A solution of `I_(2)` in aqueous Kl on reaction with an aqueous solution of `Na_(2)S_(2)O_(3)` gets decolourised. The reaction taking place here is

Options

A. `Na_(2)S_(2)O_(3)+H_(2)O+I_(2)rarrNa_(2)S_(2)O_(4)+2HI`
B. `2Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2NaI`
C. `Na_(2)S_(2)O_(3)+2H_(2)O+2I_(2)rarrNa_(2)S_(2)O_(5)+4HI`
D. `Na_(2)S_(2)O_(3)+2H_(2)O+2I_(2)rarrNa_(2)S_(4)O_(8)+4HI`

Answer

To solve the problem of the reaction between iodine (I₂) in aqueous potassium iodide (KI) and sodium thiosulfate (Na₂S₂O₃), we can follow these steps:

### Step 1: Identify the Reactants
We have iodine (I₂) and sodium thiosulfate (Na₂S₂O₃) in the reaction. Iodine is known to act as an oxidizing agent.

### Step 2: Determine the Oxidation States
- In I₂, the oxidation state of iodine is 0.
- In the thiosulfate ion (S₂O₃²⁻), the oxidation state of sulfur can be calculated. The average oxidation state of sulfur in thiosulfate is +2.

### Step 3: Write the Half-Reactions
1. **Reduction Half-Reaction**: I₂ is reduced to iodide ions (I⁻).
\[ I_2 + 2e^- \rightarrow 2I^- \]
(Here, iodine goes from an oxidation state of 0 to -1, gaining 2 electrons.)

2. **Oxidation Half-Reaction**: The thiosulfate ion (S₂O₃²⁻) is oxidized to tetrathionate (S₄O₆²⁻).
\[ 2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^- \]
(In this reaction, the sulfur in the thiosulfate is oxidized, losing electrons.)

### Step 4: Balance the Overall Reaction
Now, we can combine the half-reactions. We need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

From the reduction half-reaction, we see that 2 electrons are gained, and from the oxidation half-reaction, 2 electrons are lost. Thus, we can add the two half-reactions together:

\[ I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-} \]

### Step 5: Include the Sodium Ions
Since we started with sodium thiosulfate (Na₂S₂O₃), we need to include the sodium ions in the final balanced equation:

\[ I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6 \]

### Final Reaction
The final balanced reaction is:
\[ I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6 \]

### Conclusion
The reaction taking place is the reduction of iodine by sodium thiosulfate, resulting in the formation of iodide ions and tetrathionate.

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