In reaction of `KMnO_4` and Mohr's salt, `FeSO_4` is oxidised to

Options

A. `Fe^(2+)`
B. `Fe^(3+)`
C. Fe
D. All of these

Explanation

To determine what `FeSO4` is oxidized to when it reacts with `KMnO4`, we can follow these steps:

### Step 1: Identify the Oxidation States
In `FeSO4`, iron (Fe) is in the +2 oxidation state (Fe²⁺). In `KMnO4`, manganese (Mn) is in the +7 oxidation state (MnO4²⁻).

### Step 2: Determine the Reaction
In a redox reaction, one species is oxidized (loses electrons) and another is reduced (gains electrons). Here, `KMnO4` acts as an oxidizing agent, meaning it will reduce itself while oxidizing another species.

### Step 3: Write the Half-Reaction for Manganese
The half-reaction for the reduction of manganese can be written as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
This shows that Mn is reduced from +7 to +2 by gaining 5 electrons.

### Step 4: Write the Half-Reaction for Iron
Since `FeSO4` contains Fe²⁺, the oxidation half-reaction can be written as:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \]
This indicates that Fe²⁺ is oxidized to Fe³⁺ by losing one electron.

### Step 5: Balance the Electrons
To balance the electrons transferred in the overall reaction, we need to ensure that the number of electrons lost by iron equals the number of electrons gained by manganese. Since 5 electrons are gained by Mn, we need 5 Fe²⁺ ions to provide these electrons:
\[ 5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5\text{e}^- \]

### Step 6: Combine the Half-Reactions
Now we can combine the half-reactions:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} \]

### Conclusion
In this reaction, `FeSO4` (Fe²⁺) is oxidized to `Fe³⁺`.