6L.7a cell diagram

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Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

6L.7a cell diagram

Postby Lindsey Chheng 1E » Sun Mar 01, 2020 3:54 pm

The cell diagram for this question is Ag(s) I AgBr(s) I Br-(aq) II Ag+(aq) I Ag(s). I'm confused on why there's a single line separating Ag(s) and AgBr(s) when they are both in the same state. Can someone explain?

Myka G 1l
Posts: 100
Joined: Fri Aug 30, 2019 12:17 am

Re: 6L.7a cell diagram

Postby Myka G 1l » Sun Mar 01, 2020 3:57 pm

There’s a solid line separating Ag(s) and AgBr(s) despite the fact that they are both in the same state because Ag(s) is the electrode whereas AgBr(s) is just a solid involved in the reaction.

Brittney Hun 2C
Posts: 104
Joined: Wed Sep 11, 2019 12:15 am

Re: 6L.7a cell diagram

Postby Brittney Hun 2C » Mon Mar 02, 2020 4:11 pm

How do we know that Ag is both the oxidizing and reducing agent? Doesn't Br go from Br-(oxidation of -1) to AgBr (which has an oxidation of 0)?


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