Worksheet identifying atoms oxidized and reduced, balancing equations using half-reactions.

Let's solve each redox reaction step by step using the half-reaction method. We will:

1. Identify the atom oxidized and reduced.
2. Write the oxidation half-reaction.
3. Write the reduction half-reaction.
4. Balance each half-reaction (atoms and charge).
5. Multiply to make electrons equal.
6. Add the half-reactions together.
7. Cancel species that appear on both sides.

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1. Cu + 2 Ag⁺ → Cu²⁺ + 2 Ag

#### Step 1: Assign oxidation states
- Cu (elemental) = 0 → Cu²⁺ = +2 → oxidized
- Ag⁺ = +1 → Ag (elemental) = 0 → reduced

✔ Atom oxidized: Cu
✔ Atom reduced: Ag

#### Oxidation half-reaction:
Cu → Cu²⁺ + 2e⁻
(Balanced: atoms and charge)

#### Reduction half-reaction:
Ag⁺ + e⁻ → Ag
But we have 2 Ag⁺, so multiply by 2:
2Ag⁺ + 2e⁻ → 2Ag

#### Combine:
Cu → Cu²⁺ + 2e⁻
2Ag⁺ + 2e⁻ → 2Ag
Add them:
Cu + 2Ag⁺ + 2e⁻ → Cu²⁺ + 2Ag + 2e⁻
Cancel electrons:
Cu + 2Ag⁺ → Cu²⁺ + 2Ag

✔ Already balanced!
Electrons transferred: 2 electrons

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2. Cl₂ + 2 Br⁻ → 2 Cl⁻ + Br₂

#### Oxidation states:
- Cl₂ = 0 → Cl⁻ = -1 → reduced
- Br⁻ = -1 → Br₂ = 0 → oxidized

✔ Atom oxidized: Br⁻
✔ Atom reduced: Cl₂

#### Oxidation half-reaction:
2Br⁻ → Br₂ + 2e⁻
(Br atoms balanced, charge: -2 → 0 + 2e⁻ → -2; balanced)

#### Reduction half-reaction:
Cl₂ + 2e⁻ → 2Cl⁻
(Cl atoms and charge balanced)

#### Combine:
2Br⁻ → Br₂ + 2e⁻
Cl₂ + 2e⁻ → 2Cl⁻
Add:
Cl₂ + 2Br⁻ + 2e⁻ → 2Cl⁻ + Br₂ + 2e⁻
Cancel electrons:
Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂

✔ Balanced
Electrons transferred: 2 electrons

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3. 2 I⁻ + F₂ → 2 F⁻ + I₂

#### Oxidation states:
- I⁻ = -1 → I₂ = 0 → oxidized
- F₂ = 0 → F⁻ = -1 → reduced

✔ Atom oxidized: I⁻
✔ Atom reduced: F₂

#### Oxidation half-reaction:
2I⁻ → I₂ + 2e⁻
(Atoms and charge balanced)

#### Reduction half-reaction:
F₂ + 2e⁻ → 2F⁻
(Atoms and charge balanced)

#### Combine:
2I⁻ → I₂ + 2e⁻
F₂ + 2e⁻ → 2F⁻
Add:
F₂ + 2I⁻ + 2e⁻ → 2F⁻ + I₂ + 2e⁻
Cancel electrons:
F₂ + 2I⁻ → 2F⁻ + I₂

✔ Balanced
Electrons transferred: 2 electrons

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4. Sn²⁺ + 2 Fe³⁺ → Sn⁴⁺ + 2 Fe²⁺

#### Oxidation states:
- Sn²⁺ → Sn⁴⁺ → oxidation state increases from +2 to +4 → oxidized
- Fe³⁺ → Fe²⁺ → decreases from +3 to +2 → reduced

✔ Atom oxidized: Sn²⁺
✔ Atom reduced: Fe³⁺

#### Oxidation half-reaction:
Sn²⁺ → Sn⁴⁺ + 2e⁻
(Charge: +2 → +4 + 2e⁻ → total charge +2; balanced)

#### Reduction half-reaction:
Fe³⁺ + e⁻ → Fe²⁺
But we have 2 Fe³⁺, so:
2Fe³⁺ + 2e⁻ → 2Fe²⁺

#### Combine:
Sn²⁺ → Sn⁴⁺ + 2e⁻
2Fe³⁺ + 2e⁻ → 2Fe²⁺
Add:
Sn²⁺ + 2Fe³⁺ + 2e⁻ → Sn⁴⁺ + 2Fe²⁺ + 2e⁻
Cancel electrons:
Sn²⁺ + 2Fe³⁺ → Sn⁴⁺ + 2Fe²⁺

✔ Already balanced
Electrons transferred: 2 electrons

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5. Mg + HCl → MgCl₂ + H₂

First, write the correct balanced molecular equation.

Unbalanced:
Mg + HCl → MgCl₂ + H₂

Balance:
- Mg: 1 on both sides
- Cl: 2 on right → need 2 HCl
- H: 2 on right → 2 HCl gives 2 H

So:
Mg + 2HCl → MgCl₂ + H₂

Now analyze redox:

#### Oxidation states:
- Mg (elemental) = 0 → Mg²⁺ in MgCl₂ = +2 → oxidized
- H in HCl = +1 → H₂ = 0 → reduced

✔ Atom oxidized: Mg
✔ Atom reduced: H (in H⁺)

#### Oxidation half-reaction:
Mg → Mg²⁺ + 2e⁻

#### Reduction half-reaction:
H⁺ + e⁻ → ½H₂
But we need to balance with 2H⁺ since 2HCl:

2H⁺ + 2e⁻ → H₂

#### Combine:
Mg → Mg²⁺ + 2e⁻
2H⁺ + 2e⁻ → H₂
Add:
Mg + 2H⁺ + 2e⁻ → Mg²⁺ + H₂ + 2e⁻
Cancel electrons:
Mg + 2H⁺ → Mg²⁺ + H₂

Now include spectator ions (Cl⁻). Since HCl provides H⁺ and Cl⁻, and MgCl₂ has Mg²⁺ and 2Cl⁻, we add Cl⁻ to both sides:

Mg + 2H⁺ + 2Cl⁻ → Mg²⁺ + 2Cl⁻ + H₂
→ Mg + 2HCl → MgCl₂ + H₂

✔ Balanced
Electrons transferred: 2 electrons

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✔ Final Answers Summary:

| Reaction | Atom Oxidized | Atom Reduced | Oxidation Half-Reaction | Reduction Half-Reaction | Electrons Transferred |
|--------|----------------|---------------|----------------------------|----------------------------|------------------------|
| 1 | Cu | Ag⁺ | Cu → Cu²⁺ + 2e⁻ | 2Ag⁺ + 2e⁻ → 2Ag | 2 |
| 2 | Br⁻ | Cl₂ | 2Br⁻ → Br₂ + 2e⁻ | Cl₂ + 2e⁻ → 2Cl⁻ | 2 |
| 3 | I⁻ | F₂ | 2I⁻ → I₂ + 2e⁻ | F₂ + 2e⁻ → 2F⁻ | 2 |
| 4 | Sn²⁺ | Fe³⁺ | Sn²⁺ → Sn⁴⁺ + 2e⁻ | 2Fe³⁺ + 2e⁻ → 2Fe²⁺ | 2 |
| 5 | Mg | H⁺ | Mg → Mg²⁺ + 2e⁻ | 2H⁺ + 2e⁻ → H₂ | 2 |

All equations are balanced using the half-reaction method, and each involves 2 electrons transferred.

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