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Consider the cell Ag|AgBr(s)|Br-||AgCl(s)|Cl-|Ag at 25 degree C.The solubility product constant of AgBr and AgCl are 5*10 power(-13) and 1*10 power(-10).for what ratio of Br-and Cl- ions the emf of the cell be zero?

nitish kumar , 13 Years ago
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anser 1 Answers
Wani fiza

Last Activity: 6 Years ago

At equilibrium,Ecell=0
(Ag+)LHS=Ksp(AgBr)/(Br-)=5*10-13/a
(Ag+)RHS=Ksp(AgCl)/(Cl-)=1*10-10/b
Ecell= E0Ag/Ag+ + E0Ag+/Ag + 0.059/1 log (Ag+)RHS/(Ag+)LHS
0=0+0.059/1 log 1*10-10*a/5*10-13*b
Therefor 200a/b = antilog0 
=a/b = 1/200

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