For the redox reaction: `Fe^(2+)+ Cr_2O_7^(2-) +H^+ rarrFe^(3+) + Cr^(3+) + H_2O` The correct coefficients of the reactants for the balanced reaction are

Answer

To balance the redox reaction \( \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ \rightarrow \text{Fe}^{3+} + \text{Cr}^{3+} + \text{H}_2\text{O} \), we will follow these steps:

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

### Step 2: Write the Half-Reactions
1. **Oxidation Half-Reaction**:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
(Here, one electron is lost)

2. **Reduction Half-Reaction**:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]
(Here, six electrons are gained)

### Step 3: Balance the Electrons
To balance the number of electrons transferred, we need to multiply the oxidation half-reaction by 6:
\[ 6 \text{Fe}^{2+} \rightarrow 6 \text{Fe}^{3+} + 6 e^- \]

### Step 4: Combine the Half-Reactions
Now, we can add the balanced half-reactions together:
\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 5: Final Balanced Equation
The final balanced equation is:
\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Coefficients of the Reactants
The coefficients of the reactants are:
- \( 6 \) for \( \text{Fe}^{2+} \)
- \( 1 \) for \( \text{Cr}_2\text{O}_7^{2-} \)
- \( 14 \) for \( \text{H}^+ \)