Balanced the following equations:
`H_(2)O_(2)+H^(o+)+Fe^(2+)rarr H_(2)O+Fe^(3+)`

Explanation

To balance the redox reaction given by the equation:

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

we will follow these steps:

### Step 1: Identify Oxidation and Reduction
- **Oxidation:** The iron ion (\( \text{Fe}^{2+} \)) is oxidized to \( \text{Fe}^{3+} \). This involves the loss of one electron:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

- **Reduction:** The hydrogen peroxide (\( \text{H}_2\text{O}_2 \)) is reduced to water (\( \text{H}_2\text{O} \)). The half-reaction for this process involves the gain of electrons. To balance the oxygen atoms, we will need to add water molecules and hydrogen ions:
\[ \text{H}_2\text{O}_2 + 2 \text{H}^+ + 2 e^- \rightarrow 2 \text{H}_2\text{O} \]

### Step 2: Balance the Half-Reactions
Now we have two half-reactions:
1. Oxidation:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \] (1 electron lost)

2. Reduction:
\[ \text{H}_2\text{O}_2 + 2 \text{H}^+ + 2 e^- \rightarrow 2 \text{H}_2\text{O} \] (2 electrons gained)

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

### Step 4: Combine the Half-Reactions
Now we can combine the balanced half-reactions:
\[ 2 \text{Fe}^{2+} + \text{H}_2\text{O}_2 + 2 \text{H}^+ \rightarrow 2 \text{Fe}^{3+} + 2 \text{H}_2\text{O} \]

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