Balancing redox reactions using oxidation number method in neutral, acidic, and basic environments.

Let's solve each redox reaction using the oxidation number method and identify the oxidizing and reducing agents.

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Oxidation Number Method Steps:

1. Assign oxidation numbers to all elements.
2. Identify which species is oxidized (increase in oxidation number) and which is reduced (decrease).
3. Determine the change in oxidation number for each.
4. Balance the electron transfer by making total increase = total decrease.
5. Balance atoms other than O and H.
6. Balance O and H using H₂O and H⁺ (acidic) or OH⁻ (basic).
7. Verify charge and atom balance.

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a. B₂O₃ + Mg → MgO + Mg₃B₂ (neutral)

Step 1: Assign oxidation numbers

- In B₂O₃: O = -2 ⇒ B = +3
- Mg (elemental) = 0
- In MgO: Mg = +2, O = -2
- In Mg₃B₂: Mg = +2, B = -3 (since Mg is more electropositive)

So:
- B goes from +3 → -3 → reduction (gains electrons)
- Mg goes from 0 → +2 → oxidation (loses electrons)

Change in oxidation numbers:
- Each B: +3 → -3 → change of 6 e⁻ gained per B
- Total for 2 B: 2 × 6 = 12 e⁻ gained
- Each Mg: 0 → +2 → 2 e⁻ lost per Mg

To balance electrons:
- 12 e⁻ gained → need 6 Mg atoms (6 × 2 = 12 e⁻ lost)

Now write coefficients:
- B₂O₃ has 2 B → needs 1 B₂O₃
- Mg: 6 Mg
- Products: MgO and Mg₃B₂

But Mg₃B₂ requires 3 Mg and 2 B → so one Mg₃B₂ uses 3 Mg and 2 B

We have:
- 2 B from B₂O₃ → can make 1 Mg₃B₂
- Need 3 Mg for Mg₃B₂
- But we have 6 Mg → 3 left over → form MgO

Each MgO takes 1 Mg → 3 MgO

So:
- B₂O₃ + 6Mg → 3MgO + Mg₃B₂

Check atoms:
- B: 2 = 2 ✔
- O: 3 = 3 ✔
- Mg: 6 = 3 (in MgO) + 3 (in Mg₃B₂) = 6 ✔

✔ Balanced.

Oxidizing agent: B₂O₃ (B is reduced)
Reducing agent: Mg (oxidized)

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b. Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (acidic)

Oxidation numbers:
- Cr in Cr₂O₇²⁻: O = -2, total O = -14, overall charge = -2 ⇒ 2Cr = +12 ⇒ Cr = +6
- Cr³⁺: +3
- Fe²⁺: +2
- Fe³⁺: +3

So:
- Cr: +6 → +3 → reduction (gains 3 e⁻ per Cr)
- Fe: +2 → +3 → oxidation (loses 1 e⁻ per Fe)

Total Cr atoms: 2 → gain 2 × 3 = 6 e⁻
Each Fe loses 1 e⁻ → need 6 Fe²⁺

So:
- Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺

Now balance charge and atoms.

Left: Cr₂O₇²⁻ (-2) + 6Fe²⁺ (6×+2 = +12) → total charge = +10
Right: 2Cr³⁺ (+6) + 6Fe³⁺ (+18) → total = +24 → not balanced

Need to add H⁺ and H₂O since acidic.

Balance O: 7 O on left → add 7 H₂O on right
Add 14 H⁺ on left to balance H

So:
- Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Check charge:
Left: -2 + 6×(+2) + 14×(+1) = -2 + 12 + 14 = +24
Right: 2×(+3) + 6×(+3) = 6 + 18 = +24 ✔

Atoms:
- Cr: 2 = 2 ✔
- Fe: 6 = 6 ✔
- O: 7 = 7 ✔
- H: 14 = 14 ✔

✔ Balanced.

