Electrolysis of Water: Calculations & Theory

In summary, the author is attempting to do some calculations based upon measurements he has made, and he is also estimating what the expected minimum Joules per Litre figure would be based on known theory of electrolysis of water. He has both a theoretical and a measured figure, but is not sure of his workings, so he is posting them here in the hopes that someone will be able to point out if he has made any mistakes. The calculations appear to be correct, but he is using 6.8 times as much energy as he should have to.
  • #1
tdaffin
8
0
Hi, I have an apparatus which I'm using to electrolyse water into a mixture of hydrogen and oxygen.

I'm attempting to do some calculations based upon measurements I've made. What I want to do is come up with a Joules per Litre figure for energy consumed to evolve a volume of the mixed gas.

I'd also like to do some theoretical calculations in order to estimate what the expected minimum J/l would be based on known theory of electrolysis of water.

At present I have both a theoretical and a measured figure, but I'm not too sure of my workings, so I'm posting them here in the hopes that someone will be able to point out if I've got it wrong.

The theory:
Assume electrolysis of water requires 237.13 kJ/mol. Assume 1 mol of Hydrogen gas occupies a volume of 22.4 litres at STP. Also assume 1 mol of Oxygen gas occupies the same volume at STP. Assume the electrolysis of 1 mol of water produces 1 mol of hydrogen and 1/2 a mol of oxygen. So, we expect that we'll use 237.13 kJ to produce 33.6 litres of the mixed gas. My calculations show this implies 7057 Joules per litre.

The experiment:
The electrolyser draws 3.73 Amps of current at 10.76 Volts, for a power total of 40.1 Watts. While consuming this power it produces 50 ml of gas in 60 seconds. This implies 2406 Joules were used to produce 50 ml of gas. This comes out to 48,120 Joules per litre.

So, does all that look to be correct? I wouldn't be surprised if I've made some fundamental mistake, especially with the theoretical side, as I'm unsure of the figures I have there anyway. At present it looks like I'm using 6.8 times as much energy as I should have to, which seems a little excessive. I think this implies an efficiency of only 15%?
 
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  • #2
The calculations appear to be correct. Mind you I only have up to grade 11 chemistry so I'm assuming everything you've assumed in the first paragraph.

To make it more efficient have you tried adding a catalyst to the water? I'm guessing the lost joules are from heat generated by the resistance of the water.
 
  • #3
tdaffin said:
So, does all that look to be correct? I wouldn't be surprised if I've made some fundamental mistake, especially with the theoretical side, as I'm unsure of the figures I have there anyway.
Not that you are wrong in anyway but I thought that this website would be quiet good: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html

I will let you see if you are right or not.

The Bob (2004 ©)
 

1. What is the purpose of electrolysis of water?

The purpose of electrolysis of water is to split water molecules into hydrogen gas and oxygen gas using an electrical current. This process is important for producing clean and sustainable energy, as hydrogen gas can be used as a fuel source.

2. How is the electrical current applied in electrolysis of water?

The electrical current is applied through two electrodes, typically made of a conductive material like platinum or graphite, that are placed in a container of water. The positive electrode (anode) attracts the negatively charged oxygen ions, while the negative electrode (cathode) attracts the positively charged hydrogen ions.

3. What is the theoretical yield for hydrogen and oxygen in electrolysis of water?

The theoretical yield for hydrogen and oxygen in electrolysis of water is a 2:1 ratio, meaning that for every 2 moles of hydrogen gas produced, 1 mole of oxygen gas is also produced. This is based on the chemical equation for the reaction, which is 2H2O → 2H2 + O2.

4. How is the efficiency of electrolysis of water calculated?

The efficiency of electrolysis of water is calculated by dividing the actual yield of hydrogen gas by the theoretical yield, and then multiplying by 100 to get a percentage. This calculation can help determine the effectiveness of the process and identify any potential issues or improvements.

5. Can the products of electrolysis of water be used for other purposes?

Yes, the products of electrolysis of water can be used for other purposes. Hydrogen gas can be used as a clean fuel source for vehicles and in fuel cells to produce electricity. Oxygen gas can be used for medical purposes, such as in respiratory treatments or in industrial processes.

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