Let's tackle each part of your questions step by step.
Question 1: Balancing the Reaction
- Given Reaction:
\[ \text{Pb (NO}_3)_2 (aq) + \text{Al (s)} \rightarrow \text{Al (NO}_3)_3 (aq) + \text{Pb (s)} \]
Balancing the Reaction: The balanced reaction is: \[ 3 \text{Pb(NO}_3)_2 (aq) + 2 \text{Al (s)} \rightarrow 2 \text{Al(NO}_3)_3 (aq) + 3 \text{Pb (s)} \]
a. The Ionic Reaction
Ionic Reaction: \[ 3 \text{Pb}^{2+}(aq) + 6 \text{NO}_3^{-}(aq) + 2 \text{Al (s)} \rightarrow 2 \text{Al}^{3+}(aq) + 6 \text{NO}_3^{-}(aq) + 3 \text{Pb (s)} \]
b. Net Ionic Reaction
Net Ionic Reaction: Remove the spectator ions (i.e., \(6 \text{NO}_3^{-}\)): \[ 3 \text{Pb}^{2+}(aq) + 2 \text{Al (s)} \rightarrow 2 \text{Al}^{3+}(aq) + 3 \text{Pb (s)} \]
c. Reduced Half-Reaction
Reduced Half-Reaction (Reduction of \( \text{Pb}^{2+} \)): \[ \text{Pb}^{2+} + 2 \text{e}^- \rightarrow \text{Pb (s)} \]
d. Oxidized Half-Reaction
Oxidized Half-Reaction (Oxidation of \( \text{Al} \)): \[ \text{Al (s)} \rightarrow \text{Al}^{3+} + 3 \text{e}^- \]
e. Identify the Oxidation and Reduction Reactants
- Oxidation Reactant: Aluminum (Al) - oxidized to \( \text{Al}^{3+} \).
- Reduction Reactant: Lead ions (\( \text{Pb}^{2+} \)) - reduced to lead (Pb).
Question 2: Redox Reaction Justification
Reaction:
\[ \text{S (s)} + \text{HNO}_3 (aq) \rightarrow \text{SO}_2 (g) + \text{NO (g)} + \text{H}_2\text{O (l)} \]
Oxidation States:
- \( S(s): 0 \) (elemental state)
- \( H: +1 \), \( N: +5 \), \( O: -2 \) in \( \text{HNO}_3 \)
- In \( SO_2 \): \( S: +4 \), \( O: -2 \)
- In \( NO \): \( N: +2 \), \( O: -2 \)
- In \( H_2O \): \( H: +1 \), \( O: -2 \)
Change in Oxidation Numbers:
- Sulfur (S): \( 0 \to +4 \) (oxidized)
- Nitrogen from \( +5 \) in \( \text{HNO}_3 \) to \( +2 \) in \( \text{NO} \) (reduced)
Conclusion: Yes, it is a redox reaction because there is both oxidation (S) and reduction (N).
Question 3: Disproportionation Reaction Justification
Reaction:
\[ ClO^{-} \rightarrow Cl^{-} + ClO_3^{-} \]
Oxidation States:
- In \( ClO^{-} \): \( Cl: +1 \), \( O: -2 \)
- In \( Cl^{-} \): \( Cl: -1 \)
- In \( ClO_3^{-} \): \( Cl: +5 \), \( O: -2 \)
Changes in Oxidation Numbers:
- Chlorine in \( ClO^{-} \) is reduced from \( +1 \) to \( -1 \) in \( Cl^{-} \).
- Chlorine in \( ClO^{-} \) is oxidized from \( +1 \) to \( +5 \) in \( ClO_3^{-} \).
Conclusion: This is a disproportionation reaction because the same element (chlorine) is undergoing both oxidation and reduction.
Question 4: Balancing Half Reactions
(a) Reaction in Basic Solutions
Reaction:
\[ \text{ClO}_2 (aq) + \text{SbO}_2 (aq) \rightarrow \text{ClO}_2^{-} (aq) + \text{Sb(OH)}_6^{-} (aq) \]
Let's split into half-reactions:
- Reduction:
\[ \text{ClO}_2 \rightarrow \text{ClO}_2^- \] - Oxidation: \[ \text{SbO}_2 \rightarrow \text{Sb(OH)}_6^{-} \]
Balancing: For reduction:
- Add electrons:
\[ \text{ClO}_2 + 2 \text{e}^- \rightarrow \text{ClO}_2^{-} \]
For oxidation:
- Add water and hydroxide to balance:
\[ \text{SbO}_2 + 4 \text{H}_2\text{O} \rightarrow \text{Sb(OH)}_6^{-} + 4 \text{e}^- \]
Combine:
Multiply reduction by 2 and add:
\[ 2 \text{ClO}_2 + \text{SbO}_2 + 4 \text{H}_2\text{O} \rightarrow 2 \text{ClO}_2^{-} + \text{Sb(OH)}_6^{-} + 4 \text{e}^- + 4 \text{e}^- \]
Reduced Reactant: \( \text{SbO}_2 \)
Oxidized Reactant: \( \text{ClO}_2 \)
(b) Reaction in Acidic Solutions
Reaction: \[ \text{Hg (l)} + \text{NO}_3^{-} (aq) + \text{Cl}^{-} (aq) \rightarrow \text{HgCl}_4^{2-} + \text{NO}_2 (g) \]
Half-Reactions:
-
Oxidation:
\[ \text{Hg} \rightarrow \text{HgCl}_4^{2-} + 4 \text{e}^- \] -
Reduction:
\[ 2 \text{NO}_3^{-} + 4 \text{H}^+ + 2 \text{e}^- \rightarrow 2 \text{NO}_2 + 2 \text{H}_2\text{O} \]
Combining: Multiply oxidation by 2 to match electrons: \[ 2 \text{Hg} + 2 \text{NO}_3^{-} + 8 \text{H}^+ + 4 \text{Cl}^- \rightarrow 2 \text{HgCl}_4^{2-} + 2 \text{NO}_2 + 4 \text{H}_2\text{O} \]
Reduced Reactant: \( \text{NO}_3^{-} \)
Oxidized Reactant: \( \text{Hg} \)
This covers the problems you presented! If you have any further questions or topics, feel free to ask!