Worksheet_Redox.pdf - Free Printable
Educational worksheet: Worksheet_Redox.pdf. Download and print for classroom or home learning activities.
JPG
495×640
25.7 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1534196
⭐
Show Answer Key & Explanations
Step-by-step solution for: Worksheet_Redox.pdf
▼
Show Answer Key & Explanations
Step-by-step solution for: Worksheet_Redox.pdf
Since I can’t view the image you uploaded, I’ll solve all problems from Chapter 20 Worksheet: Redox as written in your text. This includes Problems 1–14.
---
## 🔢 Problem 1: Determine what is oxidized and reduced; identify oxidizing and reducing agents.
In redox reactions:
- Oxidation = loss of electrons → increase in oxidation number
- Reduction = gain of electrons → decrease in oxidation number
- Reducing agent = substance that gets oxidized (donates electrons)
- Oxidizing agent = substance that gets reduced (accepts electrons)
Let’s analyze each:
---
- Sr: 0 → +2 → oxidized
- O: 0 → -2 → reduced
✔ Oxidized: Sr
✔ Reduced: O₂
✔ Reducing agent: Sr
✔ Oxidizing agent: O₂
---
- Li: 0 → +1 → oxidized
- S: 0 → -2 → reduced
✔ Oxidized: Li
✔ Reduced: S
✔ Reducing agent: Li
✔ Oxidizing agent: S
---
- Cs: 0 → +1 → oxidized
- Br: 0 → -1 → reduced
✔ Oxidized: Cs
✔ Reduced: Br₂
✔ Reducing agent: Cs
✔ Oxidizing agent: Br₂
---
- Mg: 0 → +2 → oxidized
- N: 0 → -3 → reduced
✔ Oxidized: Mg
✔ Reduced: N₂
✔ Reducing agent: Mg
✔ Oxidizing agent: N₂
---
- Fe: 0 → +3 → oxidized
- O: 0 → -2 → reduced
✔ Oxidized: Fe
✔ Reduced: O₂
✔ Reducing agent: Fe
✔ Oxidizing agent: O₂
---
This is a single displacement reaction.
- Cl: 0 → -1 → reduced
- Br: -1 → 0 → oxidized
✔ Oxidized: Br⁻ (in NaBr)
✔ Reduced: Cl₂
✔ Reducing agent: NaBr
✔ Oxidizing agent: Cl₂
---
- Si: 0 → +4 → oxidized
- F: 0 → -1 → reduced
✔ Oxidized: Si
✔ Reduced: F₂
✔ Reducing agent: Si
✔ Oxidizing agent: F₂
---
- Ca: 0 → +2 → oxidized
- O: 0 → -2 → reduced
✔ Oxidized: Ca
✔ Reduced: O₂
✔ Reducing agent: Ca
✔ Oxidizing agent: O₂
---
- Mg: 0 → +2 → oxidized
- H: +1 → 0 → reduced
✔ Oxidized: Mg
✔ Reduced: H⁺ (in HCl)
✔ Reducing agent: Mg
✔ Oxidizing agent: HCl
---
- Na: 0 → +1 → oxidized
- H: +1 → 0 → reduced
✔ Oxidized: Na
✔ Reduced: H₂O (specifically H⁺)
