From the following information, calculate the solubility product of `AgBr.`
`AgBr(s)+e^(-) rarr Ag(s) +Br^(c-)(aq), " "E^(c-)=0.07V`
`Ag^(o+)(aq)+e^(-) rarr Ag(s)," "E^(c-)=0.080V`

Options

A. `(4xx10^(-133)`
B. `(4xx10^-10)`
C. `(4xx10^(-17)`
D. `(4xx10^(-7)`

Answer

A

Explanation

Cathode : `AgBr(s) + e(-) rightarrow Ag(s) + Br^(-) (aq)` `E^(@) = 0.07V`
Anode : `underline(Ag(s) rightarrow Ag^(+) (aq) + e^(-))` `E^(@) -0.8 V`
At equilibrium : `E_(cell)^(@) =0.07 - 0.8 = (0.059)/(1) log _(10) K_(sp)`
`Rightarrow log _(10) K_(sp) ~~ -12.37` `Rightarrow K_(sp) = 4xx10^(-13)`