Determining efficiency of fuel cells

In summary, solid and liquid fuel cells operate on the same basic principle of converting chemical energy into electrical energy through redox reactions. The difference lies in the type of electrolyte and operating temperature, which can affect the specific reactions and coefficients involved. I hope this helps to clarify your questions. Thank you for reading.
  • #1
boddhisattva
10
0
Hi

I need to compare solid working with
[itex]
\frac{1}{2}O_{2}+H_{2}O+2e^{-} \rightarrow 2OH^{-}

[/itex]

and
[itex]H_{2}+2OH^{-} \rightarrow 2H_{2}O+2e^{-}[/itex]

and liquid fuel cells working with

[itex]
H_{2}\rightarrow 2H^{+}+2e^{-}
[/itex]
and
[itex]
\frac{1}{2}O_{2}+2H^{+}+2e^{-} \rightarrow H_{2}O
[/itex]

My idea was to compare the required electrical energy (which I know by [itex] dt U I [/itex] and the chemical energy given by [itex] Q \cdot n_{H_{2}O}=Q\cdot n_{H_{2}O}=Q\cdot\nu_{gas}\cdot n_{O_{2}}=Q\cdot\nu_{O_{2}}\cdot\frac{PV_{O_{2}}}{n_{O_{2}}RT}
[/itex]

where [itex] \nu _{O_{2}} [itex] is a coefficient that is equal to 1/2 in the case of solid fluel cells. My question is simple: what is [itex] \nu _{O_{2}} [itex] equal to in the case of liquid cell since there is a double equation and that we don't know which portion of OH^{-} of the reductive equation is used in the oxydative equation.

Any help would be helpful, thanks
 
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  • #2


Hello,

Thank you for your post. Comparing solid and liquid fuel cells can be a complex task, as there are many factors that can affect their performance. However, I will try to address your specific questions and provide some insight into these types of fuel cells.

Firstly, it is important to understand the basic principles of fuel cells. Both solid and liquid fuel cells involve the conversion of chemical energy into electrical energy through a redox reaction. In the case of solid fuel cells, the reaction occurs on the surface of a solid electrode, while in liquid fuel cells, the reaction takes place in a liquid electrolyte solution.

In terms of the specific reactions you have mentioned, the first two reactions you listed are part of the overall reaction for a solid oxide fuel cell (SOFC). This type of fuel cell uses a solid ceramic electrolyte and operates at high temperatures (typically around 800-1000°C). The reaction on the anode side is the oxidation of hydrogen to produce protons and electrons, while the reaction on the cathode side is the reduction of oxygen to produce hydroxide ions and electrons. The overall reaction is the combination of these two reactions, with the protons and hydroxide ions reacting to form water and release energy in the form of electricity.

The coefficient [itex] \nu _{O_{2}} [itex] in this case is equal to 1/2 because only half of the oxygen molecules are involved in the overall reaction. The other half is consumed in the reaction on the anode side to produce water. This is due to the fact that the anode reaction only requires one oxygen molecule for every two hydrogen molecules, while the cathode reaction requires one oxygen molecule for every one hydrogen molecule. Therefore, the overall reaction requires only half the amount of oxygen molecules compared to the individual reactions.

In the case of liquid fuel cells, there are several types, such as proton exchange membrane (PEM) fuel cells and alkaline fuel cells. The reactions you have listed are for an alkaline fuel cell, which uses a liquid electrolyte (typically potassium hydroxide) and operates at lower temperatures (around 60-80°C). In this type of fuel cell, the overall reaction is the same as the second reaction you listed, where hydrogen is oxidized to produce protons and electrons, and the protons react with oxygen to form water and release energy. The coefficient [itex] \nu _{O_{
 

1. What is a fuel cell?

A fuel cell is a device that converts the chemical energy of a fuel, such as hydrogen, into electricity through an electrochemical reaction. It operates like a battery, but unlike a battery, it requires a continuous supply of fuel and oxygen to sustain the reaction.

2. How is the efficiency of a fuel cell determined?

The efficiency of a fuel cell is determined by the amount of electrical energy it produces compared to the amount of fuel it consumes. This is known as the fuel cell's efficiency ratio and is typically expressed as a percentage. The higher the efficiency ratio, the more efficient the fuel cell is at converting fuel into electricity.

3. What factors affect the efficiency of a fuel cell?

There are several factors that can affect the efficiency of a fuel cell, including the type of fuel used, the design of the fuel cell, and the operating conditions. For example, the temperature and pressure at which the fuel cell operates can have a significant impact on its efficiency.

4. How can the efficiency of a fuel cell be improved?

There are several ways in which the efficiency of a fuel cell can be improved. One approach is to optimize the design of the fuel cell to minimize energy losses and maximize the conversion of fuel to electricity. Another approach is to use more efficient and cleaner fuels, such as hydrogen produced from renewable sources.

5. Why is it important to determine the efficiency of fuel cells?

Determining the efficiency of fuel cells is important for several reasons. Firstly, it allows for the comparison of different fuel cell designs and technologies to determine which is the most efficient. This can help drive advancements and improvements in fuel cell technology. Additionally, knowing the efficiency of a fuel cell is crucial for accurately estimating its performance and potential applications in various industries, such as transportation and energy production.

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