Which of the following is/are correct about the redox reaction?
`MnO_(4)^(ɵ)+S_(2)O_(3)^(2-)+H^(o+)toMn^(+2)+S_(4)O_(6)^(2-)`

Options

A. 1 " mol of "`S_(2)O_(3)^(2-)` is oxidised by 8 " mol of "`MnO_(4)^(ɵ)`
B. The above redox reaction with the change of pH from 4 to 10 will have an effect on the stoichiometry of the reaction.
C. Change of pH form 4 to 7 will change the nature of the product.
D. At `pH=7,S_(2)O_(3)^(2-)` ions are oxidised to `HSO_(4)^(ɵ)`

Explanation

To determine which statements about the given redox reaction are correct, we will analyze the reaction step by step.

### Given Reaction:
\[ \text{MnO}_4^{-} + \text{S}_2\text{O}_3^{2-} + \text{H}^{+} \rightarrow \text{Mn}^{2+} + \text{S}_4\text{O}_6^{2-} \]

### Step 1: Identify Oxidation States
1. **For MnO4⁻:**
- Let the oxidation state of Mn be \( x \).
- Oxygen has an oxidation state of -2.
- Thus, the equation is:
\[ x + 4(-2) = -1 \]
\[ x - 8 = -1 \]
\[ x = +7 \]
- Therefore, the oxidation state of Mn in MnO4⁻ is +7.

2. **For Mn²⁺:**
- The oxidation state of Mn is +2.

3. **For S2O3²⁻:**
- Let the oxidation state of S be \( y \).
- Thus, the equation is:
\[ 2y + 3(-2) = -2 \]
\[ 2y - 6 = -2 \]
\[ 2y = 4 \]
\[ y = +2 \]
- Therefore, the oxidation state of S in S2O3²⁻ is +2.

4. **For S4O6²⁻:**
- Let the oxidation state of S be \( z \).
- Thus, the equation is:
\[ 4z + 6(-2) = -2 \]
\[ 4z - 12 = -2 \]
\[ 4z = 10 \]
\[ z = +2.5 \]
- Therefore, the oxidation state of S in S4O6²⁻ is +2.5.

### Step 2: Determine Changes in Oxidation States
- **Mn** changes from +7 to +2 (reduction).
- **S** changes from +2 to +2.5 (oxidation).

### Step 3: Balance the Half-Reactions
1. **For MnO4⁻ to Mn²⁺:**
- The change in oxidation state is 5 (from +7 to +2).
- To balance this, add 5 electrons:
\[ \text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

2. **For S2O3²⁻ to S4O6²⁻:**
- The change in oxidation state is 0.5 per sulfur atom (from +2 to +2.5).
- Since there are 2 sulfur atoms in S2O3²⁻, we need to add 2 electrons:
\[ 2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2e^{-} \]

### Step 4: Determine the Stoichiometry
- From the half-reactions, we see that:
- 1 mole of S2O3²⁻ reacts with 5 moles of MnO4⁻.
- Therefore, the stoichiometry is:
\[ 1 \text{ mole of } \text{S}_2\text{O}_3^{2-} \text{ reacts with } 5 \text{ moles of } \text{MnO}_4^{-} \]

### Step 5: Analyze the Options
1. **Option 1:** 1 mole of S2O3²⁻ meets 8 moles of MnO4⁻. **(Incorrect)**
2. **Option 2:** Change of pH from 4 to 10 affects stoichiometry. **(Correct)**
3. **Option 3:** Change of pH from 4 to 7 changes the nature of the product. **(Correct)**
4. **Option 4:** At pH 7, S2O3²⁻ ions are oxidized to HSO4⁻. **(Correct)**

### Conclusion
The correct statements about the redox reaction are options 2, 3, and 4.

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