Question on 6L.7

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Elizabeth Bowen 1J
Posts: 53
Joined: Wed Nov 14, 2018 12:20 am

Question on 6L.7

Postby Elizabeth Bowen 1J » Wed Mar 04, 2020 7:07 pm

In 6L.7, the overall reaction given is AgBr(s) <--> Ag+ (aq) + Br-(aq)

The cathode reaction given in the solutions manual is Ag+ (aq) + e- --> Ag(s), which makes sense,

but the anode reaction given is AgBr(s) + e- --> Ag(s) + Br- (aq)
Why it wouldn't just be Br(s) + e- --> Br- (aq), similar to the cathode reaction?

Sebastian Lee 1L
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Joined: Fri Aug 09, 2019 12:15 am
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Re: Question on 6L.7

Postby Sebastian Lee 1L » Thu Mar 05, 2020 11:50 am

To study the reaction given, you have to use both oxidation and reduction half-reactions to end up with AgBr (s) from Ag+ and Br-. If your Ag+ is being reduced to form Ag (s), you know that there must be another oxidation half-reaction that uses Br- to form your desired AgBr(s) (not just Br). If you look at the appendix chart, you will see that the reaction to form AgBr requires Ag (s) from your reduction reaction and Br- to form AgBr while losing electrons. By balancing these equations you can get the overall desired reaction.

Jessa Maheras 4F
Posts: 121
Joined: Fri Aug 02, 2019 12:16 am

Re: Question on 6L.7

Postby Jessa Maheras 4F » Sat Mar 07, 2020 3:47 pm

A shorter version of putting the above, in order for AgBr to show up in the balanced redox rxn, you have to include it in one of your half runs. :)

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