Answer
Standard emf= 2.71 V
Equation of the cell reaction is:
$Cu^{2+}+Mg(s)\rightarrow Mg^{2+}+Cu(s)$
Work Step by Step
From the standard reduction potentials in table 18.1, we write
$E^{\circ}_{Mg^{2+}/Mg}=-2.37\,V$
$E^{\circ}_{Cu^{2+}/Cu}=0.34\,V$
$E^{\circ}_{Cu^{2+}/Cu}$ is greater than $E^{\circ}_{Mg^{2+}/Mg}$ and so greater is the tendency of $Cu^{2+}$ to be reduced.
Therefore, the reaction at the cathode (reduction) is:
$Cu^{2+}+2e^{-}\rightarrow Cu(s)$
The reaction at the anode (oxidation) is:
$Mg(s)\rightarrow Mg^{2+}+2e^{-}$
The overall reaction is:
$Cu^{2+}+Mg(s)\rightarrow Mg^{2+}+Cu(s)$
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=0.34\,V-(-2.37\,V)=2.71\,V$
