Answer:
[tex]k_{sp}=4.7 \times 10^{-13}.[/tex]
Explanation:
Now equation of tqo halves are:
Oxidation : [tex]Ag(s)-->Ag^+(aq)+e^-[/tex]
Reduction : [tex]AgBr(s)+e^--->Ag(s)+Br^-(aq)[/tex]
We know,
[tex]E^o_{cell}=0.071-(0.8)=-0.729\ V.[/tex]
[tex]\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.[/tex]
[tex]k_{sp}=4.7 \times 10^{-13}.[/tex]
Hence, this is the required solution.
