## Endgame 1a

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

Michael Du 1E
Posts: 117
Joined: Sun Sep 22, 2019 12:16 am

### Endgame 1a

In the review packet, 1a, what determines the fact that the lower concentration of the Ag+ (0.10M) is on the left anode side and the higher concentration (2.0 M) is in the cathode side of the concentration cell? or it doesn't matter, thanks!

MingdaH 3B
Posts: 133
Joined: Thu Jul 11, 2019 12:17 am

### Re: Endgame 1a

anode is always oxidized, and the side with lower concentration will always be oxidized.

KnarGeghamyan1B
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

### Re: Endgame 1a

Since cells flow from anode to cathode, it makes sense that the side with the lower concentration of cations will oxidize and the higher concentration will be reduced until both sides are balanced.

AlyssaYeh_1B
Posts: 100
Joined: Sat Aug 17, 2019 12:16 am

### Re: Endgame 1a

I believe we discussed in lecture that we should always make the lower concentration the product, so that "it is in solution", which means that the lower concentration will always be on the anode side

ValerieChavarin 4F
Posts: 99
Joined: Wed Sep 18, 2019 12:18 am

### Re: Endgame 1a

the anode is the side with the lower concentration as it is the side where oxidation occurs.

CNourian2H
Posts: 106
Joined: Fri Aug 30, 2019 12:16 am

### Re: Endgame 1a

yes, just remember that the lower concentration is always in the products.

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