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

Explanation

The electrode of Mg dipping in the solution of its own.ions `Mg^2+` is. connected to Standard Hydrgen Electrode (SHE) through a śalt-bridge internally and through voltmeter in the outer circuit to complete the cell. `Mg(s)] Mg^2+ | dotH^+(aq) mid H_2(~g): Pt`
At anode: `Mg(s) long Mg^2++2 e^--` oxidation
At čathode : `2 H^+(aq)+2 e^- H_2(~g)` - Reduction
Cell .reaction is `: Mg(s)+2 H^*(aq) long`
`Mg^2+(aq)+H_2(~g)`
The e.m.f. of the cell is determined whịch is `2.36 ~V`
`2.36=0-E_M e^0 cdot or E_M_e^circ=-2.36 ~V`