Answer :
To balance the redox reaction [tex]\( \text{MnO}_4^- (aq) + \text{NO}_2^- (aq) \rightarrow \text{MnO}_2 (s) + \text{NO}_3^- (aq) \)[/tex] in an alkaline medium, follow these detailed steps:
### Step 1: Separate the reaction into two half-reactions
1. Half-reaction for [tex]\( \text{MnO}_4^- \rightarrow \text{MnO}_2 \)[/tex]
2. Half-reaction for [tex]\( \text{NO}_2^- \rightarrow \text{NO}_3^- \)[/tex]
### Step 2: Balance each half-reaction for all elements except hydrogen and oxygen
- For the manganese half-reaction:
[tex]\[ \text{MnO}_4^- \rightarrow \text{MnO}_2 \][/tex]
Manganese (Mn) is already balanced with 1 Mn atom on each side.
- For the nitrogen half-reaction:
[tex]\[ \text{NO}_2^- \rightarrow \text{NO}_3^- \][/tex]
Nitrogen (N) is already balanced with 1 N atom on each side.
### Step 3: Balance oxygen atoms by adding [tex]\( \text{H}_2\text{O} \)[/tex]
- For the manganese half-reaction:
There are 4 oxygen atoms on the left side and 2 oxygen atoms on the right side. Add 2 [tex]\( \text{H}_2\text{O} \)[/tex] molecules to the right side:
[tex]\[ \text{MnO}_4^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} \][/tex]
Now, there are 4 oxygen atoms on each side.
- For the nitrogen half-reaction:
There are 2 oxygen atoms on the left side and 3 oxygen atoms on the right side. The oxygen atoms are already balanced, no [tex]\( \text{H}_2\text{O} \)[/tex] is needed.
### Step 4: Balance hydrogen atoms by adding [tex]\( \text{OH}^- \)[/tex]
- For the manganese half-reaction:
On the right side, we have 4 hydrogen atoms (from 2 [tex]\( \text{H}_2\text{O} \)[/tex]). Add 4 [tex]\( \text{OH}^- \)[/tex] ions to the left side:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} \][/tex]
- For the nitrogen half-reaction:
Hydrogen atoms are already balanced as there are no hydrogen atoms in [tex]\( \text{NO}_2^- \)[/tex] and [tex]\( \text{NO}_3^- \)[/tex].
### Step 5: Balance the charge by adding electrons ([tex]\( \text{e}^- \)[/tex])
- For the manganese half-reaction:
The left side has a charge of [tex]\(-1 + 4(-1) = -5\)[/tex]. The right side has no charge (neutral). Add 3 electrons to the left side to balance the charge:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- + 3 \text{e}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} \][/tex]
- For the nitrogen half-reaction:
The left side has a charge of [tex]\(-1\)[/tex]. The right side has a charge of [tex]\(-1\)[/tex]. Add 1 electron to the right side to balance the charge:
[tex]\[ \text{NO}_2^- \rightarrow \text{NO}_3^- + \text{e}^- \][/tex]
### Step 6: Combine the half-reactions and simplify
Multiply the nitrogen half-reaction by 3 so that the number of electrons in both half-reactions are equal. This gives:
[tex]\[ 3 (\text{NO}_2^- \rightarrow \text{NO}_3^- + \text{e}^-) \rightarrow 3 \text{NO}_2^- \rightarrow 3 \text{NO}_3^- + 3 \text{e}^- \][/tex]
Now, we combine the two half-reactions:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- + 3 \text{NO}_2^- + 3 \text{e}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} + 3 \text{NO}_3^- + 3 \text{e}^- \][/tex]
The electrons on both sides cancel out. The final balanced equation is:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- + 3 \text{NO}_2^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} + 3 \text{NO}_3^- \][/tex]
Thus, this is the balanced redox reaction in an alkaline medium.
### Step 1: Separate the reaction into two half-reactions
1. Half-reaction for [tex]\( \text{MnO}_4^- \rightarrow \text{MnO}_2 \)[/tex]
2. Half-reaction for [tex]\( \text{NO}_2^- \rightarrow \text{NO}_3^- \)[/tex]
### Step 2: Balance each half-reaction for all elements except hydrogen and oxygen
- For the manganese half-reaction:
[tex]\[ \text{MnO}_4^- \rightarrow \text{MnO}_2 \][/tex]
Manganese (Mn) is already balanced with 1 Mn atom on each side.
- For the nitrogen half-reaction:
[tex]\[ \text{NO}_2^- \rightarrow \text{NO}_3^- \][/tex]
Nitrogen (N) is already balanced with 1 N atom on each side.
### Step 3: Balance oxygen atoms by adding [tex]\( \text{H}_2\text{O} \)[/tex]
- For the manganese half-reaction:
There are 4 oxygen atoms on the left side and 2 oxygen atoms on the right side. Add 2 [tex]\( \text{H}_2\text{O} \)[/tex] molecules to the right side:
[tex]\[ \text{MnO}_4^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} \][/tex]
Now, there are 4 oxygen atoms on each side.
- For the nitrogen half-reaction:
There are 2 oxygen atoms on the left side and 3 oxygen atoms on the right side. The oxygen atoms are already balanced, no [tex]\( \text{H}_2\text{O} \)[/tex] is needed.
### Step 4: Balance hydrogen atoms by adding [tex]\( \text{OH}^- \)[/tex]
- For the manganese half-reaction:
On the right side, we have 4 hydrogen atoms (from 2 [tex]\( \text{H}_2\text{O} \)[/tex]). Add 4 [tex]\( \text{OH}^- \)[/tex] ions to the left side:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} \][/tex]
- For the nitrogen half-reaction:
Hydrogen atoms are already balanced as there are no hydrogen atoms in [tex]\( \text{NO}_2^- \)[/tex] and [tex]\( \text{NO}_3^- \)[/tex].
### Step 5: Balance the charge by adding electrons ([tex]\( \text{e}^- \)[/tex])
- For the manganese half-reaction:
The left side has a charge of [tex]\(-1 + 4(-1) = -5\)[/tex]. The right side has no charge (neutral). Add 3 electrons to the left side to balance the charge:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- + 3 \text{e}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} \][/tex]
- For the nitrogen half-reaction:
The left side has a charge of [tex]\(-1\)[/tex]. The right side has a charge of [tex]\(-1\)[/tex]. Add 1 electron to the right side to balance the charge:
[tex]\[ \text{NO}_2^- \rightarrow \text{NO}_3^- + \text{e}^- \][/tex]
### Step 6: Combine the half-reactions and simplify
Multiply the nitrogen half-reaction by 3 so that the number of electrons in both half-reactions are equal. This gives:
[tex]\[ 3 (\text{NO}_2^- \rightarrow \text{NO}_3^- + \text{e}^-) \rightarrow 3 \text{NO}_2^- \rightarrow 3 \text{NO}_3^- + 3 \text{e}^- \][/tex]
Now, we combine the two half-reactions:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- + 3 \text{NO}_2^- + 3 \text{e}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} + 3 \text{NO}_3^- + 3 \text{e}^- \][/tex]
The electrons on both sides cancel out. The final balanced equation is:
[tex]\[ \text{MnO}_4^- + 4 \text{OH}^- + 3 \text{NO}_2^- \rightarrow \text{MnO}_2 + 2 \text{H}_2\text{O} + 3 \text{NO}_3^- \][/tex]
Thus, this is the balanced redox reaction in an alkaline medium.
