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 \( KMnO_4 \) are required to oxidize one mole of \( SnCl_2 \) in acidic medium, we can follow these steps:

### Step 1: Identify the oxidation states
- In \( SnCl_2 \), tin (Sn) has an oxidation state of +2.
- In the reaction, tin is oxidized to \( Sn^{4+} \) (oxidation state +4).
- Manganese in \( KMnO_4 \) has an oxidation state of +7 and is reduced to \( Mn^{2+} \) (oxidation state +2).

### Step 2: Write the half-reactions
- The oxidation half-reaction for tin can be written as:
\[ Sn^{2+} \rightarrow Sn^{4+} + 2e^- \]
- The reduction half-reaction for permanganate ion can be written as:
\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]

### Step 3: Determine the number of electrons transferred
- From the oxidation half-reaction, 1 mole of \( Sn^{2+} \) loses 2 electrons.
- From the reduction half-reaction, 1 mole of \( MnO_4^- \) gains 5 electrons.

### Step 4: Relate the moles of \( KMnO_4 \) to moles of \( SnCl_2 \)
- To balance the electrons, we need to find a common multiple. The least common multiple of 2 (from tin) and 5 (from permanganate) is 10.
- Therefore, to balance the reaction:
- We need 5 moles of \( Sn^{2+} \) (which will lose a total of 10 electrons).
- We need 2 moles of \( KMnO_4 \) (which will gain a total of 10 electrons).

### Step 5: Calculate the moles of \( KMnO_4 \) required for 1 mole of \( SnCl_2 \)
- Since 5 moles of \( Sn^{2+} \) require 2 moles of \( KMnO_4 \), for 1 mole of \( Sn^{2+} \), the moles of \( KMnO_4 \) required can be calculated as:
\[ \text{Moles of } KMnO_4 = \frac{2 \text{ moles of } KMnO_4}{5 \text{ moles of } Sn^{2+}} \times 1 \text{ mole of } Sn^{2+} = \frac{2}{5} \text{ moles of } KMnO_4 \]

### Final Answer
Thus, the number of moles of \( KMnO_4 \) required to oxidize one mole of \( SnCl_2 \) in acidic medium is \( \frac{2}{5} \) moles.

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