Redox chemistry worksheet with problems on oxidation states and electron transfer.
Chapter 20 Worksheet on Redox reactions, including questions on identifying oxidized and reduced substances, oxidizing and reducing agents, oxidation numbers, and electron transfer in chemical reactions.
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Step-by-step solution for: Chapter 20 Worksheet Redox
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Step-by-step solution for: Chapter 20 Worksheet Redox
You’ve uploaded a worksheet titled “Chapter 20 Worksheet: Redox”, which contains several redox reaction problems. Since I can’t view images directly, I’ll solve the problems based on the text you’ve provided — which appears to be the full content of the worksheet.
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---
## Problem 1–10: Determine what is oxidized and reduced; identify oxidizing and reducing agents
- Oxidation: Loss of electrons → increase in oxidation number.
- Reduction: Gain of electrons → decrease in oxidation number.
- Oxidizing agent: Causes oxidation → gets reduced.
- Reducing agent: Causes reduction → gets oxidized.
---
- Sr: 0 → +2 (oxidized)
- O: 0 → -2 (reduced)
✔ Oxidized: Sr
✔ Reduced: O₂
✔ Oxidizing agent: O₂
✔ Reducing agent: Sr
---
- Li: 0 → +1 (oxidized)
- S: 0 → -2 (reduced)
✔ Oxidized: Li
✔ Reduced: S
✔ Oxidizing agent: S
✔ Reducing agent: Li
---
- Cs: 0 → +1 (oxidized)
- Br: 0 → -1 (reduced)
✔ Oxidized: Cs
✔ Reduced: Br₂
✔ Oxidizing agent: Br₂
✔ Reducing agent: Cs
---
- Mg: 0 → +2 (oxidized)
- N: 0 → -3 (reduced)
✔ Oxidized: Mg
✔ Reduced: N₂
✔ Oxidizing agent: N₂
✔ Reducing agent: Mg
---
- Fe: 0 → +3 (oxidized)
- O: 0 → -2 (reduced)
✔ Oxidized: Fe
✔ Reduced: O₂
✔ Oxidizing agent: O₂
✔ Reducing agent: Fe
---
- Cl: 0 → -1 (reduced)
- Br: -1 → 0 (oxidized)
✔ Oxidized: Br⁻ (in NaBr)
✔ Reduced: Cl₂
✔ Oxidizing agent: Cl₂
✔ Reducing agent: NaBr (specifically Br⁻)
---
- Si: 0 → +4 (oxidized)
- F: 0 → -1 (reduced)
✔ Oxidized: Si
✔ Reduced: F₂
✔ Oxidizing agent: F₂
✔ Reducing agent: Si
---
- Ca: 0 → +2 (oxidized)
- O: 0 → -2 (reduced)
✔ Oxidized: Ca
✔ Reduced: O₂
✔ Oxidizing agent: O₂
✔ Reducing agent: Ca
---
- Mg: 0 → +2 (oxidized)
- H: +1 → 0 (reduced)
✔ Oxidized: Mg
✔ Reduced: H⁺ (in HCl)
✔ Oxidizing agent: HCl
✔ Reducing agent: Mg
---
- Na: 0 → +1 (oxidized)
- H: +1 → 0 (reduced)
✔ Oxidized: Na
✔ Reduced: H⁺ (in H₂O)
✔ Oxidizing agent: H₂O
✔ Reducing agent: Na
---
## Problem 11: Give the oxidation number of each kind of atom or ion
a. sulfate (SO₄²⁻):
→ O = -2 (each), total for O = -8. Charge = -2 → S + (-8) = -2 → S = +6
b. Sn (elemental tin) → 0
c. S²⁻ → -2
d. Fe³⁺ → +3
e. Sn⁴⁺ → +4
f. nitrate (NO₃⁻):
→ O = -2 × 3 = -6. Total charge = -1 → N + (-6) = -1 → N = +5
g. ammonium (NH₄⁺):
→ H = +1 × 4 = +4. Total charge = +1 → N + 4 = +1 → N = -3
✔ Answers:
a. S = +6
b. Sn = 0
c. S²⁻ = -2
d. Fe³⁺ = +3
e. Sn⁴⁺ = +4
f. N = +5
g. N = -3
---
## Problem 12: Calculate oxidation number of chromium
a. Cr₂O₃
→ O = -2 × 3 = -6. Total = 0 → 2Cr = +6 → Cr = +3
b. Na₂Cr₂O₇
→ Na = +1 × 2 = +2; O = -2 × 7 = -14. Total = 0 → 2Cr + 2 -14 = 0 → 2Cr = +12 → Cr = +6
c. CrSO₄
→ SO₄²⁻ = -2 → Cr must be +2
d. chromate (CrO₄²⁻)
