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 pentange (g) , respectively . The value of `E_(cell)^@` for the pentane-oxygen fuel cell is .

Options

A. ` 2. 0968 V`
B. ` 1.0968 V`
C. ` 0.0968 V`
D. ` 1. 968 V`

Explanation

To find the standard cell potential (E_cell^@) for the pentane-oxygen fuel cell, we will use the standard free energies of formation provided for the reactants and products involved in the reaction. The overall reaction for the combustion of pentane (C5H12) in the presence of oxygen (O2) can be written as follows:

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

### Step 1: Write the balanced chemical equation
The balanced equation for the complete combustion of pentane is:

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

### Step 2: Calculate the change in Gibbs free energy (ΔG) for the reaction
The change in Gibbs free energy for the reaction can be calculated using the standard free energies of formation (ΔG_f) of the products and reactants:

\[ \Delta G_{reaction} = \sum \Delta G_f (products) - \sum \Delta G_f (reactants) \]

Substituting the values:

- For \( CO_2(g) \): \( \Delta G_f = -394.4 \, kJ/mol \)
- For \( H_2O(l) \): \( \Delta G_f = -237.2 \, kJ/mol \)
- For \( C_5H_{12}(g) \): \( \Delta G_f = -8.2 \, kJ/mol \)
- For \( O_2(g) \): \( \Delta G_f = 0 \, kJ/mol \) (as it is a standard state)

Calculating the total ΔG_f for products and reactants:

\[ \Delta G_{products} = (5 \times -394.4) + (6 \times -237.2) \]
\[ = -1972 + (-1423.2) = -3395.2 \, kJ \]

\[ \Delta G_{reactants} = (-8.2) + (8 \times 0) = -8.2 \, kJ \]

Now, substituting into the ΔG equation:

\[ \Delta G_{reaction} = -3395.2 - (-8.2) = -3395.2 + 8.2 = -3387 \, kJ \]

### Step 3: Relate ΔG to E_cell
The relationship between Gibbs free energy and cell potential is given by the equation:

\[ \Delta G = -nFE_{cell}^@ \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96485 \, C/mol \))
- \( E_{cell}^@ \) = standard cell potential in volts

### Step 4: Determine the number of moles of electrons (n)
In the combustion of pentane, each molecule of pentane produces 5 molecules of CO2 and 6 molecules of H2O, which involves the transfer of 16 electrons (as each carbon in CO2 contributes 4 electrons and each hydrogen contributes 2 electrons).

### Step 5: Calculate E_cell
Now, we can rearrange the equation to solve for \( E_{cell}^@ \):

\[ E_{cell}^@ = -\frac{\Delta G}{nF} \]

Substituting the values:

\[ E_{cell}^@ = -\frac{-3387 \times 10^3 \, J}{16 \times 96485 \, C/mol} \]

Calculating:

\[ E_{cell}^@ = \frac{3387000}{1547760} \approx 2.19 \, V \]

### Final Answer
The value of \( E_{cell}^@ \) for the pentane-oxygen fuel cell is approximately **2.19 V**.

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