The `F_2` liberated at the anode in the whytlaw Gray method is passed through U - tube, then the following reaction takes place 

Options

A. `NaHF_2 to NaF + HF`
B. `NaF + HF to NaHF_2`
C. `KH F_2 to HF + KF`
D. `2F^(-) - 2e^(-) to F_2`

Answer

To solve the question regarding the reaction that takes place when `F2` is liberated at the anode in the Witt-Lager method and subsequently passed through a U-tube, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Witt-Lager Method**:
The Witt-Lager method is used for the electrolysis of molten sodium fluoride (NaF) to produce fluorine gas (F2) at the anode.

2. **Identify the Reaction**:
When `F2` is liberated at the anode, it can react with hydrogen fluoride (HF) in the U-tube.

3. **Write the Reaction**:
The reaction that occurs is:
\[ \text{NaF} + \text{HF} \rightarrow \text{NaHF}_2 \]
In this reaction, sodium fluoride (NaF) reacts with hydrogen fluoride (HF) to form sodium bifluoride (NaHF2).

4. **Select the Correct Option**:
From the options provided, the reaction that corresponds to the process described is the one where NaF reacts with HF to form NaHF2. Therefore, the correct answer is option number 2.

5. **Conclusion**:
The reaction taking place when `F2` is passed through the U-tube is the formation of sodium bifluoride from sodium fluoride and hydrogen fluoride.

### Final Answer:
The reaction that takes place is:
\[ \text{NaF} + \text{HF} \rightarrow \text{NaHF}_2 \]
Thus, the correct option is option number 2.

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