Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: Electrolysis and Redox equations (Diagrams provided)  (Read 5808 times)

0 Members and 1 Guest are viewing this topic.

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Electrolysis and Redox equations (Diagrams provided)
« on: August 21, 2008, 10:38:15 PM »

Hello everyone,

I had some time today and drew this pic in paint to help explain my question. The questions are bit wordy but I'll really appreciate if you can answer atleast one of them  ;)  ;)  ;)  ;)  ;)



1. What is the purpose of the salt bridge?
    Is it because when electrons flow Cu side becomes positive and KI side becomes negative. To avoid this NO3- goes to the anode to try to regain the original negative charge and K+ vise versa. What happens if there was no salt bridge? Would the current flow both ways. Is the reason why this doesn't happen is because only Cu produces electrons.So electrons must flow from anode to cathode. Can reactants flow through. I don't think so because salt bridge is not a tube. It is paper in most cases isn't it.

2. How does this really work?
    Do the electrons from Cu go to the Cu electrode and then through the wire to the carbon electrode. Then Iodine recieves these electrons. Is this what happens.

3. My dislike towards reduction
   Ok the reaction in KI is I2 + 2e- -------- 2I- .  In the KI flask there is only I- present.  However the equation implies I2 is originally present in the flask. Is it just the nature of the equation. If what happens is I2 gains electrons. Where is I2 to begin with there is only I- in the flask.

4. I thought oxidation and reduction can not happen separately. I mean in all redox equations balanced there have been 2 half equations. One for oxidation and one for reduction. I thought for oxidation to occur reduction must also occur and reverse. Also when we use electrochemical series to work out the cathode and anode we treat this as though oxidant and reductant are present in the same flask. Can oxidation and reduction happen separately?

Thank you  :)  :)  :)  :)  :)


   


Logged

cliverlong

  • Full Member
  • ****
  • Mole Snacks: +60/-14
  • Offline Offline
  • Posts: 611
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #1 on: August 22, 2008, 03:34:33 AM »

Hello everyone,

I had some time today and drew this pic in paint to help explain my question. The questions are bit wordy but I'll really appreciate if you can answer atleast one of them  ;)  ;)  ;)  ;)  ;)

Good diagram
Quote

1. What is the purpose of the salt bridge?
    Is it because when electrons flow Cu side becomes positive and KI side becomes negative. To avoid this NO3- goes to the anode to try to regain the original negative charge and K+ vise versa. What happens if there was no salt bridge? Would the current flow both ways. Is the reason why this doesn't happen is because only Cu produces electrons.So electrons must flow from anode to cathode. Can reactants flow through. I don't think so because salt bridge is not a tube. It is paper in most cases isn't it.
I would state it a bit different to that (but similar basic ideas)
An electric current flows because of a potential difference across a conductor. The set up of the cell creates the p.d. (more on that later)
An electric current requires a charge carrier in the medium
In the metal wire the charge carrier is only electrons
In the solutions any mobile, dissolved ion can carry charge and will physically move to or from the electrodes depending on charge
So as well ions in solution, ions are needed in the salt bridge to complete the circuit.

Now, I have never actually seen a description of what actually happens at the interface between the salt bridge and the two solutions. I have thought about it and found nothing - so I forget about it until someone asks about the salt bridge. Sorry can't help on that bit
Quote
2. How does this really work?
    Do the electrons from Cu go to the Cu electrode and then through the wire to the carbon electrode. Then Iodine recieves these electrons. Is this what happens.
The diagram in the following post
http://www.chemicalforums.com/index.php?topic=26468.msg99853#msg99853

Gives a really good image of what is happening at the electrode/solution boundary - basically what you are describing goes on
Quote
3. My dislike towards reduction
   Ok the reaction in KI is I2 + 2e- -------- 2I- .  In the KI flask there is only I- present.  However the equation implies I2 is originally present in the flask. Is it just the nature of the equation. If what happens is I2 gains electrons. Where is I2 to begin with there is only I- in the flask.
This is interesting. As you say for this reaction to occur there needs to be a source of I2 - which there isn't in this situation.
But is this the only reaction that can take place?

Another possible would be K+ + e- > K(s)

But I guess this deposition of potassium would not take place because the equlibrium would lie well over the the left due to the relatively high, positive electrode potential of K+ | K.