✔ Reducing agent: Na
✔ Oxidizing agent: H₂O
---
## 🧪 Problem 11: Oxidation number of each kind of atom or ion.
Rules:
- Free elements = 0
- Monatomic ions = charge
- Oxygen = -2 (except peroxides)
- Hydrogen = +1 (except with metals = -1)
- Sum of oxidation numbers in compound = 0; in polyatomic ion = ion charge
a. Sulfate (SO₄²⁻):
Let S = x
x + 4(-2) = -2 → x - 8 = -2 → x = +6
b. Sn (elemental tin) → 0
c. S²⁻ → -2
d. Fe³⁺ → +3
e. Sn⁴⁺ → +4
f. Nitrate (NO₃⁻):
Let N = x
x + 3(-2) = -1 → x - 6 = -1 → x = +5
g. Ammonium (NH₄⁺):
Let N = x
x + 4(+1) = +1 → x + 4 = 1 → x = -3
✔ Answers:
a. +6
b. 0
c. -2
d. +3
e. +4
f. +5
g. -3
---
## 🧮 Problem 12: Calculate oxidation number of Cr in each.
a. Cr₂O₃
Let Cr = x
2x + 3(-2) = 0 → 2x = 6 → x = +3
b. Na₂Cr₂O₇
Na = +1, O = -2
2(+1) + 2x + 7(-2) = 0 → 2 + 2x - 14 = 0 → 2x = 12 → x = +6
c. CrSO₄
SO₄²⁻ has -2 charge → Cr must be +2
d. Chromate (CrO₄²⁻)
x + 4(-2) = -2 → x - 8 = -2 → x = +6
e. Dichromate (Cr₂O₇²⁻)
2x + 7(-2) = -2 → 2x - 14 = -2 → 2x = 12 → x = +6
✔ Answers:
a. +3
b. +6
c. +2
d. +6
e. +6
---
## 🔄 Problem 13: Use oxidation numbers to determine what is oxidized/reduced.
*(Note: We don’t need balanced equations — just track changes)*
---
Assign oxidation numbers:
- C: 0 → +4 (in CO₂) → oxidized
- S in H₂SO₄: +6 → +4 (in SO₂) → reduced
✔ Oxidized: C
✔ Reduced: S (in H₂SO₄)
---
- N in HNO₃: +5 → +2 (in NO) → reduced
- I in HI: -1 → 0 (in I₂) → oxidized
✔ Oxidized: I⁻ (in HI)
✔ Reduced: N (in HNO₃)
---
- Mn in KMnO₄: +7 → +2 (in MnCl₂) → reduced
- Cl in HCl: -1 → 0 (in Cl₂) → oxidized
✔ Oxidized: Cl⁻ (in HCl)
✔ Reduced: Mn (in KMnO₄)
---
- Sb: 0 → +5 (in Sb₂O₅) → oxidized
- N in HNO₃: +5 → +2 (in NO) → reduced
✔ Oxidized: Sb
✔ Reduced: N (in HNO₃)
---
## ⚖️ Problem 14: Identify oxidizing and reducing agents for Problem 13.
Recall:
- Reducing agent = gets oxidized
- Oxidizing agent = gets reduced
---
- C oxidized → reducing agent: C
- S reduced → oxidizing agent: H₂SO₄
✔ Reducing agent: C
✔ Oxidizing agent: H₂SO₄
---
- I⁻ oxidized → reducing agent: HI
- N reduced → oxidizing agent: HNO₃
✔ Reducing agent: HI
✔ Oxidizing agent: HNO₃
---
- Cl⁻ oxidized → reducing agent: HCl
- Mn reduced → oxidizing agent: KMnO₄
✔ Reducing agent: HCl
✔ Oxidizing agent: KMnO₄
---
- Sb oxidized → reducing agent: Sb
- N reduced → oxidizing agent: HNO₃
✔ Reducing agent: Sb
✔ Oxidizing agent: HNO₃
---
## ✔ Final Summary
All problems solved! You now have complete answers for:
- Identifying oxidized/reduced species and agents (Problems 1–10)
- Oxidation numbers of ions/compounds (Problem 11)
- Chromium oxidation states (Problem 12)
- Redox changes using oxidation numbers (Problem 13)
- Oxidizing/reducing agents for those reactions (Problem 14)
Let me know if you want these formatted into a printable worksheet or explained further!
