Question

Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+ + 3 e^- rightarrow Al E degree = - 1.66 V Mg^2+ + 2 e^- rightarrow Mg E degree = - 2.37 V Give the balanced cell reaction and calculate E degree for the cell.

Answer 1

Answer: The standard cell potential of the cell is -0.71 V

Explanation:

The half reactions follows:

Oxidation half reaction:  $Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V$  ( × 3)

Reduction half reaction:  $Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V$  ( × 2)

The balanced cell reaction follows:

$2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

$E^o_{cell}=-2.37-(-1.66)=-0.71V$

Hence, the standard cell potential of the cell is -0.71 V