electrochemistry

@avisikta:is K_sp = 1.02 x 10⁵....pls verify????

@debosmit, na re, noi eta answer.........cud u please say how u did it????

The cell is Ag|Ag⁺||Ag⁺|Ag, so when the concentrations are unity the potential of cathode and anode will be equal and opposite, so E⁰_cell = 0

By Nernst equation

E_cell = E⁰_cell - RTnFln([Ag⁺]in satd. Ag₂CrO₄ solution[Ag⁺])

We know that E_cell = 0.164 V, T = 298K, n= 1, [Ag⁺] (that in the denominator) = 0.1 mol and E⁰_cell = 0

So 0.164 = - 0.059 log ₁₀(10([Ag⁺]in satd. Ag₂CrO₄ solution))

Gives me [Ag⁺]in satd. Ag₂CrO₄ solution = 1.66 * 10^-4

Ag₂CrO₄(aq) <---> 2 Ag⁺ + CrO₄^2- (*)

So , solubility product is = [Ag⁺]²[ CrO₄^2-]

from (*), [Ag⁺]= 1.66 * 10^-4, [CrO₄^2-] = 8.3 *10^-5

so solubility product = 2.287 * 10^-12

Is this the answer ? what i have done seems logical :D

yes dat is it...d concentrations were where i got confused....:P thanks...:)