When the redox reaction.
`Cr_(2)O_(7)^(--)+C_(2)H_(4)OtoC_(2)H_(4)O_(2)+Cr^(++)` is balanced by ion electron method in acidic medium.
what is correct co-efficient of `H^(+)` ions.

Answer

To balance the redox reaction \( \text{Cr}_2\text{O}_7^{2-} + \text{C}_2\text{H}_4\text{O} \rightarrow \text{C}_2\text{H}_4\text{O}_2 + \text{Cr}^{2+} \) using the ion-electron method in acidic medium, we will follow these steps:

### Step 1: Determine the oxidation states
1. **Chromium in \( \text{Cr}_2\text{O}_7^{2-} \)**:
- Let the oxidation state of Cr be \( x \).
- The equation becomes \( 2x + 7(-2) = -2 \).
- Solving gives \( 2x - 14 = -2 \) → \( 2x = 12 \) → \( x = +6 \).
- Therefore, Cr is in the +6 oxidation state.

2. **Carbon in \( \text{C}_2\text{H}_4\text{O} \)**:
- Let the oxidation state of C be \( y \).
- The equation becomes \( 2y + 4(1) + (-2) = 0 \).
- Solving gives \( 2y + 4 - 2 = 0 \) → \( 2y + 2 = 0 \) → \( y = -1 \).
- Therefore, C is in the -1 oxidation state.

3. **Carbon in \( \text{C}_2\text{H}_4\text{O}_2 \)**:
- Let the oxidation state of C be \( z \).
- The equation becomes \( 2z + 4(1) + 2(-2) = 0 \).
- Solving gives \( 2z + 4 - 4 = 0 \) → \( 2z = 0 \) → \( z = 0 \).
- Therefore, C is in the 0 oxidation state.

### Step 2: Identify the oxidation and reduction half-reactions
- **Oxidation half-reaction**:
- \( \text{C}_2\text{H}_4\text{O} \rightarrow \text{C}_2\text{H}_4\text{O}_2 \)
- Here, each carbon goes from -1 to 0, which means 2 electrons are lost.

- **Reduction half-reaction**:
- \( \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{2+} \)
- Each Cr goes from +6 to +2. For 2 Cr, a total of 6 electrons are gained.

### Step 3: Balance the half-reactions
1. **For the reduction half-reaction**:
- \( \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{2+} + 7 \text{H}_2\text{O} \)

2. **For the oxidation half-reaction**:
- Multiply the oxidation half-reaction by 3 to balance the electrons:
- \( 3 \text{C}_2\text{H}_4\text{O} \rightarrow 3 \text{C}_2\text{H}_4\text{O}_2 + 6 \text{e}^- \)

### Step 4: Combine the balanced half-reactions
- Combining both half-reactions gives:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- + 3 \text{C}_2\text{H}_4\text{O} \rightarrow 2 \text{Cr}^{2+} + 7 \text{H}_2\text{O} + 3 \text{C}_2\text{H}_4\text{O}_2 + 6 \text{e}^- \]
- The electrons cancel out, leading to:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 3 \text{C}_2\text{H}_4\text{O} \rightarrow 2 \text{Cr}^{2+} + 3 \text{C}_2\text{H}_4\text{O}_2 + 7 \text{H}_2\text{O} \]

### Final Answer
The coefficient of \( \text{H}^+ \) ions is **14**.

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