Two half cells are `Al^(3+)|Al" and "Mg^(2+)| Mg`. The reduction potentials of these half cells are ` -1.66 " and "-2.36 V`. Construct the galvanic cell, write the cell reaction and calculate the emf of the cell.

Explanation

For construction of a galvanic cell,
= (Electrode with lesser value of reduction potential)
- (Electrode with higher value of reduction potential)
i.e., `{:(Mg|Mg^(2+)||Al^(3+)|Al),(" anode cathode"),("oxidation reduction"):}``{:("Ox.half reaction:"," "Mg to Mg^(2+) + 2e xx 3),("Red. half reaction :",Al^(3+)+3e^(-) to Al xx 2),("Overall cell reaction :",ulbar(3Mg + 2Al^(3+) to 3Mg^(2+) + 2Al)):}`
emf of the cell ` = E_(Al^(3+)//Al)^(0) - E_(Mg^(2+)//Mg)^(0) `
` = - 1.66 - (-2.36) = 0.70 V`