Step 1: Understanding the reaction.
The given reaction is:
\[
2 \text{Ag}^+ (aq) \rightarrow \text{Ag} (s) + \text{Ag}^{2+} (aq)
\]
The standard electrode potentials are:
\[
\text{Ag}^+ (aq) + e^- \rightarrow \text{Ag} (s), E^o = 0.62 \, \text{V}
\]
and
\[
\text{Ag}^{2+} (aq) + e^- \rightarrow \text{Ag}^+ (aq), E^o = 0.12 \, \text{V}
\]
Step 2: Calculating the standard emf.
The standard emf for the reaction is:
\[
E^o = E^o (\text{cathode}) - E^o (\text{anode})
\]
Here, \( \text{Ag}^+ \) is reduced (cathode) and \( \text{Ag}^{2+} \) is oxidized (anode). So:
\[
E^o = 0.62 - 0.12 = 0.50 \, \text{V}
\]
Step 3: Conclusion.
The standard emf of the cell is 0.50 V.