Two half cells are Al³⁺(aq)/Al and Mg²⁺(aq)/Mg

The reduction potentials of these half cells are:
Al³⁺/Al = −1.66 V
Mg²⁺/Mg = −1.36 V

Calculate the cell potential. Write the cell reactions also.

Since, Mg²⁺(aq)/Mg electrode = −2.36 V is at a lower potential than Al³⁺(aq)/Al electrode = −1.66 V, therefore, Mg²⁺(aq)/Mg electrode acts as the anode and Al³⁺(aq)/Al acts as the cathode.
In other words, Mg loses electrons and Al³⁺ ion accepts electrons.

Cell reaction:
3Mg + 2Al³⁺ ⟶ 3Mg²⁺ + 2Al

Standard cell potential:
E°_cell = E°(Al³⁺/Al) − E°(Mg²⁺/Mg)
= −1.66 − (−2.36)
= +0.70 V