## 14.23A

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Elizabeth Ignacio 1C
Posts: 39
Joined: Fri Sep 29, 2017 7:07 am

### 14.23A

One of half reactions needed to calculate the overall cell potential in 14.23A is 2Hg -> Hg22+ + 2e- and the cell potential for the reverse reaction in the textbook is listed as +0.79, which the solutions manual uses in the calculations for Eocell. I'm confused why they didn't flip the cell potential to get -0.79 if the positive version was the reverse of the reaction that was actually happening.

manasa933
Posts: 72
Joined: Fri Sep 29, 2017 7:04 am

### Re: 14.23A

I'm not really sure why but my TA said that we don't flip potentials because the redox couple is the same.

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