Discover new perspectives and gain insights with IDNLearn.com. Ask your questions and get detailed, reliable answers from our community of experienced experts.
Sagot :
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.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.
