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### 6M.1 reversed anode and cathode?

Posted: Sun Mar 01, 2020 9:17 pm
the given cell diagram is M(s)|M1(aq)||Cu21(aq)|Cu(s), so in using the Ecell = E(right)- E(left) i set the copper as the cathode/right electrode and calculated as usual. i got -1.029 V instead of the answer the book gives.

the solutions manual reverses the anode and cathode roles--is this because the Ecell value is negative? in which case, is it better to think of the Ecell equation as Ecell = E(cathode) - E(anode), rather than "right" and "left"?

### Re: 6M.1 reversed anode and cathode?

Posted: Sun Mar 01, 2020 10:46 pm
Yeah I think the book reversed it because the value was negative and you typically shouldn't use negative values for the cell. And I think it is also more accurate to use E(cathode) - E(anode)

### Re: 6M.1 reversed anode and cathode?

Posted: Sun Mar 01, 2020 10:55 pm
even though the book wrote the cell diagram like that, you have to look at what is being reduced/ oxidized and then rewrite the cell diagram based on that. Also since the given Ecell is negative, you know that the given isn't spontaneous so you need to switch the Cu and M to make it spontaneous.

### Re: 6M.1 reversed anode and cathode?

Posted: Sun Mar 01, 2020 11:19 pm
so in general, I think it's a really good idea to always think of it in anode/cathode terms over left/right terms, just in case a question switches it on you. In this case, they actually do flip the cell diagram so you have to be a tad careful. so the standard reduction potential of Cu2+/Cu is +.34V, and in this case, you have this presented as the reduction. However, they're specifically asking for the Cu/Cu2+ cell, so you have to make this .34 negative (but place it in the cathode term, because in the cell diagram, it would be the cathode). So you'll have the equation:

-.649 = -.34 - Eanode. Solving for Eanode, you get +.349.
However, this value is in terms of M/M2+. You're looking for the potential in terms of M2+/M, so you have to make +.349 negative to get your answer of -.349!