## 6th edition 14.13 d [ENDORSED]

Eshwar Venkat 1F
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### 6th edition 14.13 d

For 14.13 d), I was confused as to why the cell diagram was written as is. I thought that the Au+ ion was being reduced to Au (s) and oxidized to Au3+. So why isn't the cell diagram Au+(aq) | Au3+(aq) || Au+(aq) | Au(s). Does it have something to do with the fact that Au (s) is also a pure conductor?

Chem_Mod
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### Re: 6th edition 14.13 d

Can you please post the cell diagram from the book for reference? They are writing it in this manner because they would need to add an electrode, and solid gold can serve as that electrode in the solution.

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### Re: 6th edition 14.13 d

The cell diagram from the book is Au (s) | Au +3 (aq) || Au+ (aq) | Au (s). So the only reason that this is written this way is because the Au (s) can serve as the conductor? For the half reactions in the appendix there is Au 3+ --> Au (s) and Au + --> Au (s) but no Au+ --> Au 3+. Does this have something to do with why the cell cannot be written as shown above?

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### Re: 6th edition 14.13 d  [ENDORSED]

Yes, because you do not have a half reaction using Ag+, and you would need a conducting electrode, you would simply use Ag(s) as your conducting electrode.

Eshwar Venkat 1F
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Joined: Fri Sep 28, 2018 12:22 am

### Re: 6th edition 14.13 d

I understand that Au (s) is serving as a conducting electrode in this case, but isn't the oxidation reaction that's occurring in this problem
Au+(aq) --> Au3+ (aq) + 2e- ? Is it not necessary to include Au+ when writing out the cell diagram for the anode side?

Jeannine 1I
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### Re: 6th edition 14.13 d

Eshwar Venkat 1F wrote:I understand that Au (s) is serving as a conducting electrode in this case, but isn't the oxidation reaction that's occurring in this problem
Au+(aq) --> Au3+ (aq) + 2e- ? Is it not necessary to include Au+ when writing out the cell diagram for the anode side?

You base off your cell diagram from the half-reactions found in the back of the book, we don't treat these problems the same way we would when balancing out redox reactions!

The oxidized reaction (anode) would then be Au(s) ------> Au$^{3+}$(aq) + 3e$^{-}$

You can see that there is no Au+ on this side.

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