→ O = -2 × 4 = -8. Charge = -2 → Cr + (-8) = -2 → Cr = +6
e. dichromate (Cr₂O₇²⁻)
→ O = -2 × 7 = -14. Charge = -2 → 2Cr + (-14) = -2 → 2Cr = +12 → Cr = +6
✔ Answers:
a. +3
b. +6
c. +2
d. +6
e. +6
---
## Problem 13: Use oxidation numbers to determine which elements are oxidized and reduced
*(Note: Not necessary to balance equations)*
---
Assign oxidation numbers:
- C: 0 → +4 (in CO₂) → oxidized
- S: +6 (in H₂SO₄) → +4 (in SO₂) → reduced
✔ Oxidized: C
✔ Reduced: S
---
- I: -1 (in HI) → 0 (in I₂) → oxidized
- N: +5 (in HNO₃) → +2 (in NO) → reduced
✔ Oxidized: I
✔ Reduced: N
---
- Mn: +7 (in KMnO₄) → +2 (in MnCl₂) → reduced
- Cl: -1 (in HCl) → 0 (in Cl₂) → oxidized
✔ Oxidized: Cl
✔ Reduced: Mn
---
- Sb: 0 → +5 (in Sb₂O₅) → oxidized
- N: +5 (in HNO₃) → +2 (in NO) → reduced
✔ Oxidized: Sb
✔ Reduced: N
---
## Problem 14: For each reaction in Problem 13, identify oxidizing and reducing agents
Recall:
- Oxidizing agent → gets reduced
- Reducing agent → gets oxidized
---
- Reduced: S → in H₂SO₄ → oxidizing agent: H₂SO₄
- Oxidized: C → reducing agent: C
✔ Oxidizing agent: H₂SO₄
✔ Reducing agent: C
---
- Reduced: N → in HNO₃ → oxidizing agent: HNO₃
- Oxidized: I → in HI → reducing agent: HI
✔ Oxidizing agent: HNO₃
✔ Reducing agent: HI
---
- Reduced: Mn → in KMnO₄ → oxidizing agent: KMnO₄
- Oxidized: Cl → in HCl → reducing agent: HCl
✔ Oxidizing agent: KMnO₄
✔ Reducing agent: HCl
---
- Reduced: N → in HNO₃ → oxidizing agent: HNO₃
- Oxidized: Sb → reducing agent: Sb
✔ Oxidizing agent: HNO₃
✔ Reducing agent: Sb
---
## ✔ Final Summary:
All problems solved with clear explanations using oxidation number changes.
Let me know if you’d like this formatted as a printable answer key or need help with balancing these equations too!
Let’s go problem by problem.
---
## Problem 1–10: Determine what is oxidized and reduced; identify oxidizing and reducing agents
General Rules:
- Oxidation: Loss of electrons → increase in oxidation number.
- Reduction: Gain of electrons → decrease in oxidation number.
- Oxidizing agent: Causes oxidation → gets reduced.
- Reducing agent: Causes reduction → gets oxidized.
---
1. 2Sr + O₂ → 2SrO
- Sr: 0 → +2 (oxidized)
- O: 0 → -2 (reduced)
✔ Oxidized: Sr
✔ Reduced: O₂
✔ Oxidizing agent: O₂
✔ Reducing agent: Sr
---
2. 2Li + S → Li₂S
- Li: 0 → +1 (oxidized)
- S: 0 → -2 (reduced)
✔ Oxidized: Li
✔ Reduced: S
✔ Oxidizing agent: S
✔ Reducing agent: Li
---
3. 2Cs + Br₂ → 2CsBr
- Cs: 0 → +1 (oxidized)
- Br: 0 → -1 (reduced)
✔ Oxidized: Cs
✔ Reduced: Br₂
✔ Oxidizing agent: Br₂
✔ Reducing agent: Cs
---
4. 3Mg + N₂ → Mg₃N₂
- Mg: 0 → +2 (oxidized)
- N: 0 → -3 (reduced)
✔ Oxidized: Mg
✔ Reduced: N₂
✔ Oxidizing agent: N₂
✔ Reducing agent: Mg
---
5. 4Fe + 3O₂ → 2Fe₂O₃
- Fe: 0 → +3 (oxidized)
- O: 0 → -2 (reduced)
✔ Oxidized: Fe
✔ Reduced: O₂
✔ Oxidizing agent: O₂
✔ Reducing agent: Fe
---
6. Cl₂ + 2NaBr → 2NaCl + Br₂
- Cl: 0 → -1 (reduced)
- Br: -1 → 0 (oxidized)
✔ Oxidized: Br⁻ (in NaBr)
✔ Reduced: Cl₂
✔ Oxidizing agent: Cl₂
✔ Reducing agent: NaBr (specifically Br⁻)
---
7. Si + 2F₂ → SiF₄
- Si: 0 → +4 (oxidized)
- F: 0 → -1 (reduced)
✔ Oxidized: Si
✔ Reduced: F₂
✔ Oxidizing agent: F₂
✔ Reducing agent: Si
---
8. 2Ca + O₂ → 2CaO
- Ca: 0 → +2 (oxidized)
- O: 0 → -2 (reduced)
