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

Explanation

Set up a cell consisting of `Mg|MgSO_(4)(1M)` as one electrode `(` by dipping a magnesium rod in `1M MgSO_(4)` solution `)` and standard hydrogen electrode `Pt,(1 atm) H^(o+)|(1M)` as the second electrode and measure the `EMF` of the cell. Also note the direction of deflection in the voltmeter. 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. Hence, the cell may be represented as follows `:`
`Mg|Mg^(2+)(1M)||H^(o+)(1M)||H^(o+)(1M)|H_(2),(1 atm),Pt`
`E_(cell)^(c-)=E^(c-)_((H^(o+)|H_(2)))-E_((Mg^(2+)|Mg))^(c-)`
Put `E_(H^(o+)|H_(2))^(c-)=0 `
Hence, `E_(Mg^(2+)|Mg)^(c-)=-E_(cell)^(c-)`