A silver electrode is immersed in saturated `Ag_(2)SO_(4(aq))`. The potential difference between the silver and the standard hydrogen electrode is found to be `0.711V` Determine `K_(SP)(AgSO_(4))`. Given `E_(Ag^(+)//Ag)^(@)=0799V`.

Explanation

The given all is `H_(2)|{:(H^(+)),( 1M):}||{:(Ag_(2)SO_(4(aq.))),("saturated"):}|Ag`
The reaction are, `H_(2)rarr2H^(+)+2e`
`2A^(+)+2e^(-)rarr2Ag`
Thus, `E_(cell)=E_(OP_(H))+H_(RP_(Ag))`
`0.711=0.799+(0.059)/(2)log[Ag^(+)]^(2)`
`thereforelog.[(1)/([Ag^(+)]^(2))]=([0.799-0.711]xx2)/(0.059)=3`
`therefore[Ag^(+)]^(2)=10^(-3)`
`therefore[Ag^(+)]=3.2xx10^(-2)`
Now the solubility equilibrium is,
`Ag_(2)SO_(4)hArr2Ag^(+)+SO_(4)^(2-)`
`K_(SP)=(Ag^(+))^(2)(SO_(4)^(2-))`
`=(3.2xx10^(-2))^(2)((3.2xx10^(-2))/(2))=1.6xx10^(-5)`
`("Note that if "[Ag^(+)]=3.2xx10^(-2)"then")`
`[SO_(4)^(2-)]=(1)/(2)xx3.2xx10^(-2)`