Certainly! Here’s a **step-by-step solution** for the problem: --- ## Problem Restatement **Calculate the standard cell voltage of the PEMFC at 80°C using the temperature dependence of the Gibbs free energy:** \[ E^\circ(T) = E^\circ(298) - \frac{\Delta S}{nF}(T - 298) \] Where: - \(\Delta S\) = Entropy change of reaction = **–163.2 J/mol·K** (given) - \(E^\circ(298)\) = Standard cell voltage at 25°C (298 K) - \(T\) = Temperature in Kelvin (80°C = 353 K) - \(n\) = Number of electrons transferred (for PEMFC, **n = 2**) - \(F\) = Faraday constant = **96485 C/mol** --- ## Step 1: List All Given Data - \(\Delta S = -163.2 \text{ J/mol·K}\) - \(n = 2\) - \(F = 96485 \text{ C/mol}\) - \(T = 80^\circ \text{C} = 353 \text{ K}\) - \(E^\circ(298) \approx 1.229 \text{ V}\) (for PEMFC reaction at standard conditions) --- ## Step 2: Substitute Values into the Formula \[ E^\circ(353) = E^\circ(298) - \frac{\Delta S}{nF}(353 - 298) \] \[ E^\circ(353) = 1.229 - \frac{(-163.2)}{2 \times 96485}(353 - 298) \] --- ## Step 3: Calculate Each Term - \(T - 298 = 353 - 298 = 55\) - \(nF = 2 \times 96485 = 192970\) - \(\frac{\Delta S}{nF} = \frac{-163.2}{192970} \approx -.0008456\) - \(\frac{\Delta S}{nF}(T - 298) = (-.0008456) \times 55 = -.04651\) *But since there are two negative signs, this term becomes positive:* \[ - \frac{\Delta S}{nF}(T - 298) = -(-.04651) = +.04651 \] --- ## Step 4: Plug Into Final Equation \[ E^\circ(353) = 1.229 + .04651 = \boxed{1.276 \text{ V}} \] --- ## **Final Answer** \[ \boxed{E^\circ(80^\circ\text{C}) = 1.28 \ \text{V} \ (\text{rounded to two decimal places})} \] --- ### **Summary Table** | Variable | Value | |------------------|---------------------| | \(E^\circ(298)\) | 1.229 V | | \(\Delta S\) | –163.2 J/mol·K | | \(n\) | 2 | | \(F\) | 96485 C/mol | | \(T\) | 353 K (80°C) | | \(E^\circ(353)\) | 1.28 V | --- **Explanation:** The standard cell voltage increases with temperature for this reaction because the entropy change is negative, and the formula yields a positive correction term.