What is the delta G naught for the cell in question 1 at 298 K?

Question 1: Which is spontaneous, the oxidation of copper to Cu2+ or the reduction of Cu2+ to Cu?

OX: Cu --> Cu2+ + 2e-
RE: 2e- + Cu2+ --> Cu
Overall: Cu + Cu2+ --> Cu + Cu2+

Delta G knaught = -RTln(Keq)
= -(8.314)(298)ln(1)
= 0?
R=8.314 J/Kmol
T=298 K
Keq= [Cu2+]/[Cu2+] = 1

The problem I am having is with the Keq. I think its one, but that would make the entire delta G knaught 0? I am just a bit confused.

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asked by Kait
  1. You have no concns listed so you can't get Keq that way.
    Use dGo = -nEoF
    You will need to look up the value of Eo for the reaction. Obviously the one with Eo a positive number will give you a negative dGo and that is the one that is spontaneous. If you want Keq, you can get that AFTER you know dGo by
    dGo = -RTlnK.

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