Question

Standard free energies of formation (in kj/mol) at 298 K are -237.2, -394.4 and -8.2 for $H_2O(l),CO_2(g) $ and pentane (g), respectively. The value of $E^0_{cell} $ for the pentane-oxygen fuel cell is

• A. 1.968 V
• B. 2.0968 V
• C. 1.0968 V
• D. 0.0968 V

Answer 1

The Correct Option is C

To solve this question, we need to determine the standard cell potential $E^0_{cell}$ for the pentane-oxygen fuel cell. This involves using the given standard free energies of formation for the reactants and products involved in the fuel cell reaction.

The combustion reaction for pentane (\(C_5H_{12}\)) with oxygen (\(O_2\)) can be written as:

2C_5H_{12}(g) + 15O_2(g) \rightarrow 10CO_2(g) + 12H_2O(l)

We will use the formula for the change in Gibbs free energy \(\Delta G^0\) for the reaction:

\Delta G^0 = \sum \Delta G^0_{f, \text{products}} - \sum \Delta G^0_{f, \text{reactants}}

Given standard free energies of formation are:

• \Delta G^0_f (H_2O(l)) = -237.2 \, \text{kJ/mol}
• \Delta G^0_f (CO_2(g)) = -394.4 \, \text{kJ/mol}
• \Delta G^0_f (C_5H_{12}(g)) = -8.2 \, \text{kJ/mol}

For the products, calculate the total standard free energy as follows:

10 \times (-394.4) + 12 \times (-237.2)

For the reactants (only pentane contributes as \(\Delta G^0_f\) for O₂ is zero):

2 \times (-8.2)

Now, calculate \Delta G^0_{\text{reaction}}:

\Delta G^0_{\text{reaction}} = [10 \times (-394.4) + 12 \times (-237.2)] - [2 \times (-8.2)]

\Delta G^0_{\text{reaction}} = -3944 - 2846.4 + 16.4

\Delta G^0_{\text{reaction}} = -6774 \, \text{kJ/mol}

Convert this value from kJ to J:

\Delta G^0_{\text{reaction}} = -6774000 \, \text{J/mol}

Using the relation between Gibbs free energy and cell potential:

\Delta G^0 = -nFE^0_{cell}

where \(n\) is the number of moles of electrons exchanged (16 moles of electrons for complete oxidation of 2 moles of pentane), and \(F\) is the Faraday constant (96485 C/mol).

-6774000 = -16 \times 96485 \times E^0_{cell}

Solve for E^0_{cell}:

E^0_{cell} = \frac{6774000}{16 \times 96485} = 1.0968 \, \text{V}

Therefore, the correct answer is 1.0968 V.