Redox practice worksheet featuring questions on oxidation states, identifying oxidized and reduced species, and analyzing chemical reactions.

Let's go through each question on the Redox Practice Worksheet and solve them step by step with explanations.

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1. In which substance is the oxidation number of nitrogen zero?

Options:
A. NH₃
B. N₂
C. NO₂
D. N₂O

Answer: B. N₂

Explanation:
The oxidation number of an element in its elemental form is zero. Nitrogen gas (N₂) is a diatomic molecule in its elemental state, so each nitrogen atom has an oxidation number of 0.

- In NH₃: H is +1, so N = -3
- In NO₂: O is -2, so N = +4
- In N₂O: O is -2, so N = +1

✔ So, only N₂ has oxidation number 0.

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2. What is the oxidation number of carbon in NaHCO₃?

Options:
A. +6
B. +2
C. -4
D. +4

Answer: D. +4

Explanation:
Break down the compound: NaHCO₃

- Na = +1 (alkali metal)
- H = +1
- O = -2 (each)
- There are 3 O atoms → total = 3 × (-2) = -6
- Let oxidation number of C be x

Total charge = 0 (neutral compound)

So:
(+1) + (+1) + x + (-6) = 0
→ 2 + x - 6 = 0
→ x = +4

✔ Carbon has oxidation number +4

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3. In the reaction Al⁰ + Cr³⁺ → Al³⁺ + Cr⁰, the reducing agent is

Options:
A. Al⁰
B. Cr³⁺
C. Al³⁺
D. Cr⁰

Answer: A. Al⁰

Explanation:
- Reducing agent is the species that gets oxidized (loses electrons).
- Al⁰ goes from 0 to +3 → oxidized
- Cr³⁺ goes from +3 to 0 → reduced

So, Al⁰ donates electrons → it's the reducing agent

✔ Answer: Al⁰

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4. In the reaction 2K + Cl₂ → 2KCl, the species oxidized is

Options:
A. Cl₂
B. Cl⁻
C. K
D. K⁺

Answer: C. K

Explanation:
- Potassium (K) starts as 0 (elemental form), ends as K⁺ in KCl → oxidized
- Chlorine (Cl₂) starts as 0, ends as Cl⁻ → reduced

So, K is oxidized.

✔ Answer: K

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5. As an S²⁻ ion is oxidized to an S⁰ atom, the number of protons in its nucleus

Options:
A. decreases
B. increases
C. remains the same

Answer: C. remains the same

Explanation:
Oxidation involves loss of electrons, not change in nucleus.

- S²⁻ has gained 2 electrons, but the number of protons (atomic number) stays constant.
- When it becomes S⁰, it loses 2 electrons, but nucleus unchanged.

✔ Proton count remains the same.

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**6. Given the probable reaction for the nickel-cadmium battery:

2Ni(OH)₃ + Cd → 2Ni(OH)₂ + Cd(OH)₂
Which species is oxidized during the discharge of the battery?**

Options:
A. Ni³⁺
B. Ni²⁺
C. Cd⁰
D. Cd²⁺

Answer: C. Cd⁰

Explanation:
Look at oxidation states:

- In Cd: elemental form → oxidation state = 0
- In Cd(OH)₂: Cd is +2 → oxidized

So Cd⁰ → Cd²⁺ → oxidized

Now check Ni:
- In Ni(OH)₃: Ni is +3
- In Ni(OH)₂: Ni is +2 → reduced

So Cd⁰ is oxidized.

✔ Answer: Cd⁰

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7. If element X forms the oxides XO and X₂O₃, the oxidation numbers of element X are

Options:
A. +1 and +2
B. +2 and +3
C. +1 and +3
D. +2 and +4

Answer: B. +2 and +3

Explanation:
- In XO: Oxygen is -2 → X must be +2 to balance
- In X₂O₃: Total O = 3 × (-2) = -6 → two X atoms = +6 → each X = +3

So oxidation states are +2 and +3

✔ Answer: B

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8. Oxygen has a positive oxidation number in the compound

Options:
A. H₂O
B. H₂O₂
C. OF₂
D. IO₂

Answer: C. OF₂

Explanation:
Normally oxygen is -2, but when bonded to fluorine (more electronegative), oxygen can have positive oxidation state.

- In OF₂: F is -1 (each), so 2F = -2 → O must be +2
- In H₂O: O = -2
- In H₂O₂: O = -1
- In IO₂: Iodine is less electronegative than O → O = -2

✔ Only in OF₂ does oxygen have a positive oxidation number.

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9. What is the oxidation number of sulfur in H₂SO₄?

Options:
A. 0
B. -2
C. +6
D. +4

Answer: C. +6

Explanation:
H₂SO₄:

- H = +1 each → 2 × (+1) = +2
- O = -2 each → 4 × (-2) = -8
- Let S = x

Total = 0 →
+2 + x + (-8) = 0
x = +6

✔ Sulfur has oxidation number +6

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10. In the equation Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), the oxidizing agent is

Options:
A. Cu⁰
B. Ag⁺
C. Cu²⁺
D. Ag⁰

Answer: B. Ag⁺

Explanation:
- Oxidizing agent gets reduced
- Ag⁺ → Ag⁰ → gains electrons → reduced
- So Ag⁺ is the oxidizing agent

Cu⁰ → Cu²⁺ → oxidized → it’s the reducing agent

✔ Answer: Ag⁺

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11. In the reaction Al + Cr³⁺ → Al³⁺ + Cr, the reducing agent is

Options:
A. Al
B. Cr³⁺
C. Al³⁺
D. Cr

Answer: A. Al

Explanation:
- Al goes from 0 → +3 → oxidized
- Cr³⁺ goes from +3 → 0 → reduced

So Al is the reducing agent

✔ Answer: Al

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12. In the compound Na₂HPO₄, which element has a negative oxidation number?

Options:
A. H
B. O
C. P
D. Na

Answer: B. O

Explanation:
Let’s assign oxidation numbers:

- Na = +1 → 2 × (+1) = +2
- H = +1
- O = usually -2 → 4 × (-2) = -8
- P = ?

Total charge = 0

So:
+2 (Na) + +1 (H) + x (P) + (-8) (O) = 0
→ 3 + x - 8 = 0
→ x = +5 → P = +5

Now check signs:
- Na = +1 → positive
- H = +1 → positive
- P = +5 → positive
- O = -2 → negative

✔ Only oxygen has a negative oxidation number.

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✔ Final Answers Summary:

| Q# | Answer |
|----|--------|
| 1 | B. N₂ |
| 2 | D. +4 |
| 3 | A. Al⁰ |
| 4 | C. K |
| 5 | C. remains the same |
| 6 | C. Cd⁰ |
| 7 | B. +2 and +3 |
| 8 | C. OF₂ |
| 9 | C. +6 |
| 10 | B. Ag⁺ |
| 11 | A. Al |
| 12 | B. O |

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