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How would you determine the standard reduction potential of the system Mg2+∣Mg ? Select the correct answer from above options

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Set up a cell consisting of Mg∣MgSO4(1M) as one electrode ( by dipping a magnesium rod in 1MMgSO4 solution ) and standard hydrogen electrode Pt,(1atm)H⊕∣(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|Mg2+(1M)||H⊕(1M)||H⊕(1M)|H2,(1atm),Pt E c- cell =Ec-_((H⊕∣H2))-E c- (Mg2+∣Mg) Put E c- H⊕∣H2 =0 Hence, E c- Mg2+∣Mg =-E c- cell

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