Respuesta :
Answer: The cell potential of the above reaction is 1.97 V
Explanation:
The given chemical reaction follows:
[tex]Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)[/tex]
Oxidation half reaction: [tex]Mg(s)\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V[/tex]
Reduction half reaction: [tex]Fe^{2+}+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.44V[/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-(-2.37)=1.93V[/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{[Mg^{2+}]}{[Fe^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ? V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 1.93 V
n = number of electrons exchanged = 2
[tex][Mg^{2+}]=0.210M[/tex]
[tex][Fe^{2+}]=3.60M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=1.93-\frac{0.059}{2}\times \log(\frac{0.210}{3.60})\\\\E_{cell}=1.97V[/tex]
Hence, the cell potential of the above reaction is 1.97 V