Oxidizing agent: Cr₂O₇²⁻ (Cr reduced)
Reducing agent: Fe²⁺ (oxidized)

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c. I₂ + NO₃⁻ → IO₃⁻ + NO₂ (acidic)

Oxidation numbers:
- I₂: 0
- IO₃⁻: O = -2, 3×(-2) = -6, charge = -1 ⇒ I = +5
- NO₃⁻: O = -2, 3×(-2) = -6, charge = -1 ⇒ N = +5
- NO₂: neutral molecule, O = -2, 2×(-2) = -4 ⇒ N = +4

So:
- I: 0 → +5 → oxidation (loses 5 e⁻ per I)
- N: +5 → +4 → reduction (gains 1 e⁻ per N)

I₂ has 2 I atoms → lose 2 × 5 = 10 e⁻
Each N gains 1 e⁻ → need 10 N atoms

So:
- I₂ + 10NO₃⁻ → 2IO₃⁻ + 10NO₂

Now balance O and H.

Left: 10 NO₃⁻ → 30 O
Right: 2IO₃⁻ → 6 O, 10NO₂ → 20 O → total 26 O → missing 4 O

Add 4 H₂O on right? No — better to use H⁺ and H₂O.

We need to balance oxygen and hydrogen.

Let’s write:

I₂ + 10NO₃⁻ → 2IO₃⁻ + 10NO₂

Oxygen:
Left: 10 × 3 = 30 O
Right: 2×3 + 10×2 = 6 + 20 = 26 O → need 4 more O on right → add 4 H₂O?

Wait — actually, NO₂ is gas, but in solution, it may be formed with water.

Better approach: use H⁺ and H₂O.

We are in acidic medium.

Try balancing:

I₂ → 2IO₃⁻
→ I: 0 → +5 → loss of 10 e⁻

NO₃⁻ → NO₂
N: +5 → +4 → gain of 1 e⁻ per N → need 10 NO₃⁻

So:
I₂ + 10NO₃⁻ → 2IO₃⁻ + 10NO₂

Now balance O and H.

Left: 30 O
Right: 2×3 (from IO₃⁻) + 10×2 (NO₂) = 6 + 20 = 26 O → 4 O missing → add 4 H₂O on right?

No — that adds O. Instead, we need to add H⁺ to react with extra O.

Actually, NO₃⁻ → NO₂ involves:
NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O

So for each NO₃⁻ → NO₂: needs 2H⁺ and produces 1 H₂O

For 10 NO₃⁻ → 10 NO₂: need 20 H⁺ and produce 10 H₂O

Now look at I₂ → 2IO₃⁻
I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻
(Standard half-reaction)

But we already have electron balance.

So combine:

Oxidation: I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻
Reduction: 10NO₃⁻ + 20H⁺ + 10e⁻ → 10NO₂ + 10H₂O

Add them:

I₂ + 6H₂O + 10NO₃⁻ + 20H⁺ + 10e⁻ → 2IO₃⁻ + 12H⁺ + 10e⁻ + 10NO₂ + 10H₂O

Cancel:
- 10e⁻
- 6H₂O and 10H₂O → net 4H₂O on right
- 20H⁺ and 12H⁺ → net 8H⁺ on left

So:
I₂ + 10NO₃⁻ + 8H⁺ → 2IO₃⁻ + 10NO₂ + 4H₂O

Check atoms:
- I: 2 = 2 ✔
- N: 10 = 10 ✔
- O: 30 + 8 (from H⁺? no) — wait, H⁺ has no O

Left: 10 NO₃⁻ → 30 O; 8H⁺ → no O
Right: 2IO₃⁻ → 6 O; 10NO₂ → 20 O; 4H₂O → 4 O → total 30 O ✔
H: 8H⁺ → 8H; right: 4H₂O → 8H ✔

Charge:
Left: I₂ (0), 10NO₃⁻ (10×-1 = -10), 8H⁺ (+8) → total = -2
Right: 2IO₃⁻ (2×-1 = -2), 10NO₂ (0), 4H₂O (0) → total = -2 ✔

✔ Balanced.

