## 14.97

Anna Goldberg 2I
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### 14.97

What does it mean when it asks to calculate Ka?
As well, why does the solutions manual choose the half reaction with the largest potential to be the anode? I thought the half reaction with the largest potential was the cathode, seeing as it has the most oxidizing power and is thus most likely to be reduced.

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

### Re: 14.97

Ka is the acidic constant, which means it is for acids dissociation reaction.
in acid dissociation reaction, H+ is dissociated. Therefore, H+ will always be on the product side. In this way, you write down the whole reaction happens.
For example, in practice exam, $HF\Leftrightarrow H^{+}+F^{-}$. Ka = $\frac{[H^{+}][F^{-}]}{[HF]}$.

Therefore, the half-reaction given in the question should be reversed. The E value would be -3.03V
And in the appendix, F2+2e- --- 2F-, E0 = +2.87V.
You add them up, get the Ecell value equals to -0.16V. This is not a favorable reaction.
So the largest potential in this reaction is anode and smaller potential is cathode.