Let's solve the problem step by step: ### **Step 1: Identify the Half-Reactions** Given reaction: \[ \mathrm{H_2(g) + 2OH^-(aq) + Zn^{2+}(aq) \rightarrow 2H_2O(l) + Zn(s)} \] **Half-reactions:** - **Zn²⁺ + 2e⁻ → Zn(s)**  (reduction) - **H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻**  (oxidation) ### **Step 2: Standard Reduction Potentials (from ALEKS Data Tab)** - \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s) \)  **E° = -0.76 V** - \( \text{O}_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) \)  **E° = +0.40 V** (But we have H₂ oxidation in base, so let's find: \( \text{H}_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^- \) The reduction half-reaction is: \( 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq), \) E° = -0.83 V) ### **Step 3: Calculate E°\(_{cell}\)** - **Reduction (at cathode):** \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s) \)  E° = -0.76 V - **Oxidation (at anode):** \( 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \)  E° = -0.83 V (reverse for oxidation: +0.83 V) \[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \] But since the reduction potential for H₂ is -0.83 V, and it's being oxidized, the cell potential is: \[ E^\circ_\text{cell} = (-0.76) - (-0.83) = +0.07\,\text{V} \] ### **Step 4: Calculate Standard Free Energy Change (\( \Delta G^\circ \))** \[ \Delta G^\circ = -nFE^\circ_\text{cell} \] where: - \( n = 2 \) (electrons) - \( F = 96485 \) C/mol e⁻ - \( E^\circ_\text{cell} = 0.07 \) V \[ \Delta G^\circ = -2 \times 96485 \times 0.07 \] \[ \Delta G^\circ = -2 \times 6754 = -13508\,\text{J} \] \[ \Delta G^\circ = -14\,\text{kJ} \quad \text{(rounded to 2 significant digits)} \] --- ## **Final Answer** \[ \boxed{-1.4 \times 10^4\ \text{J}} \] or \[ \boxed{-14\ \text{kJ}} \] (Rounded to 2 significant digits)

Certainly! Here's a concise explanation for each step: **Step 1: Identify the Half-Reactions** Break down the overall reaction into two half-reactions: one reduction (gain of electrons) and one oxidation (loss of electrons). In this case, zinc ions gain electrons to form zinc metal (reduction), and water (or H₂) is involved in oxidation. **Step 2: Find Standard Reduction Potentials** Use the ALEKS Data tab to find the standard reduction potentials (E°) for each half-reaction. For zinc, it's given as -0.76 V; for the H₂/H₂O couple, the reduction potential is about -0.83 V. **Step 3: Determine the Cell Potential (E°ₙ)** Calculate the standard cell potential by subtracting the anode (oxidation) potential from the cathode (reduction) potential. Since H₂ is being oxidized, reverse its reduction potential to get its oxidation potential (+0.83 V). Then, subtract: E°ₙ = E°(cathode) - E°(anode) = (-0.76 V) - (-0.83 V) = +0.07 V. **Step 4: Calculate Standard Free Energy Change (ΔG°)** Use the relation ΔG° = -nFE°ₙ, where n = 2 electrons, F = 96485 C/mol. Plug in the values: ΔG° = -2 × 96485 × 0.07 ≈ -13508 J, which rounds to **-14 kJ** for two significant digits. **Final Note:** The negative ΔG° indicates the reaction is spontaneous under standard conditions.