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Sagot :
Sure! To balance the redox reaction in an acidic solution, we will follow these steps: identify the oxidation states, write the half-reactions for reduction and oxidation, balance each half-reaction for atoms and charges, and then combine the half-reactions.
Step 1: Assign Oxidation States
First, let's identify the oxidation states of the elements in the given reaction:
[tex]\[ PbO_2 + I_2 \rightarrow Pb^{2+} + IO_3^- \][/tex]
- In [tex]\(PbO_2\)[/tex], the oxidation state of Pb is +4 (since each O is -2).
- In [tex]\(I_2\)[/tex], iodine has an oxidation state of 0.
- In [tex]\(Pb^{2+}\)[/tex], the oxidation state of Pb is +2.
- In [tex]\(IO_3^-\)[/tex], the oxidation state of I is +5 (since each O is -2 and the overall charge on the ion is -1).
Step 2: Write the Half-Reactions
Oxidation Half-Reaction:
Iodine is oxidized:
[tex]\[ I_2 \rightarrow IO_3^- \][/tex]
Reduction Half-Reaction:
Lead is reduced:
[tex]\[ PbO_2 \rightarrow Pb^{2+} \][/tex]
Step 3: Balance Each Half-Reaction
Oxidation Half-Reaction:
1. Balance iodine atoms:
[tex]\[ I_2 \rightarrow 2 IO_3^- \][/tex]
2. Balance oxygen atoms by adding water:
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- \][/tex]
3. Balance hydrogen atoms by adding [tex]\(H^+\)[/tex]:
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- + 12 H^+ \][/tex]
4. Balance charges by adding electrons ([tex]\(e^-\)[/tex]):
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- + 12 H^+ + 10 e^- \][/tex]
(Since [tex]\(I_2\)[/tex] is neutral and [tex]\(2 IO_3^-\)[/tex] has a combined charge of -2, we add 10 electrons to balance the +12 charge from [tex]\(12 H^+\)[/tex]).
Reduction Half-Reaction:
1. Balance lead atoms:
[tex]\[ PbO_2 \rightarrow Pb^{2+} \][/tex]
2. Balance oxygen atoms by adding water:
[tex]\[ PbO_2 \rightarrow Pb^{2+} + 2 H_2O \][/tex]
3. Balance hydrogen atoms by adding [tex]\(H^+\)[/tex]:
[tex]\[ PbO_2 + 4 H^+ \rightarrow Pb^{2+} + 2 H_2O \][/tex]
4. Balance charges by adding electrons ([tex]\(e^-\)[/tex]):
[tex]\[ PbO_2 + 4 H^+ + 2 e^- \rightarrow Pb^{2+} + 2 H_2O \][/tex]
(Since [tex]\(PbO_2\)[/tex] is neutral and [tex]\(Pb^{2+}\)[/tex] has a charge of +2, adding 2 electrons balances the equation).
Step 4: Combine the Half-Reactions
To combine the half-reactions, the electrons lost in the oxidation half-reaction must equal the electrons gained in the reduction half-reaction.
From the oxidation half-reaction, 10 electrons are lost. From the reduction half-reaction, 2 electrons are gained. To equalize, we multiply the reduction half-reaction by 5:
[tex]\[ 5 (PbO_2 + 4 H^+ + 2 e^- \rightarrow Pb^{2+} + 2 H_2O) \][/tex]
[tex]\[ 5 PbO_2 + 20 H^+ + 10 e^- \rightarrow 5 Pb^{2+} + 10 H_2O \][/tex]
Now, we can add the balanced half-reactions together:
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- + 12 H^+ + 10 e^- \][/tex]
[tex]\[ 5 PbO_2 + 20 H^+ + 10 e^- \rightarrow 5 Pb^{2+} + 10 H_2O \][/tex]
Combining, we get:
[tex]\[ I_2 + 6 H_2O + 5 PbO_2 + 20 H^+ + 10 e^- \rightarrow 2 IO_3^- + 12 H^+ + 10 e^- + 5 Pb^{2+} + 10 H_2O \][/tex]
Finally, we simplify by canceling out the electrons and combining like terms:
[tex]\[ I_2 + 5 PbO_2 + 14 H^+ \rightarrow 2 IO_3^- + 5 Pb^{2+} + 3 H_2O \][/tex]
So the balanced redox reaction in acidic solution is:
[tex]\[ I_2 + 5 PbO_2 + 14 H^+ \rightarrow 2 IO_3^- + 5 Pb^{2+} + 3 H_2O \][/tex]
Step 1: Assign Oxidation States
First, let's identify the oxidation states of the elements in the given reaction:
[tex]\[ PbO_2 + I_2 \rightarrow Pb^{2+} + IO_3^- \][/tex]
- In [tex]\(PbO_2\)[/tex], the oxidation state of Pb is +4 (since each O is -2).
