emf of the cell ag/agcl//0.01MAgno3 /again at 298 k is 0.169 v .calculate the solubility and solubility product of agcl at 298 k

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Solution

Dear student

The electrochemical cell can be represented as below:
Ag -------- Ag+(ag) +Cl-(aq)
Let the solubility of Ag is s so the concentration in the reaction will be
Cl- + Ag -------- Ag+(ag) +Cl-(aq)
or Ag -------- Ag+(ag)
- s
Also the another reaction can be written as:
Ag+ (cathode)---------Ag(s)
Thus the overall reaction can be written as:
Ag + Ag+ (cathode) (0.01) ------------- Ag+(anode)(s)+ Ag(s)
so using the nernst equation, one gets, $E c e l l = E o c e l l - \frac{0.059}{1} \log \frac{(s)}{0.01} T h e E o c e l l c a n b e c a l c u l a t e d a s r e d u c t i o n p o t e n t i a l o f e l e c t r o d e w h e r e r e d u c t i o n i s t a k i n g p l a c e - r e d u c t i o n p o t e n t i a l o f e l e c t r o d e w h e r e o x i d a t i o n i s t a k i n g p l a c e T h u s E o c e l l = 0.222 - 0.80 = - 0.578 T h u s t h e n e r n s t e q u a t i o n w i l l b e 0.169 = - 0.578 - 0.06 \log 100 s 0.747 = - 0.06 \log 100 s - 12.45 = \log 100 + \log s - 14.45 = \log s s = 3.5 * 10^{- 15} s o t h e s o l u b i l i t y p r o d u c t w i l l b e s^{2} = 1.25 * 10^{- 29}$
Regards

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