In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of `CO_(2)` is

Explanation

To determine the number of electrons involved in producing one molecule of \( CO_2 \) in the reaction of oxalate with permanganate in acidic medium, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are oxalate ion (\( C_2O_4^{2-} \)) and permanganate ion (\( MnO_4^{-} \)).

### Step 2: Determine the Oxidation and Reduction Processes
- **Oxidation**: The oxalate ion (\( C_2O_4^{2-} \)) is oxidized to carbon dioxide (\( CO_2 \)).
- **Reduction**: The permanganate ion (\( MnO_4^{-} \)) is reduced to manganese ion (\( Mn^{2+} \)).

### Step 3: Write the Half-Reactions
1. **Oxidation Half-Reaction**:
\[ C_2O_4^{2-} \rightarrow 2 CO_2 + 2 e^{-} \]
This indicates that for every molecule of \( C_2O_4^{2-} \) oxidized, 2 electrons are released.

2. **Reduction Half-Reaction**:
\[ MnO_4^{-} + 8 H^+ + 5 e^{-} \rightarrow Mn^{2+} + 4 H_2O \]
This indicates that for every permanganate ion reduced, 5 electrons are consumed.

### Step 4: Balance the Overall Reaction
To balance the overall reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction matches the number of electrons gained in the reduction half-reaction.

- The oxidation half-reaction produces 2 electrons for every \( C_2O_4^{2-} \).
- The reduction half-reaction consumes 5 electrons for every \( MnO_4^{-} \).

To balance, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
- Oxidation:
\[ 5 C_2O_4^{2-} \rightarrow 10 CO_2 + 10 e^{-} \]
- Reduction:
\[ 2 MnO_4^{-} + 16 H^+ + 10 e^{-} \rightarrow 2 Mn^{2+} + 8 H_2O \]

### Step 5: Combine the Half-Reactions
Combining these balanced half-reactions gives us the overall reaction:
\[ 2 MnO_4^{-} + 5 C_2O_4^{2-} + 16 H^+ \rightarrow 2 Mn^{2+} + 10 CO_2 + 8 H_2O \]

### Step 6: Determine Electrons per Molecule of \( CO_2 \)
From the balanced equation, we see that 10 \( CO_2 \) molecules are produced with a total of 10 electrons involved. Therefore, the number of electrons involved in producing one molecule of \( CO_2 \) is:
\[ \frac{10 \text{ electrons}}{10 \text{ CO}_2} = 1 \text{ electron per } CO_2 \]

### Final Answer
The number of electrons involved in producing one molecule of \( CO_2 \) is **1**.

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