What is the voltage for a cell constructed of Ni(s) in 1.0 x 10-7M Ni+2 and Ag(s) in 1.0 x 10-5 M Ag+?

The Nernst Equation: Ecell = Estandard - (RT/nF)lnQ
at 298 k ( assuming T = 25 deg. ) : Ecell= Estandard - (0.05916 /n)ln Q

n: number of electrons in half cell and Q is reaction quotient of initial concentrations.

for you question:
Ni2+ (aq) + 2e– —> Ni(s) Ehalf-cell = –0.257 V oxidation
2Ag+(aq) + 2e– —> 2Ag(s) Ehalf-cell = +0.80 V reduction

==> Ecell = (0.8 - (-0.257)) - (0.05916 / 2) * ln ([Ni2+]/[Ag+]^2)
Ecell = (1.057) - (0.05916/2) * ln (1.0 x 10-7/(1.0 x 10-5)^2) = 0.8526
Ecell = 0.8526 volts