Calculate equilibrium constant for the reaction :
`Mg(s) |Mg^(2+)(0.001 M) || Cu^(2+)(0.0001 M)|Cu(s)`
Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V " , E_((Cu^(2+)//Cu))^(@)=0.34" V "`

Explanation

The electrode reactions and cell reactions are :

Since the reaction involves 2 moles of electrons and therefore , n =2 and the Nernst equation for the cell at 298 K is :
or `E_(cell) = E_(cell)^(Theta) - (0.059)/(2) "log" ([Mg^(2+) (aq)])/([Cu^(2+) (aq)]) " " ... (i)`
`E_(cell)^(Theta) = E^(Theta) (Cu^(2+) |Cu)- E^(Theta) (Mg^(2+) | Mg)`
`E^(Theta) (Cu^(2+) | Cu) = 0.34 V , E^(Theta) (Mg^(2+) |Mg) = -2.37 V`
`E_(cell)^(Theta) = 0.34 - (-2.37) = 2.71 V`
Substituting the values in Eq. (i)
`E_(cell) = 2.71 - (0.059)/(2) "log" (0.001)/(0.0001)`
`E_(cell) = 2 . 71 - 0.0295 = 2.6805 V`