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Cu^(2+)+2e^(-)hArrCu,E^(0)=+0.34V Ag^(...

`Cu^(2+)+2e^(-)hArrCu,E^(0)=+0.34V`
`Ag^(+)+e^(-)hArr Ag, E^(0)=+0.80V`
For what concentration of `Ag^(+)` ions will the emf of the cell be zero at `25^(@)C.` The concentration of `Cu^(2+)` is 0.1M. `(log 3.919 =0.539).`

Text Solution

Verified by Experts

Cell is `Cu|Cu^(+2)||Ag^(+2)|Ag`
`E_(of cell)^(0) =E_(RHS)=E_(LHS) =0.80 -0.34 =0.46V`
Nerest equation is `E=E^(0)-(0.059)/(2)log""([Cu^(+2)])/([Ag^(+)]^(2))`
EMF of cell = 0
`E^(0) =(0.059)/(2)log ""(0.1)/([Ag^(+)]^(2))`
`log ""(0.1)/([Ag^(+)]^(2))=(0.46xx2)/(0.059)=15.5668`
`(0.1)/([Ag^(+)]^(2))=3.688xx10^(15)`
`[Ag^(+)]^(2) =(0.1)/(3.688xx10^(15))=0.027xx10^(-15)`
`[Ag^(+)]=sqrt(0.271xx10^(-16))=0.5207xx10^(-8)`
`[Ag^(+)]=0.5207xx10^(-8)M`
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