---
---
## 🔢 Problem 1: Determine what is oxidized and reduced; identify oxidizing and reducing agents.
In redox reactions:
- Oxidation = loss of electrons → increase in oxidation number
- Reduction = gain of electrons → decrease in oxidation number
- Reducing agent = substance that gets oxidized (donates electrons)
- Oxidizing agent = substance that gets reduced (accepts electrons)
Let’s analyze each:
---
1. 2Sr + O₂ → 2SrO
- Sr: 0 → +2 → oxidized
- O: 0 → -2 → reduced
✔ Oxidized: Sr
✔ Reduced: O₂
✔ Reducing agent: Sr
✔ Oxidizing agent: O₂
---
2. 2Li + S → Li₂S
- Li: 0 → +1 → oxidized
- S: 0 → -2 → reduced
✔ Oxidized: Li
✔ Reduced: S
✔ Reducing agent: Li
✔ Oxidizing agent: S
---
3. 2Cs + Br₂ → 2CsBr
- Cs: 0 → +1 → oxidized
- Br: 0 → -1 → reduced
✔ Oxidized: Cs
✔ Reduced: Br₂
✔ Reducing agent: Cs
✔ Oxidizing agent: Br₂
---
4. 3Mg + N₂ → Mg₃N₂
- Mg: 0 → +2 → oxidized
- N: 0 → -3 → reduced
✔ Oxidized: Mg
✔ Reduced: N₂
✔ Reducing agent: Mg
✔ Oxidizing agent: N₂
---
5. 4Fe + 3O₂ → 2Fe₂O₃
- Fe: 0 → +3 → oxidized
- O: 0 → -2 → reduced
✔ Oxidized: Fe
✔ Reduced: O₂
✔ Reducing agent: Fe
✔ Oxidizing agent: O₂
---
6. Cl₂ + 2NaBr → 2NaCl + Br₂
This is a single displacement reaction.
- Cl: 0 → -1 → reduced
- Br: -1 → 0 → oxidized
✔ Oxidized: Br⁻ (in NaBr)
✔ Reduced: Cl₂
✔ Reducing agent: NaBr
✔ Oxidizing agent: Cl₂
---
7. Si + 2F₂ → SiF₄
- Si: 0 → +4 → oxidized
- F: 0 → -1 → reduced
✔ Oxidized: Si
✔ Reduced: F₂
✔ Reducing agent: Si
✔ Oxidizing agent: F₂
---
8. 2Ca + O₂ → 2CaO
- Ca: 0 → +2 → oxidized
- O: 0 → -2 → reduced
✔ Oxidized: Ca
✔ Reduced: O₂
✔ Reducing agent: Ca
✔ Oxidizing agent: O₂
---
9. Mg + 2HCl → MgCl₂ + H₂
- Mg: 0 → +2 → oxidized
- H: +1 → 0 → reduced
✔ Oxidized: Mg
✔ Reduced: H⁺ (in HCl)
✔ Reducing agent: Mg
✔ Oxidizing agent: HCl
---
10. 2Na + 2H₂O → 2NaOH + H₂
- Na: 0 → +1 → oxidized
- H: +1 → 0 → reduced
✔ Oxidized: Na
✔ Reduced: H₂O (specifically H⁺)
✔ Reducing agent: Na
✔ Oxidizing agent: H₂O
---
## 🧪 Problem 11: Oxidation number of each kind of atom or ion.
Rules:
- Free elements = 0
- Monatomic ions = charge
- Oxygen = -2 (except peroxides)
- Hydrogen = +1 (except with metals = -1)
- Sum of oxidation numbers in compound = 0; in polyatomic ion = ion charge
a. Sulfate (SO₄²⁻):
Let S = x
x + 4(-2) = -2 → x - 8 = -2 → x = +6
b. Sn (elemental tin) → 0
c. S²⁻ → -2
d. Fe³⁺ → +3
e. Sn⁴⁺ → +4
f. Nitrate (NO₃⁻):
Let N = x
x + 3(-2) = -1 → x - 6 = -1 → x = +5
g. Ammonium (NH₄⁺):
Let N = x
x + 4(+1) = +1 → x + 4 = 1 → x = -3
✔ Answers:
