • #24
Ok stay safe.
 
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Related to Cu(s) vs Cu2+([H2O]6) | Who Can Explain?

1. What is the difference between Cu(s) and Cu2+([H2O]6)?

Cu(s) refers to solid copper, while Cu2+([H2O]6) refers to the hydrated copper ion, which is a positively charged ion surrounded by six water molecules.

2. How does Cu(s) and Cu2+([H2O]6) behave in a chemical reaction?

Cu(s) is a reducing agent, meaning it loses electrons in a chemical reaction, while Cu2+([H2O]6) is an oxidizing agent, meaning it gains electrons in a chemical reaction.

3. What is the significance of the subscript in Cu2+([H2O]6)?

The subscript indicates the number of water molecules surrounding the Cu2+ ion. In this case, there are six water molecules attached to each Cu2+ ion.

4. Can you explain the role of Cu(s) and Cu2+([H2O]6) in electrochemical cells?

Cu(s) is used as the anode (negative electrode) in electrochemical cells, while Cu2+([H2O]6) is used as the cathode (positive electrode). This allows for the flow of electrons and the production of electricity.

5. How do Cu(s) and Cu2+([H2O]6) affect the color of a solution?

Cu(s) is a reddish-brown color, while Cu2+([H2O]6) is a blue-green color. When dissolved in a solution, the presence of either Cu(s) or Cu2+([H2O]6) can change the color of the solution, depending on the concentration and other factors.

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