14.13 part d

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Joshua Hughes 1L
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14.13 part d

Postby Joshua Hughes 1L » Sun Feb 25, 2018 9:00 pm

I really don't understand why one of the half-reactions seems flipped for this problem. I looked at a recent post and still didn't understand it.
I thought the half reactions would be slit so that Au+(aq) goes to Au3+ and Au respectively but for some reason one of the half-reactions is Au goes to Au3+ which doesn't make sense to me because Au wasn't a reactant but a product that was formed from Au+. Au did not form Au3+ right?

Mitch Mologne 1A
Posts: 74
Joined: Fri Sep 29, 2017 7:04 am

Re: 14.13 part d

Postby Mitch Mologne 1A » Sun Feb 25, 2018 9:02 pm

Au becomes a reactant because you flip the anode half reaction.

Sabah Islam 1G
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: 14.13 part d

Postby Sabah Islam 1G » Sun Feb 25, 2018 9:58 pm

One of the half reactions is flipped because one half reaction is the anode and the other is the cathode. Au goes to Au3+ because the standard reduction potential of that is lower than that of the other reaction, and whichever has the lower reduction potential is the anode. Therefore, to make it an anode in which electrons are lost, the reaction has to be flipped.


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