✔ Oxidized: Ca
✔ Reduced: O₂
✔ Oxidizing agent: O₂
✔ Reducing agent: Ca
---
9. Mg + 2HCl → MgCl₂ + H₂
- Mg: 0 → +2 (oxidized)
- H: +1 → 0 (reduced)
✔ Oxidized: Mg
✔ Reduced: H⁺ (in HCl)
✔ Oxidizing agent: HCl
✔ Reducing agent: Mg
---
10. 2Na + 2H₂O → 2NaOH + H₂
- Na: 0 → +1 (oxidized)
- H: +1 → 0 (reduced)
✔ Oxidized: Na
✔ Reduced: H⁺ (in H₂O)
✔ Oxidizing agent: H₂O
✔ Reducing agent: Na
---
## Problem 11: Give the oxidation number of each kind of atom or ion
a. sulfate (SO₄²⁻):
→ O = -2 (each), total for O = -8. Charge = -2 → S + (-8) = -2 → S = +6
b. Sn (elemental tin) → 0
c. S²⁻ → -2
d. Fe³⁺ → +3
e. Sn⁴⁺ → +4
f. nitrate (NO₃⁻):
→ O = -2 × 3 = -6. Total charge = -1 → N + (-6) = -1 → N = +5
g. ammonium (NH₄⁺):
→ H = +1 × 4 = +4. Total charge = +1 → N + 4 = +1 → N = -3
✔ Answers:
a. S = +6
b. Sn = 0
c. S²⁻ = -2
d. Fe³⁺ = +3
e. Sn⁴⁺ = +4
f. N = +5
g. N = -3
---
## Problem 12: Calculate oxidation number of chromium
a. Cr₂O₃
→ O = -2 × 3 = -6. Total = 0 → 2Cr = +6 → Cr = +3
b. Na₂Cr₂O₇
→ Na = +1 × 2 = +2; O = -2 × 7 = -14. Total = 0 → 2Cr + 2 -14 = 0 → 2Cr = +12 → Cr = +6
c. CrSO₄
→ SO₄²⁻ = -2 → Cr must be +2
d. chromate (CrO₄²⁻)
→ O = -2 × 4 = -8. Charge = -2 → Cr + (-8) = -2 → Cr = +6
e. dichromate (Cr₂O₇²⁻)
→ O = -2 × 7 = -14. Charge = -2 → 2Cr + (-14) = -2 → 2Cr = +12 → Cr = +6
✔ Answers:
a. +3
b. +6
c. +2
d. +6
e. +6
---
## Problem 13: Use oxidation numbers to determine which elements are oxidized and reduced
*(Note: Not necessary to balance equations)*
---
a. C + H₂SO₄ → CO₂ + SO₂ + H₂O
Assign oxidation numbers:
- C: 0 → +4 (in CO₂) → oxidized
- S: +6 (in H₂SO₄) → +4 (in SO₂) → reduced
✔ Oxidized: C
✔ Reduced: S
---
b. HNO₃ + HI → NO + I₂ + H₂O
- I: -1 (in HI) → 0 (in I₂) → oxidized
- N: +5 (in HNO₃) → +2 (in NO) → reduced
✔ Oxidized: I
✔ Reduced: N
---
c. KMnO₄ + HCl → MnCl₂ + Cl₂ + H₂O + KCl
- Mn: +7 (in KMnO₄) → +2 (in MnCl₂) → reduced
- Cl: -1 (in HCl) → 0 (in Cl₂) → oxidized
✔ Oxidized: Cl
✔ Reduced: Mn
---
d. Sb + HNO₃ → Sb₂O₅ + NO + H₂O
- Sb: 0 → +5 (in Sb₂O₅) → oxidized
- N: +5 (in HNO₃) → +2 (in NO) → reduced
✔ Oxidized: Sb
✔ Reduced: N
---
## Problem 14: For each reaction in Problem 13, identify oxidizing and reducing agents
Recall:
- Oxidizing agent → gets reduced
- Reducing agent → gets oxidized
---
a. C + H₂SO₄ → ...
- Reduced: S → in H₂SO₄ → oxidizing agent: H₂SO₄
- Oxidized: C → reducing agent: C
✔ Oxidizing agent: H₂SO₄
✔ Reducing agent: C
---
b. HNO₃ + HI → ...
- Reduced: N → in HNO₃ → oxidizing agent: HNO₃
- Oxidized: I → in HI → reducing agent: HI
✔ Oxidizing agent: HNO₃
✔ Reducing agent: HI
---
c. KMnO₄ + HCl → ...
- Reduced: Mn → in KMnO₄ → oxidizing agent: KMnO₄
- Oxidized: Cl → in HCl → reducing agent: HCl
✔ Oxidizing agent: KMnO₄
✔ Reducing agent: HCl
---
d. Sb + HNO₃ → ...
- Reduced: N → in HNO₃ → oxidizing agent: HNO₃
- Oxidized: Sb → reducing agent: Sb
✔ Oxidizing agent: HNO₃
✔ Reducing agent: Sb
---
## ✔ Final Summary:
All problems solved with clear explanations using oxidation number changes.
Let me know if you’d like this formatted as a printable answer key or need help with balancing these equations too!
Parent Tip: Review the logic above to help your child master the concept of redox worksheet.
