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

`-0.037 V`

Explanation

At anode:
`Ag +CI^(-) rarr AgCI +e^(-)`
At anode:
`AgBr +e^(-) rarr Ag +Br^(-)`
`E = E_(Ag//AgCI//CI^(-))^(@) +E_(Br^(-)//AgBr//Ag)^(@) - 0.0591 log. ([Br^(-)])/([CI^(-)])` ..(i)
For reaction:
`Ag +CI^(-) rarr AgCI -e^(-)`
`Ag^(+) +e^(-) rarr Ag`
`E = O = E_(Ag//AgCr//CI^(-))^(@) +E_(Ag^(+)//Ag)^(@) - 0.0591`
`log.(1)/([Ag^(+)][CI^(-)])`
`E_(Ag//AgCI//CI^(-))^(@) = E_(Ag//Ag)^(@) - 0.0591 log K_(sp) (AgCI)` ...(ii)
for reaction
`AgBr +e^(-) rarr Ag +Br^(-)`
`Ag rarr Ag^(+) +e^(-)`
`E = 0 = E_(Br^(-)//AgBr//Ag)^(@) +E_(Ag//Ag^(+))^(@) - 0.0591 log [Ag^(+)] [Br^(-)]`
`E_(Br^(-)//AgBr//Ag)^(@) = E_(Ag^(+)//Ag)^(@) +0.0591 log K_(sp) (AgBr)` ...(iii)
from equation (1),(2) & (3)
`E = 0.0591 log.(K_(sp)(AgBr))/(K_(sp)(AgCI)) - 0.0591 log.([Br^(-)])/([CI^(-)])`
`E = 0.0591 log.(3.3 xx 10^(-13) xx 0.2)/(2.8 xx 10^(10) xx 0.001) =- 0.037 V`