Redox is a reaction in which both oxidation and reduction will take place simultaneously . It is obvious that if one substance gives electron there must be another substance to accept these electrons . In some reactions, same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions.
For calculating equivalent mass in redox reaction change in oxidtaion number is realted to n-factor which is reciprocal of molar ratio.
How many moles of `KMnO_(4)` are reacted with one mole of ferrous oxalate in acidic medium ?

Options

A. `(2)/(5)`
B. `1/5`
C. `3/5`
D. `5/3`

Answer

To solve the problem of how many moles of KMnO4 react with one mole of ferrous oxalate (FeC2O4) in acidic medium, we will follow these steps:

### Step 1: Identify the oxidation states
- In ferrous oxalate (FeC2O4), iron (Fe) is in the +2 oxidation state (Fe²⁺).
- In KMnO4, manganese (Mn) is in the +7 oxidation state (MnO4⁻).

### Step 2: Write the half-reactions
1. **Oxidation half-reaction**:
- Fe²⁺ → Fe³⁺ + e⁻ (loses 1 electron)
- The oxalate ion (C2O4²⁻) is oxidized to CO2:
- C2O4²⁻ → 2 CO2 + 2 e⁻ (loses 2 electrons)

Overall oxidation reaction:
\[ \text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Fe}^{3+} + 2 \text{CO}_2 + 3 e^- \]

2. **Reduction half-reaction**:
- MnO4⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H2O (gains 5 electrons)

### Step 3: Balance the electrons
To balance the overall reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction.

- From the oxidation half-reaction, we have a total of 3 electrons lost (from Fe²⁺ and C2O4²⁻).
- From the reduction half-reaction, we have 5 electrons gained.

To balance the electrons, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 3.

### Step 4: Write the balanced equation
1. Multiply the oxidation half-reaction by 5:
\[ 5 \text{Fe}^{2+} + 5 \text{C}_2\text{O}_4^{2-} \rightarrow 5 \text{Fe}^{3+} + 10 \text{CO}_2 + 15 e^- \]

2. Multiply the reduction half-reaction by 3:
\[ 3 \text{MnO}_4^{-} + 24 \text{H}^+ + 15 e^- \rightarrow 3 \text{Mn}^{2+} + 12 \text{H}_2\text{O} \]

### Step 5: Combine the half-reactions
Now, we can combine both half-reactions:
\[ 5 \text{Fe}^{2+} + 5 \text{C}_2\text{O}_4^{2-} + 3 \text{MnO}_4^{-} + 24 \text{H}^+ \rightarrow 5 \text{Fe}^{3+} + 10 \text{CO}_2 + 3 \text{Mn}^{2+} + 12 \text{H}_2\text{O} \]

### Step 6: Determine the mole ratio
From the balanced equation, we see that:
- 5 moles of ferrous oxalate react with 3 moles of KMnO4.

To find out how many moles of KMnO4 react with 1 mole of ferrous oxalate:
\[ \text{If } 5 \text{ moles of FeC}_2\text{O}_4 \text{ require } 3 \text{ moles of KMnO}_4, \text{ then } 1 \text{ mole of FeC}_2\text{O}_4 \text{ requires } \frac{3}{5} \text{ moles of KMnO}_4. \]

### Final Answer
Thus, 1 mole of ferrous oxalate reacts with **0.6 moles (or 3/5 moles)** of KMnO4 in acidic medium.

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