Oxidizing agent: NO₃⁻ (N reduced)
Reducing agent: I₂ (oxidized)

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d. PbSO₄ → Pb + PbO₂ + SO₄²⁻ (acidic)

This is a disproportionation of Pb²⁺ in PbSO₄.

PbSO₄ is the source of Pb²⁺ and SO₄²⁻.

We assume PbSO₄ dissociates: PbSO₄ → Pb²⁺ + SO₄²⁻

Then Pb²⁺ disproportionates:
- Some Pb²⁺ → Pb (0) → reduction
- Some Pb²⁺ → Pb⁴⁺ in PbO₂ → oxidation

So:
- Pb²⁺ → Pb: +2 → 0 → gain 2 e⁻
- Pb²⁺ → Pb⁴⁺: +2 → +4 → loss 2 e⁻

So equal amounts.

So: 2Pb²⁺ → Pb + PbO₂

But PbO₂ requires O — so need water and H⁺.

Also, SO₄²⁻ is spectator.

Start with:
2PbSO₄ → Pb + PbO₂ + 2SO₄²⁻

But check O and H.

Left: 2PbSO₄ → 8 O
Right: PbO₂ → 2 O, 2SO₄²⁻ → 8 O → total 10 O → too many

Wait: PbO₂ has 2 O, 2SO₄²⁻ has 8 O → total 10 O
Left: 2PbSO₄ → 2×4 = 8 O → need 2 more O → add 2 H₂O on left?

Better: use H⁺ and H₂O.

Write half-reactions.

Reduction: Pb²⁺ + 2e⁻ → Pb
Oxidation: Pb²⁺ + 2H₂O → PbO₂ + 4H⁺ + 2e⁻

Add:
Pb²⁺ + Pb²⁺ + 2H₂O → Pb + PbO₂ + 4H⁺

So: 2Pb²⁺ + 2H₂O → Pb + PbO₂ + 4H⁺

Now include SO₄²⁻ as spectator.

Each Pb²⁺ comes from PbSO₄ → so 2PbSO₄ → 2Pb²⁺ + 2SO₄²⁻

So:
2PbSO₄ → Pb + PbO₂ + 2SO₄²⁻ + 4H⁺ + 2H₂O

Wait — but we need to account for water.

From above: 2Pb²⁺ + 2H₂O → Pb + PbO₂ + 4H⁺

So add 2H₂O on left and 4H⁺ on right.

So full equation:
2PbSO₄ + 2H₂O → Pb + PbO₂ + 2SO₄²⁻ + 4H⁺

Check atoms:
- Pb: 2 = 1 + 1 ✔
- S: 2 = 2 ✔
- O: 8 (from PbSO₄) + 2 (H₂O) = 10
Right: PbO₂ → 2, 2SO₄²⁻ → 8, total 10 ✔
- H: 4 = 4 ✔

Charge:
Left: 0
Right: 2SO₄²⁻ → -4, 4H⁺ → +4 → total 0 ✔

✔ Balanced.

Oxidizing agent: Pb²⁺ (in PbSO₄) — part of it oxidizes others
Reducing agent: Pb²⁺ — part of it reduces others

So Pb²⁺ is both oxidizing and reducing agent (disproportionation)

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e. Cl⁻ + CrO₄²⁻ → ClO⁻ + CrO₂⁻ (basic)

Oxidation numbers:
- Cl⁻: -1
- ClO⁻: O = -2 ⇒ Cl = +1
- CrO₄²⁻: O = -2, 4×(-2) = -8, charge = -2 ⇒ Cr = +6
- CrO₂⁻: O = -2, 2×(-2) = -4, charge = -1 ⇒ Cr = +3

So:
- Cl: -1 → +1 → oxidation (loss of 2 e⁻)
- Cr: +6 → +3 → reduction (gain of 3 e⁻)

LCM of 2 and 3 is 6 → need 3 Cl⁻ and 2 CrO₄²⁻

So:
3Cl⁻ + 2CrO₄²⁻ → 3ClO⁻ + 2CrO₂⁻

Now balance O and H in basic medium.

