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`

Given the data at `25^(@)C` `Ag+I^(-)rarrAgl+e^(-)" "E^(@)=0.153V` `Ar

1.	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`
Answer» Correct Answer - 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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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`

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