What is the coefficient of oxalate ion in the following reaction ?
`MnO_(4)^(-)+C_(2)O_(4)^(-)+H^(+) to Mn^(2+)+CO_(2)H_(2)O`

Options

A. 4
B. 2
C. 3
D. 5

Explanation

To find the coefficient of the oxalate ion (C₂O₄²⁻) in the given redox reaction, we can follow these steps:

### Step 1: Write the half-reactions
The first step in balancing a redox reaction is to separate it into two half-reactions: one for oxidation and one for reduction.

1. **Reduction half-reaction**:
\[ \text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

2. **Oxidation half-reaction**:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-} \]

### Step 2: Balance the electrons
Next, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

- The reduction half-reaction involves 5 electrons.
- The oxidation half-reaction involves 2 electrons.

To balance the electrons, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:

- **Oxidation half-reaction (multiplied by 5)**:
\[ 5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10\text{e}^{-} \]

- **Reduction half-reaction (multiplied by 2)**:
\[ 2\text{MnO}_4^{-} + 16\text{H}^{+} + 10\text{e}^{-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \]

### Step 3: Combine the half-reactions
Now we can combine the two half-reactions:

\[ 2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \]

### Step 4: Identify the coefficient of the oxalate ion
From the balanced equation, we can see that the coefficient of the oxalate ion (C₂O₄²⁻) is **5**.

### Final Answer
The coefficient of the oxalate ion (C₂O₄²⁻) in the reaction is **5**.

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