Calculating K for reaction using half reactions [ENDORSED]

[Julia_Gordon_2A]

In the textbook example 14.8 on page 586, the question asks to calculate the equilibrium constant at 25 degrees C for AgCl(s)--->Ag+(aq)+Cl-(aq). The two half reactions that are required to use are
R: AgCl(s) + e- ---->Ag(s) +Cl-(aq) E=+0.22V
L: Ag+(aq) +e- -----Ag(s) E=+0.80V

It says that the second equation is reversed. Therefore, wouldn't it make that equation the anode? I thought that the more positive E value was the cathode? Could someone please explain how to know which equation to reverse and how the value of E ties into that?

Thank you!

[Beata_Vayngortin_3L]

you flip the equation where flipping the E value will give you the largest most positive E value for your overall redox reaction.

[Beata_Vayngortin_3L]

you flip the equation where flipping the E value will give you the largest most positive E value for your overall redox reaction.

[Julia_Gordon_2A]

But this would mean that Ecell=0.22-0.80=-0.58V

Why is the negative value accepted in this case?

[Zulfiqar Lokhandwala 1H]

In the explanation for the question it says that we should expect E to be negative because the K to be very small, and in order of that to happen the the value of E(naught) must be negative