In the titration of `FeSO_(4)` against `KMnO_(4)` in acidic medium, one mole of `KMnO_(4)` reacts with x moles of `FeSO_(4)`. The value of x is

Options

A. 10
B. 8
C. 5
D. 3

Answer

To solve the problem of how many moles of `FeSO4` react with one mole of `KMnO4` in acidic medium, we need to follow these steps:

### Step 1: Write the unbalanced chemical equation
In acidic medium, `KMnO4` reacts with `FeSO4`. The unbalanced reaction can be written as:
\[ \text{KMnO}_4 + \text{FeSO}_4 \rightarrow \text{Fe}_2(SO_4)_3 + \text{MnSO}_4 + \text{K}_2SO_4 + \text{H}_2O \]

### Step 2: Identify the oxidation states
In this reaction:
- The oxidation state of Mn in `KMnO4` is +7.
- The oxidation state of Fe in `FeSO4` is +2.
- In the product `Fe2(SO4)3`, the oxidation state of Fe is +3.
- Mn is reduced from +7 to +2 in `MnSO4`.

### Step 3: Balance the half-reactions
1. **Reduction half-reaction**:
\[ \text{Mn}^{7+} + 8 \text{e}^- \rightarrow \text{Mn}^{2+} \]
(1 mole of `KMnO4` gains 5 electrons)

2. **Oxidation half-reaction**:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \]
(1 mole of `FeSO4` loses 1 electron)

### Step 4: Balance the overall reaction
To balance the number of electrons transferred, we need to multiply the oxidation half-reaction by 5 (since 5 moles of `FeSO4` will provide 5 electrons):
\[ 5 \text{Fe}^{2+} \rightarrow 5 \text{Fe}^{3+} + 5 \text{e}^- \]

Now, we can combine the balanced half-reactions:
\[ \text{2 KMnO}_4 + 10 \text{FeSO}_4 + 8 \text{H}_2SO_4 \rightarrow 5 \text{Fe}_2(SO_4)_3 + 2 \text{MnSO}_4 + \text{K}_2SO_4 + 8 \text{H}_2O \]

### Step 5: Determine the mole ratio
From the balanced equation:
- 2 moles of `KMnO4` react with 10 moles of `FeSO4`.
- Therefore, 1 mole of `KMnO4` will react with 5 moles of `FeSO4`.

### Final Answer
The value of \( x \) is 5.

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