How would you determine the standard electrode potential of the system `Mg^(2+) Mg`?

Explanation

A cell consisting of `Mg | MgSO_(4)` (1 M) as one electrode (by dipping a magnesium wire in 1 M `MgSO_4` solution) and standard hydrogen electrode Pt, `H_(2)` (1 atm) | `H^(+)` (1 M) as the second electrode is set up and emf of the cell and the direction of deflection in voltmeter are noted. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. The cell may be represented as follows :
`Mg | Mg^(2+) (1 M) || H^(+) (1 M) | H_(2)`, (1 atm), Pt
or `E_("cell")^(@) = E_(H^(+)//1//2H_(2)) -E_(Mg^(2+)//Mg)^(@)`
Put `E_(H^(+)//1//2H_(2)O) =0`
Hence, `E_(Mg^(2+)//Mg)^(@) =-E_("cell")^(@)`