IDNLearn.com makes it easy to find precise answers to your specific questions. Discover in-depth answers to your questions from our community of experienced professionals.
Sagot :
Sure! Let's balance the redox reaction in an acidic solution step-by-step.
Given reaction:
[tex]\[ \mathrm{PbO_2} + \mathrm{I_2} \rightarrow \mathrm{Pb^{2+}} + \mathrm{IO_7^{-}} \][/tex]
### Step 1: Split into half-reactions
First, we split the reaction into oxidation and reduction half-reactions.
1. Oxidation half-reaction: [tex]\( \mathrm{I_2} \rightarrow \mathrm{IO_7^{-}} \)[/tex]
2. Reduction half-reaction: [tex]\( \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} \)[/tex]
### Step 2: Balance atoms in each half-reaction except for oxygen and hydrogen
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} \rightarrow \mathrm{IO_7^{-}} \][/tex]
The iodine atoms are already balanced.
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} \][/tex]
The lead atoms are already balanced.
### Step 3: Balance oxygen atoms by adding [tex]\( \mathrm{H_2O} \)[/tex]
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} \rightarrow \mathrm{IO_7^{-}} \][/tex]
On the right side, there are 7 oxygen atoms, so we add 7 [tex]\( \mathrm{H_2O} \)[/tex] on the left side:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} \][/tex]
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} \][/tex]
On the left side, there are 2 oxygen atoms, so we add 2 [tex]\( \mathrm{H_2O} \)[/tex] on the right side:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
### Step 4: Balance hydrogen atoms by adding [tex]\( \mathrm{H^+} \)[/tex]
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} \][/tex]
There are 14 hydrogen atoms on the left side, so we add 14 [tex]\( \mathrm{H^+} \)[/tex] on the right side:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} \][/tex]
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
There are 4 hydrogen atoms on the right side, so we add 4 [tex]\( \mathrm{H^+} \)[/tex] on the left side:
[tex]\[ \mathrm{PbO_2} + 4\mathrm{H^+} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
### Step 5: Balance charges by adding electrons ([tex]\( \mathrm{e^-} \)[/tex])
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} \][/tex]
The left side has no charge, and the right side has a total charge of [tex]\( -1 + 14 = 13 \)[/tex]. Therefore, we need 13 electrons on the right side to balance the charges:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} + 13\mathrm{e^-} \][/tex]
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} + 4\mathrm{H^+} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
The left side has a charge of [tex]\( 4\mathrm{H^+} = 4 \)[/tex] and the right side has a charge of [tex]\( 2 \)[/tex]. Therefore, we need 2 electrons on the left side to balance the charges:
[tex]\[ \mathrm{PbO_2} + 4\mathrm{H^+} + 2\mathrm{e^-} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
### Step 6: Combine the half-reactions
Make sure the electron loss and gain are balanced. Multiply the half-reactions by appropriate coefficients so that the number of electrons lost in oxidation equals the number gained in reduction:
[tex]\[ 2 \times \left( \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} + 13\mathrm{e^-} \right) \][/tex]
[tex]\[ 13 \times \left( \mathrm{PbO_2} + 4\mathrm{H^+} + 2\mathrm{e^-} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \right) \][/tex]
This results in:
[tex]\[ 2\mathrm{I_2} + 14\mathrm{H_2O} \rightarrow 2\mathrm{IO_7^{-}} + 28\mathrm{H^+} + 26\mathrm{e^-} \][/tex]
[tex]\[ 13\mathrm{PbO_2} + 52\mathrm{H^+} + 26\mathrm{e^-} \rightarrow 13\mathrm{Pb^{2+}} + 26\mathrm{H_2O} \][/tex]
Summing these gives:
[tex]\[ 2\mathrm{I_2} + 13\mathrm{PbO_2} + 14\mathrm{H_2O} + 52\mathrm{H^+} + 26\mathrm{e^-} \rightarrow 2\mathrm{IO_7^{-}} + 28\mathrm{H^+} + 26\mathrm{e^-} + 13\mathrm{Pb^{2+}} + 26\mathrm{H_2O} \][/tex]
### Step 7: Simplify the equation by canceling out common terms:
Cancel out the electrons ([tex]\( \mathrm{e^-} \)[/tex]), reduce the number of [tex]\( H_2O \)[/tex] and [tex]\( H^+ \)[/tex]:
[tex]\[ 2\mathrm{I_2} + 13\mathrm{PbO_2} + 26\mathrm{H^+} \rightarrow 2\mathrm{IO_7^{-}} + 13\mathrm{Pb^{2+}} + 12\mathrm{H_2O} \][/tex]
Thus, the balanced equation in acidic solution is:
[tex]\[ 2\mathrm{I_2} + 13\mathrm{PbO_2} + 26\mathrm{H^+} \rightarrow 2\mathrm{IO_7^{-}} + 13\mathrm{Pb^{2+}} + 12\mathrm{H_2O} \][/tex]
Given reaction:
[tex]\[ \mathrm{PbO_2} + \mathrm{I_2} \rightarrow \mathrm{Pb^{2+}} + \mathrm{IO_7^{-}} \][/tex]
### Step 1: Split into half-reactions
First, we split the reaction into oxidation and reduction half-reactions.