Left: 2CrO₄²⁻ → 8 O
Right: 3ClO⁻ → 3 O, 2CrO₂⁻ → 4 O → total 7 O → need 1 more O → add H₂O?

Use H₂O and OH⁻.

Oxidation: Cl⁻ → ClO⁻
Cl⁻ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻
(standard)

Reduction: CrO₄²⁻ → CrO₂⁻
CrO₄²⁻ + 2H₂O + 3e⁻ → CrO₂⁻ + 4OH⁻
(Cr: +6 → +3, gain 3e⁻)

Multiply:
- Oxidation: 3×[Cl⁻ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻] → 3Cl⁻ + 6OH⁻ → 3ClO⁻ + 3H₂O + 6e⁻
- Reduction: 2×[CrO₄²⁻ + 2H₂O + 3e⁻ → CrO₂⁻ + 4OH⁻] → 2CrO₄²⁻ + 4H₂O + 6e⁻ → 2CrO₂⁻ + 8OH⁻

Add:
3Cl⁻ + 6OH⁻ + 2CrO₄²⁻ + 4H₂O + 6e⁻ → 3ClO⁻ + 3H₂O + 6e⁻ + 2CrO₂⁻ + 8OH⁻

Cancel:
- 6e⁻
- 6OH⁻ and 8OH⁻ → net 2OH⁻ on right
- 4H₂O and 3H₂O → net 1H₂O on left

So:
3Cl⁻ + 2CrO₄²⁻ + H₂O → 3ClO⁻ + 2CrO₂⁻ + 2OH⁻

Check atoms:
- Cl: 3 = 3 ✔
- Cr: 2 = 2 ✔
- O: left: 2×4 + 1 = 9; right: 3×1 + 2×2 + 2 = 3 + 4 + 2 = 9 ✔
- H: 2 = 2 ✔

Charge:
Left: 3(-1) + 2(-2) = -3 -4 = -7
Right: 3(-1) + 2(-1) + 2(-1) = -3 -2 -2 = -7 ✔

✔ Balanced.

Oxidizing agent: CrO₄²⁻ (Cr reduced)
Reducing agent: Cl⁻ (oxidized)

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f. Ni + MnO₄⁻ → NiO + MnO₂ (basic)

Oxidation numbers:
- Ni: 0 → NiO: O = -2 ⇒ Ni = +2 → oxidation (lose 2 e⁻)
- Mn in MnO₄⁻: O = -2, 4×(-2) = -8, charge = -1 ⇒ Mn = +7
- Mn in MnO₂: O = -2, 2×(-2) = -4 ⇒ Mn = +4 → reduction (gain 3 e⁻)

So:
- Ni → Ni²⁺ → lose 2 e⁻
- Mn: +7 → +4 → gain 3 e⁻

LCM: 6 → 3 Ni and 2 MnO₄⁻

So:
3Ni + 2MnO₄⁻ → 3NiO + 2MnO₂

Now balance O and H in basic medium.

Left: 2MnO₄⁻ → 8 O
Right: 3NiO → 3 O, 2MnO₂ → 4 O → total 7 O → need 1 more O → add H₂O?

Use standard half-reactions.

Oxidation: Ni → NiO
Ni + 2OH⁻ → NiO + H₂O + 2e⁻

Reduction: MnO₄⁻ → MnO₂
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Multiply:
- Oxidation: 3×[Ni + 2OH⁻ → NiO + H₂O + 2e⁻] → 3Ni + 6OH⁻ → 3NiO + 3H₂O + 6e⁻
- Reduction: 2×[MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻] → 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻

Add:
3Ni + 6OH⁻ + 2MnO₄⁻ + 4H₂O + 6e⁻ → 3NiO + 3H₂O + 6e⁻ + 2MnO₂ + 8OH⁻

Cancel:
- 6e⁻
- 6OH⁻ and 8OH⁻ → 2OH⁻ on right
- 4H₂O and 3H₂O → 1H₂O on left

So:
3Ni + 2MnO₄⁻ + H₂O → 3NiO + 2MnO₂ + 2OH⁻

Check:
- Atoms: Ni: 3=3, Mn:2=2, O: 8+1=9; right: 3+4+2=9, H:2=2 ✔
- Charge: left: 0 + 2(-1) = -2; right: 2(-1) = -2 ✔

✔ Balanced.

