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Chapter 19: Problem 36

Discuss the advantages and disadvantages of fuel cells over conventional power plants in producing electricity.

Short Answer

Expert verified
Fuel cells offer a number of advantages over conventional power plants, including higher efficiency, environmental friendliness, quiet operation, and scalability. However, they also have some disadvantages like high costs, the need for hydrogen as fuel, limited durability, and slow start-up.

Step by step solution

01

Understand the Basics

Start by identifying the fundamental differences between fuel cells and traditional power plants. Fuel cells convert the chemical energy from a fuel into electricity through a chemical reaction with oxygen or another oxidizing agent. Conventional power plants, on the contrary, typically generate electricity from the thermal energy to produce steam, which turns a turbine connected to an electricity generator.
02

Explore the Advantages of Fuel Cells

Now, investigate on the benefits of fuel cells. Some of these advantages include: (a) High Efficiency: Fuel cells are generally much more efficient than traditional power plants, as they convert chemical energy directly into electrical energy. (b) Environmentally Friendly: Since fuel cells only emit water and heat, they do not contribute to air pollution. (c) Quiet Operation: Fuel cells operate very quietly, which is suitable for use in populated areas. (d) Scalability: Fuel cells can be small enough to power a single home, or large enough to power a major facility.
03

Identify the Disadvantages of Fuel Cells

Despite the many advantages, fuel cells also have drawbacks such as: (a) High Costs: Today, fuel cells are expensive compared to traditional power generators. (b) Hydrogen Fuel: Most fuel cells use hydrogen, which is difficult to store and transport. (c) Durability: Fuel cells have a limited lifespan and may require regular maintenance. (d) Slow Start-Up: Many fuel cells cannot provide instant power but require a warm-up period.

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Most popular questions from this chapter

Steel hardware, including nuts and bolts, is often coated with a thin plating of cadmium. Explain the function of the cadmium layer.

Explain why most useful galvanic cells give voltages of no more than 1.5 to \(2.5 \mathrm{~V}\). What are the prospects for developing practical galvanic cells with voltages of \(5 \mathrm{~V}\) or more?

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.

Tarnished silver contains \(\mathrm{Ag}_{2} \mathrm{~S}\). The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as \(\mathrm{NaCl}\). Explain the electrochemical principle for this procedure. [The standard reduction potential for the half- cell reaction \(\mathrm{Ag}_{2} \mathrm{~S}(s)+2 e^{-} \rightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)\) is \(-0.71 \mathrm{~V} .]\)
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