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?

Short Answer

Expert verified
The maximum work obtainable from a hydrogen-oxygen fuel cell that produces 1.00 kg of water at 25°C is 13,161.6 kJ. This is calculated using the standard Gibbs free energy change and considering the reaction occurs at a constant temperature and pressure. The advantages of using fuel cells over combustion reactions for electricity production include higher energy efficiency, lower environmental impact, and quiet operation. However, fuel cells have disadvantages such as higher cost, limited fuel sources, and limited infrastructure for fuel distribution.

Step by step solution

01

1. Write the chemical equation for the reaction

The chemical reaction occurring in a hydrogen-oxygen fuel cell follows this equation: \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
02

2. Determine the number of moles of water produced

Given that 1.00 kg of water is produced, we can determine the number of moles of water formed. Note that the molar mass of water (H_2O) is approximately 18.02 g/mol: \[ \text{moles of } H_2O = \frac{\text{mass of }H_2O}{\text{molar mass of }H_2O} = \frac{1.00 \text{ kg}}{18.02 \text{ g/mol}} \times \frac{1000 \text{ g}}{1 \text{ kg}} = 55.5 \text{ mol} \]
03

3. Calculate the standard Gibbs free energy change

The standard Gibbs free energy change, ΔG°, for the formation of water from hydrogen and oxygen at 25°C is given by: \[ \Delta G^\circ = -237.2\, \text{kJ/mol} \] This value is negative, implying that the reaction is spontaneous under standard conditions.
04

4. Calculate the maximum work obtained

The maximum work that can be obtained from a fuel cell is the negative of the standard Gibbs free energy change times the number of moles of water produced. This can be expressed as: \[ W_\text{max} = - n \Delta G^\circ \] Using the values we determined in Steps 2 and 3, we can find the maximum work: \[ W_\text{max} = - (55.5\, \text{mol}) \times (-237.2\,\text{kJ/mol}) = 13161.6\, \text{kJ} \] So, the maximum work obtainable from this hydrogen-oxygen fuel cell is 13,161.6 kJ.
05

5. Explain why this is the maximum work and discuss the advantages and disadvantages

The maximum work is determined by the Gibbs free energy change of the reaction, which considers both enthalpy (heat) and entropy (randomness) changes. In a fuel cell, the reaction occurs at constant temperature and pressure, making the Gibbs free energy equivalent to the maximum work obtainable. In reality, due to inefficiencies and overpotential losses, actual work obtained is usually less than the maximum work. Advantages of using fuel cells over combustion reactions for electricity production: 1. Higher energy efficiency: Fuel cells can convert chemical energy to electrical energy directly, bypassing the need for thermal engines and generators used in conventional power systems, resulting in less energy loss. 2. Lower environmental impact: Fuel cells produce water as the only byproduct, which significantly reduces greenhouse gas emissions compared to combustion reactions that produce CO2. 3. Quiet operation: Fuel cells operate quietly with no combustion noise or vibrations. Disadvantages of using fuel cells over combustion reactions for electricity production: 1. Higher cost: Fuel cells, particularly those using hydrogen, have a higher upfront cost due to their complex components and materials. 2. Limited fuel sources: Currently, hydrogen production mainly relies on natural gas, which still creates greenhouse gas emissions, and renewable methods are not yet widely available. 3. Infrastructure: The current infrastructure for hydrogen distribution is limited, so widespread deployment of fuel cells faces challenges in terms of fuel availability and transport.

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Most popular questions from this chapter

Balance the following oxidation–reduction reactions that occur in basic solution using the half-reaction method. a. $\mathrm{PO}_{3}^{3-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{PO}_{4}^{3-}(a q)+\mathrm{MnO}_{2}(s)$ b. $\operatorname{Mg}(s)+\mathrm{OCl}^{-}(a q) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Cl}^{-}(a q)$ c. $\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+(a q) \rightarrow$ $$\mathrm{HCO}_{3}(a q)+\mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$$

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