Answer:
1.99V
Explanation:
Balanced redox reaction equation:
3CU2+(aq) + 2Al(s) ------> 3Cu(s) + 2Al3+(aq)
E°cell= E°cathode- E°anode
E°cell= 0.34-(-1.66)
E°cell= 2.0V
From Nernst equation:
E= E°cell - 0.0592/n logQ
E= 2.0 - 0.0592/6 log [3.43]/[1.63]
E= 2.0- 0.0032
E= 1.99V
Answer:
The cell voltage of the given cell is 2.01 V
Explanation:
[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]
Reduction half reaction:
[tex]3Cu^{2+}(aq,3.43M)+2e^-\rightarrow 3Cu(s);E^o_{Cu^{2+}/Cu}=0.34V[/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}=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[/tex]=[42+273]K=315K
F = Faraday's constant = 96500
n = number of electrons exchanged = 6
[tex][Cu^{2+}]=3.43M[Al^{3+}]=1.63M[/tex]
[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]
Thus, the cell voltage of the given cell is 2.01
