14.13 d

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sofiakavanaugh
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Joined: Thu Jul 13, 2017 3:00 am
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14.13 d

Postby sofiakavanaugh » Thu Mar 15, 2018 12:14 pm

Write the 1/2 rxn, balanced equation, calculate E cell and write the cell diagram for the chemical rxn: Au+(aq) --> Au(s) + Au3+

I got this for the cell diagram: Pt(s)| Au+(aq), Au 3+(aq) || 2Au+(aq) | 2Au(s)

Can someone please explain to me why this is wrong and how to get the correct diagram?

Thanks!

Hubert Tang-1H
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

Re: 14.13 d

Postby Hubert Tang-1H » Thu Mar 15, 2018 1:02 pm

We split the reaction into the anode and cathode reactions:
Anode: Au(s)--->Au3+ (aq) + 3e-
Cathode: Au+(aq)+e- ----> Au(s)

We see that both the anode and cathode have solid Au, so there shouldn't be Pt in the cell diagram.

Thus, from our anode and cathode, our cell diagram should be

Au(s)| Au3+(aq)||Au+(aq)|Au(s)

Angela 1K
Posts: 80
Joined: Fri Sep 29, 2017 7:05 am

Re: 14.13 d

Postby Angela 1K » Sat Mar 17, 2018 5:32 pm

How are we supposed to know that the oxidation reaction is Au3+(aq) + 3e- --> Au(s) if, in the reaction given Au and Au3+ are on the same side (products)?


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