For the reaction: ; . If for and are -280.0 and -156.25 kJ/mol, respectively, the magnitude of for ( — Electrochemistry Chemistry Question
Question
For the reaction: $4Al(s) + 3O_2(g) + 6H_2O(l) + 4OH^-(aq) \rightleftharpoons 4Al(OH)_4^-(aq)$; $E^\circ_{cell} = 2.5 \text{ V}$. If $\Delta_f G^\circ$ for $H_2O(l)$ and $OH^-(aq)$ are -280.0 and -156.25 kJ/mol, respectively, the magnitude of $\Delta_f G^\circ$ for $Al(OH)_4^-(aq)$ (in kJ/mol) is
💡 Solution & Explanation
The reaction transfers 12 electrons ($4 \times 3$). $\Delta G^\circ_{rxn} = -nFE^\circ = -12 \times 96500 \times 2.5 = -2,895,000 \text{ J} = -2895 \text{ kJ}$. Also, $\Delta G^\circ_{rxn} = \Sigma \Delta_f G^\circ(\text{products}) - \Sigma \Delta_f G^\circ(\text{reactants})$. Let $x$ be $\Delta_f G^\circ(Al(OH)_4^-)$. $-2895 = 4x - [4(0) + 3(0) + 6(-280) + 4(-156.25)]$. $-2895 = 4x - [-1680 - 625] = 4x + 2305$. $4x = -2895 - 2305 = -5200$. $x = -1300$ kJ/mol. The magnitude is 1300. Therefore, correct answer is 1300.