The number of mole of oxalate ions oxidised by one mole of `MnO_(4)^(-)` is :

Options

A. `1//5`
B. `2//5`
C. `5//2`
D. `5`

Explanation

To solve the problem of determining the number of moles of oxalate ions (C2O4^2-) oxidized by one mole of permanganate ions (MnO4^-), we can follow these steps:

### Step 1: Write the half-reactions
The first step is to write the half-reactions for both the reduction of permanganate ions and the oxidation of oxalate ions.

1. **Reduction of MnO4^-**:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
In this reaction, MnO4^- is reduced to Mn^2+, and the oxidation state of manganese changes from +7 to +2, which involves the gain of 5 electrons.

2. **Oxidation of C2O4^2-**:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 \text{e}^- \]
In this reaction, the oxalate ion (C2O4^2-) is oxidized to carbon dioxide (CO2), and the oxidation state of carbon changes from +3 to +4, which involves the loss of 2 electrons.

### Step 2: Determine the stoichiometry of the reaction
Next, we need to determine how many moles of oxalate ions are oxidized by one mole of permanganate ions.

From the half-reactions:
- 1 mole of MnO4^- requires 5 electrons.
- Each mole of C2O4^2- loses 2 electrons.

### Step 3: Set up the relationship
Let \( x \) be the number of moles of C2O4^2- oxidized by 1 mole of MnO4^-.

From the stoichiometry:
\[ 5 \text{ e}^- \text{ (from MnO4^-)} = 2x \text{ e}^- \text{ (from C2O4^{2-})} \]

### Step 4: Solve for x
Now we can solve for \( x \):
\[ 5 = 2x \implies x = \frac{5}{2} \]

### Conclusion
Thus, the number of moles of oxalate ions oxidized by one mole of MnO4^- is:
\[ \frac{5}{2} \text{ moles of C2O4^{2-}} \]

### Final Answer
The number of moles of oxalate ions oxidized by one mole of MnO4^- is \( \frac{5}{2} \) or 2.5 moles.

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