E = Eo - (0.0592/n)log[(Cu)/(Cu^+2)]
Substitute 2 for n and 0.05 M for Cu^+2 and 0.337 for Eo. Solve.
Calculate the electrode potential of a copper half-call containing 0.05M Cu(NO3).
E^0 Cu2+/Cu = 0.337 V
3 answers
Am I solving for Cu or do I plug a value in for that also?
Answered above.