Solved Oxidation-Reduction Worksheet For each reaction | Chegg.com - Free Printable

To solve the problems in the "Oxidation-Reduction Worksheet," we need to analyze each reaction step by step. Here's how to approach each problem:

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General Steps for Solving Oxidation-Reduction Reactions:

1. Identify the oxidation states of all atoms in the reactants and products.
2. Determine which atoms are oxidized and reduced:
- Oxidation: Increase in oxidation state (loss of electrons).
- Reduction: Decrease in oxidation state (gain of electrons).
3. Identify the oxidizing agent and reducing agent:
- The oxidizing agent causes oxidation and gets reduced.
- The reducing agent causes reduction and gets oxidized.
4. Write the half-reactions:
- Oxidation half-reaction: Loss of electrons.
- Reduction half-reaction: Gain of electrons.
5. Balance the equation using the oxidation-reduction method:
- Balance atoms other than O and H.
- Balance oxygen atoms by adding \( \text{H}_2\text{O} \).
- Balance hydrogen atoms by adding \( \text{H}^+ \) (or \( \text{OH}^- \) in basic solutions).
- Balance charge by adding electrons.
- Combine the half-reactions, ensuring the number of electrons is equal.

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Solutions for Each Reaction:

#### 1. \( \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \)

- Oxidation States:
- Mg: 0 → +2
- H: +1 → 0
- Cl: -1 (unchanged)

- Oxidation: Mg (0 → +2)
- Reduction: H (+1 → 0)
- Oxidizing Agent: HCl (reduces H)
- Reducing Agent: Mg (oxidizes itself)

- Half-Reactions:
- Oxidation: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \)
- Reduction: \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \)

- Balanced Equation: Already balanced.

#### 2. \( 2\text{Fe} + 3\text{V}_2\text{O}_3 \rightarrow \text{Fe}_2\text{O}_3 + 6\text{VO} \)

- Oxidation States:
- Fe: 0 → +3
- V: +3 → +4
- O: -2 (unchanged)

- Oxidation: Fe (0 → +3)
- Reduction: V (+3 → +4)
- Oxidizing Agent: \( \text{V}_2\text{O}_3 \) (reduces V)
- Reducing Agent: Fe (oxidizes itself)

- Half-Reactions:
- Oxidation: \( 2\text{Fe} \rightarrow \text{Fe}_2\text{O}_3 + 6e^- \)
- Reduction: \( 3\text{V}_2\text{O}_3 + 6e^- \rightarrow 6\text{VO} \)

- Balanced Equation: Already balanced.

#### 3. \( \text{KMnO}_4 + \text{KNO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow \text{MnSO}_4 + 2\text{H}_2\text{O} + \text{KNO}_3 + \text{K}_2\text{SO}_4 \)

- Oxidation States:
- Mn: +7 → +2
- N: +3 → +5
- S: +6 (unchanged)
- O: -2 (unchanged)
- K: +1 (unchanged)
- H: +1 (unchanged)

- Oxidation: N (+3 → +5)
- Reduction: Mn (+7 → +2)
- Oxidizing Agent: \( \text{KMnO}_4 \) (reduces Mn)
- Reducing Agent: \( \text{KNO}_2 \) (oxidizes N)

- Half-Reactions:
- Oxidation: \( \text{KNO}_2 \rightarrow \text{KNO}_3 + e^- \)
- Reduction: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 5.
- Combine and balance.

#### 4. \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{SnCl}_2 + \text{HCl} \rightarrow \text{CrCl}_3 + \text{SnCl}_4 + \text{KCl} + \text{H}_2\text{O} \)

- Oxidation States:
- Cr: +6 → +3
- Sn: +2 → +4
- Cl: -1 (unchanged)
- K: +1 (unchanged)
- H: +1 (unchanged)
- O: -2 (unchanged)

- Oxidation: Sn (+2 → +4)
- Reduction: Cr (+6 → +3)
- Oxidizing Agent: \( \text{K}_2\text{Cr}_2\text{O}_7 \) (reduces Cr)
- Reducing Agent: \( \text{SnCl}_2 \) (oxidizes Sn)

- Half-Reactions:
- Oxidation: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \)
- Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 3.
- Combine and balance.

