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

Explanation

`E^(0)` vlaue of `Mg^(2+)//Mg` electrode is determined by setting up an electrochemical cell. For this purpose, a Mg electrode is dipped in `1 M Mg SO_(4)` solution, which acts as one half cell i.e., oxidation half-cell.
In the same way, the standard hydrogen electrode acts as the other half-cell i.e., reduction half cell. The deflection of voltmeter placed in the cell circuit is towards the Mg electrode which indicates the direction of flow of current.
The cell may be represented as : `Mg|Mg^(2+)(1M)||H^(+)(M)|H_(2) (1 atm).` Pt The reading as given by boltmeter gives `E_(cell)^(0)`
`E_(cell)^(0) =E_((H^(+)//1//2H_(2)))^(0) -E_((Mg^(2+)//Mg))^(0) =0-E_((Mg^(2+)//Mg))^(0)`