## Question 14.41

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Arjun Sharma 1D
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

### Question 14.41

For question 14.41 part (a), why is the Cu2+ with lower concentration the anode and Cu2+ with higher concentration the cathode? I thought you would want it the other way around since it would flow from high concentration to low concentration and you would want that to be anode to cathode.

Kellina Tran 2I
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Joined: Sat Jul 22, 2017 3:00 am

### Re: Question 14.41

I’m also confused, since our notes say that we should always make the lower concentration the product

Arjun Sharma 1D
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

### Re: Question 14.41

Yeah it's weird too cause the lower concentration ends up in the product according to the solution manual. Maybe it's because you have to flip the sign of the anode?

Justin Folk 3I
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Joined: Wed Sep 21, 2016 2:56 pm

### Re: Question 14.41

Arjun Sharma 1D
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

### Re: Question 14.41

In a concentration cell, why would you want your lower concentration reactant on the anode and the higher one on the cathode?

melissa carey 1f
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Joined: Fri Sep 29, 2017 7:06 am

### Re: Question 14.41

You want e- to flow anode to cathode. At the anode Cu(s) > Cu2+ + 2e-. At the cathode Cu2+ + 2e- > Cu(s). So the anode [] is a product and cathode [] is a reactant. Since you want postive E and Q=P/R, so, using the eq, you can see that anode should always have lower [].
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Yu Chong 2H
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Joined: Fri Sep 29, 2017 7:05 am

### Re: Question 14.41

The flow of electrons is not the same as the "flow" of the Cu2+. The electrons will flow towards the side with higher concentration of Cu2+ so that the copper ions there can get reduced to form Cu(s) and therefore reducing the concentration of Cu2+ in that cell.

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