Answer:
a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)
Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)
b. ΔE° = + 0.715 V
c. It's an electrolytic cell, because it's a nonspontaneous reaction.
d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)
Explanation:
a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:
Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)
Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)
b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:
Al(s) = -1.66 V
Mg(s) = -2.375 V
ΔE° = -1.66 - (-2.375)
ΔE° = + 0.715 V
c. It's an electrolytic cell.
A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.
d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)
The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.
