A galvanic cell based on $2 Al (s) + 3 Cu X^{2 +} (aq) 2 Al X^{3 +} (aq) + 3 Cu (s)$ operates at a steady EMF of $2.00 V$ . The electrical energy (in kJ, two decimals) delivered while $0.54 g$ of aluminium is consumed is (Al = 27, $F = 96500 C mo l^{- 1}$ ).

Numerical answer type, answer in kJ, tolerance 0.02.

Answer: 11.58 kJ

Moles of Al

= 0.54/27 = 0.020 mol

. Each Al releases 3 electrons, so moles of electrons

= 0.060

. Charge

= 0.060 \times 96500 = 5790 C

. Energy

= QE = 5790 \times 2.00 = 11580 J = 11.58 kJ

. Note that the energy depends on the electrons passed, not on the stoichiometric coefficient 6 in the balanced equation.