Answer :
To solve the given chemical equation and balance it in a basic solution, we can follow these steps in detail:
1. Write down the unbalanced chemical equation:
[tex]\[ \text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) \][/tex]
2. Determine the oxidation states of the elements involved in the reaction:
- For [tex]\(\text{Mn}\)[/tex] in [tex]\(\text{MnO}_4^-\)[/tex]: The oxidation state is [tex]\(+7\)[/tex].
- For [tex]\(\text{Mn}\)[/tex] in [tex]\(\text{MnO}_2\)[/tex]: The oxidation state is [tex]\(+4\)[/tex].
- For [tex]\(\text{Br}\)[/tex] in [tex]\(\text{Br}^-\)[/tex]: The oxidation state is [tex]\(-1\)[/tex].
- For [tex]\(\text{Br}\)[/tex] in [tex]\(\text{BrO}_3^-\)[/tex]: The oxidation state is [tex]\(+5\)[/tex].
3. Identify the reduction and oxidation half-reactions:
- Reduction half-reaction (MnO[tex]\(_4^-\)[/tex] to MnO[tex]\(_2\)[/tex]):
[tex]\[ \text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) \][/tex]
- Oxidation half-reaction (Br[tex]\(^-\)[/tex] to BrO[tex]\(_3^-\)[/tex]):
[tex]\[ \text{Br}^-(\text{aq}) \rightarrow \text{BrO}_3^-(\text{aq}) \][/tex]
4. Balance atoms other than oxygen and hydrogen in each half-reaction:
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) \][/tex]
The manganese atoms are already balanced.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) \rightarrow \text{BrO}_3^-(\text{aq}) \][/tex]
The bromine atoms are also balanced.
5. Balance oxygen atoms by adding H[tex]\(_2\)[/tex]O:
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \][/tex]
There are 4 oxygen atoms on the left and 2 on the right. We add 2 H[tex]\(_2\)[/tex]O to the right.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) \][/tex]
There are 3 oxygen atoms on the right, so we add 3 H[tex]\(_2\)[/tex]O to the left.
6. Balance hydrogen atoms by adding H[tex]\(^+\)[/tex]:
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \][/tex]
We added 4 H[tex]\(^+\)[/tex] to balance the hydrogen from the 2 H[tex]\(_2\)[/tex]O.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) \][/tex]
We added 6 H[tex]\(^+\)[/tex] to balance the hydrogen.
7. Balance the charge by adding electrons (e[tex]\(^-\)[/tex]):
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) + 3e^- \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \][/tex]
The left side has a charge of [tex]\(-1 + 4 = +3\)[/tex], so we add 3 electrons to the left to balance the charge.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 6e^- \][/tex]
The right side has a charge of [tex]\(-1 + 0 = -1\)[/tex], and to balance with 6 H[tex]\(^+\)[/tex], we need 6 electrons on the right side.
8. Combine the half-reactions ensuring electrons cancel out:
We multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 to make the electron count the same:
[tex]\[ 2(\text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) + 3e^- \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) ) \][/tex]
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 6e^- \][/tex]
Combining these, we get:
[tex]\[ 2\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 6e^- + \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + 4\text{H}_2\text{O}(\text{l}) + \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 6e^- \][/tex]
Simplify:
[tex]\[ 2\text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) + 8\text{H}^+(\text{aq}) \][/tex]
Thus, the balanced chemical equation is:
[tex]\[ \text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) + 8\text{H}^+(\text{aq}) \][/tex]
This equation has the following phases:
- [tex]\(\text{MnO}_4^-\)[/tex]: aqueous (aq)
- [tex]\(\text{Br}^-\)[/tex]: aqueous (aq)
- [tex]\(\text{H}_2\text{O}\)[/tex]: liquid (l)
- [tex]\(\text{MnO}_2\)[/tex]: solid (s)
- [tex]\(\text{BrO}_3^-\)[/tex]: aqueous (aq)
- [tex]\(\text{H}^+\)[/tex]: aqueous (aq)
Therefore, the final, balanced chemical equation in a basic solution is:
[tex]\[ \text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) + 8\text{H}^+\text{(aq)} \][/tex]
1. Write down the unbalanced chemical equation:
[tex]\[ \text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) \][/tex]
2. Determine the oxidation states of the elements involved in the reaction:
- For [tex]\(\text{Mn}\)[/tex] in [tex]\(\text{MnO}_4^-\)[/tex]: The oxidation state is [tex]\(+7\)[/tex].
