6L.3 e

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Rafsan Rana 1A
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Joined: Sat Aug 24, 2019 12:16 am

6L.3 e

Postby Rafsan Rana 1A » Mon Mar 02, 2020 6:07 pm

for 6L.3 part e, the for the half reaction in the anode the solutions manual says
Sn4+ + 2e- --> Sn2+, but the book problem doesn't show Sn2+, just Sn2 so aren't we supposed to assume it is two molecules of Sn instead a charge of 2+ so the balanced equation should be 2Sn4+ + 8e- -->Sn2?

Charlyn Ghoubrial 2I
Posts: 50
Joined: Wed Nov 13, 2019 12:26 am

Re: 6L.3 e

Postby Charlyn Ghoubrial 2I » Mon Mar 02, 2020 11:45 pm

If I’m looking at the correct question, your half reaction for oxidation (at anode) should be Sn^(2+) —> Sn^(4+) +2e- and your half reaction for reduction (at cathode) should be Hg2Cl2 + 2e- —> 2Hg + 2Cl-
So then your overall balanced equation after you cancel out the 2e- will be Hg2Cl2 + Sn ^(2+) —> Sn^(4+) + 2Hg + 2Cl-


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