in which form is the k+ ion released from khf2 during whytlaw-grays method?

please clearly explain the proceedure with any pictures and material if available... thank u

Dear Venkata,

Whytlaw-Gray Method:
The cell consists of a cylindrical copper vessel that acts as the cathode. A graphite anode is fixed in the center of a coaxial copper tube, which is suspended in the copper vessel.
KHF₂ is electrolysed at 523 K. A current of 12-15 amperes is passed at 15 volts. H₂ is liberated at the cathode and F₂ at the anode. F₂ is contaminated with HF and CF₄. HF is removed by passing the contaminated gas through NaF absorption tubes and CF₄ by passing the contaminated gas through liquid O₂. Pure F₂ is then stored in suitable cylinders.
Advantages of the Whytlaw-Gray Method over Dennis Method:
1. Since the anode and the cathode are separated by a perforated diaphragm, mixing of H₂ and F₂ is prevented.
2. Current efficiency is about 80%.
3. The exit tubes are wide hence they do not get blocked.
4. The cell is cylindrical, hence KHF₂ is kept at a low level to prevent its creeping.

Reactions:
Dissociation:
KHF₂ KF + HF
KF ? K⁺ + F^-

At Cathode:
2K⁺ + 2e^- ? 2K
2K + 2HF ? 2KF + H₂

At Anode:
2F- ? 2F + 2e^-
2F ? F₂

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