Respuesta :
Answer: The voltage of the cell is 0.80 V.
Explanation:
The given cell is:
[tex]Cu(s)/Cu^{2+}(0.0257M)||Br_2(l)/2Br^-(0.392M)[/tex]
Half reactions for the given cell follows:
Oxidation half reaction: [tex]Cu(s)\rightarrow Cu^{2+}(0.0257M)+2e^-;E^o_{Cu^{2+}/Cu}=0.34V[/tex]
Reduction half reaction: [tex]Br_2(l)+2e^-\rightarrow 2Br^-(0.392M);E^o_{Br_2^/2Br-}=1.07V[/tex]
Net reaction: [tex]Cu(s)+Br_2(l)\rightarrow Cu^{2+}(0.0257M)+2Br^-(0.392M)[/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}=1.07-(0.34)=0.73V[/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 ([Cu^{2+}]\times [Br^{-}]^2)[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.73 V
n = number of electrons exchanged = 2
[tex][Cu^{2+}]=0.0257M[/tex]
[tex][Br^{-}]=0.392M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.73-\frac{0.059}{2}\times \log((0.0257)\times (0.392)^2)\\\\E_{cell}=0.80V[/tex]
Hence, the voltage of the cell is 0.80 V.
