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 determine the standard Gibbs free energy change (ΔG°) for the fuel cell reaction and then use it to find the standard entropy change (ΔS°). Here's a step-by-step solution:

### Step 1: Write down the given data
- The fuel cell reaction is:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
- The standard enthalpy of formation for water is given as:
\[ \Delta_f H_{298}^\circ (H_2O, l) = -285.5 \, \text{kJ/mol} \]
- The standard electric potential for the reaction is:
\[ E^\circ = 1.23 \, \text{V} \]

### Step 2: Calculate ΔG° using the formula
The Gibbs free energy change can be calculated using the formula:
\[ \Delta G^\circ = -nFE^\circ \]
where:
- \( n \) = number of moles of electrons transferred (4 for this reaction)
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E^\circ \) = standard electric potential (1.23 V)

Substituting the values:
\[ \Delta G^\circ = -4 \times 96500 \, \text{C/mol} \times 1.23 \, \text{V} \]
\[ \Delta G^\circ = -474480 \, \text{J/mol} \quad (\text{or } -47.448 \, \text{kJ/mol}) \]

### Step 3: Relate ΔG° to ΔH° and ΔS°
We can use the Gibbs free energy equation:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
Rearranging gives:
\[ \Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} \]

### Step 4: Substitute the values
- We need to convert the enthalpy of formation for 2 moles of water:
\[ \Delta H^\circ = 2 \times (-285.5 \, \text{kJ/mol}) = -571.0 \, \text{kJ/mol} \]
- Convert this to Joules:
\[ \Delta H^\circ = -571000 \, \text{J/mol} \]
- Temperature \( T = 298 \, \text{K} \)

Now substituting into the equation:
\[ \Delta S^\circ = \frac{-571000 \, \text{J/mol} - (-474480 \, \text{J/mol})}{298 \, \text{K}} \]
\[ \Delta S^\circ = \frac{-571000 + 474480}{298} \]
\[ \Delta S^\circ = \frac{-96520}{298} \approx -323.5 \, \text{J/K/mol} \]

### Step 5: Convert to kJ
\[ \Delta S^\circ \approx -0.3235 \, \text{kJ/K/mol} \]

### Final Answer
The standard entropy change for the given fuel cell reaction is approximately:
\[ \Delta S^\circ \approx -0.324 \, \text{kJ/K/mol} \]