Respuesta :
From the reduction standard potentials;
The emf of Zinc = -0.76 V
and the emf of Aluminium = -1.66 V
In a galvanic cell the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment.
Therefore. the voltage of a galvanic cell made with zinc and aluminium will be;
Voltage =Ered- Eoxd
= -0.76 - (-1.66)
= 0.9 V
The emf of Zinc = -0.76 V
and the emf of Aluminium = -1.66 V
In a galvanic cell the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment.
Therefore. the voltage of a galvanic cell made with zinc and aluminium will be;
Voltage =Ered- Eoxd
= -0.76 - (-1.66)
= 0.9 V
Answer : The voltage of galvanic cell made with zinc and aluminum is, 0.90V
Explanation :
We are taking the value of standard reduction potential form the standard table.
[tex]E^0_{[Al^{3+}/Al]}=-1.66V[/tex]
[tex]E^0_{[Zn^{2+}/Zn]}=-0.76V[/tex]
In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode compartment. The second forms the cathode compartment.
The half oxidation-reduction reaction will be :
Oxidation : [tex]Al+3e^-\rightarrow Al^{3+}[/tex] [tex]E^0_{[Al/Al^{3+}]}=1.66V[/tex]
Reduction : [tex]Zn^{2+}+2e^-\rightarrow Zn[/tex] [tex]E^0_{[Zn^{2+}/Zn]}=-0.76V[/tex]
The expression for standard cell is,
[tex]E^0{cell}=E^0{Anode}+E^0{Cathode}[/tex]
[tex]E^0=E^0_{[Al/Al^{3+}}+E^0_{[Zn^{2+}/Zn]}[/tex]
[tex]E^0=1.66V+(-0.76V)[/tex]
[tex]E^0=0.90V[/tex]
Therefore, the voltage of galvanic cell made with zinc and aluminum is, 0.90 V
