Answer : The cell emf for this cell is 0.077 V
Solution :
The balanced cell reaction will be,
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/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{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = ?
[tex][Zn^{2+}{diluted}][/tex] = 0.0111 M
[tex][Zn^{2+}{concentrated}][/tex] = 4.50 M
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}[/tex]
[tex]E_{cell}=0.077V[/tex]
Therefore, the cell emf for this cell is 0.077 V
