Respuesta :
Answer: The potential of the given cell is 1.551 V
Explanation:
The given chemical cell follows:
[tex]Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)[/tex]
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V[/tex]
Reduction half reaction: [tex]Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V[/tex] ( × 2)
Net cell reaction: [tex]Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(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}=0.799-(-0.762)=1.561V[/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+}]}{[Ag^{+}]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ? V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.561 V
n = number of electrons exchanged = 2
[tex][Zn^{2+}]=0.125M[/tex]
[tex][Ag^{+}]=0.240M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})[/tex]
[tex]E_{cell}=1.551V[/tex]
Hence, the potential of the given cell is 1.551 V
