Question

Given the data at `25^(@)C`.
`Ag+I^(-) to Agl+e^(-) , E^(@)=-0.152" V "`
`Ag to Ag^(+)+e^(-) , E^(@)=-0.800" V "`.
The value of log `K_(sp)` for AgI is :
A. -8.12
B. 8.612
C. -37.83
D. -16.13

Answer 1

Correct Answer - D
(d) `Agl+e^(-) to Ag+I^(-) , E^(@)=-0.152" V "`
`Ag to Ag^(+)+e^(-) , E^(@)=-0.800" V "`
`Agl to Ag^(+)+I^(-) , E^(@)=-0.952" V "`
`E_(cell)^(@)=(0.059)/(n)logK=(0.059)/(n)logK_(sp)`
`logK_(sp)=(E_(cell)^(@)xxn)/(0.059)=((-0.952)xx1)/((0.059))=-16.135`