Colour of `I_2` solution is discharged , when solution of 'X' is added . 'X' is

Options

A. `H_2SO_4`
B. `Na_2SO_4`
C. `Na_2S_2O_3`
D. `S_8`

Answer

To determine the solution 'X' that can discharge the color of iodine (I2) solution, we need to analyze the reaction that occurs when iodine is reduced.

### Step-by-Step Solution:

1. **Understanding Iodine Color**: Iodine (I2) in solution appears violet due to its molecular structure. When we want to discharge this color, we need a reducing agent that can convert I2 into a colorless species.

2. **Identifying the Reduction Reaction**: The reduction of iodine involves the conversion of I2 to iodide ions (I-). The half-reaction for this process can be represented as:
\[ I_2 + 2e^- \rightarrow 2I^- \]
This indicates that iodine is gaining electrons (reduction) and forming iodide ions, which are colorless.

3. **Choosing the Reducing Agent**: A common reducing agent that can achieve this is sodium thiosulfate (Na2S2O3). When sodium thiosulfate is added to iodine, it reacts as follows:
\[ I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6 \]
In this reaction, iodine is reduced to iodide ions, and the sodium tetrathionate (Na2S4O6) formed is colorless.

4. **Conclusion**: Therefore, the solution 'X' that can discharge the color of iodine solution is sodium thiosulfate (Na2S2O3).

### Final Answer:
The solution 'X' is sodium thiosulfate (Na2S2O3).

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