The emf of the given cell is 0.788 V at 25∘C.Ag|AgI in KI(0.05 M)|| – InterviewSolution

1.	The emf of the given cell is 0.788 V at 25∘C.Ag\|AgI in KI(0.05 M)\|\|AgNO3(0.05 M)\|AgCalculate the value of solubility product of AgI at 25∘CTake:10−14.63≈2.34×10−15
Answer» The emf of the given cell is $0.788 V$ at $25^{\circ} C .$ $A g \| A g I in K I (0.05 M) \| \| A g N O_{3} (0.05 M) \| A g$ Calculate the value of solubility product of $A g I$ at $25^{\circ} C$ Take: $10^{- 14.63} \approx 2.34 \times 10^{- 15}$

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