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For the Galvanic cell, Ag|AgCl(s) |KCl...

For the Galvanic cell,
`Ag|AgCl(s) |KCl (0.2 M)|| KBr (0.001 M) |AgBr(s) |Ag`
Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell at `25^(@)` C.
Given `K_(sp)(AgCl) = 2.8 xx 10^(-10), K_(sp)(AgBr) = 3.3 xx 10^(-13)`

Text Solution

Verified by Experts

`AgCl `K_(sp)(AgCl) `therefore [Ag^(+)]_(LHS) = (K_(sp)[AgCl])/[Cl^(-)] = (2.8 xx 10^(-10))/0.2 = 1.4 xx 10^(-9)` M
`AgBr `K_(sp)(AgBr) = [Ag^(+)][Br^(-)]`
`[Ag^(+)]_(LHS) = (K_(sp)(AgBr))/[Br^(-)] = (3.3 xx 10^(-13))/(0.001) = 3.3 xx 10^(-10)` M
`{:("Cell", "Cell Reaction", E^(@)),(LHS,Ag to Ag^(+)+e^(-),E_(Ag//Ag^(+))^(@) =xv),(RHS, Ag^(+) + e^(-) to Ag, E_(Ag^(+)//Ag)=-E_(Ag//Ag^(+))^(@) =-xV):}`
`K= ([Ag^(+)]_(LHS))/([Ag^(+)]_(RHS))`
`E_("cell") = E_("cell"^(@)) -(0.0591)/n log K = 0-(0.0591)/1 log (1.4 xx 10^(-9))/(3.3 xx 10^(-10)) = -0.0371` V
To make `E_("cell")` Positive, LHS cell should be cathode (+ve half cell).
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