jonnyboy118
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Hello,
So
The redox equilibrium in two half cells is shown:
Zn2+ + 2e- <--> Zn (s)

Cu2+ (aq) + 2e- <---> Cu (s)

Apparently the Zn/Zn2+ equilibrium releases electrons more readily than the Cu/Cu2+ equilibrium. So electrons flow from zinc to copper.
Then my book says the Zn2+/Zn equililbrium moves to the left.
this is the part i do not understand. Does it move to the left because the equilibrium releases electrons more readily so there concentration will decrease shifting the equilibrium to the left?
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univ4464
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(Original post by jonnyboy118)
Hello,
So
The redox equilibrium in two half cells is shown:
Zn2+ + 2e- <--> Zn (s)

Cu2+ (aq) + 2e- <---> Cu (s)

Apparently the Zn/Zn2+ equilibrium releases electrons more readily than the Cu/Cu2+ equilibrium. So electrons flow from zinc to copper.
Then my book says the Zn2+/Zn equililbrium moves to the left.
this is the part i do not understand.
Since the zn more readily releases electrons this reaction occurs: Zn -> Zn2 + 2e- more than the Cu version and Zn backwards reaction, so there are more Zn2 ions lying around, and so the position of equilibrium is shifted to the left (of the original equation) as this concentration has increased. This stuff is always confusing
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jonnyboy118
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Thanks, one last thing.
By Le Chatelier priniciple increasing the concentration moves the equilibrium position to the other side but...

if A <--> B

And i increase the concentration of a, equilibrium shifts to the right but the position of equilbrium does not actually change. is this true?
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Reality Check
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(Original post by alfredholmes)
Since the zn more readily releases electrons this reaction occurs: Zn -> Zn2 + 2e- more than the Cu version and Zn backwards reaction, so there are more Zn2 ions lying around, and so the position of equilibrium is shifted to the left (of the original equation) as this concentration has increased. This stuff is always confusing
Hi, I was wondering if you'd be able to help this person out with a similar question? The thread is here:

https://www.thestudentroom.co.uk/sho...886&highlight=

No worries if you can't, but you seem a helpful sort of person
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univ4464
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(Original post by jonnyboy118)
Thanks, one last thing.
By Le Chatelier priniciple increasing the concentration moves the equilibrium position to the other side but...

if A <--> B

And i increase the concentration of a the position of equilibrium just returns to the same place. Why doesnt this work?
Its easier to explain this in terms of Kc = [B] / [A] for this reaction. Initially when you increase the concentration of A the position of 'eqm' is towards A and so it moves back to where is was because for the two equilibriums the Kc value is the same (same ratio of products to reactants).

I put eqm in quotes because the system is not actually in equilibrium since when you increase the concentration of A the rate of A -> B has increased so the system is not in equilibrium. So in restoring the eqm the sort of composition of the system moves towards B
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jonnyboy118
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(Original post by alfredholmes)
Its easier to explain this in terms of Kc = [B] / [A] for this reaction. Initially when you increase the concentration of A the position of 'eqm' is towards A and so it moves back to where is was because for the two equilibriums the Kc value is the same (same ratio of products to reactants).

I put eqm in quotes because the system is not actually in equilibrium since when you increase the concentration of A the rate of A -> B has increased so the system is not in equilibrium. So in restoring the eqm the sort of composition of the system moves towards B
Thanks,
So am i right in saying, for this particular equilibrium, if you increase the concentration of a the position lies to a. equilibrium shifts to the right because the quotient does not equal kc but this shift does not actually produce a change in the position of equilibrium (as the ratio's have not actually changed). Thanks for your help. and by 'shift in equilibrium i guess i mean a--> B is favoured.
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univ4464
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(Original post by jonnyboy118)
Thanks,
So am i right in saying, for this particular equilibrium, if you increase the concentration of a the position lies to a. equilibrium shifts to the right because the quotient does not equal kc but this shift does not actually produce a change in the position of equilibrium (as the ratio's have not actually changed). Thanks for your help.
Yeah, that's basically it. The problem is being equilibrium means the rate of the forwards and backwards reactions are equal. If you increase the concentration then the system is not in equilibrium so (at least to me anyway) it doesn't make much sense to talk about the position of equilibrium. I'd argue that it'd make more sense to define the position of equilibrium as the composition after you leave it for a long time, so you definitely don't change the equilibrium by increasing the concentration of A. The change of concentration is useful for things like the Harber process where to increase yield you remove product as it is produced in order to shift the position of equilibrium towards the products. Hope this makes sense.
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jonnyboy118
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(Original post by alfredholmes)
Yeah, that's basically it. The problem is being equilibrium means the rate of the forwards and backwards reactions are equal. If you increase the concentration then the system is not in equilibrium so (at least to me anyway) it doesn't make much sense to talk about the position of equilibrium. I'd argue that it'd make more sense to define the position of equilibrium as the composition after you leave it for a long time, so you definitely don't change the equilibrium by increasing the concentration of A. The change of concentration is useful for things like the Harber process where to increase yield you remove product as it is produced in order to shift the position of equilibrium towards the products. Hope this makes sense.
Thanks For your time
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