What about other ions present in solution? Could something else in the KI solution be reacting (reducing) at the cathode? How could they behave and what are the electrode potentials?

Is the carbon significant? Is it just conducting electricity or reacting? I'm not sure. I did read something once about using carbon electrodes in a cell (and I think that's what is in standard batteries) but I can't find the link

Most cell setups I have studied have the electrode being the same material as the +ve ion in the salt solution. I suppose this makes the discussion and voltage calculation easier.
Quote
4. I thought oxidation and reduction can not happen separately. I mean in all redox equations balanced there have been 2 half equations. One for oxidation and one for reduction. I thought for oxidation to occur reduction must also occur and reverse.
Yes. Yes. Yes.
Quote
Also when we use electrochemical series to work out the cathode and anode we treat this as though oxidant and reductant are present in the same flask.
You have the key to the answer in your question.
The cell is set up in such a way to separate the redox two processes into two flasks. The setup uses the pd to drive a current through the wire and hence produce a usable and useful electric current when you complete the circuit. The whole purpose of doing this activity (apart from setting electrochemistry questions)

Indeed, as you write, if you mixed the reactants in the same container, the reactions would occur, but you wouldn't derive any usable electric current
Quote
Thank you  :)  :)  :)  :)  :)
Logged

Astrokel

  • Full Member
  • ****
  • Mole Snacks: +65/-10
  • Offline Offline
  • Gender: Male
  • Posts: 989
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #2 on: August 22, 2008, 07:40:57 AM »

hey sam,  ;D ;D

Quote
1. What is the purpose of the salt bridge?

Your ideas are pretty correct. The functions of salt bridge are

1) It allows ions to flow between the two half-cells so as to help to maintain the electrical neutrality in each half-cell. (as cliverlong mentioned)
2) It allows electrical contact between the solutions.
3) It prevents mixing of the solutions

In your example, the release of Cu2+ ions into the Cu(NO3)2 solution causes a surplus of Cu2+ cations, hence upsetting the electrical neutrality of the solution. Hence the anions NO3- from the salt bridge would be released into the Cu(NO3)2 solution to balance the surplus of Cu2+ ions. Same principle applies to the other half cell.

Quote
Where is I2 to begin with there is only I- in the flask

This is actually a very good question, i'll just give my two cents worth which could be incorrect.. If you already have noticed it, the iodine electrode cannot be iodine because it is a solid under rtp. I would think that the electrode solution should be 'compatible' to both Iodine and iodide. A good solution would be KI, since  I2 + I-   <---> I3- (brown) at equilibrium. Perhaps experts on this site could give the better or right answer.
Logged
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #3 on: August 22, 2008, 05:54:38 PM »

Thanks for coming for the rescue as usual clive long and astrokel ;D

You guys are doing a great job here  ;)  ;)  ;)  ;)  ;)

Quote
I would state it a bit different to that (but similar basic ideas)
An electric current flows because of a potential difference across a conductor. The set up of the cell creates the p.d. (more on that later)
An electric current requires a charge carrier in the medium
In the metal wire the charge carrier is only electrons
In the solutions any mobile, dissolved ion can carry charge and will physically move to or from the electrodes depending on charge
So as well ions in solution, ions are needed in the salt bridge to complete the circuit.

Now, I have never actually seen a description of what actually happens at the interface between the salt bridge and the two solutions. I have thought about it and found nothing - so I forget about it until someone asks about the salt bridge. Sorry can't help on that bit

So is it the potential difference (one side postive and one side negative) that makes the electrons flow (just to clarify). From your explanations it got to be the P.E as you have emphasized. So if there is no salt bridge and polarity of the sides changes would the electrons  flow the other way. Is the reason why this doesn't happen is electron can only from anode to cathode because only anode produces electrons. This question is bit stuck in me head. Thanks a lot for your help I understood everything else really well.  ;)

hey sam,  ;D ;D
Quote
1. What is the purpose of the salt bridge?
1) It allows ions to flow between the two half-cells so as to help to maintain the electrical neutrality in each half-cell. (as cliverlong mentioned)

When you say electrical neutrality do you mean no charge or to maintain the orginal polarity. Anode negative and cathode positive. I mean you don't want zero charge right. I think I'm just hooked up in the word but just to clarify. Thanks again for your help it was very helpful  ;)

Quote
You have the key to the answer in your question.
The cell is set up in such a way to separate the redox two processes into two flasks. The setup uses the pd to drive a current through the wire and hence produce a usable and useful electric current when you complete the circuit. The whole purpose of doing this activity (apart from setting electrochemistry questions)

So basically the reaction takes place but this is a clever arrangement that makes the reaction occur with creating current as well.

