What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{~kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

The balanced chemical equation for a hydrogen-oxygen fuel cell is as follows: \[ 2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l) \]

As we are given that 1.00 kg of water is produced, we can calculate the number of moles by dividing the mass by the molar mass of water: \[n_{H_2O} = \frac{m_{H_2O}}{M_{H_2O}}\] Where \(n_{H_2O}\) is the number of moles of water, \(m_{H_2O}\) is the mass of water (1.00 kg), and \(M_{H_2O}\) is the molar mass of water (18.015 g/mol). Converting mass to grams: \[n_{H_2O} = \frac{1000 g}{18.015 g/mol} = 55.56 \ \text{mol}\]

The Gibbs free energy change of the reaction can be calculated using the standard Gibbs free energies of formation of the products and reactants: \[\Delta G^\circ = \sum n_{\text{products}} G_f^\circ(\text{products}) - \sum n_{\text{reactants}} G_f^\circ(\text{reactants})\] For this reaction, the standard Gibbs free energy of formation of liquid water is -237.13 kJ/mol, while the Gibbs free energy of formation of hydrogen and oxygen gases are zero (since they are elements in their standard states). \[\Delta G^\circ = 2 \times (-237.13 \ \text{kJ/mol}) - 0 = -474.26 \ \text{kJ/mol}\]

Now, we can calculate the maximum work that can be obtained from this reaction using the number of moles of water produced and the molar Gibbs free energy change: \[W_{\text{max}} = n_{H_2O} \times \Delta G^\circ = 55.56 \ \text{mol} \times (-474.26 \ \mathrm{kJ/mol})\] \[W_{\text{max}} = -26354 \ \text{kJ}\]

This value represents the maximum work that can be obtained because the Gibbs free energy change represents the theoretical maximum reversible work achievable from a reaction. Any potential inefficiencies or energy losses in the reaction will result in less than this amount of work being obtained. Fuel cells have some advantages over combustion reactions for electricity production. They are generally more efficient, as they convert chemical energy directly into electrical energy, while combustion reactions require a mechanical power cycle, resulting in energy losses. Fuel cells also typically have lower emissions than combustion processes, particularly if pure hydrogen is used. However, fuel cells also have some disadvantages. They can be more expensive to produce, and the hydrogen fuel used in fuel cells is not as widely available or easily transported as fossil fuels. Additionally, production, storage, and transportation of hydrogen can be energy-intensive, which could negate some of the emission advantages of fuel cells if not done in an environmentally friendly manner. In conclusion, the maximum work that can be obtained from a hydrogen-oxygen fuel cell producing 1.00 kg of water at 25°C is -26354 kJ. This is considered the maximum work because the Gibbs free energy change represents the theoretical maximum reversible work achievable from a reaction. Fuel cells offer some efficiency and emission advantages over combustion reactions for electricity production, but also have challenges related to cost and hydrogen infrastructure.