Respuesta :
Answer: The voltage of the cell is 1.05 V.
Explanation:
The given cell is:
[tex]Zn(s)/Zn^{2+}(10M)||Cu^{2+}(0.20M)/Cu(s)[/tex]
Half reactions for the given cell follows:
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}(10M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V[/tex]
Reduction half reaction: [tex]Cu^{2+}(0.02M)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V[/tex]
Net reaction: [tex]Zn(s)+Cu^{2+}(0.20M)\rightarrow Zn^{2+}(10M)+Cu(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.34-(-0.76)=1.1V[/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+}]=0.02M[/tex]
[tex][Zn^{2+}]=10M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=1.1-\frac{0.059}{2}\times \log(\frac{10}{0.2})\\\\E_{cell}=1.05V[/tex]
Hence, the voltage of the cell is 1.05 V.
