Answer: The cell potential of the above reaction is 0.52 V
Explanation:
The given chemical equation follows:
[tex]Pb^{2+}(aq.)+Zn(s)\rightarrow Zn^{2+}(aq.)+Pb(s)[/tex]
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-[/tex]
Reduction half reaction: [tex]Pb^{2+}(aq.)+2e^-\rightarrow Pb(s)[/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+}]}{[Pb^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ? V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.63 V
n = number of electrons exchanged = 2
[tex][Zn^{2+}]=1.0M[/tex]
[tex][Pb^{2+}]=2.0\times 10^{-4}M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.63-\frac{0.059}{2}\times \log(\frac{1.0}{2.0\times 10^{-4}})\\\\E_{cell}=0.52V[/tex]
Hence, the cell potential of the above reaction is 0.52 V
The potential of the cell under the given conditions is -0.46 V.
The equation of the reaction is;
Pb2+(aq) + Zn(s) ---> Zn2+(aq) + Pb(s)
We need to use Nernst equation;
Ecell = E°cell - 0.592/n log Q
E°cell = +0.63 V
n = 2
To find Q;
Q = [Zn2+]/[Pb2+]
Q = [1.0]/[2.0 x 10-4]
Q = 5000
Substituting values;
Ecell = 0.63 V - 0.592/2 log(5000)
Ecell = -0.46 V
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