For Ag(s) | AgCl(s) | Cl-(aq) | Br- / AgBr(s) | Ag(s) , E° cell shall be :
A) - 138.91 log10 Ksp AgBrKsp AgCl
A) - 138.91 log10 Ksp AgClKsp AgBr
A) - 138.91 loge Ksp AgBrKsp AgCl
A) - 138.91 loge Ksp AgClKsp AgBr
My Doubt:
If it is a concentration cell, the E° should be zero ? Isn't it? However E would not be zero, But what the problem seeks is weird. Please Correct me here.
Otherwise, C is the answer which I get for E of cell!!
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2 Answers
21E°=E at conc=1 and standard conditions.
At equilibrium Amt of solute dissolved =Ksp
Here AgCl and AgBr have diff. Ksp and to achieve equilibrium, they will change their conc to diff values(Ksp) from original conc 1.
Thus dissociation of Ag into Ag2+will be diff. in the two half cells and this is responsible for creating a net potential diff.
Hence ans is true for E°.
