Textbook 6.57 Question about Favorability, choosing Cathode, anode

[Sam Wentzel 1F 14B]

E* = 3.03 V for F2(g) + 2 H+(AQ) + 2 e- -> 2 HF (aq), E* = 3.03 V
Find Ka of HF

the E* value for F2 + 2 e- -> 2 F- is 2.87 V.

If we want the cell to be favorable, we want a positive E*cell value right?

So why in the solutions manual do they do E*cell = 2.87 V - 3.03 V? = -0.16 V
(Why do they choose HF reaction to be the anode, and the 2F- reaction to be the cathode?)

Additionally, how do we find Ka after calculating K?

[Chem_Mod]

Normally if you are determining the cell potential of a Galvanic cell, you want the most positive cell potential. However, in this problem, you are not constructing a Galvanic cell; instead, you are simply trying to find Ka for HF.

The K that you find using the Nernst equation is for double the reaction you want. Therefore, to get K for the single reaction (dividing the overall reaction by 2), you need to take the square root of K (raise K to the 1/2 power).

[FionaHunter21]

Why doesn't the 3.03V become negative when the equation gets flipped?

[Alisa Nagashima 1B]

Why doesn't the 3.03V become negative when the equation gets flipped?

I believe the 3.03V becomes negative.
E°cell = E°cathode - E°anode
-0.16V = +2.87V - 3.03V