Respuesta :
Answer: The [tex]E^o_{cell}\text{ and }K_{eq}[/tex] of the reaction is 0.78 V and [tex]2.44\times 10^{26}[/tex] respectively.
Explanation:
For the given half reactions:
Oxidation half reaction: [tex]Fe(s)\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]
Reduction half reaction: [tex]Cu^{2+}+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V[/tex]
Net reaction: [tex]Fe(s)+Cu^{2+}\rightarrow Fe^{2+}+Cu(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.34-(-0.44)=0.78V[/tex]
To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:
[tex]\Delta G^o=-nfE^o_{cell}[/tex]
and,
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
Equating these two equations, we get:
[tex]nfE^o_{cell}=RT\ln K_{eq}[/tex]
where,
n = number of electrons transferred = 2
F = Faraday's constant = 96500 C
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.78 V
R = Gas constant = 8.314 J/K.mol
T = temperature of the reaction = [tex]25^oC=[273+25]=298K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant of the reaction = ?
Putting values in above equation, we get:
[tex]2\times 96500\times 0.78=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=2.44\times 10^{26}[/tex]
Hence, the [tex]E^o_{cell}\text{ and }K_{eq}[/tex] of the reaction is 0.78 V and [tex]2.44\times 10^{26}[/tex] respectively.
