Respuesta :
Answer: The cell voltage of the given cell is 2.01
Explanation:
The given chemical cell equation follows:
[tex]3Cu^{2+}(aq.)+2Al(s)\rightarrow 3Cu(s)+2Al^{3+}(aq.)[/tex]
Oxidation half reaction: [tex]2Al(s)\rightarrow 2Al^{3+}(aq,1.63M)+3e^-;E^o_{Al^{3+}/Al}=-1.66V[/tex] ( × 2)
Reduction half reaction: [tex]3Cu^{2+}(aq,3.43M)+2e^-\rightarrow 3Cu(s);E^o_{Cu^{2+}/Cu}=0.34V[/tex] ( × 3)
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-(-1.66)=2.00V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ? V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +2.00 V
R = Gas constant = 8.314 J/mol.K
T = temperature = [tex]42^oC=[42+273]K=315K[/tex]
F = Faraday's constant = 96500
n = number of electrons exchanged = 6
[tex][Cu^{2+}]=3.43M[/tex]
[tex][Al^{3+}]=1.63M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=2.00-\frac{2.303\times 8.314\times 315}{6\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})[/tex]
[tex]E_{cell}=2.01[/tex]
Hence, the cell voltage of the given cell is 2.01
