Question #44774

A \ce{H2-O2} fuel cell operates at 298 K. The overall cell reaction is:

$\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$

Given thermodynamic data at 298 K: $\Delta G^\circ_{\text{overall}} = -237 \text{ kJ/mol}$, $\Delta H^\circ_{\text{overall}} = -286 \text{ kJ/mol}$.

The anodic oxidation of \ce{H2} occurs via a two-step mechanism on the Pt catalyst: Step 1: $\ce{H2(g) + 2Pt(s) -> 2Pt-H(ads)}$ (Irreversible, Rate-Determining Step) Step 2: $\ce{2Pt-H(ads) <=> 2Pt(s) + 2H+(aq) + 2e-}$ (Fast, Reversible)

The standard free energy change for Step 1 is $\Delta G^\circ_1 = -45 \text{ kJ/mol}$. Because Step 1 is strictly irreversible, it produces no electrical work. The subsequent electrochemical steps (Step 2 and cathodic reduction) operate reversibly.

The fuel cell is connected to a closed test chamber of constant volume initially containing 2.0 moles of \ce{H2(g)} at 298 K, with \ce{O2(g)} supplied in excess. The cell operates until the output current drops to 25% of its initial value.

Assuming standard state thermodynamic values apply throughout the process, calculate the maximum operating thermodynamic efficiency ($\eta = \frac{|W_{\text{elec}}|}{|\Delta H^\circ_{\text{overall}}|}$) of the cell, and the total heat dissipated by the cell during this period.