Answer :
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]
