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
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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.
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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
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Reduction half-reaction: \[ \text{MnO}_4^- + 3 \text{e}^- \rightarrow \text{MnO}_2^- \]
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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).
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Adjusted Reduction half-reaction: \[ 4 \text{MnO}_4^- + 12 \text{e}^- \rightarrow 4 \text{MnO}_2^- \]
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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⁻.
- 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: \[ 4 \text{MnO}_4^- + 3 \text{N}_2\text{H}_4 \rightarrow 4 \text{MnO}_2^- + 3 \text{N}_2 + 8 \text{H}_2\text{O} \]
- 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:
\[ 4 \text{MnO}_4^- + 3 \text{N}_2\text{H}_4 + 8 \text{OH}^- \rightarrow 4 \text{MnO}_2^- + 3 \text{N}_2 + 8 \text{H}_2\text{O} \]
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