Worksheet for balancing redox reactions using the oxidation number method, including neutral, acidic, and basic conditions.

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, charge = -2 ⇒ 2Cr = +12 ⇒ Cr = +6
- Cr³⁺ = +3 → reduction
- Fe²⁺ = +2, Fe³⁺ = +3 → oxidation

Changes:
- Cr: +6 → +3 → gain of 3 e⁻ per Cr → 2 Cr → 6 e⁻ gained
- Fe: +2 → +3 → loss of 1 e⁻ per Fe

So need 6 Fe²⁺ to lose 6 e⁻

Balance:
- Cr₂O₇²⁻ → 2Cr³⁺
- 6Fe²⁺ → 6Fe³⁺

Now balance O and H (acidic):

Left: Cr₂O₇²⁻ has 7 O → add 7H₂O on right?
Wait — better to use standard steps:

Half-reactions (but using oxidation number):

Since we’re using oxidation number method:

- Cr₂O₇²⁻ → 2Cr³⁺: 2 Cr go from +6 to +3 → total gain of 6 e⁻
- 6 Fe²⁺ → 6 Fe³⁺: loss of 6 e⁻

So:
- Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺

Now balance O and H.

Left: 7 O in Cr₂O₇²⁻
Right: none → add 7 H₂O on right?

No — we must balance with H⁺ and H₂O.

In acidic solution:
- Add 14 H⁺ to left (to combine with O)
- Add 7 H₂O to right

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

Check charge:
- Left: -2 + 6×(+2) + 14×(+1) = -2 + 12 + 14 = +22
- Right: 2×(+3) + 6×(+3) = 6 + 18 = +24 ✘

Wait — mistake:

Fe²⁺ is +2, so 6Fe²⁺ = +12
Cr₂O₇²⁻ = -2
14H⁺ = +14
Total left: -2 + 12 + 14 = +24

Right: 2Cr³⁺ = +6, 6Fe³⁺ = +18 → total +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)

Assign oxidation numbers:

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

So N: +5 → +4 → reduction

I: 0 → +5 → loses 5 e⁻ per I
N: +5 → +4 → gains 1 e⁻ per N

So for every I₂ (2 I), lose 10 e⁻ → need 10 N atoms to gain 10 e⁻

So:
- I₂ → 2IO₃⁻ (lose 10 e⁻)
- 10NO₃⁻ → 10NO₂ (gain 10 e⁻)

Now balance:

I₂ + 10NO₃⁻ → 2IO₃⁻ + 10NO₂

Now balance O and H.

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

Wait — actually:

- Left: 10 NO₃⁻ → 30 O
- Right: 2IO₃⁻ → 6 O, 10NO₂ → 20 O → 26 O → difference of 4 O → add 4 H₂O to right? No.

Better: add H⁺ and H₂O.

We are in acidic solution.

I₂ → 2IO₃⁻ → adds O → needs water and H⁺

For I₂ → 2IO₃⁻:
- Add 6H₂O to left (to supply O), add 12H⁺ to right (for H)

Standard way:

I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻ (oxidation)

NO₃⁻ → NO₂:
- NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O

So for 10 NO₃⁻: 10NO₃⁻ + 20H⁺ + 10e⁻ → 10NO₂ + 10H₂O

Add both half-reactions:

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

Simplify:
- H⁺: 20H⁺ - 12H⁺ = 8H⁺ on left
- H₂O: 6H₂O - 10H₂O → move 4H₂O to right

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

Check atoms:
- I: 2 = 2 ✔
- N: 10 = 10 ✔
- O: 30 (left) + 8H⁺ has no O → total O: 30
Right: 2IO₃⁻ → 6 O, 10NO₂ → 20 O, 4H₂O → 4 O → 30 ✔
- H: 8 = 8 ✔

Charge:
- Left: 0 + 10×(-1) + 8×(+1) = -10 + 8 = -2
- Right: 2×(-1) + 10×(0) + 0 = -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 Pb in PbSO₄.

PbSO₄: S = +6, O = -2, SO₄²⁻ = -2 ⇒ Pb = +2

Products:
- Pb (elemental) = 0 → reduction
- PbO₂: O = -2, 2×(-2)= -4 ⇒ Pb = +4 → oxidation

So Pb²⁺ both oxidized and reduced.

Let’s say:
- Some Pb²⁺ → Pb (0) → gain of 2 e⁻ per Pb
- Some Pb²⁺ → Pb⁴⁺ → loss of 2 e⁻ per Pb

So equal amounts.

Let x moles of Pb²⁺ reduce to Pb, x moles oxidize to Pb⁴⁺.

So total PbSO₄ used: 2x

Products:
- Pb: x
- PbO₂: x
- SO₄²⁻: 2x

But SO₄²⁻ comes from PbSO₄, so 2x SO₄²⁻ produced.