- In [tex]\(I_2\)[/tex], iodine has an oxidation state of 0.
- In [tex]\(Pb^{2+}\)[/tex], the oxidation state of Pb is +2.
- In [tex]\(IO_3^-\)[/tex], the oxidation state of I is +5 (since each O is -2 and the overall charge on the ion is -1).
Step 2: Write the Half-Reactions
Oxidation Half-Reaction:
Iodine is oxidized:
[tex]\[ I_2 \rightarrow IO_3^- \][/tex]
Reduction Half-Reaction:
Lead is reduced:
[tex]\[ PbO_2 \rightarrow Pb^{2+} \][/tex]
Step 3: Balance Each Half-Reaction
Oxidation Half-Reaction:
1. Balance iodine atoms:
[tex]\[ I_2 \rightarrow 2 IO_3^- \][/tex]
2. Balance oxygen atoms by adding water:
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- \][/tex]
3. Balance hydrogen atoms by adding [tex]\(H^+\)[/tex]:
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- + 12 H^+ \][/tex]
4. Balance charges by adding electrons ([tex]\(e^-\)[/tex]):
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- + 12 H^+ + 10 e^- \][/tex]
(Since [tex]\(I_2\)[/tex] is neutral and [tex]\(2 IO_3^-\)[/tex] has a combined charge of -2, we add 10 electrons to balance the +12 charge from [tex]\(12 H^+\)[/tex]).
Reduction Half-Reaction:
1. Balance lead atoms:
[tex]\[ PbO_2 \rightarrow Pb^{2+} \][/tex]
2. Balance oxygen atoms by adding water:
[tex]\[ PbO_2 \rightarrow Pb^{2+} + 2 H_2O \][/tex]
3. Balance hydrogen atoms by adding [tex]\(H^+\)[/tex]:
[tex]\[ PbO_2 + 4 H^+ \rightarrow Pb^{2+} + 2 H_2O \][/tex]
4. Balance charges by adding electrons ([tex]\(e^-\)[/tex]):
[tex]\[ PbO_2 + 4 H^+ + 2 e^- \rightarrow Pb^{2+} + 2 H_2O \][/tex]
(Since [tex]\(PbO_2\)[/tex] is neutral and [tex]\(Pb^{2+}\)[/tex] has a charge of +2, adding 2 electrons balances the equation).
Step 4: Combine the Half-Reactions
To combine the half-reactions, the electrons lost in the oxidation half-reaction must equal the electrons gained in the reduction half-reaction.
From the oxidation half-reaction, 10 electrons are lost. From the reduction half-reaction, 2 electrons are gained. To equalize, we multiply the reduction half-reaction by 5:
[tex]\[ 5 (PbO_2 + 4 H^+ + 2 e^- \rightarrow Pb^{2+} + 2 H_2O) \][/tex]
[tex]\[ 5 PbO_2 + 20 H^+ + 10 e^- \rightarrow 5 Pb^{2+} + 10 H_2O \][/tex]
Now, we can add the balanced half-reactions together:
[tex]\[ I_2 + 6 H_2O \rightarrow 2 IO_3^- + 12 H^+ + 10 e^- \][/tex]
[tex]\[ 5 PbO_2 + 20 H^+ + 10 e^- \rightarrow 5 Pb^{2+} + 10 H_2O \][/tex]
Combining, we get:
[tex]\[ I_2 + 6 H_2O + 5 PbO_2 + 20 H^+ + 10 e^- \rightarrow 2 IO_3^- + 12 H^+ + 10 e^- + 5 Pb^{2+} + 10 H_2O \][/tex]
Finally, we simplify by canceling out the electrons and combining like terms:
[tex]\[ I_2 + 5 PbO_2 + 14 H^+ \rightarrow 2 IO_3^- + 5 Pb^{2+} + 3 H_2O \][/tex]
So the balanced redox reaction in acidic solution is:
[tex]\[ I_2 + 5 PbO_2 + 14 H^+ \rightarrow 2 IO_3^- + 5 Pb^{2+} + 3 H_2O \][/tex]
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