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 cell is `PtH_(2)|(H^(+)),(1M)||(Ag_(2)SO_(4(aq.))),("saturated")|Ag`
The reaction are, `H_(2) rarr 2H^(+) + 2e`
`2Ag^(+) + 2e rarr 2Ag`
Thus, `E_(cell) = E_(OP_(H)) + E_(RP_(H))`
`= 0.711 = 0.799 + (0.059)/(2)log[Ag^(+)]^(2)`
`:. log[(1)/([Ag^(+)]^(2))] = ([0.799 - 0.711] xx 2)/(0.059) = 3`
`:. [Ag^(+)]^(2) = 10^(-3)`
`:. [Ag^(+)]^(2) = 3.2 xx 10^(-2)`
Now the solubility equilibrium is,
`Ag_(2)SO_(4) hArr 2Ag^(+) + SO_(4)^(2-)`
`K_(SP) = (Ag^(+))^(2)(SO_(4)^(2-))`
`= (3.2 xx 10^(-2))^(2)((3.2 xx 10^(-2))/(2))`
`= 1.6 xx 10^(-5)`
(Note that if `[Ag^(+)] = 3.2 xx 10^(-2)` then `[SO_(4)^(2-)] = (1)/(2) xx 3.2 xx 10^(-2))`