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A half cell reaction Ag(2)S(s)+2e^(-) ra...

A half cell reaction `Ag_(2)S(s)+2e^(-) rarr 3Ag(s)+S^(2-)` is carried out in a half cell `Pt_(Ag_(2)S|Ag,H_(2)S(0.1M)),` at`[H^(o+)]=10^(-3),` The `EMF` of the half cell is
`[E^(c-)._(Ag^(o+)|Ag)=0.80V,K_(a(H_(2)S))=10^(-21),`and `K_(sp)` of `Ag_(2)S=10^(-49)]`

A

`-0.1735V`

B

`-0.19V`

C

`+0.1735V`

D

`+0.19V`

Text Solution

Verified by Experts

The correct Answer is:
a


`K_(a)=([H^(o+)]^(2)[S^(2-)])/([H_(2)S])=((10^(-3))^(2)xx[S^(2-)])/(0.1)`
`:.[S^(2-)]=(10^(-21)xx0.1)/(10^(-6))=10^(-16)`
Since `[Ag^(o+)]=sqrt(K_(sp)/([S^(2-)]))=sqrt((10^(-49))/(10^(-16)))=sqrt(10^(-33))`
`2Ag^(o+)+2e^(-)rarr Ag`
`E_(S^(2-)|Ag_(2)S|Ag)=E_(ag^(o+)|Ag)`
`E_(S^(2-)|Ag_(2)S|Ag)=E^(c-)._(Ag^(o+)|Ag)+(0.059)/(2)log[Ag^(o+)]^(2)`
`=0.80+(0.059)/(2)log10^(-33)=-0.1735V`
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