Answer:
[tex]E^o_{cell}=0\ V[/tex]
Explanation:
Concentration of silver cathode = [tex]0.10\ M \ AgNO_3.[/tex]
Concentration of silver anode = [tex]0.00003\ M \ AgNO_3.[/tex]
It is given that , [tex]E^o_{red}=+0.80 \ V.[/tex]
We know in a standard cell,
Reduction:
Oxidation : [tex]Ag^++e^-->Ag \ \ \ \ \ \ \ E^o=+0.80\ V \\ Ag\ ->Ag^++e^-\ \ \ \ \ \ E^o=-0.80\ V[/tex]
We know, [tex]E^o_{cell}=E_{red}+ E_{ox}\\=-0.80\ +0.80 \ V\\=0\ V.[/tex]
Therefore, [tex]E^o_{cell}=0\ V[/tex]
Hence this is the required solution.