Aluminum is a highly reducing metal. — Thermodynamics Chemistry Question
Problem Context
Aluminum is a highly reducing metal.
The standard reduction potentials for Al3+(aq) and Fe3+(aq) are
Al3+ (aq) + 3 e– → Al (s) E°red = –1.66 V
Fe3+ (aq) + 3 e– → Fe (s) E°red = –0.04 V
and the standard S˚ values for these species are
i. Calculate ∆G˚ for the reduction of Fe3+(aq) by Al at 25 ˚C.
Model Answer
E0 = E0red - E0ox E0 = -0.04V − (-1.66V) E0 = 1.62V
∆G0 = -nFE0 ∆G0 = -3×(96500J/mol V)×1.62V ∆G0 = -468990J
ii. Calculate ∆S˚ for the reduction of Fe3+(aq) by Al.
Model Answer
∆S0 = S0(Fe) + S0(Al3+) − [S0(Al) + S0(Fe3+)]
∆S0 = -293.3 + 27 – [28-313.8] ∆S0 = -266.3 − [-285.8] ∆S0 = 19.5J/K
iii. Calculate ∆H˚ for the reduction of Fe3+(aq) by Al at 25 ˚C.
Model Answer
∆G0 = ∆H0 − T∆S0 ∆H0 = -468990J + 298×(19.5J/K) ∆H0 = -463.2kJ
i. Construct a Born-Haber cycle for the formation of Al2O3 from its elements, showing each step in the process
4 Al(s) + 3 O2 (g) → 2 Al2O3(s)
ii. Use the data below to calculate the lattice enthalpy of Al2O3(s) in kJ/mol.
Model Answer
2 ∆Hf = 4 ∆Hsub + 4(IE1+IE2+IE3) + 3BDE − 6(EA1+EA2) − LE
2(-1675.7kJ) = 4(330.0) + 4(5139.1) + 3(493.6) − 6(141) + 6(1779.6) – LE
-3351.4 = 1320 + 20556.4 + 1480.8 – 846 + 10677.6 – 2LE
LE = 3351.4 + 1320 + 20556.4 + 1480.8 + 9831.6
= 36540.2kJ / 2mol = 18270kJ/mol