Answer:
0.48 V
Explanation:
Usually in the cell notation, the left side shows oxidation. So,
Oxidation half reaction:
[tex]Sn_{(s)}\rightarrow Sn^{2+}_{(aq)}+2e^-[/tex] [tex]E^o_{ox}=-E^o_{red}=-(-0.14\ V)=0.14\ V[/tex]
Reduction half reaction:
[tex]Cu^{2+}_{(aq)}+2e^-\rightarrow Cu_{(s)}[/tex] [tex]E^o_{red}=0.34\ V[/tex]
[tex]E^o_{cell}=E^o_{ox}+E^o_{red}= (0.14+0.34)\ V=0.48\ V[/tex]