1. Oxidation half-reaction: [tex]\( \mathrm{I_2} \rightarrow \mathrm{IO_7^{-}} \)[/tex]
2. Reduction half-reaction: [tex]\( \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} \)[/tex]
### Step 2: Balance atoms in each half-reaction except for oxygen and hydrogen
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} \rightarrow \mathrm{IO_7^{-}} \][/tex]
The iodine atoms are already balanced.
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} \][/tex]
The lead atoms are already balanced.
### Step 3: Balance oxygen atoms by adding [tex]\( \mathrm{H_2O} \)[/tex]
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} \rightarrow \mathrm{IO_7^{-}} \][/tex]
On the right side, there are 7 oxygen atoms, so we add 7 [tex]\( \mathrm{H_2O} \)[/tex] on the left side:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} \][/tex]
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} \][/tex]
On the left side, there are 2 oxygen atoms, so we add 2 [tex]\( \mathrm{H_2O} \)[/tex] on the right side:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
### Step 4: Balance hydrogen atoms by adding [tex]\( \mathrm{H^+} \)[/tex]
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} \][/tex]
There are 14 hydrogen atoms on the left side, so we add 14 [tex]\( \mathrm{H^+} \)[/tex] on the right side:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} \][/tex]
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
There are 4 hydrogen atoms on the right side, so we add 4 [tex]\( \mathrm{H^+} \)[/tex] on the left side:
[tex]\[ \mathrm{PbO_2} + 4\mathrm{H^+} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
### Step 5: Balance charges by adding electrons ([tex]\( \mathrm{e^-} \)[/tex])
1. Oxidation half-reaction:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} \][/tex]
The left side has no charge, and the right side has a total charge of [tex]\( -1 + 14 = 13 \)[/tex]. Therefore, we need 13 electrons on the right side to balance the charges:
[tex]\[ \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} + 13\mathrm{e^-} \][/tex]
2. Reduction half-reaction:
[tex]\[ \mathrm{PbO_2} + 4\mathrm{H^+} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
The left side has a charge of [tex]\( 4\mathrm{H^+} = 4 \)[/tex] and the right side has a charge of [tex]\( 2 \)[/tex]. Therefore, we need 2 electrons on the left side to balance the charges:
[tex]\[ \mathrm{PbO_2} + 4\mathrm{H^+} + 2\mathrm{e^-} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \][/tex]
### Step 6: Combine the half-reactions
Make sure the electron loss and gain are balanced. Multiply the half-reactions by appropriate coefficients so that the number of electrons lost in oxidation equals the number gained in reduction:
[tex]\[ 2 \times \left( \mathrm{I_2} + 7\mathrm{H_2O} \rightarrow \mathrm{IO_7^{-}} + 14\mathrm{H^+} + 13\mathrm{e^-} \right) \][/tex]
[tex]\[ 13 \times \left( \mathrm{PbO_2} + 4\mathrm{H^+} + 2\mathrm{e^-} \rightarrow \mathrm{Pb^{2+}} + 2\mathrm{H_2O} \right) \][/tex]
This results in:
[tex]\[ 2\mathrm{I_2} + 14\mathrm{H_2O} \rightarrow 2\mathrm{IO_7^{-}} + 28\mathrm{H^+} + 26\mathrm{e^-} \][/tex]
[tex]\[ 13\mathrm{PbO_2} + 52\mathrm{H^+} + 26\mathrm{e^-} \rightarrow 13\mathrm{Pb^{2+}} + 26\mathrm{H_2O} \][/tex]
Summing these gives:
[tex]\[ 2\mathrm{I_2} + 13\mathrm{PbO_2} + 14\mathrm{H_2O} + 52\mathrm{H^+} + 26\mathrm{e^-} \rightarrow 2\mathrm{IO_7^{-}} + 28\mathrm{H^+} + 26\mathrm{e^-} + 13\mathrm{Pb^{2+}} + 26\mathrm{H_2O} \][/tex]
### Step 7: Simplify the equation by canceling out common terms:
Cancel out the electrons ([tex]\( \mathrm{e^-} \)[/tex]), reduce the number of [tex]\( H_2O \)[/tex] and [tex]\( H^+ \)[/tex]:
[tex]\[ 2\mathrm{I_2} + 13\mathrm{PbO_2} + 26\mathrm{H^+} \rightarrow 2\mathrm{IO_7^{-}} + 13\mathrm{Pb^{2+}} + 12\mathrm{H_2O} \][/tex]
Thus, the balanced equation in acidic solution is:
[tex]\[ 2\mathrm{I_2} + 13\mathrm{PbO_2} + 26\mathrm{H^+} \rightarrow 2\mathrm{IO_7^{-}} + 13\mathrm{Pb^{2+}} + 12\mathrm{H_2O} \][/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