With that I2 equation I'll ask the teacher and see.

Great work guys. Thanks again  ;)


Logged

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #4 on: August 22, 2008, 09:06:07 PM »

Too late to modify  :-\

I was thinking too much again and went blank a bit again with the salt bridge  ;)

This is what I understood. Please tell me if each statement is right

*To begin with both flasks have neutral charge. Flask 1 (Cu 2+ and NO3-) making it neutral. Flask 2 (K+ and I-) making it neutral.
*So no reaction can occur because electrons don't have a potential difference to get to the other side.
*When you add the salt bridge flask 1 becomes negative due to extra NO3- and flask 2 positive due to extra K+
*Now there is a potential difference and a path for electrons to flow hence oxidation and reduction can occur.
*So is it fair to say that whole purpose of the salt bridge is to create the  polarity in each flask.
*So no salt bridge no reaction to begin with.

Is what I have said right or can a reaction happen atleast for a little while without a salt bridge. I mean lets say a reaction occur

Cu ------ Cu2+ + 2e-

The charge is still neutral isn't it. So only way to create a negative charge is to add more anions.

So to summarise the salt bridge is required to start the reaction and also when the reaction proceeds because the electrons are flowing flask 1 becomes more positive and to maintain the negative charge it is required as well.

So it is required to start and when the reaction proceeds as well.

Very interesting how the ions in the salt bridge knows which side it must go to  ;)




Logged

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #5 on: August 22, 2008, 10:03:26 PM »

Another question too late to modify again!!

Why should the electrolyte solution must be the same as the electrode. I mean in the case of copper why can't the electrolyte be something else. It is only copper solid that is taking part in the reaction. What happens if you only leave the cu electrode. Would a reaction still occur?
Logged

Astrokel

  • Full Member
  • ****
  • Mole Snacks: +65/-10
  • Offline Offline
  • Gender: Male
  • Posts: 989
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #6 on: August 23, 2008, 04:36:58 AM »

Quote
When you say electrical neutrality do you mean no charge or to maintain the orginal polarity. Anode negative and cathode positive. I mean you don't want zero charge right. I think I'm just hooked up in the word but just to clarify. Thanks again for your help it was very helpful
 

Electrical neutrality means that the concentration of Cu2+ and NO3- are balanced in each half cell, so neither one is in excess.

and do update me about the iodine electrodes. Thanks!

ps: i wouldn't link anode to negative and cathode to positive because it is reversed in the case of electrolysis, however anode will always be oxidation and cathode is reduction.
« Last Edit: August 23, 2008, 04:50:49 AM by Astrokel »
Logged
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

cliverlong

  • Full Member
  • ****
  • Mole Snacks: +60/-14
  • Offline Offline
  • Posts: 611
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #7 on: August 23, 2008, 09:29:36 AM »

Another question too late to modify again!!

Why should the electrolyte solution must be the same as the electrode. I mean in the case of copper why can't the electrolyte be something else. It is only copper solid that is taking part in the reaction. What happens if you only leave the cu electrode. Would a reaction still occur?
Well think of the simple example of placing a Zinc rod into a copper sulphate solution.

The solution goes from blue to clear and a red deposit is seen.

Why?

Because of the difference in electrode potential or reactivity between zinc and copper. If the metal and metal ion are different it is possible or likely one of the electrodes will dissolve . I can't think at the moment whether you could have a cell setup like this yet still create a voltage.


Clive
Logged

cliverlong

  • Full Member
  • ****
  • Mole Snacks: +60/-14
  • Offline Offline
  • Posts: 611
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #8 on: August 23, 2008, 09:41:17 AM »

Too late to modify  :-\

I was thinking too much again and went blank a bit again with the salt bridge  ;)

This is what I understood. Please tell me if each statement is right

*To begin with both flasks have neutral charge. Flask 1 (Cu 2+ and NO3-) making it neutral. Flask 2 (K+ and I-) making it neutral.
*So no reaction can occur because electrons don't have a potential difference to get to the other side.
*When you add the salt bridge flask 1 becomes negative due to extra NO3- and flask 2 positive due to extra K+
*Now there is a potential difference and a path for electrons to flow hence oxidation and reduction can occur.
*So is it fair to say that whole purpose of the salt bridge is to create the  polarity in each flask.
*So no salt bridge no reaction to begin with.
I wouldn't really think of it like that.

