How many moles of `KMnO_(4)` are required to oxidise one mole of `SnCl_(2)` in acidic medium?

Options

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

Answer

To determine how many moles of `KMnO4` are required to oxidize one mole of `SnCl2` in acidic medium, we need to analyze the redox reaction that occurs between potassium permanganate (KMnO4) and tin(II) chloride (SnCl2) in an acidic environment.

### Step-by-step Solution:

1. **Identify the Oxidation States:**
- In `SnCl2`, tin (Sn) has an oxidation state of +2.
- In `KMnO4`, manganese (Mn) has an oxidation state of +7.

2. **Determine the Reaction:**
- In acidic medium, `KMnO4` acts as an oxidizing agent and will reduce from Mn(VII) to Mn(II) while oxidizing Sn(II) to Sn(IV).
- The half-reaction for the reduction of MnO4^- in acidic medium is:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
- The half-reaction for the oxidation of Sn(II) to Sn(IV) is:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^- \]

3. **Balance the Electrons:**
- From the half-reaction of MnO4^-, we see that 5 electrons are needed to reduce one MnO4^- ion.
- From the half-reaction of Sn(II), 2 electrons are released when one Sn(II) is oxidized to Sn(IV).
- To balance the electrons, we need to find a common multiple. The least common multiple of 5 and 2 is 10.
- Therefore, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
\[ 5\text{Sn}^{2+} \rightarrow 5\text{Sn}^{4+} + 10\text{e}^- \]
\[ 2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \]

4. **Combine the Half-Reactions:**
- Now we can combine both half-reactions:
\[ 2\text{MnO}_4^- + 5\text{Sn}^{2+} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{Sn}^{4+} + 8\text{H}_2\text{O} \]

5. **Determine Moles of KMnO4 Required:**
- From the balanced equation, we see that 2 moles of `KMnO4` are required to oxidize 5 moles of `SnCl2`.
- Therefore, to find out how many moles of `KMnO4` are required to oxidize 1 mole of `SnCl2`, we set up a ratio:
\[ \text{Moles of KMnO}_4 = \frac{2 \text{ moles of KMnO}_4}{5 \text{ moles of SnCl}_2} \times 1 \text{ mole of SnCl}_2 = \frac{2}{5} \text{ moles of KMnO}_4 \]

### Final Answer:
To oxidize one mole of `SnCl2`, **0.4 moles of `KMnO4` are required.**