The correct answer is the difference in reduction potentials of the half-cells.
E(cell)=E(cathode)-E(anode).
You can also think of it as the reduction potential of the cathode plus the oxidation potential of the anode since if you reverse the sign of the reduction potential you have the oxidation potential
example E(cathode)=0.7V and E(anode)=-1.2V. E(cell)=0.7V+1.2V=1.9V
(the reduction potential for the anode is -1.2V but the oxidation potential is 1.2V)
I hope this helps. Let me know if anything is unclear.