I would say the important behaviour is what is happening at the electrode between the metal and the metal ion in solution.

Look at the half-equation and its equilibrium (and its free energy I guess). The position of the equilibrium of M | M+
relative to the other electrode (and its the relative that's important) determine which reaction will oxidize and which will reduce and hence which way the equilibriums will go and hence where the electrons will flow.

Remember the salt bridge is only there to complete the circuit because it contains charged species that can carry electrical current by them physically moving. Remember above I wrote I haven't foudn a detailed description of what happens to the salt bridge ions and I guess apart from completing the circuit their behaviour can be ignored.
Quote
Is what I have said right or can a reaction happen atleast for a little while without a salt bridge. I mean lets say a reaction occur

Cu ------ Cu2+ + 2e-
Now I have seen that

Have a look here

http://www.chemguide.co.uk/physical/redoxeqia/introduction.html

which I think is in line with what you have written
Quote
The charge is still neutral isn't it. So only way to create a negative charge is to add more anions.
Don't understand what you mean there , mate
Quote
So to summarise the salt bridge is required to start the reaction
No. What leads you to think that? I'm interested to follow your thinking.
Quote
and also when the reaction proceeds because the electrons are flowing flask 1 becomes more positive and to maintain the negative charge it is required as well.

So it is required to start and when the reaction proceeds as well.

Very interesting how the ions in the salt bridge knows which side it must go to  ;)
You apply a voltage across an aqueous solution of an ionic compound and the ions will flow according to basic electrical principles.

Why does that seem strange?
Logged

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #9 on: August 23, 2008, 02:08:29 PM »

Thanks again Clivelong and Astrokel for help  ;). But this time I'm a bit more confused. I'll try to express my self better this time. Taking this step by step.

* FIRST OF ALL REALLY SORRY I'M TALKING ABOUT DANIEL CELL  :-[ . Does electrolysis mean the reverse process of splitting into ions. I'm talking about how chemicals can be used to produce current. Not sure if my wording in the topic is right.

* Is it the + charge  in one flask and - charge in the other flask that forces the electrons to move through the wire or is it simply because the oxidant in the other flask tries to grab them.

* Why is the anode negative. Our teacher has wrote electrons are produced there as the reason.
Cu-------Cu2+ + 2e-
But cations are produced at the same time don't the charge of Cu 2+ and the electrons cancel out.

* If it is the + and - in each flask that makes the eletrons move how do we set up this potential difference.

* I mean flask 1 has neutral charge Cu2+ and N03- and flask 2 has neutral charge due to K+ and I-.

*So what I am saying is to have an inbalance of charge don't the ions in the salt bridge move to each flask to make one negative and one positive.

Logged

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #10 on: August 23, 2008, 06:26:20 PM »

IGNORE MY PREVIOUS POST  ;D

I just read clive long's chemguide link on this and it seems I have demonstrated a very poor understanding of the topic  ;D  ;D  ;D  ;D  ;D. I think we are learning this in such a basic level at school and this is the reason. Now I understand this chapter has more to do with equilibrium than I have ever imagined  ;D . I was searching for this chapter on chemguide but didn't find it. Now I know I never knew equilibrium existed in redox reactions  ;D. Thanks heaps for the link Clivelong  ;)

Still as usual I got few more problems  ;)

Ok now I realize that more electrons are deposited in the electrode in the anode side than the cathode side. This is what makes the electrons flow from anode to cathode

With the salt bridge Astrokel wrote,

1) It allows ions to flow between the two half-cells so as to help to maintain the electrical neutrality in each half-cell. (as cliverlong mentioned)
2) It allows electrical contact between the solutions.

I would really appreacite is you guys can re explain each of these points to me now that I have a much better understanding of the overall idea  ;) These are the questions I got from Astrokel's points

When the electrons flow

Cu  <--------> Cu2+ + 2e-

1)I'm assuming according to Le chatelier principle the system would try to make more electrons by going to the right. However as this is a partial change there would be more Cu2+ ions. So what would this extra positive charge do to the sytem? How does salt bridge help to stop this effect?

