Balance the following equations by ion electron method:
a. `Cr_(2)O_(7)^(2-) + C_(2)H_(4)O+H^(o+) rarr 2Cr^(3+) + C_(2)H_(4)O_(2) + H_(2)O`
b. `Cu_(2)O + H^(o+)+NO_(3)^(Ө) rarr Cu^(2+)+NO+H_(2)O`

Explanation

To balance the given equations using the ion-electron method, we will follow a systematic approach for each part.

### Part A:
**Given Reaction:**
\[ \text{Cr}_{2}\text{O}_{7}^{2-} + \text{C}_{2}\text{H}_{4}\text{O} + \text{H}^{+} \rightarrow 2\text{Cr}^{3+} + \text{C}_{2}\text{H}_{4}\text{O}_{2} + \text{H}_{2}\text{O} \]

**Step 1: Identify Oxidation and Reduction**
- Chromium in \(\text{Cr}_{2}\text{O}_{7}^{2-}\) is in the +6 oxidation state and is reduced to +3 in \(\text{Cr}^{3+}\).
- Carbon in \(\text{C}_{2}\text{H}_{4}\text{O}\) is oxidized from -1 to 0 in \(\text{C}_{2}\text{H}_{4}\text{O}_{2}\).

**Step 2: Write the Half-Reactions**
- **Oxidation Half-Reaction:**
\[ \text{C}_{2}\text{H}_{4}\text{O} \rightarrow \text{C}_{2}\text{H}_{4}\text{O}_{2} + 2e^{-} \]

- **Reduction Half-Reaction:**
\[ \text{Cr}_{2}\text{O}_{7}^{2-} + 6e^{-} + 14\text{H}^{+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O} \]

**Step 3: Equalize Electron Transfer**
- The oxidation half-reaction involves 2 electrons, while the reduction half-reaction involves 6 electrons. To equalize, multiply the oxidation half-reaction by 3:
\[ 3\text{C}_{2}\text{H}_{4}\text{O} \rightarrow 3\text{C}_{2}\text{H}_{4}\text{O}_{2} + 6e^{-} \]

**Step 4: Combine the Half-Reactions**
Now, add the two half-reactions:
\[ \text{Cr}_{2}\text{O}_{7}^{2-} + 3\text{C}_{2}\text{H}_{4}\text{O} + 14\text{H}^{+} \rightarrow 2\text{Cr}^{3+} + 3\text{C}_{2}\text{H}_{4}\text{O}_{2} + 7\text{H}_{2}\text{O} \]

**Step 5: Final Check**
- Count atoms and charges on both sides to ensure they are balanced.

### Part B:
**Given Reaction:**
\[ \text{Cu}_{2}\text{O} + \text{H}^{+} + \text{NO}_{3}^{-} \rightarrow \text{Cu}^{2+} + \text{NO} + \text{H}_{2}\text{O} \]

**Step 1: Identify Oxidation and Reduction**
- Copper is oxidized from +1 in \(\text{Cu}_{2}\text{O}\) to +2 in \(\text{Cu}^{2+}\).
- Nitrogen is reduced from +5 in \(\text{NO}_{3}^{-}\) to +2 in \(\text{NO}\).

**Step 2: Write the Half-Reactions**
- **Oxidation Half-Reaction:**
\[ \text{Cu}_{2}\text{O} \rightarrow 2\text{Cu}^{2+} + 2e^{-} \]

- **Reduction Half-Reaction:**
\[ \text{NO}_{3}^{-} + 3e^{-} + 4\text{H}^{+} \rightarrow \text{NO} + 2\text{H}_{2}\text{O} \]

**Step 3: Equalize Electron Transfer**
- The oxidation half-reaction involves 2 electrons, while the reduction half-reaction involves 3 electrons. To equalize, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
\[ 3\text{Cu}_{2}\text{O} \rightarrow 6\text{Cu}^{2+} + 6e^{-} \]
\[ 2\text{NO}_{3}^{-} + 6e^{-} + 8\text{H}^{+} \rightarrow 2\text{NO} + 4\text{H}_{2}\text{O} \]

**Step 4: Combine the Half-Reactions**
Now, add the two half-reactions:
\[ 3\text{Cu}_{2}\text{O} + 2\text{NO}_{3}^{-} + 8\text{H}^{+} \rightarrow 6\text{Cu}^{2+} + 2\text{NO} + 4\text{H}_{2}\text{O} \]

**Step 5: Final Check**
- Count atoms and charges on both sides to ensure they are balanced.