a. +6
b. 0
c. -2
d. +3
e. +4
f. +5
g. -3
---
## 🧮 Problem 12: Calculate oxidation number of Cr in each.
a. Cr₂O₃
Let Cr = x
2x + 3(-2) = 0 → 2x = 6 → x = +3
b. Na₂Cr₂O₇
Na = +1, O = -2
2(+1) + 2x + 7(-2) = 0 → 2 + 2x - 14 = 0 → 2x = 12 → x = +6
c. CrSO₄
SO₄²⁻ has -2 charge → Cr must be +2
d. Chromate (CrO₄²⁻)
x + 4(-2) = -2 → x - 8 = -2 → x = +6
e. Dichromate (Cr₂O₇²⁻)
2x + 7(-2) = -2 → 2x - 14 = -2 → 2x = 12 → x = +6
✔ Answers:
a. +3
b. +6
c. +2
d. +6
e. +6
---
## 🔄 Problem 13: Use oxidation numbers to determine what is oxidized/reduced.
*(Note: We don’t need balanced equations — just track changes)*
---
a. C + H₂SO₄ → CO₂ + SO₂ + H₂O
Assign oxidation numbers:
- C: 0 → +4 (in CO₂) → oxidized
- S in H₂SO₄: +6 → +4 (in SO₂) → reduced
✔ Oxidized: C
✔ Reduced: S (in H₂SO₄)
---
b. HNO₃ + HI → NO + I₂ + H₂O
- N in HNO₃: +5 → +2 (in NO) → reduced
- I in HI: -1 → 0 (in I₂) → oxidized
✔ Oxidized: I⁻ (in HI)
✔ Reduced: N (in HNO₃)
---
c. KMnO₄ + HCl → MnCl₂ + Cl₂ + H₂O + KCl
- Mn in KMnO₄: +7 → +2 (in MnCl₂) → reduced
- Cl in HCl: -1 → 0 (in Cl₂) → oxidized
✔ Oxidized: Cl⁻ (in HCl)
✔ Reduced: Mn (in KMnO₄)
---
d. Sb + HNO₃ → Sb₂O₅ + NO + H₂O
- Sb: 0 → +5 (in Sb₂O₅) → oxidized
- N in HNO₃: +5 → +2 (in NO) → reduced
✔ Oxidized: Sb
✔ Reduced: N (in HNO₃)
---
## ⚖️ Problem 14: Identify oxidizing and reducing agents for Problem 13.
Recall:
- Reducing agent = gets oxidized
- Oxidizing agent = gets reduced
---
a. C + H₂SO₄ → ...
- C oxidized → reducing agent: C
- S reduced → oxidizing agent: H₂SO₄
✔ Reducing agent: C
✔ Oxidizing agent: H₂SO₄
---
b. HNO₃ + HI → ...
- I⁻ oxidized → reducing agent: HI
- N reduced → oxidizing agent: HNO₃
✔ Reducing agent: HI
✔ Oxidizing agent: HNO₃
---
c. KMnO₄ + HCl → ...
- Cl⁻ oxidized → reducing agent: HCl
- Mn reduced → oxidizing agent: KMnO₄
✔ Reducing agent: HCl
✔ Oxidizing agent: KMnO₄
---
d. Sb + HNO₃ → ...
- Sb oxidized → reducing agent: Sb
- N reduced → oxidizing agent: HNO₃
✔ Reducing agent: Sb
✔ Oxidizing agent: HNO₃
---
## ✔ Final Summary
All problems solved! You now have complete answers for:
- Identifying oxidized/reduced species and agents (Problems 1–10)
- Oxidation numbers of ions/compounds (Problem 11)
- Chromium oxidation states (Problem 12)
- Redox changes using oxidation numbers (Problem 13)
- Oxidizing/reducing agents for those reactions (Problem 14)
Let me know if you want these formatted into a printable worksheet or explained further!
---
Parent Tip: Review the logic above to help your child master the concept of redox reaction worksheet.