2)This might sound stupid but I don't understand why electrical contact between the solutions is required. I mean all it matters is how electrons are transferred between the electrodes right?

Also a quick question if there was no wire in this experiment oxidation and reduction would not occur right. I mean in chemguide in the first section they took one fuel cell (Mg) at a time to explain what happens. Then they showed how Mg ------ Mg2+ + 2e- but for this reaction to occur there should be some other element right. I mean if you just leave Mg in a flask this wouldn't occur right. I think chemguide did this to teach the concept one step at a time. I hope you understood what I meant  ;) Simply Mg would only change into ion if has another element to give these electrons to. Right?


Looking forward to your replies  ;)





Logged

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #11 on: August 23, 2008, 08:52:58 PM »


Quote
chemguide says, Suppose you have a piece of magnesium in a beaker of water. There will be some tendency for the magnesium atoms to shed electrons and go into solution as magnesium ions. The electrons will be left behind on the magnesium.

I thought for oxdiation to happen reduction should also occur. So how can Mg left in water create this equilibrium. I thought ions couldn't exist by itself either. Is this just a case of Mg dissolving into ions. Where are the electrons produced going to.




Wouldn't the equilibrium position of this reaction change because there are more Zn2+ in the solution as well. Same applies to the other cell.


If the electrode is inert like carbon what creates the equilibrium. The solution? how?

Sorry I know this topic has turned into bit of a nightmare but I think I'm in the brink of getting this  ;)


« Last Edit: August 23, 2008, 09:10:01 PM by sameeralord »
Logged

Astrokel

  • Full Member
  • ****
  • Mole Snacks: +65/-10
  • Offline Offline
  • Gender: Male
  • Posts: 989
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #12 on: August 23, 2008, 09:58:36 PM »

Quote
I thought for oxdiation to happen reduction should also occur. So how can Mg left in water create this equilibrium. I thought ions couldn't exist by itself either. Is this just a case of Mg dissolving into ions. Where are the electrons produced going to.

Mg2+(aq) + 2e-  <----> Mg(s)

Isn't forward reduction and backward oxidation? Mg2+ exists as [Mg(H2O)6]2+ in water solution and magnesium is not reactive towards water, or even to small extent becomes hydroxide, but in the actual half cell, you wouldn't even use water as the electrolyte!

Quote
Wouldn't the equilibrium position of this reaction change because there are more Zn2+ in the solution as well. Same applies to the other cell.

what is the function of salt bridge?

Quote
If the electrode is inert like carbon what creates the equilibrium. The solution? how?


Do you know the setup of Fe2+ & Fe3+ half cell using an inert electrode(coupling with S.H.E)

Fe3+(aq) + e- ---> Fe2+(aq)
Mg2+(aq) + 2e-  <----> Mg(s)
Can you see the similarities and what causes the equilibrium?
Logged
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

sameeralord

  • Regular Member
  • ***
  • Mole Snacks: +12/-4
  • Offline Offline
  • Posts: 90
Re: Electrolysis and Redox equations (Diagrams provided)
« Reply #13 on: August 23, 2008, 11:33:32 PM »

Thank you for the help  ;)

 So is the function of the electrolyte is to dissolve the metal electrode into ions. So actually oxidation and reduction both occur in the equilibrium. I though only way some thing can become an ion is if another  metal like Cl comes along and an ionic compound forms like MgCl2. Is there other ways things can become ions.

Wouldn't the equilibrium position of this reaction change because there are more Zn2+ in the solution as well. Same applies to the other cell.

what is the function of salt bridge?

The salt bridge can only elimanate the charge so I don't understand.

Do you know the setup of Fe2+ & Fe3+ half cell using an inert electrode(coupling with S.H.E)

Fe3+(aq) + e- ---> Fe2+(aq)
Mg2+(aq) + 2e-  <----> Mg(s)
Can you see the similarities and what causes the equilibrium?

Can you explain this to me.  ;)

Sorry I'm bit tired today after lots of research and stuff so can't really write much anymore  ;)

« Last Edit: August 23, 2008, 11:46:35 PM by sameeralord »
Logged
Pages: [1]   Go Up
 

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.091 seconds with 23 queries.