Answer: 1.07 V
Explanation: [tex]E^0_{[Zn^{2+}/Zn}=-0.76V[/tex]
[tex]E^0_{[Cu^{2+}/Cu]}=+0.34V[/tex]
The metal with negative reduction potential will easily lose electrons and thus is oxidized and the one with positive reduction potential will easily gain electrons and thus is reduced.
[tex]Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu[/tex]
[tex]E^o_{cell}[/tex] = standard electrode potential =[tex]E^0_{cathode}- E^0_{anode}=0.34-(-0.76)=1.1V[/tex]
Using Nernst equation:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}=1.10-\frac{0.0592}{2}\log \frac{[0.1]}{[0.01]}[/tex]
[tex]E_{cell}=1.07V[/tex]
