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Step-by-Step Solution
Step 1: Identify the Relevant Data
We are given the standard free energies of formation at 298 K (in kJ/mol) as follows:
$\Delta G_f^\circ(\text{H}_2\text{O}(l)) = -237.2$ kJ/mol
$\Delta G_f^\circ(\text{CO}_2(g)) = -394.4$ kJ/mol
$\Delta G_f^\circ(\text{C}_5\text{H}_{12}(g)) = -8.2$ kJ/mol
We need to find the standard cell potential, $E^\circ_{\text{cell}}$, for the pentane-oxygen fuel cell reaction.
Step 2: Write the Half-Reactions
Anode (oxidation) reaction:
$ \text{C}_5\text{H}_{12} + 10\,\text{H}_2\text{O} \;\to\; 5\,\text{CO}_2 + 32\,\text{H}^+ + 32\,e^-
$
Cathode (reduction) reaction:
$ 8\,\text{O}_2 + 32\,\text{H}^+ + 32\,e^- \;\to\; 16\,\text{H}_2\text{O}
$
Step 3: Write the Overall Cell Reaction
By adding the anode and cathode reactions, we get:
$ \text{C}_5\text{H}_{12}(g) + 8\,\text{O}_2(g) \;\to\; 5\,\text{CO}_2(g) + 6\,\text{H}_2\text{O}(l)
$
Step 4: Calculate the Overall Change in Gibbs Free Energy ($\Delta G$)
Using the standard free energies of formation, we express the reaction’s $\Delta G$ as:
$ \Delta G = \Big[5 \times \Delta G_f^\circ(\text{CO}_2) + 6 \times \Delta G_f^\circ(\text{H}_2\text{O}(l))\Big] - \Big[\Delta G_f^\circ(\text{C}_5\text{H}_{12}(g)) + 8 \times \Delta G_f^\circ(\text{O}_2(g))\Big].
$
Since $\Delta G_f^\circ(\text{O}_2(g)) = 0$, the term involving $\text{O}_2(g)$ vanishes. Substituting the given values in kJ/mol:
$ \Delta G = \big[5 \times (-394.4) + 6 \times (-237.2)\big] - \big[-8.2\big]
= -1972 + -1423.2 + 8.2
= -3387 \text{ kJ mol}^{-1}.
$
Convert to J/mol:
$ -3387 \text{ kJ mol}^{-1} = -3387 \times 10^3 \text{ J mol}^{-1}.
$
Step 5: Relate $\Delta G$ to the Cell Potential
The relation between Gibbs free energy change and cell potential is:
$ \Delta G = -n F E^\circ_{\text{cell}},
$
where:
$n$ is the number of moles of electrons transferred in the overall reaction,
$F$ is the Faraday constant ($96500$ C/mol),
$E^\circ_{\text{cell}}$ is the standard cell potential.
From the overall reaction’s half-reactions, the total moles of electrons transferred is $n = 32$.
Step 6: Solve for $E^\circ_{\text{cell}}$
Substitute $\Delta G$ and $n$ into the relation:
$ -3387 \times 10^3 \text{ J mol}^{-1} = -32 \times 96500 \times E^\circ_{\text{cell}}.
$
Rearranging to solve for $E^\circ_{\text{cell}}$:
$ E^\circ_{\text{cell}}
= \frac{-3387 \times 10^3}{-32 \times 96500}
\approx 1.0968 \text{ V}.
$
Final Answer
$\boxed{1.0968 \text{ V}}$
