Question

1. This is the complete reaction. (5 Marks)
The reaction is not balanced. You should balance first.
Pb (NO3)2 (aq) + Al (s) → Al(NO3 )3 (aq) + Pb (s)
Please write
a. The Ionic reaction
b. Net Ionic reaction
c. Reduced half- reaction
d. Oxidized half- reaction
e. Identify the oxidation and reduction reactant.


2. Is the following reaction a redox reaction? Justify your answer by showing the oxidation numbers for each element and show your work. (5 Marks)

S(s) + HNO3 (aq ) → SO2 (g) + NO (g) + H2O (l)



3. The following reaction is DISPROPORTIONATION reactions.
Please justify your answer by showing oxidation numbers for each element and identifying the reduced and oxidized elements.
(5 Marks)
DISPROPORTIONATION REACTION: A reaction in which some atoms of an element are oxidized, and other atoms of the same element are reduced. (Page 598 Textbook)

ClO - (aq) → Cl - (aq) + ClO3 – (aq)




4. For the following reactions:
Balance each equation using half-reactions in the given conditions and find the reduced and Oxidized reactants.
(Each question has 5 Marks) You MUST show your work.


ClO2 (aq) + SbO2 (aq) → ClO2 - (aq) + Sb(OH)6 - (aq)

(basic solutions)

a. Hg (l) + NO3^- (aq ) + Cl^- (aq) → HgCl4^2- + NO2 (g )
_
(Acidic solutions)

Answer 1

Let's tackle each part of your questions step by step.

### Question 1: Balancing the Reaction

1. **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**:
1. \( S(s): 0 \) (elemental state)
2. \( H: +1 \), \( N: +5 \), \( O: -2 \) in \( \text{HNO}_3 \)
3. In \( SO_2 \): \( S: +4 \), \( O: -2 \)
4. In \( NO \): \( N: +2 \), \( O: -2 \)
5. 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**:
1. In \( ClO^{-} \): \( Cl: +1 \), \( O: -2 \)
2. In \( Cl^{-} \): \( Cl: -1 \)
3. 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**:

1. **Reduction**:
\[ \text{ClO}_2 \rightarrow \text{ClO}_2^- \]
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**:

1. **Oxidation**:
\[ \text{Hg} \rightarrow \text{HgCl}_4^{2-} + 4 \text{e}^- \]

2. **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!