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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)

1 answer

1. Balancing the Reaction

Given the unbalanced reaction: \[ \text{Pb(NO}_3\text{)}_2 (aq) + \text{Al} (s) \rightarrow \text{Al(NO}_3)_3 (aq) + \text{Pb} (s) \]

To balance the reaction:

  1. Identify the oxidation states:

    • Pb in \(\text{Pb(NO}_3\text{)}_2\) is \(+2\).
    • Al in \(\text{Al}\) is \(0\).
    • Al in \(\text{Al(NO}_3)_3\) is \(+3\).
    • Pb in \(\text{Pb}\) is \(0\).

  2. Write the half-reactions:

    • Oxidation: \[\text{Al} \rightarrow \text{Al}^{3+} + 3e^{-}\]
    • Reduction: \[\text{Pb}^{2+} + 2e^{-} \rightarrow \text{Pb}\]

  3. Balance the electrons transferred: To balance the electrons, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:

    • Oxidation: \[2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^{-}\]
    • Reduction: \[3\text{Pb}^{2+} + 6e^{-} \rightarrow 3\text{Pb}\]

  4. The balanced reaction is: \[3\text{Pb(NO}_3\text{)}_2 + 2\text{Al} \rightarrow 2\text{Al(NO}_3)_3 + 3\text{Pb}\]

a. The 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: \[ 3\text{Pb}^{2+} (aq) + 2\text{Al} (s) \rightarrow 2\text{Al}^{3+} (aq) + 3\text{Pb} (s) \]

c. Reduced Half-Reaction: \[ 3\text{Pb}^{2+} + 6e^{-} \rightarrow 3\text{Pb} \]

d. Oxidized Half-Reaction: \[ 2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^{-} \]

e. Identify the Oxidation and Reduction Reactants:

  • Oxidation Reactant: Al (is oxidized to Al\(^{3+}\))
  • Reduction Reactant: Pb\(^{2+}\) (is reduced to Pb)

2. Determine if the Following Reaction is a Redox Reaction

Reaction: \[ \text{S} (s) + \text{HNO}_3 (aq) \rightarrow \text{SO}_2 (g) + \text{NO} (g) + \text{H}_2\text{O} (l) \]

  1. Determine oxidation states:

    • S in S = \(0\)
    • H in \(\text{HNO}_3\) = \(+1\), N = \(+5\), O = \(-2\)
    • S in \(\text{SO}_2\) = \(+4\)
    • N in \(\text{NO}\) = \(+2\)
    • O in \(\text{H}_2\text{O}\) = \(-2\)

  2. Calculate changes:

    • S: \(0 \rightarrow +4\) (oxidized)
    • N: \(+5 \rightarrow +2\) (reduced)

Since there is a change in oxidation states, it confirms that the reaction is a redox reaction.

3. Disproportionation Reaction

\[ \text{ClO}^{-} (aq) \rightarrow \text{Cl}^{-} (aq) + \text{ClO}_3^{-} (aq) \]

  1. Determine oxidation states:

    • Cl in \(\text{ClO}^{-}\): \(+1\)
    • Cl in \(\text{Cl}^{-}\): \(-1\)
    • Cl in \(\text{ClO}_3^{-}\): \(+5\)

  2. Identify changes:

    • Cl is reduced from \(+1\) to \(-1\) (reduction).
    • Cl is oxidized from \(+1\) to \(+5\) (oxidation).

Since the same element (Cl) is both oxidized and reduced, this confirms it is a disproportionation reaction.

4. Balancing the Reactions

Reaction 1:

\[ \text{ClO}_2 (aq) + \text{SbO}_2 (aq) \rightarrow \text{ClO}_2^{-} (aq) + \text{Sb(OH)}_6^{-} (aq) \] (Basic solution)

  1. Half-Reactions:

    • Oxidation half-reaction: \[ \text{SbO}_2 + 2e^{-} \rightarrow \text{Sb(OH)}_6^{-} \]
    • Reduction half-reaction: \[ \text{ClO}_2 + 2e^{-} \rightarrow \text{ClO}_2^{-} \]

  2. Balancing: You would multiply each half-reaction accordingly to balance electrons exchanged if necessary and combine the two half-reactions.

  3. Identify reduced and oxidized:

    • Reduced Reactant: ClO\(_2\) (goes from +4 to -1)
    • Oxidized Reactant: SbO\(_2\) (goes from +3 to +5)

Reaction 2:

\[ \text{Hg} (l) + \text{NO}_3^{-} (aq) + \text{Cl}^{-} (aq) \rightarrow \text{HgCl}_4^{2-} + \text{NO}_2 (g) \] (Acidic solution)

  1. Half-Reactions:

    • Oxidation half-reaction: \[ \text{Hg} \rightarrow \text{HgCl}_4^{2-} + 2e^{-} \]
    • Reduction half-reaction: \[ 2\text{NO}_3^{-} + 4e^{-} \rightarrow \text{NO}_2 + 2H+ + 2H_2O \]

  2. Balancing: Make sure the overall charge and number of atoms are balanced on both sides.

  3. Identify reduced and oxidized:

    • Reduced Reactant: NO\(_3^{-}\) (is reduced)
    • Oxidized Reactant: Hg (is oxidized)

This is the systematic approach to balancing the reactions and determining whether they are redox reactions, as well as identifying the oxidation states and the oxidized/reduced species. Each step is crucial for understanding the chemistry involved.