14.15A

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Dylan Mai 1D
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Joined: Sat Jul 22, 2017 3:00 am

14.15A

Postby Dylan Mai 1D » Sun Feb 18, 2018 8:59 pm

How do you know that the Anode reaction is AgBr + e- to Ag (s) + Br - and not Br2 + 2e- to 2Br-?

sahajgill
Posts: 20
Joined: Fri Sep 29, 2017 7:06 am

Re: 14.15A

Postby sahajgill » Sun Feb 18, 2018 9:52 pm

There is no 2Br in the equation

Swetha Sundaram 1E
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Re: 14.15A

Postby Swetha Sundaram 1E » Mon Feb 19, 2018 8:47 am

For this question, I was confused in how oxidation numbers were distributed in the original question because I thought Ag was +1 and Br was -1 and they stayed the same throughout because the ions split but this does not make sense in regards to reduction/oxidation half-reactions.

Christy Lee 2H
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Re: 14.15A

Postby Christy Lee 2H » Mon Feb 19, 2018 3:59 pm

For this problem the cell diagram in the solutions manual divides Ag(s) and AgBr(s) on the anode side with a vertical line. Can someone explain why they aren't separated by a comma even though they're the same phase?

Jana Sun 1I
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Joined: Sat Jul 22, 2017 3:00 am

Re: 14.15A

Postby Jana Sun 1I » Tue Feb 20, 2018 12:47 am

Christy Lee 2H wrote:For this problem the cell diagram in the solutions manual divides Ag(s) and AgBr(s) on the anode side with a vertical line. Can someone explain why they aren't separated by a comma even though they're the same phase?


This is addressed in this thread: viewtopic.php?f=140&t=28121. I also think it might be because Ag(s) and AgBr(s) aren't ions. On page 574 of the book, it says that ions are only separated by a comma in the same state.


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