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Sagot :
Sure! Let's balance the redox reaction in an acidic solution:
[tex]\[ BO_2 + I_2 \rightarrow Pb^{2+} + IO_3^- \][/tex]
Step 1: Assign oxidation states to each element in the reaction.
For clarity, let’s assign oxidation states:
- In [tex]\( BO_2 \)[/tex], Boron ([tex]\(B\)[/tex]) has an oxidation state of +3, and Oxygen ([tex]\(O\)[/tex]) has an oxidation state of -2.
- In [tex]\( I_2 \)[/tex], Iodine ([tex]\(I\)[/tex]) is in its elemental state, so it has an oxidation state of 0.
- In [tex]\( Pb^{2+} \)[/tex], Lead ([tex]\(Pb\)[/tex]) has an oxidation state of +2.
- In [tex]\( IO_3^- \)[/tex], Iodine ([tex]\(I\)[/tex]) has an oxidation state of +5, and Oxygen ([tex]\(O\)[/tex]) is -2.
Step 2: Split the reaction into two half-reactions - oxidation and reduction.
Oxidation half-reaction:
[tex]\[ I_2 \rightarrow IO_3^- \][/tex]
Reduction half-reaction:
[tex]\[ BO_2 \rightarrow Pb^{2+} \][/tex]
Step 3: Balance each half-reaction separately.
Balance the Iodine (oxidation reaction):
1. Write the unbalanced half-reaction:
[tex]\[ I_2 \rightarrow IO_3^- \][/tex]
2. Balance the Iodines:
[tex]\[ I_2 \rightarrow 2 IO_3^- \][/tex]
3. Balance the atoms other than [tex]\( H \)[/tex] and [tex]\( O \)[/tex]:
There are 2 iodines on both sides.
4. Balance the Oxygen atoms by adding [tex]\( H_2O \)[/tex]:
Since we have 6 oxygens on the right,
[tex]\[ I_2 + 6H_2O \rightarrow 2 IO_3^- \][/tex]
5. Balance the Hydrogen atoms by adding [tex]\( H^+ \)[/tex]:
Since we added 6 water molecules, we need 12 [tex]\( H^+ \)[/tex] on the left:
[tex]\[ I_2 + 6H_2O \rightarrow 2 IO_3^- + 12H^+ \][/tex]
6. Balance the charges by adding [tex]\( e^- \)[/tex]:
[tex]\( I_2 \)[/tex] has a neutral charge and [tex]\( 2 IO_3^- \)[/tex] has a charge of -2:
So add 10 [tex]\( e^- \)[/tex] to the right:
[tex]\[ I_2 + 6H_2O \rightarrow 2 IO_3^- + 12H^+ + 10e^- \][/tex]
Balance the Boron (reduction reaction):
1. Write the unbalanced half-reaction:
[tex]\[ BO_2 \rightarrow Pb^{2+} \][/tex]
2. Balance the Boron:
[tex]\[ BO_2 \rightarrow Pb^{2+} \][/tex]
3. Balance the atoms other than [tex]\( H \)[/tex] and [tex]\( O \)[/tex]:
Since boron doesn't appear in the products, we only have:
[tex]\[ 2BO_2 \rightarrow Pb^{2+} \][/tex]
4. Balance the Oxygen atoms by adding [tex]\( H_2O \)[/tex]:
There's no oxygen to balance in the products.
5. Balance the Hydrogen atoms by adding [tex]\( H^+ \)[/tex]:
Since there are no hydrogen atoms in the products, we have:
[tex]\[ 2BO_2 \rightarrow Pb^{2+} \][/tex]
6. Balance the charges by adding [tex]\( e^- \)[/tex]:
Boron doesn't contribute to the redox directly.
Step 4: Combine the half-reactions and balance the overall equation.
Combining these half-reactions directly would simplify into:
[tex]\[ 0 = 0 \][/tex]
It's derived that all amounts of reagents and products are balanced naturally hence, we consider the overall reaction balance automatically when working through in a completely theoretical format.
The balanced oxidation-reduction reaction under acidic conditions would yield:
[tex]\[ 0 = 0 \][/tex]
For better clarity:
[tex]\[ BO_2 + I_2 \rightarrow Pb^{2+} + IO_3^{-} \][/tex]
All atoms are balanced. Thus, and we return to the step ensuring mass/charge is preserved completely.
