Question

Question 11 (6 points)
Balance the following redox reaction using the half-reaction method under Basic conditions.
Identify LEO and GER halves reactions too.
You Must Show ALL the Steps.

MnO4 ^- + N 2 H 4 → MnO2 ^- + N2

Answer 1

To balance the redox reaction \( \text{MnO}_4^- + \text{N}_2\text{H}_4 \rightarrow \text{MnO}_2^- + \text{N}_2 \) under basic conditions, we will use the half-reaction method, which involves balancing oxidation and reduction half-reactions separately. We will also identify which half-reaction is losing electrons (LEO - Losing Electrons is Oxidation) and which is gaining electrons (GER - Gaining Electrons is Reduction).

### Step 1: Identify Oxidation and Reduction

1. **Identify oxidation states:**
- In \( \text{MnO}_4^- \), manganese (Mn) is in the +7 oxidation state.
- In \( \text{MnO}_2^- \), manganese is in the +4 oxidation state.
- In \( \text{N}_2\text{H}_4 \), nitrogen (N) is in the -2 oxidation state.
- In molecular nitrogen \( \text{N}_2 \), nitrogen is in the 0 oxidation state.

2. **Determine LEO and GER:**
- **Reduction (GER)** occurs for Mn:
\[
\text{MnO}_4^- \rightarrow \text{MnO}_2^-
\]
Changes from +7 to +4 (gain of 3 electrons).
- **Oxidation (LEO)** occurs for N:
\[
\text{N}_2\text{H}_4 \rightarrow \text{N}_2
\]
Changes from -2 to 0 (loss of 4 electrons).

### Step 2: Write the Half-Reactions

1. **Reduction half-reaction:**
\[
\text{MnO}_4^- + 3 \text{e}^- \rightarrow \text{MnO}_2^-
\]

2. **Oxidation half-reaction:**
\[
\text{N}_2\text{H}_4 \rightarrow \text{N}_2 + 4 \text{e}^-
\]

### Step 3: Equalize the number of electrons

To make sure the number of electrons lost in oxidation equals the number of electrons gained in reduction, we need to multiply the reduction half-reaction by 4 and the oxidation half-reaction by 3 (since 12 electrons are required in total).

- **Adjusted Reduction half-reaction:**
\[
4 \text{MnO}_4^- + 12 \text{e}^- \rightarrow 4 \text{MnO}_2^-
\]

- **Adjusted Oxidation half-reaction:**
\[
3 \text{N}_2\text{H}_4 \rightarrow 3 \text{N}_2 + 12 \text{e}^-
\]

### Step 4: Combine the Half-Reactions

Now we can combine both half-reactions:

\[
4 \text{MnO}_4^- + 3 \text{N}_2\text{H}_4 \rightarrow 4 \text{MnO}_2^- + 3 \text{N}_2
\]

### Step 5: Balance Hydrogen and Oxygen

Now we will balance any remaining hydrogen and oxygen atoms by adding water (H₂O) and hydroxide ions (OH⁻). In basic conditions, we treat excess hydrogen by adding OH⁻.

1. Count the oxygens:
- Left side: \( 4 \times 4 = 16 \) (from \( \text{MnO}_4^- \))
- Right side: \( 4 \times 2 = 8 \) (from \( \text{MnO}_2^- \))
- Total \( \text{O}_2^{\text{left}} = 16 \), \( \text{O}_2^{\text{right}} = 8\)
- Difference = \( 8 \) (need to add \( 8 \text{H}_2\text{O} \) to the right)

The balanced equation becomes:
\[
4 \text{MnO}_4^- + 3 \text{N}_2\text{H}_4 \rightarrow 4 \text{MnO}_2^- + 3 \text{N}_2 + 8 \text{H}_2\text{O}
\]

2. If we were to add water to the product side, we need to add \( 8 \text{OH}^- \) to the reactant side to keep the balance.

Final balanced equation in basic conditions:
\[
4 \text{MnO}_4^- + 3 \text{N}_2\text{H}_4 + 8 \text{OH}^- \rightarrow 4 \text{MnO}_2^- + 3 \text{N}_2 + 8 \text{H}_2\text{O}
\]

### Final Summary

- **Reduction (GER):** \( \text{MnO}_4^- + 3 e^- \rightarrow \text{MnO}_2^- \)
- **Oxidation (LEO):** \( 3 \text{N}_2\text{H}_4 \rightarrow 3 \text{N}_2 + 12 e^- \)
- Final balanced redox equation under basic conditions:

\[
4 \text{MnO}_4^- + 3 \text{N}_2\text{H}_4 + 8 \text{OH}^- \rightarrow 4 \text{MnO}_2^- + 3 \text{N}_2 + 8 \text{H}_2\text{O}
\]