What will be e.m.f of the cell  $2{\mathrm{Ag}}^{+}+\mathrm{Cu}\to {\mathrm{Cu}}^{+2}+2\mathrm{Ag}$  , Given
$$\begin{gathered} {\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\mathrm{O}}=+0.8\mathrm{V}; \\ {\mathrm{E}}_{{\mathrm{Cu}}^{+2}/\mathrm{Cu}}^{\mathrm{O}}=+0.3\mathrm{V} \end{gathered}$$

Among the two electrodes that are combined to construct a cell, the electrode with smaller reduction potential will acts as anode and another electrode will acts as cathode. 
As the SOP of Ag electrode is -0.8V, hence its SRP is +0.8V and that of Cu is +0.3V. Therefore, Cu acts as anode and Ag electrode acts as cathode. EMF of cell = E_cathode – E_anode = E_Ag – E_Cu = 0.8 – 0.3 = +0.5