Electrode potential Mg is greater than EP of Al .
Hence Al is anode and Mg is cathode.
emf = E(cathode) - E(anode) = -0.54- (-1.66) = 1.12V
option B is correct not D
if emf is - ve the reaction is not feasible

The electrode potential of a substance is a measure of the tendency of that substance to undergo reduction or oxidation. The more negative the electrode potential, the greater the tendency for reduction to occur, and vice versa. In a galvanic cell, the electrode potential difference between the two electrodes determines the electromotive force (e.m.f.) of the cell.

In this case, the electrode potential of aluminum (Al) is -1.66 V, and the electrode potential of magnesium (Mg) is -0.54 V. The e.m.f. of the galvanic cell can be determined by subtracting the electrode potential of the anode (Mg) from the electrode potential of the cathode (Al). In this case, it would be -1.66 V - (-0.54 V) = -1.12 V.

Explanation:
1. Understand the concept of electrode potential:
- Electrode potential is a measure of the tendency of a substance to undergo reduction or oxidation.
- More negative electrode potential indicates a greater tendency for reduction to occur, and vice versa.

2. Determine the electrode potentials of Al and Mg:
- The electrode potential of Al is -1.66 V.
- The electrode potential of Mg is -0.54 V.

3. Calculate the e.m.f. of the galvanic cell:
- The e.m.f. of the cell is determined by subtracting the electrode potential of the anode (Mg) from the electrode potential of the cathode (Al).
- In this case, the e.m.f. would be -1.66 V - (-0.54 V) = -1.12 V.

4. Identify the correct answer:
- The correct answer is option 'D', which states that the e.m.f. of the galvanic cell is -1.12 V.

In summary, the e.m.f. of the galvanic cell with electrode potentials of Al equal to -1.66 V and Mg equal to -0.54 V is -1.12 V.