if the reduction potential for
Cu^2+ + 2e---> Cu is 0.00V. Then the remaining reduction potentials for the voltages between Cu[cathode]/Zn[anode] is like???
Ecell=1.043v
My answers:Zn-->Zn2+ 2e- voltage = -1.043V
Yes, that's correct as a reduction potential but you wrote the Zn ==> Zn^+ + 2e as an oxidation. The way you wrote the half cell is the way it is operating spontaneously in the cell; however, the question asked for the reduction potential.
