redox equilibrium (Is their such a thing?)

[scilearner]

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

[hermanntrude]

half-reactions cannot occur on their own. the reason for this is that electrons aren't fond of being alone. What you are missing is that there is a half-reaction for water too, so if zinc is in water, the zinc has a small tendency to lose two electrons and the water can pick them up and undergo its own half-reaction:

 

reduction: [ce]2H2O + 2e- -> H2 + 2OH-[/ce] E°=-0.828V

oxidation: [ce]Zn -> Zn^2+ + 2e-[/ce] E°=+0.763V

 

total: [ce]Zn + 2H2O -> Zn^2+ + H2 + 2OH-[/ce] E°cell = -0.065V

 

 

 

This reaction is not spontaneous, however, since the sum of the electrode potentials for the two half reactions is negative, so we don't see a reaction. The reaction can occur, but a little bit of energy needs to be applied before it will.

 

If we chose a metal which is slightly less attached to its electrons, however, we could see that it can and does react with water:

 

reduction: [ce]2H2O + 2e- -> H2 + 2OH-[/ce] E°=-0.828V

oxidation: [ce]Mg -> Mg^2+ + 2e-[/ce] E°=+2.356V

 

total: [ce]Mg + 2H2O -> Mg^2+ + H2 + 2OH-[/ce] E°cell = +1.528V

 

Since the sum of the electrode potentials for these half reactions is positive, it means that magnesium will indeed react with water, separating it into hydrogen gas and hydroxide ions, which will essentially cause magnesium hydroxide to form.

 

note that these values are only true for 1M solutions at 25°C using standard electrodes. As you study electrochemistry further you will find there are equations for predicting non-standard situations

[scilearner]

Thanks a lot!!. This is actually one of the most satisfying answers I have recieved

 

Only few problems.

 

So how is equilibrium related to all this. Is the Mg 2+ formed in equlibrium?

 

Does redox equlibrium occur only when elements are in a solution such as this!!

 

Can same be said of Mg ions in a solution. Would they try to use water's half equations and create this same equilibrium. Mg <------> Mg2+ + 2e-

 

Thanks again

[hermanntrude]

one thing which people don't learn until late (too late, in my opinion) in chemistry courses, is that ALL reactions are equilibria. When we write the one-way arrow to indicate a one-way reaction, what it REALLY indicates is that the reactions equilibrium constant is so large that there are no reactants to speak of left at the end. However, there is always a little bit of the reactants left, and often in a reaction which doesn't appear to "go", the only reason is that the equilibrium constant is so small that there are only tiny quantities of reactant at any one time. Reactions which don't appear to work can also be non-spontaneous or incredibly slow.

 

If you're doing a chemistry course you'll come across these topics later on in your curriculum.

[scilearner]

Thanks a lot Hermantrude . I have tried to get help for this question from variety of sources but I think you have understood my question better than anyone else. I especially liked the water equation that makes a lot of sense why Mg dissolves. Yeah we just finished equilibrium and totally forgot about that fact that everything is in equilibrium.

 

tiny quantities of reactant

 

I'm pretty sure you meant products. Yeah!!

 

Ok now I'm going to make a topic on the daniel cell. I want to see if my understanding on that is right and I got some problems with inert electrodes.

 

Looking forward to your help there especially!!

 

Also thanks for understanding I'm a begginer and would not be able to grasp some of the concepts immediately.

 

Thanks again!!

 

PS: Just one more question. If Mg electrode is left in a solution of Mg2+ not water there would be an equilibrium right between the ions and the solid Mg <----------> Mg 2+ + 2e- Here the two half reactions are the forward one and backone. I'm assuming here the Mg2+ can act as the oxidant!!

Edited August 26, 2008 by scilearner

[hermanntrude]

i'm not really sure how your question differs from the first one.

 

if you want to show appreciation for anyone's responses, click the scales at the top right of someone's post

[scilearner]

No probs. I didn' know there was a system like this here. I didn't think I expressed myself clearly last time. If you can please answer this question I'll sure add more scales . This is my confusion

 

chemguide says.....

 

 

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

 

 

 

In respect to your previous post "half-reactions cannot occur on their own"

 

In this case is this equlibrium caused by this equilibrium

 

I mean one cannot occur without the other

 

Would these equilbrium rise if the cells were not connected.

 

However I get the feeling chemguide is suggesting that these equilibrium would exist if these cells were not connected. I'm not sure. If my understanding is wrong please tell me how this equilibrium have occured.

 

Yes you probably have mentioned this in previous post but could you please reexplain this with this specific example. I'll be most greatful. Thanks

Edited August 27, 2008 by scilearner

[big314mp]

The way I understand it (feel free to correct me) is that there is an equilibrium on each side (when the cells are disconnected)...but it is something of a meaningless equilibrium:

 

Cu2+ + Cu <-> Cu + Cu2+ and something similar happening on the zinc side.

 

Your half reactions would be:

Cu2+ + 2e- <-> Cu

Cu <-> Cu2+ + 2e-

 

Now when you connect the cells, you can think of it like this:

The zinc can provide electrons more easily than the copper can, so on the copper side there is an "excess" of electrons. This drives the copper side (Le Chatelier) towards the metal:

Cu2+ + 2e- <-> Cu, shifts towards right because of excess electrons on left

 

On the zinc side, the copper cell acts as an electron "sink" so there is a deficit of electrons, shifting that reaction towards the zinc ion.

 

Hope that doesn't confuse you anymore

[hermanntrude]

i'm not sure if it's sensible to speak of one half-reaction being coupled with it's own reverse reaction. You could, of course, speak of the reaction with the reduction/oxidation of water, which does happen in significant quantities with the more reactive metals...

 

I don't honestly know if there is a reaction with water for the less reactive metals and there just isn't very much of it (the case where the equilibrium constant is very small), or whether it just doesn't happen because the reaction is non-spontaneous due to the cell potential being negative.

 

I think it also depends on the specfic half-cell

[big314mp]

Well the chemistry basis I used for it was that the cell voltage would be (of the half reaction coupled to itself) zero, therefore delta G would be zero, which implies an equilibrium. It was basically an accounting trick to help explain the overall principle from the perspective of Le Chatelier.

[scilearner]

Thanks both of you for your replies.

 

That was a good respone, big314mp I like to understand the way you have showed. However I can only do it if I know how this meaningless equilibrium is caused?

 

Can anyone explain me to how is these two equilibrium caused in these two cells?

[Ladeira]

I've never seen this cell, but I am very familiarized to the reduction potential.

 

i'm not really sure how your question differs from the first one.

 

if you want to show appreciation for anyone's responses, click the scales at the top right of someone's post

 

LOL!