Answer to Question #87799 in Inorganic Chemistry for Rahul bisht

Question

What were the difficulties in the preparation of flourine? Explain how these difficulties were overcome?

Answer 1

What were the difficulties in the preparation of flourine? Explain how these difficulties were overcome?

Answer:

Difficulties in the Isolation of Fluorine:

■   The following difficulties were Involved in the isolation of fluorine.

■   Its oxidation is not possible.

■   It exists as fluoride (F~) in its minerals. The preparation of it from a fluoride is an oxidation reaction.

2F -> F₂ + 2e^-

■   It is the most electronegative element and itself is a very strong oxidizing agent.    F₂ is a more powerful oxidizing agent than 0₂. Therefore oxidation of fluoride to fluorine could not be carried out by chemical oxidizing agents. Therefore, no oxidizing agent was able to separate fluorine from hydrogen.

■ When hydrofluoric acid is heated with oxidizing agents, like Mn0₂, KMnO₄, K₂Cr₂0₇ no fluorine is obtained. This is because of the High affinity of fluorine for hydrogen.

■   H-F bond Is considerably strong due to small bond length and high polarity.

■   It cannot be prepared by electrolysis of HF.

■   Electrolysis of hydrofluoric acid solution gave ozonized oxygen at the anode, instead of fluorine. This is due to the action of F₂ on water.

2H₂0 + 2F₂ -> 4HF + 0₂

3H₂0 + 3F₂ -> 6HF + 0₃

■ Then the electrolysis of anhydrous hydrogen fluoride was attempted. It was found to be a poor conductor (non-conductor) of electricity. HF is highly volatile and poisonous.

■ Proper apparatus for preparation and storage was not available.

■ As F₂ is extremely reactive, it attacks glass, platinum, carbon and other materials commonly used for the construction of the apparatus of electrolysis.

■ Aqueous HF is an active solvent. It attacks the glass and various metals. Thus the main need was to find a suitable apparatus end suitable electrolyte.

 Well known method of producing F₂ is the Dennis method. The prince of this method are including pure, dry and anhydrous potassium hydrogen fluoride (KHF₂) is electrolyzed in the molten state at 523 K when hydrogen is liberated at the cathode while F₂ is liberated at the anode. A V-shaped tube prevents fluorine liberated at the anode and hydrogen at the cathode from coming in contact with each other.

    Graphite is not suitable electrode at anode because it combines with fluorine formed

and produces CF4. Hence it is necessary to replace graphite electrode time to time. The

reaction is as follows

C + 2F₂ -> CF₄