Calculate the cell potential for this voltaic cell at 298K if it is constructed using solutions of .25M Ni(NO3)2 and .01M Cu(NO3)2

Question

Any explanation would be spectacular

Answer 1

Look up Eo for Ni^2+ + 2e ==> Ni(s) as a reduction potential.
Do the same for Cu^2+ + 2e ==> Cu(s)
Reverse the Ni equation, change sign of Eo for Ni and add Ni Eo(as an oxidation) to Cu as a reduction to arrive at Eocell and this equation.
Ni(s) + Cu^2+(0.01M) ==> Ni^2+(0.25M) + Cu(s)
Then
Ecell = Eocell - (0.0592/2)log Q
where Q = (Ni^2+)(Cu)/(Ni)(Cu^2+)