Answer:
The correct answer is "+2.37 V".
Explanation:
Given that,
Concentration of [tex]Mg^{2+}[/tex]
= 1.00 M
[tex]E^0 \ Mg^{2+}/Mg[/tex],
= -2.37 V
[tex]E^0 \ \mu^+/H_2[/tex],
= 0.00V
When S.H.E as well as anode act as cathode, then
⇒ [tex]E^0 cell=E^cathode-E^0 anode[/tex]
By putting the values, we get
⇒ [tex]=0.00V-(-2.37 V)[/tex]
⇒ [tex]=+2.37 \ V[/tex]