#### 5. \( \text{KMnO}_4 + \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Cl}_2 + \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \)

- Oxidation States:
- Mn: +7 → +2
- Cl: -1 → 0
- S: +6 (unchanged)
- O: -2 (unchanged)
- K: +1 (unchanged)
- Na: +1 (unchanged)
- H: +1 (unchanged)

- Oxidation: Cl (-1 → 0)
- Reduction: Mn (+7 → +2)
- Oxidizing Agent: \( \text{KMnO}_4 \) (reduces Mn)
- Reducing Agent: \( \text{NaCl} \) (oxidizes Cl)

- Half-Reactions:
- Oxidation: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \)
- Reduction: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 5.
- Combine and balance.

#### 6. \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{O} + \text{S} \rightarrow \text{SO}_2 + \text{KOH} + \text{Cr}_2\text{O}_3 \)

- Oxidation States:
- Cr: +6 → +3
- S: 0 → +4
- O: -2 (unchanged)
- K: +1 (unchanged)
- H: +1 (unchanged)

- Oxidation: S (0 → +4)
- Reduction: Cr (+6 → +3)
- Oxidizing Agent: \( \text{K}_2\text{Cr}_2\text{O}_7 \) (reduces Cr)
- Reducing Agent: S (oxidizes itself)

- Half-Reactions:
- Oxidation: \( \text{S} \rightarrow \text{SO}_2 + 4e^- \)
- Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 3.
- Combine and balance.

#### 7. \( \text{KClO}_3 + \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow \text{KCl} + \text{H}_2\text{O} + \text{CO}_2 \)

- Oxidation States:
- Cl: +5 → -1
- C: 0 → +4
- O: -2 (unchanged)
- K: +1 (unchanged)
- H: +1 (unchanged)

- Oxidation: C (0 → +4)
- Reduction: Cl (+5 → -1)
- Oxidizing Agent: \( \text{KClO}_3 \) (reduces Cl)
- Reducing Agent: \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) (oxidizes C)

- Half-Reactions:
- Oxidation: \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow 12\text{CO}_2 + 22\text{H}^+ + 24e^- \)
- Reduction: \( \text{ClO}_3^- + 6\text{H}^+ + 6e^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the reduction half-reaction by 4.
- Combine and balance.

#### 8. \( \text{H}_2\text{C}_2\text{O}_4 + \text{K}_2\text{MnO}_4 \rightarrow \text{CO}_2 + \text{K}_2\text{O} + \text{Mn}_2\text{O}_3 + \text{H}_2\text{O} \)

- Oxidation States:
- C: +3 → +4
- Mn: +6 → +3
- O: -2 (unchanged)
- K: +1 (unchanged)
- H: +1 (unchanged)

- Oxidation: C (+3 → +4)
- Reduction: Mn (+6 → +3)
- Oxidizing Agent: \( \text{K}_2\text{MnO}_4 \) (reduces Mn)
- Reducing Agent: \( \text{H}_2\text{C}_2\text{O}_4 \) (oxidizes C)

- Half-Reactions:
- Oxidation: \( \text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^- \)
- Reduction: \( \text{MnO}_4^{2-} + 4\text{H}^+ + 3e^- \rightarrow \text{Mn}_2\text{O}_3 + 2\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 3.
- Combine and balance.

