Calculate the EMF of the cell,   $\mathrm{Mg}/{\mathrm{Mg}}^{2+}\left(1\mathrm{M}\right)//{\mathrm{Cu}}^{2+}\left(1\mathrm{M}\right)/\mathrm{Cu}.$

(Given  ${\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^0=-2.37 \mathrm{V} ; {\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0=+0.34 \mathrm{V} ;$)

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^0=-2.37 \mathrm{V} \mathrm{and} {\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0=+0.34 \mathrm{V}.; \\ \mathrm{Both} \mathrm{are} \mathrm{S}.\mathrm{R}.\mathrm{P} \mathrm{values}.; \end{gathered}$$

From the cell notation ,

Magnesium electrode ( L.H.E ) is the anode.

Copper electrode ( R.H.E ) is the cathode.

${\mathrm{E}}_{\mathrm{cell}}^{°} ={\mathrm{E}}_{\mathrm{cathode}}^{°}-{\mathrm{E}}_{\mathrm{anode}}^{°}=+0.34 \mathrm{V}-(-2.37) \mathrm{V} =+2.71 \mathrm{V}$