## 14.15a

Kyung_Jin_Kim_1H
Posts: 53
Joined: Thu Jul 27, 2017 3:00 am

### 14.15a

I'm really confused on how to find the half reactions when given: AgBr(s) --> Ag+(aq) + Br-(aq), a solubility equilibrium.

I understand what happens at a cathode (reduction) and anode (oxidation). I thought that the half reactions would we somewhere along the lines of:
cathode: Ag+ + e- --> Ag(s)
anode: Be- --> Br(s) + e-
but the answer is totally different...
Can someone walk me through the process?

Angela G 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: 14.15a

I think it's because the oxidation number for bromine doesn't change - for the reactant AgBr, Ag has oxidation no. = +1, and Br has oxidation no. = -1 for an overall charge of 0. So in the anode, Ag(s) + Br- -> AgBr + e (Ag goes from 0 to +1); cathode, Ag+ + e -> Ag(s) (Ag goes from +1 to 0).

The cell notation is Ag(s) | AgBr(s) | Br-(aq) || Ag+(aq) | Ag(s) with the species in the order as they are formed, save for Br-(aq), which is put at the end of the anode half and out of order because it is a spectator ion.

soniatripathy
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: 14.15a

the best way to write half reactions is to first assign all the elements an oxidation number, then look at how it changes between the right and left side of the equation. Based on what is oxidized and what it reduced, you can write the half reactions and assign electrons to the correct side

Desiree1G
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 2 times

### Re: 14.15a

I am confused on this too but essentially I think it is because the oxidation number for Bromine do not change.