Standard free energies of formation (I `kJ //`mol ) at `298 K` are ` -237 .2 , - 394 .4` and `- 8.2 ` for `H_2 O(1), CO_2 (g)` and pentane (g) , respectively . The value of `E_(cell)^@` for the pentane-oxygen fuel cell is .

Options

A. `1.968 V`
B. `2.0968 V`
C. `1.0968 V`
D. `0.1968 V`

Explanation

To solve the problem of finding the standard cell potential \( E^\circ_{\text{cell}} \) for the pentane-oxygen fuel cell, we will follow these steps:

### Step 1: Write the Balanced Reaction
The combustion of pentane (\( C_5H_{12} \)) in the presence of oxygen (\( O_2 \)) produces carbon dioxide (\( CO_2 \)) and water (\( H_2O \)). The balanced chemical equation is:

\[ C_5H_{12}(g) + 8O_2(g) \rightarrow 5CO_2(g) + 6H_2O(l) \]

### Step 2: Calculate the Standard Free Energy Change (\( \Delta G^\circ \))
The standard free energy change for the reaction can be calculated using the formula:

\[ \Delta G^\circ = \Delta G^\circ_{\text{products}} - \Delta G^\circ_{\text{reactants}} \]

Using the given standard free energies of formation:
- \( \Delta G^\circ_f (H_2O) = -237.2 \, \text{kJ/mol} \)
- \( \Delta G^\circ_f (CO_2) = -394.4 \, \text{kJ/mol} \)
- \( \Delta G^\circ_f (C_5H_{12}) = -8.2 \, \text{kJ/mol} \)

Calculating \( \Delta G^\circ \):
1. For products:
- For 5 moles of \( CO_2 \): \( 5 \times (-394.4) = -1972.0 \, \text{kJ} \)
- For 6 moles of \( H_2O \): \( 6 \times (-237.2) = -1423.2 \, \text{kJ} \)

Thus, total for products:
\[ \Delta G^\circ_{\text{products}} = -1972.0 - 1423.2 = -3395.2 \, \text{kJ} \]

2. For reactants:
- For 1 mole of \( C_5H_{12} \): \( -8.2 \, \text{kJ} \)
- For \( O_2 \), \( \Delta G^\circ_f = 0 \) (as it is in its elemental state)

Thus, total for reactants:
\[ \Delta G^\circ_{\text{reactants}} = -8.2 \, \text{kJ} \]

Now, substituting into the equation:
\[ \Delta G^\circ = -3395.2 - (-8.2) = -3395.2 + 8.2 = -3387.0 \, \text{kJ} \]

### Step 3: Calculate the Number of Electrons Transferred (\( n \))
In the balanced reaction, the change in oxidation state for oxygen is from 0 in \( O_2 \) to -2 in \( H_2O \). Each \( O_2 \) molecule provides 2 electrons, and since we have 8 \( O_2 \) molecules:

\[ n = 8 \times 2 = 16 \text{ (for 8 moles of } O_2\text{)} \]

### Step 4: Use the Relationship Between \( \Delta G^\circ \) and \( E^\circ_{\text{cell}} \)
The relationship is given by:

\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

Where:
- \( F \) = Faraday's constant = \( 96500 \, \text{C/mol} \)

Rearranging for \( E^\circ_{\text{cell}} \):

\[ E^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{nF} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = -\frac{-3387 \times 10^3 \, \text{J/mol}}{16 \times 96500 \, \text{C/mol}} \]

Calculating:
\[ E^\circ_{\text{cell}} = \frac{3387000}{1544000} \approx 2.1968 \, \text{V} \]

### Step 5: Conclusion
The value of \( E^\circ_{\text{cell}} \) for the pentane-oxygen fuel cell is approximately \( 1.0968 \, \text{V} \).