Answer: The [tex]E_{cell}[/tex] is 0.63 V
Explanation:
In the given reaction :
[tex]Zn(s)+Pb^{2+}(aq)\rightarrow Zn^{2+}(aq)+Pb(s)[/tex]
Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0_{cell}=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Pb^{2+}/Pb]}= -0.13V[/tex]
[tex]E^0_{[Zn^{2+}/Zn]}=-0.76V[/tex]
[tex]E^0_{cell}=E^0_{[Pb^{2+}/Pb]}- E^0_{[Zn^{2+}/Zn]}[/tex]
[tex]E^0_{cell}=-0.13-(-0.76)=0.63V[/tex]
Thus the [tex]E_{cell}[/tex] is 0.63 V
