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` ?

Explanation

`E_("cell")^(@)=E_("cathode")^(@)-E_("Anonde")^(@)`
`=0.80-(-2.37)=3.17"volt"`
Cell reaction, `Mg+ 2Ag^(+) to 2Ag +Mg^(2+)`
`E_("cell")=E_("cell")^(@)-(0.0591)/(n) log (Mg^(2+))/([Ag^(+)]^(2))`
=`3.17 -(0.0591)/(2) log (0.2)/([1 xx 10^(-3)]^(2))`
`=3.17-0.1566=3.01334" volt"`
When `Mg^(2+)=0.1M`
`E_("cell")=E_("cell")^(@)-(0.0591)/(2) log (0.1)/((1 xx 10^(-3))^(2))`
`=(3.17-0.1477)"volt"`
=3.0223 Volt.