Balance the following equation stepwise:
`Cr_(2)O_(7)^(2-) + Fe^(2+)++H^(o+)rarrCr^(3+) + Fe^(3+) + H_(2)O`

Explanation

To balance the given redox reaction stepwise, we will follow a systematic approach. The reaction is:

\[ \text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} + \text{H}_2\text{O} \]

### Step 1: Identify Oxidation and Reduction
- **Oxidation**: Iron (\( \text{Fe}^{2+} \)) is oxidized to (\( \text{Fe}^{3+} \)).
- **Reduction**: Chromium (\( \text{Cr}_2\text{O}_7^{2-} \)) is reduced from \( +6 \) oxidation state to \( +3 \).

### Step 2: Write Half-Reactions
1. **Oxidation Half-Reaction**:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

2. **Reduction Half-Reaction**:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 3: Balance Electrons
To balance the electrons, we need to multiply the oxidation half-reaction by 6 so that the number of electrons lost equals the number of electrons gained:

\[ 6 \text{Fe}^{2+} \rightarrow 6 \text{Fe}^{3+} + 6 e^- \]

### Step 4: Combine Half-Reactions
Now we can combine the balanced half-reactions:

\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 5: Final Balanced Equation
The final balanced equation is:

\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]