Answer
2.46 V
$Al(s)+3Ag^{+}\rightarrow Al^{3+}+3Ag(s)$
Work Step by Step
From the standard reduction potentials in table 18.1, we write
$E^{\circ}_{Al^{3+}/Al}=-1.66\,V$
$E^{\circ}_{Ag^{+}/Ag}=0.80\,V$
$E^{\circ}_{Ag^{+}/Ag}$ is greater than $E^{\circ}_{Al^{3+}/Al}$ and so greater is the tendency to be reduced.
Therefore, the reaction at the cathode (reduction) is:
$3Ag^{+}+3e^{-}\rightarrow 3Ag(s)$
The reaction at the anode (oxidation) is:
$Al(s)\rightarrow Al^{3+}+3e^{-}$
The overall reaction is:
$Al(s)+3Ag^{+}\rightarrow Al^{3+}+3Ag(s)$
(We multiplied the reduction of $Ag^{+}$ by 3 to balance the overall equation. $E^{\circ}$ is not affected by this procedure.)
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=0.80\,V-(-1.66\,V)=2.46\,V$
