For the fuel cell reaction `2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(l), Deta_(f)H_(298)^(@) (H_(2)O,l) =- 285.5 kJ//mol` What is `Delta_(298)^(@)` for the given fuel cell reaction?
Given: `O_(2)(g) +4H^(+)(aQ) +4e^(-) rarr 2H_(2)O(l) E^(@) = 1.23 V`

Options

A. `-0.322 J//K`
B. `-0.635 kJ//K`
C. `3.51 kJ//K`
D. `-0.322 kJ//K`

Explanation

To solve the problem, we need to calculate the standard entropy change (ΔS₀) for the fuel cell reaction given by:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]

We are provided with the following information:
- The standard enthalpy of formation of water, \( \Delta_f H_{298}^0(H_2O, l) = -285.5 \, \text{kJ/mol} \)
- The cell reaction involves the half-reaction:
\[ O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l) \]
with a standard cell potential \( E^0 = 1.23 \, \text{V} \).

### Step 1: Calculate ΔH₀ for the reaction
Since the reaction produces 2 moles of water, we need to calculate the total enthalpy change for the formation of 2 moles of water:

\[ \Delta H^0 = 2 \times \Delta_f H_{298}^0(H_2O, l) = 2 \times (-285.5 \, \text{kJ/mol}) = -571.0 \, \text{kJ} \]

### Step 2: Convert ΔH₀ to Joules
Since we will be using Joules in our calculations, we convert ΔH₀ from kJ to J:

\[ \Delta H^0 = -571.0 \, \text{kJ} = -571000 \, \text{J} \]

### Step 3: Calculate ΔG₀ using the electrochemical relationship
Using the relationship from electrochemistry:

\[ \Delta G^0 = -nFE^0 \]

Where:
- \( n = 4 \) (number of moles of electrons transferred)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( E^0 = 1.23 \, \text{V} \)

Calculating ΔG₀:

\[ \Delta G^0 = -4 \times 96500 \, \text{C/mol} \times 1.23 \, \text{V} = -474780 \, \text{J} \]

### Step 4: Use the Gibbs free energy equation
Using the Gibbs free energy equation:

\[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \]

We can rearrange this to solve for ΔS₀:

\[ \Delta S^0 = \frac{\Delta H^0 - \Delta G^0}{T} \]

### Step 5: Substitute the values
Substituting the values we have:

- \( \Delta H^0 = -571000 \, \text{J} \)
- \( \Delta G^0 = -474780 \, \text{J} \)
- \( T = 298 \, \text{K} \)

\[ \Delta S^0 = \frac{-571000 - (-474780)}{298} \]

Calculating the numerator:

\[ \Delta S^0 = \frac{-571000 + 474780}{298} = \frac{-96220}{298} \]

Calculating ΔS₀:

\[ \Delta S^0 \approx -322.8 \, \text{J/K} \]

### Step 6: Convert ΔS₀ to kJ
To convert to kJ:

\[ \Delta S^0 \approx -0.3228 \, \text{kJ/K} \]

### Final Answer
Thus, the standard entropy change for the given fuel cell reaction is approximately:

\[ \Delta S^0 \approx -0.3228 \, \text{kJ/K} \]