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Consider the cell AG|AgBr(s)|Br^(-)||AgC...

Consider the cell `AG|AgBr(s)|Br^(-)||AgCI(s)|CI^(-)|Ag` at `25^(@)C`. The solubility product constants of `AgBr & AgCI` are respectively `5 xx 10^(-13) & 1 xx 10^(-10)`. For what ratio of the concentration of `Br^(-) & CI^(-)` ions would the emf of the cell be zero?

Text Solution

Verified by Experts

The correct Answer is:
`[Br^(-)]:[Cl^(-)]=1:200`
If cell is taken to be conc, cell `E_(cell)^(@)=0`
Anode: `AgtoAg_(a)^(+)+e^(-)`
cathode: `underline(Ag_(c)^(+)+e^(-)toAg)`
`AghArrAg_(a)^(+)`
From nearest Eq,
`E_(cell)=E_(cell)^(@)-(0.059)/(1)log(([Ag^(+)]_(a))/([Ag^(+)]_(c)))implies0=0-(0.059)/(1)log(([Ag^(+)]_(a))/([Ag^(+)]_(c)))`
`therefore[Ag^(+)]_(a)=[Ag^(+)]_(c)implies(K_(sp)" of " AgBr)/([Br^(-)])=(K_(sp)" of " AgCl)/([Cl^(-)])`
or, `(5xx10^(-13))/(10^(-10))=([Br^(-)])/([Cl^(-)])=([Br^(-)])/([Cl^(-)])=(1)/(200)`
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