Book problem 14.15

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DanielLee3H
Posts: 17
Joined: Wed Sep 21, 2016 2:55 pm

Book problem 14.15

Postby DanielLee3H » Tue Feb 14, 2017 11:11 pm

I have no idea how to even approach these.

Do we have to just know the half reactions of these formulas or that you have to add KOH to the cell in part c?

Sangita_Sub_3H
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: Book problem 14.15

Postby Sangita_Sub_3H » Wed Feb 15, 2017 1:37 am

Hi! So for part a) of this problem, the equation is AgBr --> Ag(plus) + Br (minus). So for the half reactions, you look at how the oxidation state of each reactant changes. In the case of Ag, the oxidation state changes from 0 to +1, which means that it is being oxidized since it is losing one electron. In the case of Br, the oxidation state changes from 0 to -1, which means that is it being reduced since it is gaining one electron. So, the half reactions would be Ag --> Ag(plus) + 1e- and Br + 1e- --> Br(minus). Then for the galvanic cell, you know that Ag goes in the anode and Br in the cathode, since they correspond to the oxidized and reduced state respectively.

Hope this helps!


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