Standard Electrode Potentials?

Calculate the standard free-energy change at 25°C for the following reaction. Use standard electrode potentials.





4 Al(s) + 3 O2(g) + 12 H+(aq) --%26gt; 4 Al3+(aq) + 6 H2O(l)





_________ J

Standard Electrode Potentials?
The half reactions and their standard electrode potentials are


(I) :


Al³⁺ + 3 e⁺ ⇌ Al(s)


E°(I) = -1.66 V


(II) :


O₂(g) + 4 H⁺ + 4 e⁻ ⇌ 2 H₂O(l)


E°(II) = +1.23 V





Overall reaction is three times reaction (II) forward plus 4 times reaction (I) reverse:


3·(II)-4·(I) : 4 Al(s) + 3 O₂(g) + 12 H⁺ + 4 e⁻ ⇌ 4Al³⁺ + 6 H₂O(l)


Hence:


ΔE° = 3·E°(II) - 4·E°(I)


= 3·(+1.23VI) - 4·(-1.66V)


= +10.33 V





Gibbs energy of reaction and potential difference are related as:


ΔG° = - z·F·ΔE°


where


z number of electrons exchanged


F Faraday constant





So for this reaction:


ΔG° = - 12 · 96485C/mol · +10.33V = 11960 J/mol