In the reaction `MnO_(4)^(-)+SO_(3)^(-2)+H^(+)rarrSO_(4)^(-2)+Mn^(2+)+H_(2)O`

Options

A. `MnO_(4)^(-) and H^(+)` both are reduced
B. `MnO_(4)^(-)` is reduced and `H^(+)` is oxidised
C. `MnO_(4)^(-)` is reduced and `SO_(3)^(2-)` is oxidised
D. `MnO_(4)^(-)` is oxidised and `SO_(3)^(2-)` is reduced

Answer

To determine which species is oxidized and which is reduced in the reaction:

\[ \text{MnO}_4^{-} + \text{SO}_3^{2-} + \text{H}^{+} \rightarrow \text{SO}_4^{2-} + \text{Mn}^{2+} + \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Identify the oxidation states of the elements in the reactants and products.

1. **For MnO₄⁻**:
- Let the oxidation state of Mn be \( x \).
- The oxidation state of O is \(-2\).
- The equation becomes:
\[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \]
- Thus, the oxidation state of Mn in MnO₄⁻ is \( +7 \).

2. **For SO₃²⁻**:
- Let the oxidation state of S be \( y \).
- The equation becomes:
\[ y + 3(-2) = -2 \implies y - 6 = -2 \implies y = +4 \]
- Thus, the oxidation state of S in SO₃²⁻ is \( +4 \).

### Step 2: Identify the oxidation states in the products.

1. **For SO₄²⁻**:
- Let the oxidation state of S be \( y' \).
- The equation becomes:
\[ y' + 4(-2) = -2 \implies y' - 8 = -2 \implies y' = +6 \]
- Thus, the oxidation state of S in SO₄²⁻ is \( +6 \).

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

### Step 3: Determine the changes in oxidation states.

- **For Mn**:
- Changes from \( +7 \) in MnO₄⁻ to \( +2 \) in Mn²⁺.
- This is a decrease in oxidation state, indicating reduction.

- **For S**:
- Changes from \( +4 \) in SO₃²⁻ to \( +6 \) in SO₄²⁻.
- This is an increase in oxidation state, indicating oxidation.

### Conclusion:
- **MnO₄⁻ is reduced** (gains electrons, reduction).
- **SO₃²⁻ is oxidized** (loses electrons, oxidation).

Thus, the final answer is:
- MnO₄⁻ is reduced, and SO₃²⁻ is oxidized.

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