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A silver electrode is immersed in satura...

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`.

Text Solution

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The cell may be represented as :
`Pt(H_(2)" 1 atm")|H^(+) ("1 M")||Ag^(+)" (salt)"|Ag(s)`
`E_(cell)^(@)=E_(Ag^(+)//Ag)^(@)-E_(H^(+)//H_(2))^(@)`
`0.799-0=0.799 V`
Given, emf of the cell `=0.711 V`
`H_(2)+2Ag^(+) rarr 2Ag+2H^(+)`
`Q=([Ag]^(2)[H^(+)]^(2))/([H_(2)][Ag^(+)]^(2))=(1^(2)xx1^(2))/(1xx[Ag^(+)]^(2))=1/([Ag^(+)]^(2))`
Applying Nernst equation,
`E=E^(@)-0.0591/nlog_(10) Q`
or `0.711=0.99-0.0591/2"log"_(10) 1/([Ag^(+)]^(2))`
`[Ag^(+)]=0.03243" mol L"^(-1)`
`Ag_(2)SO_(4) hArr underset(0.03243)(2Ag^(+))+underset(0.016215)(SO_(4)^(2-))`
`K_(sp)=[Ag^(+)]^(2)[SO_(4)^(2-)]=(0.03243)^(2)xx(0.016215)=1.705xx10^(-5)`.
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