[sameeralord]

Hello everyone  ,

I thought before I could move to the daniel cell I have to understand this concept.

1. If we leave  Mg solid electrode in water it says there is a tendency for Mg to dissolve into ions creating equilibrium Mg <------->  Mg2+ + 2e-
I thought one half reaction cannot occur independant of the other half reaction. I mean oxidation and reduction should always occur at the same time. I understand here that forward is oxidation and backward is reduction.   So is this true that this happens when you put Mg in water.
2. How about Mg+ ions in an solution would they create an equilibrium like this or is it only solids that can dissolve create this equilibrium?
3. So after looking at redox equilibrium this is my new understanding of redox equations. Tell me if this is right .

1. There is zinc in a glass tube
   It is in equilibrium Zn <-----> Zn2+ +  2e-
2. There is Mg in another glass tube
    It is in equilibrium Mg <-----> Mg2+ + 2e-
3. You add them together and then according to their potentials one reaction gives electrons to another reaction. The equilibriums of each reactions is upset and they eventually become oneway reactions.

So infact the halfway reactions are occuring all the time it is just that redox reactions upset the equilibrium and make them oneway reactions. Am I right?

Thank you 

[Borek]

1. If we leave  Mg solid electrode in water it says there is a tendency for Mg to dissolve into ions creating equilibrium Mg <------->  Mg2+ + 2e-

Tendency doesn't necesarilly meanreaction occurs, It will not occur before you add some 'electron sink'.

2. How about Mg+ ions in an solution would they create an equilibrium like this or is it only solids that can dissolve create this equilibrium?

Ions can create solids, the same type of equilibrium, but in the reverse direction.

So infact the halfway reactions are occuring all the time it is just that redox reactions upset the equilibrium and make them oneway reactions. Am I right?

Yes and no. Reactions to some extent are occuring all the time, but that's because there are no "other halfreaction free" environements. For example when you put solid Mg in water, it will to some extent react with H⁺ from water autodissociation.

[sameeralord]

Thanks a lot for taking your time to answer my question 

What is an electron sink. I searched in the internet didn't get much useful info.

In this case

Cl- <--------> Cl2 + 2e-

I'm interested how there is an equilibrium. Wouldn't the Cl2 gas escape to the air. Also as you have said ion can turn into solids. Can they turn into gases as well in equilibrium as in this case.

If Mg reacts with H+ what happens I don't understand how it can affect the equilibrium.

Looking forward to your replies 

[Borek]

What is an electron sink. I searched in the internet didn't get much useful info.

It is not a formal term, I meant just any oxidising agent that will be able to consume electrons produced in the oxidation reaction.

In this case

Cl- <--------> Cl2 + 2e-

I'm interested how there is an equilibrium. Wouldn't the Cl2 gas escape to the air.

It will. But any chlorine produced that will escape the solution will be in contact with the solution - so there will be some chlorine dissolved (see http://en.wikipedia.org/wiki/Henry's_law). Could be you are thinking about situation when chlorine escapes completely (that's equivalent of the situation when the air volume is infinite) - then you are just removing one of the products to move equilbrium to the right (that's Le Chatelier's principle at work).

Also as you have said ion can turn into solids. Can they turn into gases as well in equilibrium as in this case.

Yes.

If Mg reacts with H+ what happens I don't understand how it can affect the equilibrium.

If you put some solid Mg into water it will - to some extent - react with H⁺:

Mg + 2H⁺ -> Mg²⁺ + H₂

You will end with some dissolved Mg²⁺ and some gaseous H₂. This reaction occurs almost each time you put any metal into water - just depending on the metal activity it will dissolve more or less. The more active metal, the more will get dissolved. How much - depends on the metal activity. Dissolution equilibrium can be calculated, although it is probably way above your head for now.

[sameeralord]

Thanks a lot Borek for you help   

Now I think of it in a daniel cell Mg electrode is put into a solution of Mg2+ (only talking about one half cell before wire is connected) .  Here an equilibrium is caused Mg <-----------> Mg2+  +  2e-. Is Mg 2+ the electron sink in this case. I also get the impression that this a full redox reaction.

I mean the two half equations are

Mg ---------> Mg 2+  +  2e-
Mg 2+  +  2e-  -------- > Mg

I know this cancels out to zero but can we take this as  full redox equation atleast theroretically .  Mg redutant and Mg 2+ oxidant.

Only talking about the cell before wire is connected

[Borek]

Mg ---------> Mg 2+  +  2e-
Mg 2+  +  2e-  -------- > Mg

I know this cancels out to zero but can we take this as  full redox equation atleast theroretically .  Mg redutant and Mg 2+ oxidant.

