Answer: 0.94 V
Explanation:
For the given chemical reaction :
[tex]Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.10 V
[tex]E_{cell}[/tex] = emf of the cell = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}[/tex]
[tex]E_{cell}=+1.10-0.16V=0.94V[/tex]
The cell potential for this reaction is 0.94 V
