Consider the reaction:
`H_(2)SO_(3)(aq)+Sn^(4+)(aq)+H_(2)O(l)toSn^(2+)(aq)+HSO_(4)^(-)(aq)+3H^(+)(aq)`
Which of the following statements is correct?

Options

A. `Sn4^(+)` is the oxidizing agent because it undergoes oxidation.
B. `Sn4^(+)` is the reducing agent because it undergoes oxidation.
C. `H2SO_(3)` is the reducing agent because it undergoes oxidation.
D. `H2SO_(3)` is the reducing agent because it undergoes reduction.

Explanation

To analyze the given redox reaction and determine the correct statement, we will follow these steps:

### Step 1: Identify the oxidation states of the elements in the reactants and products.
- In `H₂SO₃`, the oxidation state of sulfur (S) can be calculated as follows:
- Let the oxidation state of S be x.
- The oxidation state of H is +1 and O is -2.
- The equation becomes:
\[ 2(+1) + x + 3(-2) = 0 \implies 2 + x - 6 = 0 \implies x = +4 \]
- Therefore, the oxidation state of sulfur in `H₂SO₃` is +4.

- In `Sn^(4+)`, the oxidation state of Sn is +4.

- In `Sn^(2+)`, the oxidation state of Sn is +2.

- In `HSO₄^(-)`, the oxidation state of sulfur can be calculated similarly:
- Let the oxidation state of S be y.
- The equation becomes:
\[ +1 + y + 4(-2) = -1 \implies 1 + y - 8 = -1 \implies y = +6 \]
- Therefore, the oxidation state of sulfur in `HSO₄^(-)` is +6.

### Step 2: Determine the changes in oxidation states.
- In the reaction, `H₂SO₃` (S from +4 to +6 in `HSO₄^(-)`) indicates that sulfur is oxidized (loss of electrons).
- `Sn^(4+)` (Sn from +4 to +2 in `Sn^(2+)`) indicates that tin is reduced (gain of electrons).

### Step 3: Identify the oxidizing and reducing agents.
- The reducing agent (RA) is the species that gets oxidized. Here, `H₂SO₃` is the reducing agent as it loses electrons.
- The oxidizing agent (OA) is the species that gets reduced. Here, `Sn^(4+)` is the oxidizing agent as it gains electrons.

### Step 4: Evaluate the statements provided in the options.
- Based on the analysis:
- `Sn^(4+)` is the oxidizing agent because it undergoes reduction.
- `H₂SO₃` is the reducing agent because it undergoes oxidation.

### Conclusion:
The correct statement is that `Sn^(4+)` is the oxidizing agent, and `H₂SO₃` is the reducing agent.