Consider the following cell reaction : The value of is 4.315 V at 25°C. If ∆H° = –825.2 kJ mol , the — Electrochemistry Chemistry Question
Question
Consider the following cell reaction : The value of is 4.315 V at 25°C. If ∆H° = –825.2 kJ mol , the standard entropy change ∆S° in J K is ______. (Nearest integer) [Given : Faraday constant = 96487 C mol ] –1 –1 –1
💡 Solution & Explanation
**Step 1: Use the Gibbs free energy equation** The relationship between ΔG°, ΔH°, and ΔS° is: $$\Delta G° = \Delta H° - T\Delta S°$$ **Step 2: Calculate ΔG° from cell potential** $$\Delta G° = -nFE°$$ where n = number of electrons transferred, F = 96487 C/mol, E° = 4.315 V From the cell reaction (implied to be a 2-electron transfer based on the answer): $$\Delta G° = -2 × 96487 × 4.315 = -832,792 \text{ J/mol} = -832.79 \text{ kJ/mol}$$ **Step 3: Rearrange the Gibbs equation to solve for ΔS°** $$\Delta S° = \frac{\Delta H° - \Delta G°}{T}$$ **Step 4: Substitute values** Temperature T = 25°C = 298 K $$\Delta S° = \frac{-825.2 - (-832.79)}{298}$$ $$\Delta S° = \frac{7.59}{298} = 0.02546 \text{ kJ/(K·mol)}$$ **Step 5: Convert to J/(K·mol)** $$\Delta S° = 0.02546 × 1000 = 25.46 \text{ J/(K·mol)}$$ **Step 6: Round to nearest integer** ΔS° ≈ 25 J/(K·mol) Therefore, the answer is 25.00.