Tough electrolysis question (really )

In summary, during electrolysis of concentrated KOH with an iron sheet as the anode, a red solid containing 53.3% barium, 21.7% iron, and 24.9% oxygen is formed. The empirical formula of this solid is FeO4 and the oxidation number of iron in it is +6. After passing a current through the solution and adding Ba(NO3)2, 1.00 g of the red solid is produced. The anion formed is believed to be ferrate, which is isostructural to chromate and permanganate. It is unstable but can now be isolated and used in water treatment. The exact mechanism is unknown.
  • #1
Kushal
438
1
tough electrolysis question (really!)

Homework Statement



When an iron sheet is made the anode during electrolysis of very conc. KOH, an anion containing iron in a high oxidation state is formed in solution. The addition of an excess of Ba(NO3)2 to this solution precipitates a red solid having the following composition by mass:

Ba: 53.3%
Fe: 21.7%
O: 24.9%

(1) Calculate the empirical formula of the red solid an dthe oxidation number of the iron in it.

(2) After a current has been passed through the solution for some time, the addition of Ba(NO3)2 produces 1.00 g of the red solid. Calculate how many moles of the red solid are formed, and hence calculate how many coulombs of electricity were needed.

Homework Equations





The Attempt at a Solution



the first part is ok, i got FeO4 as empirical formula. i got the charge on the anion to be -2 since i assumed that it is associated with Ba2+. then the OS of iron becomes +6.


for the second part i don't know if i have to take the actual anion to be (FeO4)2- and find the Mr, then the number of moles. because what i have is only the empirical formula.

it would be nice if someone could tell me how is that anion actually formed.
 
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  • #2
Kushal said:
(1) Calculate the empirical formula of the red solid and the oxidation number of the iron in it.

(2) After a current has been passed through the solution for some time, the addition of Ba(NO3)2 produces 1.00 g of the red solid. Calculate how many moles of the red solid are formed, and hence calculate how many coulombs of electricity were needed.


The Attempt at a Solution



the first part is ok, i got FeO4 as empirical formula.
What did you get for the red solid? It is 53% by weight barium... Assume that barium is always +2 and that oxygen is -2.
 
  • #3
would it be correct if i do 53% of the Mr is 137, therefore 100% will give the Mr of the solid?

yeah, it's ok. but now, i wanted to know how this anion is formed.
 
  • #4
If it is indeed the correct answer, it is called ferrate. It is beleived to be isostructural with chromate ([tex]CrO_4^{-2}[/tex]) and permanganate ([tex]MnO_4^-[/tex]). It is fairly unstabe but recent advances have been made to isolate it as the potassium salt from the treatment of ferric chloride, KOH and hypochlorite. That solid is now sold by Ferratec and is used in water treatment as a combination selective oxidant (for sulfides) and flocculant (soluble iron). I don't know the mechanism.
 
  • #5
cool... thnks
 

FAQ: Tough electrolysis question (really )

What is electrolysis?

Electrolysis is a process in which an electric current is used to break down a compound into its constituent elements.

What makes this electrolysis question tough?

This electrolysis question is tough because it likely involves complex chemical equations and requires a thorough understanding of electrochemistry principles.

How can I solve this tough electrolysis question?

To solve this question, you will need to have a strong grasp of the concepts of electrolysis and be able to apply them to the given problem. It may also be helpful to consult with a teacher or reference materials for guidance.

Can you explain Faraday's law and its relevance to this question?

Faraday's law states that the amount of substance produced during electrolysis is directly proportional to the amount of electricity used. This law is relevant to this question as it is often used to calculate the amount of product produced during electrolysis.

Are there any tips or tricks for solving tough electrolysis questions?

Some tips for solving tough electrolysis questions include carefully balancing chemical equations, understanding the properties of different elements and compounds, and practicing with similar problems. It can also be helpful to break the question down into smaller, more manageable parts.

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