Write correctly balanced equaitons for the following redox reaction. Using half reaction:
`H_(2)S+Fe^(3+)rarrFe^(2+)+Sdarr+H^(o+)`

Explanation

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

### Step 1: Identify the oxidation and reduction half-reactions
The given reaction is:
\[ H_2S + Fe^{3+} \rightarrow Fe^{2+} + S + H^+ \]

- **Oxidation Half-Reaction**: \( H_2S \rightarrow S \)
- **Reduction Half-Reaction**: \( Fe^{3+} \rightarrow Fe^{2+} \)

### Step 2: Assign oxidation states
- In \( H_2S \), sulfur (S) has an oxidation state of -2.
- In elemental sulfur (S), the oxidation state is 0.
- In \( Fe^{3+} \), iron (Fe) has an oxidation state of +3.
- In \( Fe^{2+} \), iron (Fe) has an oxidation state of +2.

### Step 3: Write the half-reactions
1. **Oxidation Half-Reaction**:
\[ H_2S \rightarrow S + 2e^- \]
(Sulfur is oxidized from -2 to 0, releasing 2 electrons.)

2. **Reduction Half-Reaction**:
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \]
(Iron is reduced from +3 to +2, gaining 1 electron.)

### Step 4: Balance the electrons
To balance the number of electrons transferred in both half-reactions, we need to multiply the reduction half-reaction by 2:
\[ 2(Fe^{3+} + e^- \rightarrow Fe^{2+}) \]
This gives:
\[ 2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+} \]

### Step 5: Combine the half-reactions
Now, we can add the oxidation and the modified reduction half-reactions:
1. \( H_2S \rightarrow S + 2e^- \)
2. \( 2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+} \)

Adding them together:
\[ H_2S + 2Fe^{3+} \rightarrow S + 2Fe^{2+} \]

### Step 6: Add protons to balance hydrogen
Since there are 2 hydrogen atoms in \( H_2S \), we need to add 2 \( H^+ \) ions to the product side to balance the hydrogen:
\[ H_2S + 2Fe^{3+} \rightarrow S + 2Fe^{2+} + 2H^+ \]

### Final Balanced Equation
The final balanced equation for the redox reaction is:
\[ H_2S + 2Fe^{3+} \rightarrow S + 2Fe^{2+} + 2H^+ \]