Respuesta :
Answer: The EMF of the cell is -0.100 V and the reaction is non-spontaneous
Explanation:
The half reaction for the equation follows:
Oxidation half reaction: [tex]Cd(s)\rightarrow Cd^{2+}(aq,0.00550M)+2e^-;E^o_{Cd^{2+}/Cd}=-0.40V[/tex]
Reduction half reaction: [tex]Fe^{2+}(aq,0.600M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.44V[/tex]
Net cell reaction: [tex]Cd(s)+Fe^{2+}(aq,0.600M)\rightarrow Cd^{2+}(aq,0.00550M)+Fe(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}=-0.44-(-0.40)=-0.04V[/tex]
- To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Cd^{2+}]}{[Fe^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = -0.04 V
n = number of electrons exchanged = 2
[tex][Cd^{2+}]=0.00550M[/tex]
[tex][Fe^{2+}]=0.600M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=-0.04-\frac{0.059}{2}\times \log(\frac{0.00550}{0.600})[/tex]
[tex]E_{cell}=-0.100V[/tex]
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
For a reaction to be spontaneous, the standard electrode potential must be positive.
Hence, the EMF of the cell is -0.100 V and the reaction is non-spontaneous
