W redox reactions 13-14.pdf - RHSMartinAPChem - Free Printable

Let’s solve each redox reaction step by step, identifying what is being reduced, what is being oxidized, the reducing agent, and the oxidizing agent.

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🔹 Key Concepts:

- Reduction: Gain of electrons → decrease in oxidation number.
- Oxidation: Loss of electrons → increase in oxidation number.
- Reducing agent: The species that gets oxidized (it causes reduction in another).
- Oxidizing agent: The species that gets reduced (it causes oxidation in another).

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## ✔ 1. Br⁻(aq) + MnO₄⁻(aq) → Br₂(l) + Mn²⁺(aq)

First, assign oxidation numbers:

- Br⁻: Br = -1
- MnO₄⁻: O = -2 → 4×(-2) = -8 → Mn = +7 (since ion charge is -1)
- Br₂: Br = 0 (elemental form)
- Mn²⁺: Mn = +2

Now track changes:

- Br⁻ → Br₂: -1 → 0 → increase in oxidation number → oxidized
- MnO₄⁻ → Mn²⁺: +7 → +2 → decrease in oxidation number → reduced

→ So:

- Substance being reduced: MnO₄⁻ (specifically, Mn⁷⁺)
- Substance being oxidized: Br⁻
- Reducing agent: Br⁻ (it gets oxidized, so it reduces MnO₄⁻)
- Oxidizing agent: MnO₄⁻ (it gets reduced, so it oxidizes Br⁻)

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## ✔ 2. MnO₄⁻ + H₂C₂O₄ → CO₂ + Mn²⁺

Assign oxidation numbers:

- MnO₄⁻: Mn = +7 (as before)
- H₂C₂O₄ (oxalic acid): H = +1, O = -2 → Let C = x
→ 2(+1) + 2x + 4(-2) = 0 → 2 + 2x - 8 = 0 → 2x = 6 → x = +3
- CO₂: O = -2 → C = +4
- Mn²⁺: Mn = +2

Changes:

- C in H₂C₂O₄: +3 → +4 → oxidized
- Mn in MnO₄⁻: +7 → +2 → reduced

→ So:

- Substance being reduced: MnO₄⁻
- Substance being oxidized: H₂C₂O₄
- Reducing agent: H₂C₂O₄
- Oxidizing agent: MnO₄⁻

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## ✔ 3. As₂O₃ + NO₃⁻ → NO + H₃AsO₄

Assign oxidation numbers:

- As₂O₃: O = -2 → 3×(-2) = -6 → 2As = +6 → As = +3
- NO₃⁻: O = -2 → 3×(-2) = -6 → N = +5 (since ion = -1)
- NO: O = -2 → N = +2
- H₃AsO₄: H = +1, O = -2 → 3(+1) + As + 4(-2) = 0 → 3 + As - 8 = 0 → As = +5

Changes:

- As in As₂O₃: +3 → +5 → oxidized
- N in NO₃⁻: +5 → +2 → reduced

→ So:

- Substance being reduced: NO₃⁻
- Substance being oxidized: As₂O₃
- Reducing agent: As₂O₃
- Oxidizing agent: NO₃⁻

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## ✔ 4. Cr₂O₇²⁻ + Cl⁻ → Cr³⁺ + Cl₂

Assign oxidation numbers:

- Cr₂O₇²⁻: O = -2 → 7×(-2) = -14 → 2Cr = +12 → Cr = +6
- Cl⁻: Cl = -1
- Cr³⁺: Cr = +3
- Cl₂: Cl = 0

Changes:

- Cr in Cr₂O₇²⁻: +6 → +3 → reduced
- Cl⁻: -1 → 0 → oxidized

→ So:

- Substance being reduced: Cr₂O₇²⁻
- Substance being oxidized: Cl⁻
- Reducing agent: Cl⁻
- Oxidizing agent: Cr₂O₇²⁻

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## 📝 Final Answers:

1.

- What substance is being reduced? → MnO₄⁻
- What substance is being oxidized? → Br⁻
- What is the reducing agent? → Br⁻
- What is the oxidizing agent? → MnO₄⁻

2.

- What substance is being reduced? → MnO₄⁻
- What substance is being oxidized? → H₂C₂O₄
- What is the reducing agent? → H₂C₂O₄
- What is the oxidizing agent? → MnO₄⁻

3.

- What substance is being reduced? → NO₃⁻
- What substance is being oxidized? → As₂O₃
- What is the reducing agent? → As₂O₃
- What is the oxidizing agent? → NO₃⁻

4.

- What substance is being reduced? → Cr₂O₇²⁻
- What substance is being oxidized? → Cl⁻
- What is the reducing agent? → Cl⁻
- What is the oxidizing agent? → Cr₂O₇²⁻

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✔ All done! Each answer is based on tracking changes in oxidation numbers to identify electron transfer. Let me know if you’d like the balanced equations too!