Worksheet for identifying oxidation-reduction reactions, including atom oxidized/reduced, oxidizing/reducing agents, and balancing equations.

To solve the problems in the oxidation-reduction worksheet, we need to identify the atoms that are oxidized and reduced, determine the oxidizing and reducing agents, write the half-reactions for oxidation and reduction, and balance the overall equation using the method of oxidation-reduction. Below is a detailed explanation for each reaction:

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1. Mg + 2HCl → MgCl₂ + H₂

#### Step 1: Identify oxidation and reduction
- Mg changes from 0 (in Mg) to +2 (in Mg²⁺): Oxidation
- H changes from +1 (in HCl) to 0 (in H₂): Reduction

#### Step 2: Identify oxidizing and reducing agents
- Reducing agent: Mg (loses electrons)
- Oxidizing agent: H⁺ (from HCl, gains electrons)

#### Step 3: Write half-reactions
- Oxidation half-reaction:
\[
\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-
\]
- Reduction half-reaction:
\[
2\text{H}^+ + 2e^- \rightarrow \text{H}_2
\]

#### Step 4: Balance the equation
The equation is already balanced:
\[
\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2
\]

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2. 2Fe + 3V₂O₃ → Fe₂O₃ + 6VO

#### Step 1: Identify oxidation and reduction
- Fe changes from 0 (in Fe) to +3 (in Fe₂O₃): Oxidation
- V changes from +3 (in V₂O₃) to +2 (in VO): Reduction

#### Step 2: Identify oxidizing and reducing agents
- Reducing agent: Fe (loses electrons)
- Oxidizing agent: V₂O₃ (gains electrons)

#### Step 3: Write half-reactions
- Oxidation half-reaction:
\[
2\text{Fe} \rightarrow \text{Fe}_2\text{O}_3 + 6e^-
\]
(Each Fe goes from 0 to +3, so 2 Fe give 6 electrons.)
- Reduction half-reaction:
\[
3\text{V}_2\text{O}_3 + 6e^- \rightarrow 6\text{VO}
\]
(Each V goes from +3 to +2, so 3 V₂O₃ (6 V atoms) gain 6 electrons.)

#### Step 4: Balance the equation
The equation is already balanced:
\[
2\text{Fe} + 3\text{V}_2\text{O}_3 \rightarrow \text{Fe}_2\text{O}_3 + 6\text{VO}
\]

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3. KMnO₄ + KNO₂ + 2H₂SO₄ → MnSO₄ + 2H₂O + KNO₃ + K₂SO₄

#### Step 1: Identify oxidation and reduction
- Mn changes from +7 (in KMnO₄) to +2 (in MnSO₄): Reduction
- N changes from +3 (in KNO₂) to +5 (in KNO₃): Oxidation

#### Step 2: Identify oxidizing and reducing agents
- Reducing agent: KNO₂ (loses electrons)
- Oxidizing agent: KMnO₄ (gains electrons)

#### Step 3: Write half-reactions
- Oxidation half-reaction:
\[
\text{KNO}_2 \rightarrow \text{KNO}_3 + 2e^-
\]
(N goes from +3 to +5, losing 2 electrons.)
- Reduction half-reaction:
\[
\text{KMnO}_4 + 5e^- \rightarrow \text{MnSO}_4
\]
(Mn goes from +7 to +2, gaining 5 electrons.)

#### Step 4: Balance the equation
To balance the electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
\[
5\text{KNO}_2 \rightarrow 5\text{KNO}_3 + 10e^-
\]
\[
2\text{KMnO}_4 + 10e^- \rightarrow 2\text{MnSO}_4
\]
Combine the half-reactions and balance the remaining atoms:
\[
2\text{KMnO}_4 + 5\text{KNO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + 5\text{KNO}_3 + 2\text{H}_2\text{O} + \text{K}_2\text{SO}_4
\]

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4. K₂Cr₂O₇ + SnCl₂ + HCl → CrCl₃ + SnCl₄ + KCl + H₂O

#### Step 1: Identify oxidation and reduction
- Cr changes from +6 (in K₂Cr₂O₇) to +3 (in CrCl₃): Reduction
- Sn changes from +2 (in SnCl₂) to +4 (in SnCl₄): Oxidation

#### Step 2: Identify oxidizing and reducing agents
- Reducing agent: SnCl₂ (loses electrons)
- Oxidizing agent: K₂Cr₂O₇ (gains electrons)

#### Step 3: Write half-reactions
- Oxidation half-reaction:
\[
\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^-
\]
- Reduction half-reaction:
\[
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
\]

#### Step 4: Balance the equation
To balance the electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1:
\[
3\text{Sn}^{2+} \rightarrow 3\text{Sn}^{4+} + 6e^-
\]
\[
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
\]
Combine the half-reactions and balance the remaining atoms:
\[
\text{K}_2\text{Cr}_2\text{O}_7 + 3\text{SnCl}_2 + 14\text{HCl} \rightarrow 2\text{CrCl}_3 + 3\text{SnCl}_4 + 2\text{KCl} + 7\text{H}_2\text{O}
\]

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Final Answer

\[
\boxed{1. \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2}
\]
\[
\boxed{2. 2\text{Fe} + 3\text{V}_2\text{O}_3 \rightarrow \text{Fe}_2\text{O}_3 + 6\text{VO}}
\]
\[
\boxed{3. 2\text{KMnO}_4 + 5\text{KNO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + 5\text{KNO}_3 + 2\text{H}_2\text{O} + \text{K}_2\text{SO}_4}
\]
\[
\boxed{4. \text{K}_2\text{Cr}_2\text{O}_7 + 3\text{SnCl}_2 + 14\text{HCl} \rightarrow 2\text{CrCl}_3 + 3\text{SnCl}_4 + 2\text{KCl} + 7\text{H}_2\text{O}}
\]

The same process can be applied to the remaining reactions (5–10). Let me know if you need further assistance!