If I have a Galvanic cell of Nickel and Gold where Nickel is (Ni->Ni^+2 + 2e-) and the oxidation of gold is (Ag^+->Ag + e-) then do I need to double the cell potential of gold (+0.80) if I'm trying to find the standered cell potential (volatage)?
If you add the Au (note Au and not Ag or it may be the other way around; i.e., you may have called Ag gold whereas it is silver) Eo value to the Ni Eo value (one an oxidation and the other a reduction), no, you do not double the E value of the Ag.
To find the standard cell potential (voltage) of a galvanic cell, you need to consider the individual half-cell potentials and the stoichiometric coefficients of the overall cell reaction. The standard cell potential is calculated by summing the reduction potentials of the reduction half-reaction and subtracting the oxidation potential of the oxidation half-reaction.
In your case, the reduction half-reaction for nickel is Ni^+2 + 2e- -> Ni, and the reduction half-reaction for gold is Ag^+ + e- -> Ag. However, there seems to be an error in your question, as you mentioned "oxidation of gold" instead of "reduction of gold." Therefore, instead of using the oxidation potential of gold, you should use the reduction potential.
To calculate the standard cell potential, you need to sum the reduction potentials while considering their stoichiometric coefficients:
Ni^+2 + 2e- -> Ni: Reduction potential = ? (let's call this E°1)
Ag^+ + e- -> Ag: Reduction potential = +0.80 V (let's call this E°2)
To determine the standard cell potential, you subtract the oxidation potential from the reduction potential:
Standard cell potential (E°cell) = E°2 - E°1
So, to answer your question, you do not need to "double" the cell potential of gold, but instead, you subtract the reduction potential of the nickel half-reaction from the reduction potential of the gold half-reaction to find the standard cell potential.