When the redox reaction
`Cr_(2)O_(7)^(--)+Fe^(++)+C_(2)O_(4)^(--)toCr^(+++)+Fe^(+++)+CO_(2)`
balanced by ion electron method in acid medium what is correct co-efficient of `Cr_(2)O_(7)^(--)`

Answer

To balance the given redox reaction using the ion-electron method in acidic medium, we will follow these steps:

### Step 1: Identify the half-reactions
The first step is to identify the oxidation and reduction half-reactions in the overall reaction.

1. **Oxidation half-reaction**:
- Iron (Fe) is oxidized from Fe²⁺ to Fe³⁺.
- The half-reaction can be written as:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

2. **Oxidation of oxalate (C₂O₄²⁻)**:
- The oxalate ion (C₂O₄²⁻) is oxidized to carbon dioxide (CO₂).
- The half-reaction can be written as:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 e^- \]

3. **Reduction half-reaction**:
- Dichromate ion (Cr₂O₇²⁻) is reduced to chromium (Cr³⁺).
- The half-reaction can be written as:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 2: Balance the electrons
Next, we need to balance the number of electrons lost in the oxidation half-reactions with the electrons gained in the reduction half-reaction.

- From the oxidation of Fe²⁺, we have 1 electron lost per Fe²⁺.
- From the oxidation of C₂O₄²⁻, we have 2 electrons lost per C₂O₄²⁻.
- Therefore, to balance the electrons, we need to multiply the oxidation half-reaction of Fe²⁺ by 2:
\[ 2 \text{Fe}^{2+} \rightarrow 2 \text{Fe}^{3+} + 2 e^- \]

### Step 3: Combine the half-reactions
Now we can combine the half-reactions, ensuring that the electrons cancel out.

1. The oxidation half-reactions:
\[ 2 \text{Fe}^{2+} \rightarrow 2 \text{Fe}^{3+} + 2 e^- \]
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 e^- \]

2. The reduction half-reaction:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

3. Combine all half-reactions:
- Since we have 6 electrons in the reduction half-reaction, we need to multiply the oxidation half-reactions by 3:
\[ 3 \text{C}_2\text{O}_4^{2-} \rightarrow 6 \text{CO}_2 + 6 e^- \]
\[ 2 \text{Fe}^{2+} \rightarrow 2 \text{Fe}^{3+} + 2 e^- \]

4. The overall balanced equation becomes:
\[ \text{Cr}_2\text{O}_7^{2-} + 3 \text{C}_2\text{O}_4^{2-} + 2 \text{Fe}^{2+} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 6 \text{CO}_2 + 2 \text{Fe}^{3+} + 7 \text{H}_2\text{O} \]

### Step 4: Identify the coefficient of Cr₂O₇²⁻
From the balanced equation, we can see that the coefficient of Cr₂O₇²⁻ is 1.

### Final Answer
The correct coefficient of Cr₂O₇²⁻ in the balanced equation is **1**.

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