A silver electrode is immersed in saturated `AgSO_(4(aq.))`. The potential difference between the silver and the standard hydrogen electrode is found to be `0.711 V`. Determine `K_(SP)(Ag_(2)SO_(4))`. Given `E_(Ag^(+)//Ag)^(@) = 0.799 V`.

Options

A. `1.7xx10^(-5)`
B. `1.7xx10^(-4)`
C. `2.7xx10^(-5)`
D. `2.7xx10^(-4)`

Answer

A

Explanation

Cell reaction `H_(2)+2Ag^(+)rarr2Ag+2H^(+)`
`E=E^(@)-(0.0591)/(n)log_(10)Q`
`or 0.711=0.799-(0.0591)/(2),log_(10).(1)/([Ag^(+)]^(2))`
`[Ag^(+)]=0.3243"mol L"^(-1)`
`{:(Ag_(2)SO_(4)hArr2Ag^(+)+SO_(4)^(2-)),(" 0.03243 0.0162154"):}`
`K_(sp)=[Ag^(+)]^(2)[SO_(4)^(2-)]`
`=[0.03243]^(2)xx(0.016215)=1.705xx10^(-5)`