A schematic diagram for a methane-oxygen fuel cell is shown in Figure 25.15. The equations for the half-cell reactions are given below. Anode: CH4(g) + 2 H2O(l) → CO2(g) + 8 H^+(aq) + 8 e^– Cathode: 2 O2(g) + 8 H^+(aq) + 8 e^– → 4 H2O(l) / Overall: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Use the

Question

A schematic diagram for a methane-oxygen fuel cell is shown in Figure 25.15. The equations for the half-cell reactions are given below.

Anode: [latex]CH_{4}(g) + 2\ H_{2}O(l ) → CO_2(g) + 8\ H^+(aq) + 8\ e^–[/latex]
Cathode: [latex]2\ O_2(g) + 8\ H^+(aq) + 8\ e^– → 4\ H_2O(l)[/latex]
-----------------------------------------------------------------------------------------------
Overall: [latex]CH_4(g) + 2\ O_2(g) → CO_2(g) + 2\ H_2O(l)[/latex]

Use the thermodynamic data in Appendix D to calculate the standard cell voltage produced by this methane-oxygen fuel cell.

Accepted Answer

[latex]\Delta G^{\circ}_{rxn} =\Delta G^{\circ}_{f}[CO_2(g)] + 2\ \Delta G^{\circ}_{f}[H_{2}O(l )] –\Delta G^{\circ}_{f}[CH_{4}(g)] – 2\ \Delta G^{\circ}_{f}[O_{2}(g)] \ = (–394.4\ kJ\cdot mol^{–1}) + (2)(–237.1\ kJ\cdot mol^{–1}) – (–50.5\ kJ\cdot mol^{–1}) – 0 \ = –818.1\ kJ\cdot mol^{–1}[/latex]

Using Equation 25.5 at standard conditions gives us

[latex]\Delta G_{rxn} = –ν_{e}F E_{cell}[/latex] (25.5)

[latex]E^{\circ}_{cell}=\frac{-\Delta G_{rxn}}{ν_{e}F} =\frac{818.1 × 10^3\ J·mol^{–1}}{(8)(96\ 485\ C·mol^{–1})}= 1.060\ V[/latex]

The key information that we obtain from the equations for the half reactions is the value of [latex]ν_{e}[/latex]. You really don’t need the details of the half reactions, however, if you realize that the oxidation state of the carbon atom goes from –4 in [latex]CH_{4}(g)[/latex] to +4 in [latex]CO_{2}(g)[/latex], for a total change of 8. The following Practice Problem utilizes this approach.

Question 25.11: A schematic diagram for a methane-oxygen fuel cell is shown ......

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