6M.5 part c 7th edition

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inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

6M.5 part c 7th edition

Postby inlovewithchemistry » Wed Feb 27, 2019 8:47 pm

How do you determine which is the cathode and which is the anode in this question? Im confused because it looks to me that both half reactions are losing electrons.
thank u!

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 6M.5 part c 7th edition

Postby Matthew Tran 1H » Wed Feb 27, 2019 11:00 pm

I think that you assigned your oxidation numbers incorrectly. Hg(l) has an oxidation state of 0 while Hg in Hg2 2+ has an oxidation state of +1. Since Hg is oxidized, this half reaction takes place at the anode. N in NO3- has an oxidation state of +5 (-1 = 3*(-2) + 5) while N in NO has an oxidation state of +2. N is reduced so this half reaction takes place at the cathode.

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

Re: 6M.5 part c 7th edition

Postby inlovewithchemistry » Wed Mar 06, 2019 7:01 pm

This makes so much more sense now, thank you!


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