1mol of `MnO_(4)^(-)` will oxidise x mole of ferric oxalate in acidic medium, x is -

Options

A. `( 5)/( 6)`
B. `( 6)/( 5)`
C. 5
D. 6

Answer

To solve the problem of how many moles of ferric oxalate (Fe(C2O4)3) can be oxidized by 1 mole of permanganate ion (MnO4^-), we will follow these steps:

### Step 1: Write the balanced redox reaction
The reaction between permanganate ion and ferric oxalate in acidic medium can be represented as follows:

\[ \text{MnO}_4^- + \text{Fe(C}_2\text{O}_4)_3 \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{CO}_2 \]

### Step 2: Determine the oxidation states and n-factors
1. **For MnO4^-**:
- Manganese (Mn) in MnO4^- is in the +7 oxidation state.
- It gets reduced to Mn^2+, which is in the +2 oxidation state.
- The change in oxidation state is from +7 to +2, which means it gains 5 electrons. Therefore, the n-factor for MnO4^- is 5.

2. **For Fe(C2O4)3**:
- Iron (Fe) in Fe(C2O4)3 is in the +3 oxidation state.
- It remains as Fe^3+ after the reaction, so there is no change in oxidation state for iron.
- However, each oxalate ion (C2O4^2-) is oxidized to CO2, which means each oxalate loses 2 electrons.
- Since there are 3 oxalate ions in Fe(C2O4)3, the total n-factor for the oxalate is 3 * 2 = 6.

### Step 3: Set up the equivalence equation
Using the concept of equivalents, we can set up the following equation based on the n-factors:

\[ \text{Equivalent of MnO}_4^- = \text{Equivalent of Fe(C}_2\text{O}_4)_3 \]

This can be expressed as:

\[ n_{\text{MnO}_4^-} \times \text{moles of MnO}_4^- = n_{\text{Fe(C}_2\text{O}_4)_3} \times \text{moles of Fe(C}_2\text{O}_4)_3 \]

Substituting the n-factors:

\[ 5 \times 1 = 6 \times x \]

### Step 4: Solve for x
Now, we can solve for x:

\[ 5 = 6x \]

\[ x = \frac{5}{6} \]

### Conclusion
Thus, 1 mole of MnO4^- will oxidize \(\frac{5}{6}\) moles of ferric oxalate.

### Final Answer
x = \(\frac{5}{6}\)

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