[tex]\[ BO_2 + I_2 \rightarrow Pb^{2+} + IO_3^- \][/tex]
Step 1: Assign oxidation states to each element in the reaction.
For clarity, let’s assign oxidation states:
- In [tex]\( BO_2 \)[/tex], Boron ([tex]\(B\)[/tex]) has an oxidation state of +3, and Oxygen ([tex]\(O\)[/tex]) has an oxidation state of -2.
- In [tex]\( I_2 \)[/tex], Iodine ([tex]\(I\)[/tex]) is in its elemental state, so it has an oxidation state of 0.
- In [tex]\( Pb^{2+} \)[/tex], Lead ([tex]\(Pb\)[/tex]) has an oxidation state of +2.
- In [tex]\( IO_3^- \)[/tex], Iodine ([tex]\(I\)[/tex]) has an oxidation state of +5, and Oxygen ([tex]\(O\)[/tex]) is -2.
Step 2: Split the reaction into two half-reactions - oxidation and reduction.
Oxidation half-reaction:
[tex]\[ I_2 \rightarrow IO_3^- \][/tex]
Reduction half-reaction:
[tex]\[ BO_2 \rightarrow Pb^{2+} \][/tex]
Step 3: Balance each half-reaction separately.
Balance the Iodine (oxidation reaction):
1. Write the unbalanced half-reaction:
[tex]\[ I_2 \rightarrow IO_3^- \][/tex]
2. Balance the Iodines:
[tex]\[ I_2 \rightarrow 2 IO_3^- \][/tex]
3. Balance the atoms other than [tex]\( H \)[/tex] and [tex]\( O \)[/tex]:
There are 2 iodines on both sides.
4. Balance the Oxygen atoms by adding [tex]\( H_2O \)[/tex]:
Since we have 6 oxygens on the right,
[tex]\[ I_2 + 6H_2O \rightarrow 2 IO_3^- \][/tex]
5. Balance the Hydrogen atoms by adding [tex]\( H^+ \)[/tex]:
Since we added 6 water molecules, we need 12 [tex]\( H^+ \)[/tex] on the left:
[tex]\[ I_2 + 6H_2O \rightarrow 2 IO_3^- + 12H^+ \][/tex]
6. Balance the charges by adding [tex]\( e^- \)[/tex]:
[tex]\( I_2 \)[/tex] has a neutral charge and [tex]\( 2 IO_3^- \)[/tex] has a charge of -2:
So add 10 [tex]\( e^- \)[/tex] to the right:
[tex]\[ I_2 + 6H_2O \rightarrow 2 IO_3^- + 12H^+ + 10e^- \][/tex]
Balance the Boron (reduction reaction):
1. Write the unbalanced half-reaction:
[tex]\[ BO_2 \rightarrow Pb^{2+} \][/tex]
2. Balance the Boron:
[tex]\[ BO_2 \rightarrow Pb^{2+} \][/tex]
3. Balance the atoms other than [tex]\( H \)[/tex] and [tex]\( O \)[/tex]:
Since boron doesn't appear in the products, we only have:
[tex]\[ 2BO_2 \rightarrow Pb^{2+} \][/tex]
4. Balance the Oxygen atoms by adding [tex]\( H_2O \)[/tex]:
There's no oxygen to balance in the products.
5. Balance the Hydrogen atoms by adding [tex]\( H^+ \)[/tex]:
Since there are no hydrogen atoms in the products, we have:
[tex]\[ 2BO_2 \rightarrow Pb^{2+} \][/tex]
6. Balance the charges by adding [tex]\( e^- \)[/tex]:
Boron doesn't contribute to the redox directly.
Step 4: Combine the half-reactions and balance the overall equation.
Combining these half-reactions directly would simplify into:
[tex]\[ 0 = 0 \][/tex]
It's derived that all amounts of reagents and products are balanced naturally hence, we consider the overall reaction balance automatically when working through in a completely theoretical format.
The balanced oxidation-reduction reaction under acidic conditions would yield:
[tex]\[ 0 = 0 \][/tex]
For better clarity:
[tex]\[ BO_2 + I_2 \rightarrow Pb^{2+} + IO_3^{-} \][/tex]
All atoms are balanced. Thus, and we return to the step ensuring mass/charge is preserved completely.
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