[Four-digit Integer] For the reaction: 4Al(s) + 3(g) + 6(l) + 4OH^-(aq) ⇌ 4Al(OH)4^-(aq); E°_cell = — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Balance the Redox Half-Reactions and Determine the Electron Exchange ($n$) The overall balanced cell reaction is given as: $$\ce{4Al(s) + 3O2(g) + 6H2O(l) + 4OH^-(aq) <=> 4Al(OH)4^-(aq)}$$ To find the number of moles of electrons transferred ($n$) in this redox process, we break down the overall cell reaction into its corresponding oxidation and reduction half-reactions: * **Oxidation Half-Reaction (at the Anode):** Each neutral aluminum atom loses $3$ electrons to form the tetrahydroxoaluminate complex: $$\ce{Al(s) + 4OH^-(aq) -> [Al(OH)4]^-(aq) + 3e^-}$$ For the $4$ moles of aluminum present in the overall cell equation, the total electrons lost are: $$\ce{4Al(s) + 16OH^-(aq) -> 4[Al(OH)4]^-(aq) + 12e^-}$$ * **Reduction Half-Reaction (at the Cathode):** Oxygen gas undergoes reduction in the alkaline aqueous medium: $$\ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)}$$ For the $3$ moles of oxygen gas present in the overall cell equation, the total electrons gained are: $$\ce{3O2(g) + 6H2O(l) + 12e^- -> 12OH^-(aq)}$$ Summing these two balanced half-reactions yields the net cell reaction, demonstrating that the total number of moles of electrons transferred is: $$n = 12$$ Step 2 - Calculate the Standard Gibbs Free Energy Change ($\Delta_r G^\circ$) of the Cell Reaction The standard Gibbs free energy change of the reaction is related to the standard cell potential ($E^\circ_{\text{cell}}$) by the fundamental electrochemical equation: $$\Delta_r G^\circ = -n F E^\circ_{\text{cell}}$$ Substitute the calculated $n = 12$, Faraday's constant $F = 96,500\text{ C mol}^{-1}$, and the given standard cell potential $E^\circ_{\text{cell}} = 2.5\text{ V}$: $$\Delta_r G^\circ = -12 \times 96,500\text{ C mol}^{-1} \times 2.5\text{ V}$$ $$\Delta_r G^\circ = -30 \times 96,500\text{ J mol}^{-1}$$ $$\Delta_r G^\circ = -2,895,000\text{ J mol}^{-1} = -2895\text{ kJ mol}^{-1}$$ Step 3 - Formulate the Expression for $\Delta_r G^\circ$ using Gibbs Free Energy of Formation The standard Gibbs free energy change of any reaction can be expressed as the difference between the sum of the standard free energies of formation ($\Delta_f G^\circ$) of the products and those of the reactants: $$\Delta_r G^\circ = \sum \Delta_f G^\circ(\text{products}) - \sum \Delta_f G^\circ(\text{reactants})$$ For our balanced equation: $$\Delta_r G^\circ = 4\Delta_f G^\circ(\ce{Al(OH)4^-}) - \left[ 4\Delta_f G^\circ(\ce{Al, s}) + 3\Delta_f G^\circ(\ce{O2, g}) + 6\Delta_f G^\circ(\ce{H2O, l}) + 4\Delta_f G^\circ(\ce{OH^-, aq}) \right]$$ Since elemental aluminum ($\ce{Al, s}$) and oxygen gas ($\ce{O2, g}$) are in their standard elemental states, their standard Gibbs free energies of formation are zero by definition: $$\Delta_f G^\circ(\ce{Al, s}) = 0\text{ kJ mol}^{-1}$$ $$\Delta_f G^\circ(\ce{O2, g}) = 0\text{ kJ mol}^{-1}$$ Let $X$ represent the unknown standard Gibbs free energy of formation of $\ce{Al(OH)4^-(aq)}$: $$X = \Delta_f G^\circ(\ce{Al(OH)4^-})$$ Substituting the given values $\Delta_f G^\circ(\ce{H2O, l}) = -280.0\text{ kJ mol}^{-1}$ and $\Delta_f G^\circ(\ce{OH^-, aq}) = -156.25\text{ kJ mol}^{-1}$: $$\Delta_r G^\circ = 4X - \left[ 4(0) + 3(0) + 6(-280.0\text{ kJ mol}^{-1}) + 4(-156.25\text{ kJ mol}^{-1}) \right]$$ $$\Delta_r G^\circ = 4X - \left[ -1680\text{ kJ mol}^{-1} - 625\text{ kJ mol}^{-1} \right]$$ $$\Delta_r G^\circ = 4X - \left[ -2305\text{ kJ mol}^{-1} \right]$$ $$\Delta_r G^\circ = 4X + 2305\text{ kJ mol}^{-1}$$ Step 4 - Solve for $X$ and Determine the Magnitude Equating the two expressions obtained for $\Delta_r G^\circ$ in Step 2 and Step 3: $$-2895\text{ kJ mol}^{-1} = 4X + 2305\text{ kJ mol}^{-1}$$ Subtract $2305\text{ kJ mol}^{-1}$ from both sides to isolate the $4X$ term: $$4X = -2895\text{ kJ mol}^{-1} - 2305\text{ kJ mol}^{-1}$$ $$4X = -5200\text{ kJ mol}^{-1}$$ Divide by $4$ to find the standard free energy of formation: $$X = \frac{-5200}{4}\text{ kJ mol}^{-1} = -1300\text{ kJ mol}^{-1}$$ The standard Gibbs free energy of formation of $\ce{Al(OH)4^-(aq)}$ is $-1300\text{ kJ mol}^{-1}$. Taking the absolute value to find its magnitude: $$\text{Magnitude} = |X| = |-1300|\text{ kJ mol}^{-1} = 1300\text{ kJ mol}^{-1}$$ Since the question requires a four-digit integer format: $$\boxed{1300}$$