So equation:
2PbSO₄ → Pb + PbO₂ + 2SO₄²⁻

Now check O and H.

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

Wait — PbO₂ has 2 O, but SO₄²⁻ has 4 O each → 2×4=8 O → total 10 O

Left: 2×4 = 8 O → not balanced.

Need to add water or H⁺.

But this is in acidic solution.

PbSO₄ is solid, but let's treat as ions.

Actually, PbSO₄ is insoluble, but in acidic medium, it may dissolve.

Better approach: balance using oxidation numbers.

Pb²⁺ → Pb⁰: gain 2 e⁻
Pb²⁺ → Pb⁴⁺: lose 2 e⁻

So 1 Pb²⁺ reduced, 1 Pb²⁺ oxidized → total 2 Pb²⁺

So:
2PbSO₄ → Pb + PbO₂ + 2SO₄²⁻

Now balance O and H.

Left: 8 O (from 2SO₄)
Right: PbO₂ → 2 O, 2SO₄²⁻ → 8 O → 10 O → excess 2 O

So need to remove O → add H⁺ and form H₂O

Add 2H⁺ to left, add H₂O to right?

Try:

2PbSO₄ + 2H⁺ → Pb + PbO₂ + 2SO₄²⁻ + H₂O

Check O: left: 8 O, right: 2 (PbO₂) + 8 (SO₄) + 1 (H₂O) = 11 O → no

Wait — PbSO₄ has 4 O per formula → 2×4=8 O

Right: PbO₂ has 2 O, 2SO₄²⁻ has 8 O, H₂O has 1 O → 11 O → imbalance

Alternative idea: perhaps PbO₂ forms from Pb²⁺ + H₂O → PbO₂ + 2H⁺ + 2e⁻

And Pb²⁺ + 2e⁻ → Pb

So:

Pb²⁺ + 2e⁻ → Pb
Pb²⁺ + 2H₂O → PbO₂ + 4H⁺ + 2e⁻

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

So 2Pb²⁺ + 2H₂O → Pb + PbO₂ + 4H⁺

Now include SO₄²⁻: since PbSO₄ dissociates to Pb²⁺ and SO₄²⁻, but only Pb changes.

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

Now check:

Atoms:
- Pb: 2 = 1 + 1 ✔
- S: 2 = 2 ✔
- O: 8 (PbSO₄) + 2 (H₂O) = 10
Right: PbO₂ → 2 O, 2SO₄²⁻ → 8 O, H₂O → 0 → 10 ✔
- H: 4 = 4 ✔

Charge:
- Left: 0 (solids) + 0 (H₂O) = 0
- Right: 2SO₄²⁻ = -4, 4H⁺ = +4 → net 0 ✔

✔ Balanced.

Oxidizing agent: Pb²⁺ (it oxidizes itself, but technically Pb²⁺ is both oxidizing and reducing agent — disproportionation)

So PbSO₄ is both oxidizing and reducing agent.

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

Oxidation numbers:

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

So:
- Cl⁻ → ClO⁻: -1 → +1 → loss of 2 e⁻
- CrO₄²⁻ → CrO₂⁻: +6 → +3 → gain of 3 e⁻

LCM of 2 and 3 is 6.

So:
- 3Cl⁻ → 3ClO⁻ → lose 6 e⁻
- 2CrO₄²⁻ → 2CrO₂⁻ → gain 6 e⁻

So:
3Cl⁻ + 2CrO₄²⁻ → 3ClO⁻ + 2CrO₂⁻

Now balance O and H in basic solution.

Left: 2CrO₄²⁻ → 8 O
Right: 3ClO⁻ → 3 O, 2CrO₂⁻ → 4 O → 7 O → 1 O short

Also, H needed.

Add H₂O and OH⁻.

For oxidation: Cl⁻ → ClO⁻

Cl⁻ + H₂O → ClO⁻ + 2H⁺ + 2e⁻ → but in basic, convert:

Cl⁻ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻

For reduction: CrO₄²⁻ → CrO₂⁻

CrO₄²⁻ + 2H⁺ + 3e⁻ → CrO₂⁻ + H₂O → in basic:

CrO₄²⁻ + H₂O + 3e⁻ → CrO₂⁻ + 2OH⁻

Now multiply:
- Oxidation: 3Cl⁻ + 6OH⁻ → 3ClO⁻ + 3H₂O + 6e⁻
- Reduction: 2CrO₄²⁻ + 2H₂O + 6e⁻ → 2CrO₂⁻ + 4OH⁻

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

Cancel:
- 6e⁻ cancel
- 2H₂O vs 3H₂O → leave 1 H₂O on right
- 6OH⁻ vs 4OH⁻ → leave 2OH⁻ on left

