How to determine anode and cathode in 6.57?

[Tiffany Chao 2H]

6.57: 6.57 Use the data in Appendix 2B and the fact that, for the half- reaction F2(g) + 2 H+ (aq) + 2 e- --> 2HF (aq), E* = + 3.03 V, to calculate the value of Ka for HF.

Based on F2 (g) + 2 e- --> 2F-, E*= +2.87V, how am I suppose to know which one is the cathode and anode?

Thanks

[Adam Kramer 1A]

this is the half-reaction for the cathode. We know this because there are electrons being added to the reactants, reducing the Fluorine.

[JonathanS 1H]

Because you are given E^o for the whole reaction, we can use E^o = E_cathode - E_anode, all we need is the E value for the other half reaction and arrange the equation to give us the given E^o value.

[Tiffany Chao 2H]

this is the half-reaction for the cathode. We know this because there are electrons being added to the reactants, reducing the Fluorine.

Sorry I meant F2 (g) + 2 e- --> 2F- is the cathode and F2(g) + 2 H+ (aq) + 2 e- --> 2HF (aq) is the anode according to the solutions manual. It says the E*cell = -0.16V

[Chris Charton 1B]

The cathode will be the reaction with the larger E* value.

[Natalie Benitez 1E]

The cathode should be the one with the larger E value.

[Emily Burghart 1k]

The cathode should be the one with the larger E value.

Why again is the cathode the one with the larger E value?