6M.1

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Brian_Ho_2B
Posts: 221
Joined: Fri Aug 09, 2019 12:16 am

6M.1

Postby Brian_Ho_2B » Wed Feb 26, 2020 9:26 am

I cannot wrap my head around this one. It asks us to calculate E (M2+/M) for the cell diagram M|M2+||Cu2+|Cu and we are given that Cu|Cu2+ is +0.34 and that the E cell is -0.689. However, calculating E(M/M2+) using E cell = E cathode - E anode gives me positive 1.029. The answer is -0.349. This answer could only be obtained if we do E anode - E cathode, based on the cell diagram. Why is this the case? Is it because the cell diagram is reversed? If so, how would we know that the cell diagram is reversed? The answer to this problem isn't consistent with the examples that the book provides.

KBELTRAMI_1E
Posts: 108
Joined: Sat Jul 20, 2019 12:17 am

Re: 6M.1

Postby KBELTRAMI_1E » Wed Feb 26, 2020 10:44 am

Without seeing your calculations, chances are one of two things happened. Either you did E anode- Ecathode which is backwards (should be Ecathode-Eanode) or you just made a minor calculation error with the double negative. Since the anode has a negative value AND is being subtracted, it will become positive and be subtracted from both sides, rather than added.

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

Re: 6M.1

Postby Brooke Yasuda 2J » Wed Feb 26, 2020 11:05 am

I think what happened is that the equation is Ecell = Eanode - Ecathode. So you have -.698 = Eanode - (.34 V). So I think you just forgot that the species that is oxidized is subtracted from the species that is reduced.

Betania Hernandez 2E
Posts: 107
Joined: Fri Aug 02, 2019 12:15 am

Re: 6M.1

Postby Betania Hernandez 2E » Wed Feb 26, 2020 8:09 pm

We know that the cell diagram is reversed when the cell potential is negative. The cell potential was given as -0.689 which means that the cathode is the left of the cell diagram and the anode is the right side of the cell diagram. This makes the unknown metal the cathode and the copper the anode.

Megan Cao 1I
Posts: 103
Joined: Sat Sep 07, 2019 12:18 am

Re: 6M.1

Postby Megan Cao 1I » Wed Feb 26, 2020 9:47 pm

using the Ecell = Ecathode - Eanode equation,
since Ecell= -0.689 and the cell diagram M|M2+||Cu2+|Cu, where Cu|Cu2+ is +0.34, shows that the copper is the anode side:
-0.689 = Ecathode -(0.34) (all in volts, V)
so you get Ecathode = -0.349 V.

based off the diagram you have to see that the cell diagram is flipped, so the copper seems like the cathode side, but it's really the anode.

chari_maya 3B
Posts: 108
Joined: Sat Sep 07, 2019 12:18 am

Re: 6M.1

Postby chari_maya 3B » Sat Feb 29, 2020 2:40 pm

Why is the cell diagram reversed when the potential is negative? Is it because reversing is the only way to make the reaction favorable/spontaneous?


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