Calculate the cell potential of a cell having reaction `Ag_(2)S +2e^(-) hArr 2Ag +S^(2-)` in a solution buffered at `pH =3` and which is also saturated with `0.1M H_(2)S.`
For `H_(2)S: K_(1) = 10^(-8)` and `K_(2) = 1.1 xx 10^(-13), K_(sp) (Ag_(2)S) = 2xx 10^(-49), E_(Ag^(+)//Ag)^(@) =0.8`

Answer

`E^(@) = 0.71V`

Explanation

`Cu^(2+) +4NH_(3) hArr [Cu(NH_(3))_(4)]^(+2)`
`:. K_(f) = 1xx 10^(-12) =([Cu(NH_(3))_(4)]^(+2))/([Cu^(+2)][NH_(3)]^(4))=(1.0)/(x(2.0)^(4))`
`:. x =6.25 xx 10^(-14)M`
Note that due to high value of `K_(f)` almost all of the `Cu^(+2)` ions are converted to `Cu(NH_(3))_(4)^(2+)` ion
Now `E_(cell) =E_(Cell)^(@) +(0.059)/(2)log.(Cu^(2+))/(Zn^(+2))`
`=1.1 +(0.059)/(2)log_(10). [(6.25 xx10^(-14))/(1)]`
`E_(Cell) = 0.71V`