How would you determine the standard electrode potential of the system $Mg^{2+} | Mg$?

(N/A) The standard electrode potential of $Mg^{2+} | Mg$ is determined by constructing an electrochemical cell with the standard hydrogen electrode $(SHE)$.
$1$. Set up a cell with a magnesium electrode ($Mg$ rod dipped in $1 \, M \, MgSO_4$ solution) as the anode and a standard hydrogen electrode $(Pt_{(s)}, H_{2_{(g)}} (1 \, bar) | H^{+}_{(aq)} (1 \, M))$ as the cathode.
$2$. The cell representation is: $Mg | Mg^{2+} (aq, 1 \, M) || H^{+} (aq, 1 \, M) | H_2 (g, 1 \, bar), Pt_{(s)}$.
$3$. Measure the electromotive force $(emf)$ of this cell using a voltmeter.
$4$. The standard cell potential is given by $E^{\Theta}_{cell} = E^{\Theta}_{cathode} - E^{\Theta}_{anode}$.
$5$. Since the standard electrode potential of the $SHE$ is defined as $0.00 \, V$,we have $E^{\Theta}_{cell} = 0 - E^{\Theta}_{Mg^{2+}/Mg}$.
$6$. Therefore,$E^{\Theta}_{Mg^{2+}/Mg} = -E^{\Theta}_{cell}$.