Electrolysis calcul with faraday

In summary, the conversation was about an exercise at college regarding the amount of electricity needed for electrolysis on a sea surface of specific dimensions. The initial suggestion was to use nuclear facilities, but it was later clarified that the distance between the electrodes does not affect the electrochemical process. It was also mentioned that a substantial amount of current and electrode area is needed for electrolysis. Additionally, the potential harm to marine life was discussed. Finally, a link was shared for a similar problem and the college student mentioned that they would ask their professor for clarification.
  • #1
cancerman
3
0
Hello felow's my boy of 15 years old asked a question about an exercice at college, he asked me how much electricity it would take to generate electrolysis on a sea surface of 100 feet wide and half a mile long.

I told him (jocking) you may use some kind of nuclear facilities

I thought using Faraday loies could help me but I'm buging, thank you advice in any case

matt's
 
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  • #2
Hello and :welcome:

First of all you need a lot of cable :smile: .

But the electrochemical processes do not really depend on the distance between the electrodes: the sea can be supposed to be at zero potential and the processes are driven by potential difference.

If you want to electrolyze a substantial amount of sea water, you need a lot of current and thereby a lot of electrode area.

And the fish won't be all that happy with the chemical products !
 
  • #3
Hello BvU

first thank you for your answer it's nice :) i will tell him about the electrodes first.

so for your what kind of energy are we talking ?maping 100 feet wide and half a mile long of multiple electrode. Fish may won't be all that happy you right , a good way of fishing lol
 
  • #4
Problem example 2 in http://www.chem1.com/acad/webtext/elchem/ec8.html works it out: 5580 MegaJoule for 1000 kg of Cl2.
The 100 ft and the half mile don't really come into it - except that you need a lot of area and small distance between cathode and anode to get down to the cells operating at 2 Volt.

So what's the problem formulation for this exercise at college ? Sounds ambitious !
 
  • #5
example 2 look good thank you for the link BvU, yes i think the formulation is wrong somewhere, i will ask the professor

thank you
 

FAQ: Electrolysis calcul with faraday

What is electrolysis and how does it work?

Electrolysis is a chemical process that uses electricity to break down a substance into its constituent elements. It works by passing an electric current through an electrolyte solution, causing the ions in the solution to migrate to the electrodes and undergo chemical reactions.

Who is Faraday and what is his contribution to electrolysis?

Michael Faraday was a British scientist who is credited with the discovery of electrolysis in the early 19th century. He also formulated the laws of electrolysis, which describe the relationship between the amount of substance produced during electrolysis and the amount of electricity used.

How is Faraday's law used in electrolysis calculations?

Faraday's law is used to calculate the amount of substance produced during electrolysis, given the amount of electricity used. This is done by multiplying the amount of electricity in coulombs by the equivalent weight of the substance in grams per mole.

What is the Faraday constant and how is it related to Faraday's law?

The Faraday constant is a physical constant that represents the amount of electric charge carried by one mole of electrons. It is equal to the product of the Avogadro constant and the charge on an electron. It is used in Faraday's law to convert between coulombs and moles.

What are some practical applications of Faraday's law and electrolysis?

Faraday's law and electrolysis have many practical applications, including metal plating, production of metals and alloys, and purification of substances. It is also used in the production of hydrogen and chlorine gas, as well as in the extraction of reactive metals from their ores.

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