The emf of a cell involving the following re 

$2{\mathrm{Ag}}^{+}+{\mathrm{H}}_2⟶2\mathrm{Ag}+2{\mathrm{H}}^{+}$  is 0.80 volt. The

 standard oxidation potential of silver electrode is: 

For the cell reaction:

 ${\mathrm{H}}_2\left|2{\mathrm{H}}^{+}\right|2\mathrm{Ag}\left|2\mathrm{Ag}\right|$

  $$\begin{gathered} {\mathrm{H}}_2\to 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-} \\ {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag} \\ \mathrm{EMF}=0.8 \mathrm{V} \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}^{\circ }={\mathrm{E}}_{\mathrm{cathode}}^{\circ }-{\mathrm{E}}_{\mathrm{anode}}^{\circ } \\ 0.80 \mathrm{V}={\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\circ }-0 \\ 0.80 \mathrm{V}={\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\circ } \\ -0.80 \mathrm{V}={\mathrm{E}}_{\mathrm{Ag}/{\mathrm{Ag}}^{+}}^{\circ } \end{gathered}$$.