Pure water is saturated with pure solid `AgCl`, a silver electrode is placed in the solution and the potential is measured against normal calomel electrode at `25^(@)C`. This experiment is then repeated with a saturated solution of `Agl`. If the difference in potential in the two cases is `0.177V`. What is the ratio of solubility product (solubility) of `AgCl` and Agl at the temperature of the experiment?

Options

A. `10^(3)`
B. `10^(6)`
C. `10^(2)`
D. `10^(2)`

Answer

A

Explanation

`AgtoAg^(+)+e^(-)`
`E_(1)+E_("oxid")+E_("calomel")`
`=E'-(0.0591)/(1)logK_(sp_(1))+E_("calomel")`
`E_(2)=E'-(0.0591)/(1)logK_(sp_(2))+E_("calomel")`
`E_(2)-E_(1)=0.177=0.0591log((K_(sp_(1)))/(K_(sp_(2))))`
`(K_(sp_(1)))/(K_(sp_(2)))=10^(3)`