Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.
Solution:
1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
[tex]E^0=E_{cat}^0-E_{an}^0[/tex]
where [tex]E_{cat}^0[/tex] is the standard potential at the cathode, while [tex]E_{an}^0[/tex] is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: [tex]E_{Cu}^0=+0.34 V[/tex], while at the anode we have zinc: [tex]E_{Zn}^0=-0.76 V[/tex]. Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
[tex]E^0=+0.34 V-(-0.76 V)=+1.1 V[/tex]
2) To calculate [tex]E^0[/tex] at any temperature T, we should use Nerst equation:
[tex]E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]} [/tex]
where
[tex]R=8.31 J/(K mol)[/tex]
[tex]T=473.15 K[/tex] is the temperature in our problem
[tex]z=2[/tex] is the number of electrons transferred in the cell's reaction
[tex]F=9.65\cdot 10^4 C/mol[/tex] is the Faraday's constant
[tex][Zn][/tex] and [tex][Cu][/tex] are the molar concentrations of zinc and in copper, and in our problem we have [tex][Zn]=10[Cu][/tex].
Using all these data inside the equation, and using [tex]E^0=+1.1 V[/tex], in the end we find:
[tex]E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V[/tex]
