## 14.97

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

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JadeSebti1L
Posts: 35
Joined: Fri Sep 28, 2018 12:19 am

### 14.97

The two standard potentials are +2.87V and +3.03V so wouldn't the latter be the cathode, making the total cell potential positive?
The solutions manual makes +2.87V potential the cathode potential.
Also, why do you get Ka when you take the square root of K?

Kristen Kim 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:16 am

### Re: 14.97

In the half reactions, F2(g) is being reduced, so this is the cathode, and HF(aq) is being oxidized, so this is the anode.

305113590
Posts: 66
Joined: Fri Sep 28, 2018 12:28 am

### Re: 14.97

This question is unique because this is an acid-base reaction. Hence we know that for this question, the weak acid will donate an electron:
HF<--->H+(aq) + F-(aq)

This means that the Ka= [H+][F-]/[HF]. Because the half reactions must adhere to this acid base reaction, the Ecell value is negative, -0.16V with the overall equation being 2HF(aq)--->2H+(aq) +2F-(aq).

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