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

`-0.167V`

Explanation

`{:(Ag_(2)S+2e^(-)hArr2Ag+ S^(-2),E^(@) =x ..(i)),(2AghArr2Ag^(+)+2e^(-),E^(@) =- 0.8),(Ag_(2)ShArr2Ag^(+)+S^(-2),E^(@) =x -0.8):}`
`E = x - 0.8-(0.0591)/(2)log [Ag^(+)]^(2) [S^(-2)]`
`0 =x - 0.8 -(0.0591)/(2)log K_(sp)`
`x = 0.8 +(0.0591)/(2) log 2 xx 10^(-49) =- 0.639`
so for equation (i)
`E = E^(@) -(0.0591)/(2) log [S^(-2)]` ..(ii)
`H_(2)S hArr 2H^(+) +S^(-2) k = k_(1) xx k_(2)`
`1.1 xx 10^(-21) =([H^(+)]^(2)[S^(-2)])/([H_(2)S])= ((10^(-3))^(2)(S^(-2)))/((0.1))`
`[S^(-2)] = 1.1 xx 10^(-16)`
putting this in equatin (ii)
`E =- 0.639 - (0.0591)/(2)log 1.1 xx 10^(-16) =- 0.1674V`