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Given the data at 25^(@)C Ag+I^(-)rarr...

Given the data at `25^(@)C`
`Ag+I^(-)rarrAgl+e^(-)" "E^(@)=0.153V`
`Ar rarrAg^(+)+e^(-)E^(@)=0.800V`
What is the value of `log K_(sp)` for AgI?

A

`-37.83`

B

`-16.13`

C

`-8.12`

D

`+8.612`

Text Solution

Verified by Experts

The correct Answer is:
B
`Ag+I^(-)rarrAgI+e^(-),E^(@)=0.152V`
`Ag^(+)+e^(-)rarrAg,E^(@)=0.800V`
`Ag^(+)+I^(-)rarrAgI,E_("cell")=0.952V`
`k_(C)=( [AgI])/([Ag^(+)][I^(-)])=(1)/(K_(sp))`
`therefore 0.952=(2.303)/(F)RT"log"k_(C)`
`0.952=0.059"log"(1)/(k_(sp))`
`0.952=-0.059"log" k_(sp)`
or `"log" k_(sp)=-16.13`
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