Only talking about the cell before wire is connected

If there is no wire, then no - it cancels out and nothing happens. If there are two separate tanks with different concentrations of cation - than that will be called concentration cell and it will work till both concentrations are identical. But magnesium is far from being the good choice - it is too active, it reacts with water on its own. Copper is perfect for concentration cells.

[sameeralord]

Thanks again for your help Borek 

Then why is Mg electrode put into a solution of Mg2+ in a daniel cell. I thought this was done to create an equilibrium.  I don't understand how redox equilibrium is related to the daniel cell. I read about this in chemguide and now confused. If you can help me in this I'll be most greatful. Thanks 

MODIFY: Another possibility by looking at your example of water is that when Mg electrode is put into Mg2+ the equilibrium is caused by water's half reaction. I'm really confused not what you have said previously but how redox equilibrium is related to daniel cell. For the first time chemguide has made me more confused

I'll try to be more specific...

This is what chemguide says..

The two equilibria which are set up in the half cells are:

Then they go on talking about how electrons flow to one cell according to their potentials and eventually become one way reaction.

Simply how did those equilibrium they have mentioned have formed.

This is what has troubled me the most!!

[sameeralord]

I'm interested as to know that if each equlibrium that has occured is due to the other equilibrium that has occured. If the wire is connected and Tht one equilibrium cann't occur without the other.

or is that when you leave half a cell with Zn and no connection to other half cell this Zn equilibrium occurs.

Thanks again for anyone who is going to help 

[Borek]

Not sure if I understand what you are asking about. Each system is in its own equilibrium. When you add a wire, you combine two systems into one system, and two earlier equilibria have to change for the whole (new) system to rech new equilibrium.

[sameeralord]

Thank you for your reply 

Ok I understand what you are saying it does make sense. My question is how is the equilibrium caused in half a cell. Taking the Zn half cell for instance  how is this Zn + 2e-  <------->  Zn2+  equilibrium archieved. Does the Zn electrode and Zinc suplhate solution create this equilibrium  or is this the result of water acting with Zn.

Simply if you can give me an explanation how each of the equilibrium in each half cell is orginally created.

How is Zn + 2e-  <------->  Zn2+ caused and how is
Cu2+  +  2e-  <--------> Cu caused

Looking forward to your reply 

[Borek]

I have already explained details on August 25^th. Water (or H⁺) will react with solid metal, dissolving just enough to create the equilibrium. If you put salt into the solution, the same equilibrium between water and solid will exist, it will be just shifted by the presence of excess ions.

[Hunt]

Tendency doesn't necesarilly meanreaction occurs, It will not occur before you add some 'electron sink'.

What if there's a closed red/ox system ( such as Fe³ <--> Fe² ) in aqueous solution with a conductor ( like a Pt electrode ) dipped in it? There should be an exchange of electrons through the conductor until equlibrium is reached. Then the concentration of the electroactive species has changed. Can you confirm this ?   

[Borek]

What if there's a closed red/ox system ( such as Fe³ <--> Fe² ) in aqueous solution with a conductor ( like a Pt electrode ) dipped in it? There should be an exchange of electrons through the conductor until equlibrium is reached. Then the concentration of the electroactive species has changed. Can you confirm this ?  

Draw the system you are referring to. So far it sounds like you are talking about solution containing both Fe(II) and Fe(III) - sticking Pt wire into it won't change a thing.

[Hunt]

What if there's a closed red/ox system ( such as Fe³ <--> Fe² ) in aqueous solution with a conductor ( like a Pt electrode ) dipped in it? There should be an exchange of electrons through the conductor until equlibrium is reached. Then the concentration of the electroactive species has changed. Can you confirm this ?  

Draw the system you are referring to. So far it sounds like you are talking about solution containing both Fe(II) and Fe(III) - sticking Pt wire into it won't change a thing.

Yes exactly , suppose the system is simple like you have a beaker with Fe(III)/Fe(II) soln and a Pt electrode. I used to think there should be no change in concentration, but a while back while I reading about electrochemical process I remember that the writer said that the system reaches dynamic equilibrium after the electrons are transfered through the conductor. If this is so , why is there no net change in the number of particles of Fe(III) ? 

[Borek]

If you have two beakers, one containing Fe(II) and the other one Fe(III), and they are connected with salt bridge, and you connect them with Pt wire, there will be charge transfer though the wire, till both beakers contain identical concentrations of both Fe(II) and Fe(III). Note, that it will also mean exnchage of counterions thorugh the salt bridge. But if you mix both ions in one beaker, sticking Pt wire wont change anything.