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

Check atoms:
- Cl: 3 = 3 ✔
- Cr: 2 = 2 ✔
- O: 8 (CrO₄) + 2 (OH⁻) = 10
Right: 3 (ClO⁻) + 4 (CrO₂⁻) + 1 (H₂O) = 8 → wait, 3+4+1=8 → mismatch

Wait: CrO₄²⁻ has 4 O each → 2×4=8
OH⁻ has 1 O → total 9 O

Right: 3ClO⁻ → 3 O, 2CrO₂⁻ → 4 O, H₂O → 1 O → 8 O → still off

Wait — OH⁻ has 1 O and 1 H

Let’s count carefully:

Left:
- 3Cl⁻
- 2CrO₄²⁻ → 8 O
- 2OH⁻ → 2 O, 2 H

Total O: 8 + 2 = 10
H: 2

Right:
- 3ClO⁻ → 3 O
- 2CrO₂⁻ → 4 O
- H₂O → 1 O, 2 H

Total O: 3+4+1=8
H: 2

→ O mismatch

Error in half-reactions?

Recheck reduction:

CrO₄²⁻ → CrO₂⁻

CrO₄²⁻ → CrO₂⁻ + 2O²⁻ → but in basic, better:

CrO₄²⁻ + 2H₂O + 3e⁻ → CrO₂⁻ + 4OH⁻

Yes! Because CrO₄²⁻ to CrO₂⁻ removes 2 O, which combines with H⁺ to form H₂O, but in basic:

CrO₄²⁻ + 2H₂O + 3e⁻ → CrO₂⁻ + 4OH⁻

Similarly, oxidation:
Cl⁻ → ClO⁻ → needs to add O

Cl⁻ + H₂O → ClO⁻ + 2H⁺ + 2e⁻ → in basic:

Cl⁻ + H₂O → ClO⁻ + 2H⁺ + 2e⁻ → add 2OH⁻ to both sides:

Cl⁻ + H₂O + 2OH⁻ → ClO⁻ + 2H₂O + 2e⁻ → simplify:

Cl⁻ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻

Correct.

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

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

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

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

Now check atoms:

Left:
- Cl: 3
- Cr: 2
- O: 8 (CrO₄) + 1 (H₂O) = 9
- H: 2

Right:
- 3ClO⁻: 3 O
- 2CrO₂⁻: 4 O
- 2OH⁻: 2 O, 2 H
- Total O: 3+4+2=9 ✔
- H: 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
- Mn in MnO₄⁻: O = -2, 4×(-2)= -8, charge = -1 ⇒ Mn = +7
- MnO₂: O = -2, 2×(-2)= -4 ⇒ Mn = +4 → reduction

So:
- Ni: 0 → +2 → lose 2 e⁻
- Mn: +7 → +4 → gain 3 e⁻

LCM: 6

So:
- 3Ni → lose 6 e⁻
- 2MnO₄⁻ → gain 6 e⁻

So:
3Ni + 2MnO₄⁻ → 3NiO + 2MnO₂

Now balance O and H in basic.

Left: 2MnO₄⁻ → 8 O
Right: 3NiO → 3 O, 2MnO₂ → 4 O → 7 O → 1 O short

Add H₂O and OH⁻.

Use half-reactions:

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

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

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

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

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

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

Check atoms:
- Ni: 3 = 3 ✔
- Mn: 2 = 2 ✔
- O: 8 (MnO₄) + 1 (H₂O) = 9
Right: 3 (NiO) + 4 (MnO₂) + 2 (OH⁻) = 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 → oxidation
- Ce⁴⁺ → Ce³⁺ → reduction (gain of 1 e⁻)

I: -1 → +5 → loss of 6 e⁻
Ce: +4 → +3 → gain of 1 e⁻

So need 6 Ce⁴⁺ per I⁻

So:
I⁻ + 6Ce⁴⁺ → IO₃⁻ + 6Ce³⁺

Now balance O and H in basic.

Left: no O
Right: IO₃⁻ has 3 O

So add H₂O and OH⁻.

Oxidation: I⁻ → IO₃⁻

I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻ → in basic:

I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻ → add 6OH⁻ to both sides:

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

I⁻ + 6OH⁻ → IO₃⁻ + 3H₂O + 6e⁻

Reduction: Ce⁴⁺ + e⁻ → Ce³⁺ → multiply by 6:

6Ce⁴⁺ + 6e⁻ → 6Ce³⁺

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

Cancel 6e⁻:

I⁻ + 6Ce⁴⁺ + 6OH⁻ → IO₃⁻ + 6Ce³⁺ + 3H₂O

Check:
- Atoms: I, Ce, O, H — all balanced
- Charge: left: -1 + 6(+4) + 6(-1) = -1 + 24 -6 = 17
Right: -1 + 6(+3) = -1 + 18 = 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⁺ | PbSO₄ (self) | PbSO₄ (self) |
| 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⁻ |

Let me know if you'd like these formatted neatly!