Question

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`
A. `4xx10^(-13)`
B. `4xx10^(-10)`
C. `4xx10^(-17)`
D. `4xx10^(-7)`

Answer 1

Correct Answer - a
Cathode`:" "AgBr(s)+e^(-) rarr Ag(s)+Br^(c-)(aq)" "E^(c-)._(red)=0.07V`
Anode`: Ag(s) rarr Ag^(o+) rarr Ag^(o+)(aq)+e^(-)`
`ulbar(AgBr(aq)rarrAg^(o+)(aq)+Br^(c-)(aq))`
At equilibrium `:E^(c-)._(cell)=0.07-0.8=(0.059)/(1) log K_(sp)`
`implieslogK_(sp)~~_12.37impliesK_(sp)=4xx10^(-13)`