Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
[tex]AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 1
[tex]E_{cell}[/tex] = ?
[tex][Cl^{-}{diluted}][/tex] = 0.0222 M
[tex][Cl^{-}{concentrated}][/tex] = 2.22 M
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}[/tex]
[tex]E_{cell}=0.118V[/tex]
Therefore, the cell emf for this cell is 0.118 V