Given standard reduction potentials $E^{\circ}(\mathrm{Cu^{2+}/Cu}) = +0.34$ V and $E^{\circ}(\mathrm{Zn^{2+}/Zn}) = -0.76$ V, the standard EMF of the Daniell cell is:

Cu is the cathode (higher reduction potential), Zn is the anode. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (+0.34) - (-0.76) = +1.10$ V. The positive value confirms the reaction is spontaneous as written. A common slip is to drop the negative sign on Zn, giving $+0.42$ V — option A is the trap.