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?

In the hydrogen-oxygen fuel cell, hydrogen gas (H₂) is oxidized to produce protons and electrons at the anode, while oxygen gas (O₂) is reduced to form water (H₂O) at the cathode. The balanced half-cell reactions are as follows: Anode: \(H_{2} \rightarrow 2H^{+} + 2e^{-}\) Cathode: \(O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O\) Now, we can add these two half-reactions to obtain the overall balanced reaction in the fuel cell: \(2H_{2} + O_{2} \rightarrow 2H_{2}O\)

To calculate the maximum work, we need to find the Gibbs free energy change using the following relationship: \(\Delta G = -nFE\) Where: ∆G = Gibbs free energy change n = moles of electrons transferred F = Faraday's constant (96485 C/mol) E = standard cell potential For this cell, E = 1.23 V and, from the balanced equation, each mole of water produced requires 2 moles of electrons being transferred. To produce 1.00 kg of water, we must first determine how many moles of water that is: 1.00 kg H₂O × (1 mol H₂O / 18.015 g H₂O) = 55.56 mol H₂O Now, since each mole of water requires 2 moles of electrons: n = 2 × 55.56 mol = 111.1 mol Now, we can calculate the Gibbs free energy change: \(\Delta G = - (111.1 \, \text{mol}) (96485 \, \text{C/mol}) (1.23 \, \text{V}) = -1.34 \times 10^7 \, \text{J/mol}\)

The Gibbs free energy change calculated in the previous step is the maximum work that can be obtained from the fuel cell under standard conditions. This is because any real fuel cell will have some inefficiencies and energy losses, so the actual work obtained will be less than the theoretical maximum. Therefore, the maximum work that can be obtained from the fuel cell is: Maximum work = \(-\Delta G = 1.34 \times 10^7 \, \text{J/mol}\)

Now, let's briefly discuss the advantages and disadvantages of using fuel cells compared to combustion reactions for electricity generation. Advantages: 1. Fuel cells are more efficient than combustion-based systems because they convert the chemical energy of the fuel directly into electrical energy, bypassing the intermediate steps involving heat and mechanical work. 2. They emit fewer pollutants and greenhouse gases than combustion-based systems as they mainly produce water as a byproduct. 3. Fuel cells can be more reliable and have a longer life since they have fewer moving parts and lower maintenance requirements. Disadvantages: 1. The high cost of fuel cell systems, as they involve expensive materials like platinum catalysts. 2. The requirement for pure fuels, such as hydrogen, which might not be readily available or easily produced. 3. Limitations in power density and size, which can make them less suitable for certain applications.