Respuesta :
Answer: The concentration of Pb2+ in the anode compartment is [tex]1.74\times 10^{-6}M[/tex]
Explanation:
[tex]Sn^{2+}(aq)+Pb(s)\rightarrow Sn(s)+Pb^{2+}(aq)[/tex]
Here Pb undergoes oxidation by loss of electrons, thus act as anode. Sn undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Sn^{2+}/Sn]}=-0.14V[/tex]
[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]
[tex]E^0=E^0_{[Sn^{2+}/Sn]}- E^0_{[Pb^{2+}/Pb]}[/tex]
[tex]E^0=(-0.14-(-0.13)V=-0.01V[/tex]
Now using Nernst Eqn :
[tex]E=E^0-\frac{0.059}{n}\log\frac{[Pb^{2+}]}{[Sn^{2+}]}[/tex]
[tex]0.16=(-0.01)-\frac{0.059}{2}\log\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex]0.17=-0.0295\log\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex]-5.76=\log\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex]1.74\times 10^{-6}=\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex][Pb^{2+}]=1.74\times 10^{-6}[/tex]
