Effects of a non-inert electrode in a half cell reaction involving hydrogen gas as the electrolyte ion
If you started with a full cell with two inert electrodes immersed in sulfuric acid and applied a large enough voltage between them, a current would flow. At one electrode, the easiest to oxidize species would be oxidized (in this case, that would probably be 2 H2O -> O2 + 4 H+ + 4e-). At the same time, the easiest to reduce species would be reduced at the other electrode (probably 2H+ + 2e- -> H2). If you switched the cathode electrode from some inert metal to copper, the easiest reduction would still be 2H+ + 2e- -> H2 (Remember, there is no Cu2+ in the system to reduce). If, on the other hand, you changed the anode from inert metal to copper, the easiest oxidation would probably be Cu -> Cu2+ + 2e-, so the net reaction would change.