# Finding cell emf in a concentration cell

1. Apr 24, 2010

### 4LeafClover

Question:
A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: AgCl(s) + e- = Ag(s) + Cl-(aq). The two cell compartments have [Cl-]= 1.41×10−2 M and [Cl-]= 2.70 M, respectively. What is the emf of the cell for the concentrations given?

My Work:
I found the standard emf of the cell to be 0 V and then I tried to use the Nernst equation to find emf, but it came out wrong. I also answered a previous multiple choice question which told me that the compartment with the 1.41×10−2 M Cl(aq) is the cathode (I don't understand why though because I thought the more concentrated one was the cathode...) So then here is what I did: Standard EMF = 0-((8.3145*298)/(1*96485)) * ln ((1.41×10−2 M)/2.7) but the answer I am getting is incorrect. Where did I go wrong?

2. Apr 24, 2010

### Staff: Mentor

More concentrated what? Concentration cell tries to make both concentrations identical.

I don't see what can be wrong with your calculation of EMF though, equation seems correct. Unless I am too sleepy to think correctly.

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3. Apr 25, 2010

### 4LeafClover

Ya, I thought my calculations were correct too, but they're not for some reason. And what I meant by more concentrated, I wondering if, when you make a concentration cell and you start off with one solution that is more concentrated and then one that is more diluted (in order to make electrons transfer) then isn't the one that is more concentrated usually the cathode end and the one that is more diluted the anode end? I understand that in the end they will end up with the same concentration, but at the beginning is this how it works??

4. Apr 26, 2010

### Staff: Mentor

1. Solution is not cathode - electrode in solution is either anode or cathode.

2. You have not stated concentration of what.

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methods

5. Apr 26, 2010

### 4LeafClover

My chemistry book explicitly states that, in a concentration cell whose compartments both consist of a strip of nickel metal immersed in solutions of different concentrations of Ni^2+, that oxidation occurs in the half cell containing the more dilute solution and is therefore the anode and that reduction of Ni^2+ occurs in the half cell containing the more concentrated solution and is the cathode department. But when I applied this to the question above and chose the more concentrated solution to be the anode, (the one where [Cl-]=2.70M) I got the answer wrong. When I chose the more dilute solution to be the cathode, I got it correct. Why does my chem book say one thing but then my homework is the opposite? What is correct?

6. Apr 26, 2010

### Staff: Mentor

You are missing something very basic, and I have already pointed twice to the problem in your thinking. There is NO DIFFERENCE between your homework and your textbook.

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methods

7. Apr 26, 2010

### 4LeafClover

Actually my textbook is telling me that the more concentrated solution is the cathode but my homework is telling me the more dilute ([Cl-]= 1.41×10−2 M) is my cathode. Can you explain to me what you mean because just telling me I'm not understanding isn't helping me to understand.

8. Apr 26, 2010

### Staff: Mentor

Compare both examples. Higher concentration of WHAT?

What are electrode reactions?

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methods

9. Apr 26, 2010

### 4LeafClover

Ok, well, if the reaction is AgCl(s) + e- = Ag(s) + Cl-(aq) and the two cell compartments have [Cl-]= 1.41×10−2 M and [Cl-]= 2.70 M, that means that there is higher electron concentration in the first because lower Cl- concentration means that there are more free electrons. Is that what they are meaning?

10. Apr 26, 2010

### Staff: Mentor

No, there is no such thing as free electrons, nor free electron concentration.

Write electrode reactions for both Ni case and Ag case.

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11. Apr 26, 2010

### 4LeafClover

For Ni:
Anode: Ni(s) --> Ni^2+(aq) + 2e-
Cathode: Ni^2+(aq) + 2e- --> Ni(s)

For Ag:
Anode: Ag(s) + Cl-(aq) --> AgCl(aq) + 1e-
Cathode: AgCl(aq) + 1e- --> Cl-(aq) + Ag(s)

12. Apr 26, 2010

### Staff: Mentor

Have you noticed difference between these two cases? Can you rewrite second equation in terms of Ag+ cation reduction? Do you know how to calculate Ag+ concentration in solution (given Cl- concentration)?

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13. Apr 27, 2010

### 4LeafClover

Well I guess the difference between the two would be that in the second, Ag is reduced while the oxidation state of Cl- remains the same. The concentration for this equation is given in terms of Cl- instead of the species that is reduced whereas in the first instance the concentration is given in terms of the species that is reduced.

The second equation, just in terms of Ag, would be AgCl(aq) + 1e- --> Ag(s) because in the AgCl case Ag is +1 and in the solid Ag case, Ag is 0.

If I had an equilibrium constant I could calculate the exact concentration of Ag+ in terms of Cl- but it would be something like K=[Cl-]/[AgCl] and therefore as the concentration of AgCl (thus the concentration of Ag+) increased, then so would the concentration of Cl-.

14. Apr 27, 2010

### Staff: Mentor

Actually you can think about it as if AgCl dissolves before Ag+ is being reduced. And solubility product plays an important role.

15. Apr 27, 2010

### 4LeafClover

So how does a higher concentration of AgCl make for a lower concentration of Cl-? Please just explain

16. Apr 27, 2010

### Staff: Mentor

No such thing as "higher concentration of AgCl" - AgCl is a solid, so it doesn't have concentration. However, as every weakly soluble solid it is in equilibrium with its saturated solution.

What reacts on the electrode are Ag+. AgCl is in contact with solution of chlorides. How does concentration of the Ag+ cation depend on the concentration of chlorides?

17. Apr 27, 2010

### 4LeafClover

The Ag+ will react with Cl- and form AgCl. So less Cl- means more Ag+

18. Apr 27, 2010

### Staff: Mentor

And you are now on the right track. Do you see that more Cl- means less Ag+? How is that realated to cathode/anode problem?

19. Apr 27, 2010

I understand