Balancing Redox Reactions worksheet with 14 equations to balance using oxidation numbers.

Of course! I'll solve the first problem from your worksheet as an example, showing you the complete step-by-step process using oxidation numbers. Since there are 14 problems, I will demonstrate the method thoroughly on Problem 1 so you can apply it to the rest.

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## Problem 1:
H₂S + HNO₃ → S₈ + NO + H₂O

We need to:
- Balance the redox reaction using oxidation numbers.
- Underline the oxidizing agent.
- Circle the reducing agent.

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Step 1: Assign Oxidation Numbers

Let’s assign oxidation numbers to all atoms involved.

#### Reactants:
- H₂S: H is usually +1 → 2×(+1) = +2 → So S must be -2 to make molecule neutral.
- HNO₃: H = +1, O = -2 (×3 = -6). Let N = x → +1 + x - 6 = 0 → x = +5

#### Products:
- S₈: Elemental form → 0
- NO: O = -2 → N = +2 (since +2 - 2 = 0)
- H₂O: H = +1, O = -2

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Step 2: Identify What is Oxidized and Reduced

- Sulfur: from -2 in H₂S → 0 in S₈ → Oxidation (loss of electrons)
- Nitrogen: from +5 in HNO₃ → +2 in NO → Reduction (gain of electrons)

✔ So:
- Reducing agent: H₂S (it gets oxidized → causes reduction of others)
- Oxidizing agent: HNO₃ (it gets reduced → causes oxidation of others)

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Step 3: Write Half-Reactions

#### Oxidation Half-Reaction (S⁻² → S⁰):
Each S atom loses 2 electrons. But product is S₈, so we need 8 sulfur atoms.

So:
8 H₂S → S₈ + 16 e⁻
(because 8 S atoms × 2 e⁻ each = 16 e⁻ lost)

But we also need to balance H and O — we’ll do that later after balancing electrons.

#### Reduction Half-Reaction (N⁺⁵ → N⁺²):
Each N gains 3 electrons.

So:
HNO₃ + 3 e⁻ → NO

But again, we need to balance atoms and charge properly.

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Step 4: Balance Atoms and Electrons

Let’s write proper half-reactions with atoms balanced.

#### Oxidation: H₂S → S₈
Start with:
8 H₂S → S₈

Now balance H: 16 H on left → add 16 H⁺ on right? But this is not acidic medium specified — actually, since H₂O is a product, we can use H⁺/H₂O method or just balance by inspection.

Actually, let’s balance atoms and then electrons.

We’ll balance the entire equation by finding the least common multiple of electrons transferred.

From above:

- Each S atom loses 2 e⁻ → for 8 S atoms: 16 e⁻ lost
- Each N atom gains 3 e⁻ → for 1 N atom: 3 e⁻ gained

LCM of 16 and 3 is 48.

So multiply oxidation by 3 → 3 × 16e⁻ = 48e⁻ lost
Multiply reduction by 16 → 16 × 3e⁻ = 48e⁻ gained

So:

Oxidation (×3):
3 × [8 H₂S → S₈ + 16 e⁻] → 24 H₂S → 3 S₈ + 48 e⁻

Reduction (×16):
16 × [HNO₃ + 3 e⁻ → NO] → 16 HNO₃ + 48 e⁻ → 16 NO

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Step 5: Combine Half-Reactions

Add them together:

24 H₂S + 16 HNO₃ → 3 S₈ + 16 NO + 48 e⁻ - 48 e⁻

Cancel electrons:

24 H₂S + 16 HNO₃ → 3 S₈ + 16 NO

Now balance H and O using H₂O.

Left side:
H: 24×2 + 16×1 = 48 + 16 = 64 H
O: 16×3 = 48 O

Right side:
H: 0
O: 16×1 = 16 O (from NO) → need more O and H → add H₂O

Let’s add x H₂O to right.

Then right side H = 2x, O = 16 + x

Set equal to left:

H: 64 = 2x → x = 32

O: 48 = 16 + x → x = 32 ✔

Perfect!

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Final Balanced Equation:

24 H₂S + 16 HNO₃ → 3 S₈ + 16 NO + 32 H₂O

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Step 6: Identify Agents

- Oxidizing agent: HNO₃ → *underline it*
- Reducing agent: H₂S → *circle it*

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## ✔ Final Answer for Problem 1:

Balanced Equation:
24 H₂S + 16 HNO₃ → 3 S₈ + 16 NO + 32 H₂O

Oxidizing Agent: <u>HNO₃</u>
Reducing Agent: ⭕ H₂S

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## 🔄 How to Apply This to Other Problems

For each problem:

1. Assign oxidation numbers to all elements.
2. Identify what is oxidized and reduced (change in oxidation number).
3. Write half-reactions for oxidation and reduction.
4. Balance atoms (except H and O), then balance electrons.
5. Multiply half-reactions so electrons lost = electrons gained.
6. Combine and balance H and O using H₂O or H⁺ if needed (usually H₂O here).
7. Identify agents:
- Oxidizing agent = species that gets reduced.
- Reducing agent = species that gets oxidized.

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Would you like me to solve the next one (Problem 2) in the same detail? Or would you prefer to try it yourself and I check your work?

Let me know how you’d like to proceed!