The theoretical efficiency of the fuel cell is given by — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Recall the formula for theoretical efficiency of a fuel cell.** The theoretical efficiency of a fuel cell is defined as the ratio of the maximum electrical work (Gibbs free energy) to the total heat content (enthalpy) of the fuel: $$\eta = \frac{|\Delta G^\circ|}{|\Delta H^\circ|} \times 100\%$$ **Step 2: Identify the given values for the H₂–O₂ fuel cell.** From the passage: - Overall reaction: $\ce{2H2(g) + O2(g) -> 2H2O(l)}$ - $\Delta H^\circ = -285.8\ \text{kJ/mol}$ (per mole of $\text{H}_2\text{O}$) - $\Delta G^\circ = -237.39\ \text{kJ/mol}$ (per mole of $\text{H}_2\text{O}$) **Step 3: Calculate the theoretical efficiency.** $$\eta = \frac{237.39}{285.8} \times 100\% = 83.06\%$$ This is significantly higher than the Carnot efficiency of a heat engine operating between the same temperatures, demonstrating a key advantage of fuel cells. $$\boxed{\text{Answer: A — } \eta = 83.06\%}$$