6L.5 (d) half reactions

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inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

6L.5 (d) half reactions

Postby inlovewithchemistry » Tue Feb 26, 2019 7:24 pm

can someone help me understand how to write the half reactions for part (d) of 6L.5? the equation is as follows:

Au+(aq)-->Au(s) + Au3+(aq)

The sol'n manual says that the half reactions are :
Au+ + e- --> Au(s)
Au3+ + 3e- -->Au(s)

but i dont understand how to get there. thank u!

Gracie Ge 2E
Posts: 28
Joined: Wed Nov 21, 2018 12:19 am

Re: 6L.5 (d) half reactions

Postby Gracie Ge 2E » Tue Feb 26, 2019 9:30 pm

i think for this problem you'll have to look at the standard reduction potentials page. the two possible reactions are
Au+ + e- = Au (+1.69v)
and
Au3+ + 3e- = Au (+1.40v)
from the reduction potentials we know that the first reaction is the cathode and the second one is the anode. so we invert the second one.
so the reaction as a whole is basically forming Au(s) from Au+(aq), then forming Au3+(aq) from Au (s).

204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

Re: 6L.5 (d) half reactions

Postby 204929947 » Tue Feb 26, 2019 9:37 pm

how do you tell the difference between a cathode and an anode?????

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

Re: 6L.5 (d) half reactions

Postby inlovewithchemistry » Tue Feb 26, 2019 10:53 pm

204929947 wrote:how do you tell the difference between a cathode and an anode?????

A cathode is what is being reduced, or gaining electrons. An anode is what is being oxidized, or losing electrons.


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