To solve the problem, we need to find the values of \( \Delta S_{298}^\circ \) and \( \Delta G_{298}^\circ \) for the given fuel cell reaction:
\[ 2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l) \]
Given data:
- \( \Delta_f H_{298}^\circ (H_{2}O, l) = -285.5 \, \text{kJ/mol} \)
- Standard cell potential \( E^\circ = 1.23 \, V \)
- Number of electrons transferred \( n = 4 \)
### Step 1: Calculate \( \Delta H_{298}^\circ \)
Since the enthalpy change is given for the formation of 1 mole of water, we need to multiply it by 2 for the reaction involving 2 moles of water:
\[
\Delta H_{298}^\circ = 2 \times (-285.5 \, \text{kJ/mol}) = -571.0 \, \text{kJ}
\]
### Step 2: Calculate \( \Delta G_{298}^\circ \)
Using the relationship between Gibbs free energy, enthalpy, and entropy:
\[
\Delta G_{298}^\circ = \Delta H_{298}^\circ - T \Delta S_{298}^\circ
\]
We also know that:
\[
\Delta G_{298}^\circ = -nFE^\circ
\]
Where:
- \( F \) is Faraday's constant \( \approx 96500 \, \text{C/mol} \)
- \( E^\circ = 1.23 \, V \)
Substituting the values:
\[
\Delta G_{298}^\circ = -4 \times 96500 \, \text{C/mol} \times 1.23 \, V
\]
Calculating this gives:
\[
\Delta G_{298}^\circ = -474780 \, \text{J} = -474.78 \, \text{kJ}
\]
### Step 3: Substitute \( \Delta G_{298}^\circ \) into the equation
Now we can substitute \( \Delta G_{298}^\circ \) and \( \Delta H_{298}^\circ \) into the Gibbs free energy equation to find \( \Delta S_{298}^\circ \):
\[
-474.78 \, \text{kJ} = -571.0 \, \text{kJ} - 298 \Delta S_{298}^\circ
\]
Rearranging gives:
\[
298 \Delta S_{298}^\circ = -571.0 \, \text{kJ} + 474.78 \, \text{kJ}
\]
Calculating the right side:
\[
298 \Delta S_{298}^\circ = -96.22 \, \text{kJ}
\]
Now, divide by 298 to find \( \Delta S_{298}^\circ \):
\[
\Delta S_{298}^\circ = \frac{-96.22 \, \text{kJ}}{298} \approx -0.322 \, \text{kJ/K} = -322 \, \text{J/K}
\]
### Final Results
- \( \Delta G_{298}^\circ = -474.78 \, \text{kJ} \)
- \( \Delta S_{298}^\circ = -0.322 \, \text{kJ/K} \)