#### 9. \( \text{Mn(NO}_3)_2 + \text{NaBiO}_3 + \text{HNO}_3 \rightarrow \text{HMnO}_4 + \text{Bi(NO}_3)_3 + \text{NaNO}_3 + \text{H}_2\text{O} \)

- Oxidation States:
- Mn: +2 → +7
- Bi: +5 → +3
- N: +5 (unchanged)
- O: -2 (unchanged)
- H: +1 (unchanged)
- Na: +1 (unchanged)

- Oxidation: Mn (+2 → +7)
- Reduction: Bi (+5 → +3)
- Oxidizing Agent: \( \text{NaBiO}_3 \) (reduces Bi)
- Reducing Agent: \( \text{Mn(NO}_3)_2 \) (oxidizes Mn)

- Half-Reactions:
- Oxidation: \( \text{Mn}^{2+} \rightarrow \text{MnO}_4^- + 8\text{H}^+ + 5e^- \)
- Reduction: \( \text{BiO}_3^- + 3e^- \rightarrow \text{Bi}^{3+} + 3\text{OH}^- \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 3.
- Combine and balance.

#### 10. \( \text{H}_2\text{C}_2\text{O}_4 + \text{KMnO}_4 \rightarrow \text{CO}_2 + \text{K}_2\text{O} + \text{Mn}_2\text{O}_3 + \text{H}_2\text{O} \)

- Oxidation States:
- C: +3 → +4
- Mn: +7 → +3
- O: -2 (unchanged)
- K: +1 (unchanged)
- H: +1 (unchanged)

- Oxidation: C (+3 → +4)
- Reduction: Mn (+7 → +3)
- Oxidizing Agent: \( \text{KMnO}_4 \) (reduces Mn)
- Reducing Agent: \( \text{H}_2\text{C}_2\text{O}_4 \) (oxidizes C)

- Half-Reactions:
- Oxidation: \( \text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^- \)
- Reduction: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

- Balanced Equation:
- Multiply the oxidation half-reaction by 5.
- Combine and balance.

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Final Answer:

Each reaction has been analyzed for oxidation states, oxidizing/reducing agents, half-reactions, and balanced equations. The detailed steps for balancing can be followed as outlined above.

For brevity, the final balanced equations are:

1. \( \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \)
2. \( 2\text{Fe} + 3\text{V}_2\text{O}_3 \rightarrow \text{Fe}_2\text{O}_3 + 6\text{VO} \)
3. \( 2\text{KMnO}_4 + 3\text{KNO}_2 + 8\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + 4\text{H}_2\text{O} + 3\text{KNO}_3 + \text{K}_2\text{SO}_4 \)
4. \( \text{K}_2\text{Cr}_2\text{O}_7 + 3\text{SnCl}_2 + 14\text{HCl} \rightarrow 2\text{CrCl}_3 + 3\text{SnCl}_4 + 2\text{KCl} + 7\text{H}_2\text{O} \)
5. \( 2\text{KMnO}_4 + 10\text{NaCl} + 8\text{H}_2\text{SO}_4 \rightarrow 5\text{Cl}_2 + \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 8\text{H}_2\text{O} + 5\text{Na}_2\text{SO}_4 \)
6. \( \text{K}_2\text{Cr}_2\text{O}_7 + 3\text{H}_2\text{O} + 3\text{S} \rightarrow 3\text{SO}_2 + 2\text{KOH} + \text{Cr}_2\text{O}_3 \)
7. \( 8\text{KClO}_3 + \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow 8\text{KCl} + 11\text{H}_2\text{O} + 12\text{CO}_2 \)
8. \( 3\text{H}_2\text{C}_2\text{O}_4 + 2\text{K}_2\text{MnO}_4 \rightarrow 6\text{CO}_2 + \text{K}_2\text{O} + \text{Mn}_2\text{O}_3 + 3\text{H}_2\text{O} \)
9. \( 3\text{Mn(NO}_3)_2 + 2\text{NaBiO}_3 + 10\text{HNO}_3 \rightarrow 3\text{HMnO}_4 + \text{Bi(NO}_3)_3 + 2\text{NaNO}_3 + 5\text{H}_2\text{O} \)
10. \( 5\text{H}_2\text{C}_2\text{O}_4 + 2\text{KMnO}_4 \rightarrow 10\text{CO}_2 + \text{K}_2\text{O} + \text{Mn}_2\text{O}_3 + 8\text{H}_2\text{O} \)

\boxed{\text{See detailed steps above for each reaction.}}