The number of moles of `KMnO_(4)` needed to react with one mole of `SO_(3)^(2-)` in acidic solution is

Options

A. `4//5`
B. `2//5`
C. 1
D. `3//5`

Explanation

To determine the number of moles of KMnO₄ needed to react with one mole of SO₃²⁻ in acidic solution, we can follow these steps:

### Step 1: Identify the half-reactions
In acidic solution, KMnO₄ (which contains MnO₄⁻) is reduced to Mn²⁺, and SO₃²⁻ is oxidized to SO₄²⁻.

### Step 2: Write the reduction half-reaction
The reduction half-reaction for MnO₄⁻ to Mn²⁺ in acidic medium is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
This shows that 1 mole of MnO₄⁻ gains 5 electrons.

### Step 3: Write the oxidation half-reaction
The oxidation half-reaction for SO₃²⁻ to SO₄²⁻ is:
\[ \text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 2 \text{H}^+ + 2 \text{e}^- \]
This shows that 1 mole of SO₃²⁻ loses 2 electrons.

### Step 4: Balance the electrons
To balance the overall reaction, 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 1 mole of MnO₄⁻ requires 5 electrons.
- From the oxidation half-reaction, 1 mole of SO₃²⁻ produces 2 electrons.

To balance the electrons, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
- 5 SO₃²⁻ will produce \( 5 \times 2 = 10 \) electrons.
- 2 MnO₄⁻ will consume \( 2 \times 5 = 10 \) electrons.

### Step 5: Write the balanced overall reaction
Combining the balanced half-reactions gives us:
\[ 2 \text{MnO}_4^- + 16 \text{H}^+ + 5 \text{SO}_3^{2-} \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + 5 \text{SO}_4^{2-} \]

### Step 6: Determine the moles of KMnO₄ needed
From the balanced equation, we see that:
- 5 moles of SO₃²⁻ react with 2 moles of MnO₄⁻.

To find out how many moles of KMnO₄ are needed to react with 1 mole of SO₃²⁻, we set up a proportion:
\[ \frac{2 \text{ moles of KMnO}_4}{5 \text{ moles of SO}_3^{2-}} = x \text{ moles of KMnO}_4 \text{ for } 1 \text{ mole of SO}_3^{2-} \]

Solving for \( x \):
\[ x = \frac{2}{5} \text{ moles of KMnO}_4 \]

### Final Answer
The number of moles of KMnO₄ needed to react with one mole of SO₃²⁻ in acidic solution is \( \frac{2}{5} \) moles.

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