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Pure water is saturated with pure solid ...

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 AgI. If the difference in potential in the two cases is 0.177V, what is the ratio of solubilities of AgCl and AgI at the temperature of experiment ?

A

`10^(3)`

B

`10^(6)`

C

`10^(4)`

D

`10^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Assume solubility of AgCl in pure water is `S_(1)`, solubility of AgBr in pure water is `S_(2)`. The cell is `Ag|Ag^(+)||Cl^(-)|Hg_(2)Cl_(2)|Hg`
`E_(1)= E^(@)+ (0.059)/(1) log S_(2), E_(2)= E^(@)+ (0.059)/(1) log S_(2)`
`E_(1)- E_(2)= 0.177 = 0.059 "log" (S_(1))/(S_(2)) therefore (S_(1))/(S_(2))= 10^(3)`
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