Answer: The standard cell potential of the cell is -0.71 V
Explanation:
The half reactions follows:
Oxidation half reaction: [tex]Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V[/tex] ( × 3)
Reduction half reaction: [tex]Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V[/tex] ( × 2)
The balanced cell reaction follows:
[tex]2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)[/tex]
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]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation, we get:
[tex]E^o_{cell}=-2.37-(-1.66)=-0.71V[/tex]
Hence, the standard cell potential of the cell is -0.71 V