Oxidizing agent: MnO₄⁻
Reducing agent: Ni

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g. I⁻ + Ce⁴⁺ → IO₃⁻ + Ce³⁺ (basic)

Oxidation numbers:
- I⁻: -1
- IO₃⁻: O = -2, 3×(-2) = -6, charge = -1 ⇒ I = +5
- Ce⁴⁺ → Ce³⁺: +4 → +3 → reduction (gain 1 e⁻)

I: -1 → +5 → loss of 6 e⁻

So:
- I⁻ → IO₃⁻: lose 6 e⁻
- Ce⁴⁺ → Ce³⁺: gain 1 e⁻

So need 6 Ce⁴⁺ per I⁻

So:
I⁻ + 6Ce⁴⁺ → IO₃⁻ + 6Ce³⁺

Now balance in basic medium.

Oxidation: I⁻ → IO₃⁻
I⁻ + 6OH⁻ → IO₃⁻ + 3H₂O + 6e⁻

Reduction: Ce⁴⁺ + e⁻ → Ce³⁺
6×: 6Ce⁴⁺ + 6e⁻ → 6Ce³⁺

Add:
I⁻ + 6OH⁻ + 6Ce⁴⁺ + 6e⁻ → IO₃⁻ + 3H₂O + 6e⁻ + 6Ce³⁺

Cancel 6e⁻:
I⁻ + 6Ce⁴⁺ + 6OH⁻ → IO₃⁻ + 6Ce³⁺ + 3H₂O

Check:
- Atoms: I:1=1, Ce:6=6, O:6+6=12; right: 3+3=6? Wait — IO₃⁻ has 3 O, 3H₂O has 3 O → total 6 O? No!

Left: 6OH⁻ → 6 O, I⁻, 6Ce⁴⁺
Right: IO₃⁻ → 3 O, 3H₂O → 3 O → total 6 O → but left has 6 O from OH⁻ → ok

H: left: 6H; right: 6H → 3H₂O → 6H ✔

Charge:
Left: I⁻ (-1) + 6Ce⁴⁺ (+24) + 6OH⁻ (-6) → total = +17
Right: IO₃⁻ (-1) + 6Ce³⁺ (+18) → total = +17 ✔

✔ Balanced.

Oxidizing agent: Ce⁴⁺
Reducing agent: I⁻

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✔ Final Answers:

| Reaction | Balanced Equation | Oxidizing Agent | Reducing Agent |
|--------|-------------------|------------------|----------------|
| a | B₂O₃ + 6Mg → 3MgO + Mg₃B₂ | B₂O₃ | Mg |
| b | Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O | Cr₂O₇²⁻ | Fe²⁺ |
| c | I₂ + 10NO₃⁻ + 8H⁺ → 2IO₃⁻ + 10NO₂ + 4H₂O | NO₃⁻ | I₂ |
| d | 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2SO₄²⁻ + 4H⁺ | Pb²⁺ (in PbSO₄) | Pb²⁺ (in PbSO₄) |
| e | 3Cl⁻ + 2CrO₄²⁻ + H₂O → 3ClO⁻ + 2CrO₂⁻ + 2OH⁻ | CrO₄²⁻ | Cl⁻ |
| f | 3Ni + 2MnO₄⁻ + H₂O → 3NiO + 2MnO₂ + 2OH⁻ | MnO₄⁻ | Ni |
| g | I⁻ + 6Ce⁴⁺ + 6OH⁻ → IO₃⁻ + 6Ce³⁺ + 3H₂O | Ce⁴⁺ | I⁻ |

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