Calculate the emf of the cell.
`Mg(s)|Mg^(2+) (0.2 M)||Ag^(+) (1xx10^(-3))|Ag`
`E_(Ag^(+)//Ag)^(@)=+0.8` volt, `E_(Mg^(2+)//Mg)^(@)=-2.37` volt
What will be the effect on emf If concentration of `Mg^(2+)` ion is decreased to `0.1 M` ?

Answer

3.0134`" V "`; increases

Explanation

Step I. Calculation of EMF of the cell
Cell reaction : `Mg(s)+2Ag^(+)(1xx10^(-3)" M")to Mg^(2+)(0.2" M") +2Ag(s)`
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)])/([Ag^(+)]^(2))=[0.8-(-2.37)]-(0.0591)/(2)"log"(0.1)/((1xx10^(-4))^(2)`
`=3.17-0.02955" log "(2xx10^(5))=3.17-0.02955xx5.3010`
`=3.17-0.15664=3.0134" V "`
Step II. Effect of decreasing concentration of `Mg^(2+)` ions on emf of cell
The EMF of cell wil increase on decreasing the concentration of `Mg^(2+)` ions in solution.