Respuesta :
Answer:
Cell potential is 0.03 V
Explanation:
Anode: [tex]Zn\rightarrow Zn^{2+}(0.01M)+2e^{-}[/tex]
Cathode: [tex]Zn^{2+}(0.10M)+2e^{-}\rightarrow Zn[/tex]
Overall: [tex]Zn^{2+}(0.10M)\rightarrow Zn^{2+}(0.01M)[/tex]
Nenrst equation for this cell reaction at [tex]25^{0}\textrm{C}[/tex]:
Cell potential, [tex]E_{cell}=\frac{-0.059}{n}log{\frac{[Zn^{2+}]_{0.01M}}{[Zn^{2+}]_{0.10M}}}[/tex]
Where n is number of electron exchanged
Here n=2
So, [tex]E_{cell}=\frac{-0.059}{2}log{\frac{0.01}{0.10}}=0.03V[/tex]
Answer: The cell potential of the cell is +0.03 V
Explanation:
The half reactions for the cell is:
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] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[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] = 0.010 M
[tex][Zn^{2+}{concentrated}][/tex] = 0.10 M
Putting values in above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0101M}{0.10M}[/tex]
[tex]E_{cell}=0.03V[/tex]
Hence, the cell potential of the cell is +0.03 V
