Balance the following equations by ion electron method:
`Cr_(2)O_(7)^(2-) + C_(2)H_(4)O+H^(o+) rarr 2Cr^(3+) + C_(2)H_(4)O_(2) + H_(2)O`

Explanation

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

### Step 1: Identify the half-reactions
The given reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{C}_2\text{H}_4\text{O} + \text{H}^+ \rightarrow 2\text{Cr}^{3+} + \text{C}_2\text{H}_4\text{O}_2 + \text{H}_2\text{O} \]

We can split this into two half-reactions:
1. Oxidation half-reaction:
\[ \text{C}_2\text{H}_4\text{O} \rightarrow \text{C}_2\text{H}_4\text{O}_2 \]
2. Reduction half-reaction:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \]

### Step 2: Balance the oxidation half-reaction
For the oxidation half-reaction:
\[ \text{C}_2\text{H}_4\text{O} \rightarrow \text{C}_2\text{H}_4\text{O}_2 \]

- The oxidation state of carbon changes from -1 in \(\text{C}_2\text{H}_4\text{O}\) to 0 in \(\text{C}_2\text{H}_4\text{O}_2\).
- This requires 2 electrons to balance:
\[ \text{C}_2\text{H}_4\text{O} \rightarrow \text{C}_2\text{H}_4\text{O}_2 + 2e^- \]

- Now, balance the oxygen atoms:
- There is 1 oxygen on the left and 2 on the right, so we add 1 water molecule to the left:
\[ \text{C}_2\text{H}_4\text{O} + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_4\text{O}_2 + 2e^- \]

- Now, balance the hydrogen atoms:
- We have 4 hydrogen on the left and 6 on the right, so we add 2 protons (\(2\text{H}^+\)) to the left:
\[ \text{C}_2\text{H}_4\text{O} + \text{H}_2\text{O} + 2\text{H}^+ \rightarrow \text{C}_2\text{H}_4\text{O}_2 + 2e^- \]

### Step 3: Balance the reduction half-reaction
For the reduction half-reaction:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \]

- The oxidation state of chromium changes from +6 to +3, which means each chromium atom gains 3 electrons. Therefore, for 2 chromium atoms, we need 6 electrons:
\[ \text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} \]

- Now, balance the oxygen atoms:
- There are 7 oxygen atoms on the left, so we add 7 water molecules to the right:
\[ \text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

- Now, balance the hydrogen atoms:
- We have 14 hydrogen on the right, so we add 14 protons (\(14\text{H}^+\)) to the left:
\[ \text{Cr}_2\text{O}_7^{2-} + 6e^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 4: Equalize the number of electrons
Now we have:
1. Oxidation half-reaction:
\[ \text{C}_2\text{H}_4\text{O} + \text{H}_2\text{O} + 2\text{H}^+ \rightarrow \text{C}_2\text{H}_4\text{O}_2 + 2e^- \]
2. Reduction half-reaction:
\[ \text{Cr}_2\text{O}_7^{2-} + 6e^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

To equalize the number of electrons, we multiply the oxidation half-reaction by 3:
\[ 3\text{C}_2\text{H}_4\text{O} + 3\text{H}_2\text{O} + 6\text{H}^+ \rightarrow 3\text{C}_2\text{H}_4\text{O}_2 + 6e^- \]

### Step 5: Combine the half-reactions
Now we can add the two half-reactions:
\[ 3\text{C}_2\text{H}_4\text{O} + 3\text{H}_2\text{O} + 6\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 3\text{C}_2\text{H}_4\text{O}_2 + 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 6: Simplify the equation
Now, we can simplify by canceling out the water molecules:
- We have 3 water on the left and 7 on the right, so we cancel 3 from both sides:
\[ 3\text{C}_2\text{H}_4\text{O} + \text{Cr}_2\text{O}_7^{2-} + 6\text{H}^+ \rightarrow 3\text{C}_2\text{H}_4\text{O}_2 + 2\text{Cr}^{3+} + 4\text{H}_2\text{O} \]

### Final Balanced Equation
The final balanced equation is:
\[ \text{Cr}_2\text{O}_7^{2-} + 3\text{C}_2\text{H}_4\text{O} + 8\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{C}_2\text{H}_4\text{O}_2 + 4\text{H}_2\text{O} \]