Respuesta :
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
[tex]Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)[/tex]
The half-cell reactions are:
Oxidation half reaction (anode): [tex]Zn\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Pb^{2+}+2e^-\rightarrow Pb[/tex]
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]25^oC=273+25=298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.63 V
[tex]E_{cell}[/tex] = cell potential for the reaction = ?
[tex][Zn^{2+}][/tex] = 3.5 M
[tex][Pb^{2+}][/tex] = [tex]2.0\times 10^{-4}M[/tex]
Now put all the given values in the above equation, we get:
[tex]E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}[/tex]
[tex]E_{cell}=0.50V[/tex]
Therefore, the cell potential for this reaction is 0.50 V
