The emf of a galvanic cell, with electrode potentials of silver $=+0.80\mathrm{V}$ and that of copper $=+0.34\mathrm{V}$, is 

Reduction potential is higher for silver, thereby it is at cathode.

$$\begin{gathered} {\mathrm{E}}^{\mathrm{o}}={\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{o}}-{\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}} \\ =0.80\mathrm{V}-0.34 \mathrm{V}=+0.46\mathrm{V} \end{gathered}$$.