- For [tex]\(\text{Mn}\)[/tex] in [tex]\(\text{MnO}_2\)[/tex]: The oxidation state is [tex]\(+4\)[/tex].
- For [tex]\(\text{Br}\)[/tex] in [tex]\(\text{Br}^-\)[/tex]: The oxidation state is [tex]\(-1\)[/tex].
- For [tex]\(\text{Br}\)[/tex] in [tex]\(\text{BrO}_3^-\)[/tex]: The oxidation state is [tex]\(+5\)[/tex].
3. Identify the reduction and oxidation half-reactions:
- Reduction half-reaction (MnO[tex]\(_4^-\)[/tex] to MnO[tex]\(_2\)[/tex]):
[tex]\[ \text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) \][/tex]
- Oxidation half-reaction (Br[tex]\(^-\)[/tex] to BrO[tex]\(_3^-\)[/tex]):
[tex]\[ \text{Br}^-(\text{aq}) \rightarrow \text{BrO}_3^-(\text{aq}) \][/tex]
4. Balance atoms other than oxygen and hydrogen in each half-reaction:
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) \][/tex]
The manganese atoms are already balanced.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) \rightarrow \text{BrO}_3^-(\text{aq}) \][/tex]
The bromine atoms are also balanced.
5. Balance oxygen atoms by adding H[tex]\(_2\)[/tex]O:
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \][/tex]
There are 4 oxygen atoms on the left and 2 on the right. We add 2 H[tex]\(_2\)[/tex]O to the right.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) \][/tex]
There are 3 oxygen atoms on the right, so we add 3 H[tex]\(_2\)[/tex]O to the left.
6. Balance hydrogen atoms by adding H[tex]\(^+\)[/tex]:
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \][/tex]
We added 4 H[tex]\(^+\)[/tex] to balance the hydrogen from the 2 H[tex]\(_2\)[/tex]O.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) \][/tex]
We added 6 H[tex]\(^+\)[/tex] to balance the hydrogen.
7. Balance the charge by adding electrons (e[tex]\(^-\)[/tex]):
For the reduction half-reaction:
[tex]\[ \text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) + 3e^- \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \][/tex]
The left side has a charge of [tex]\(-1 + 4 = +3\)[/tex], so we add 3 electrons to the left to balance the charge.
For the oxidation half-reaction:
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 6e^- \][/tex]
The right side has a charge of [tex]\(-1 + 0 = -1\)[/tex], and to balance with 6 H[tex]\(^+\)[/tex], we need 6 electrons on the right side.
8. Combine the half-reactions ensuring electrons cancel out:
We multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 to make the electron count the same:
[tex]\[ 2(\text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) + 3e^- \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) ) \][/tex]
[tex]\[ \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 6e^- \][/tex]
Combining these, we get:
[tex]\[ 2\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 6e^- + \text{Br}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + 4\text{H}_2\text{O}(\text{l}) + \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) + 6e^- \][/tex]
Simplify:
[tex]\[ 2\text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) + 8\text{H}^+(\text{aq}) \][/tex]
Thus, the balanced chemical equation is:
[tex]\[ \text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) + 8\text{H}^+(\text{aq}) \][/tex]
This equation has the following phases:
- [tex]\(\text{MnO}_4^-\)[/tex]: aqueous (aq)
- [tex]\(\text{Br}^-\)[/tex]: aqueous (aq)
- [tex]\(\text{H}_2\text{O}\)[/tex]: liquid (l)
- [tex]\(\text{MnO}_2\)[/tex]: solid (s)
- [tex]\(\text{BrO}_3^-\)[/tex]: aqueous (aq)
- [tex]\(\text{H}^+\)[/tex]: aqueous (aq)
Therefore, the final, balanced chemical equation in a basic solution is:
[tex]\[ \text{MnO}_4^-(\text{aq}) + \text{Br}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^-(\text{aq}) + 8\text{H}^+\text{(aq)} \][/tex]
