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14.15a

Posted: Wed Feb 21, 2018 9:06 pm
by Lucia H 2L
AgBr(s) <--> Ag+(aq) + Br-(aq)

anode: AgBr(s) + e- --> Ag(s) + Br-(aq)
cathode: Ag+(aq) + e- --> Ag(s)

Where did these half reactions come from? Shouldn't the anode be losing electrons? Also, where is the AgBr in the cathode equation?

Re: 14.15a  [ENDORSED]

Posted: Thu Feb 22, 2018 8:46 am
by Chem_Mod
First, identify which component is oxidized and which is reduced by assigning oxidation numbers.

Then, look up the relevant half-reactions in Appendix 2B that match the half-reactions of oxidation and reduction you just identified.

If the cell is to be galvanic (AKA voltaic, spontaneous) then E°cell must be positive. Assign the half-reaction with the lowest (most negative) standard reduction potential to the cathode.

Re: 14.15a

Posted: Fri Feb 23, 2018 1:22 am
by 904676178
Chem_Mod wrote:First, identify which component is oxidized and which is reduced by assigning oxidation numbers.

Then, look up the relevant half-reactions in Appendix 2B that match the half-reactions of oxidation and reduction you just identified.

If the cell is to be galvanic (AKA voltaic, spontaneous) then E°cell must be positive. Assign the half-reaction with the lowest (most negative) standard reduction potential to the cathode.

I am somewhat confused about this question as well, isn't the half-reaction with the lowest most negative standard reduction potential supposed to be assigned to the anode? or are we using the (most negative) lowest standard reduction potential and assigning it to the cathode because it is a non-spontaneous reaction?

Re: 14.15a

Posted: Mon Feb 26, 2018 10:45 am
by Chem_Mod
Vasiliki gives a good explanation on choosing your anode and cathode for this problem. See the answer here: viewtopic.php?f=140&t=28135&p=86714&hilit=agbr&sid=d0dd8e815892f0d08cafc4fba831ff83#p86714