Answer: The cell potential of the above reaction is 1.2 V
Explanation:
The given chemical reaction follows:
[tex]Zn(s)+Cu^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Cu(s)[/tex]
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction: [tex]Cu^{2+}+2e^-\rightarrow Cu(s)[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ? V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 1.1 V
n = number of electrons exchanged = 2
[tex][Cu^{2+}]=1.0M[/tex]
[tex][Zn^{2+}]=1.0\times 10^{-3}M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=1.1-\frac{0.059}{2}\times \log(\frac{1.0\times 10^{-3}}{1.0})\\\\E_{cell}=1.2V[/tex]
Hence, the cell potential of the above reaction is 1.2 V
