Solved Balancing Redox Reactions Worksheet Balancing Redox | Chegg.com - Free Printable

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 electron transfer by making total increase = total decrease.
5. Balance atoms other than O and H.
6. Balance O and H using H₂O, H⁺ (acidic), or OH⁻ (basic).
7. Check charge balance.

We'll go through each one.

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a. B₂O₃ + Mg → MgO + Mg₃B₂ (neutral)

Step 1: Assign oxidation numbers

- In B₂O₃: Oxygen is -2, so total for O = -6.
Let oxidation state of B be x:
2x + (-6) = 0 → x = +3 → B = +3

- In Mg (elemental): 0

- In MgO: O = -2 → Mg = +2

- In Mg₃B₂: This is a compound with Mg²⁺ and B³⁻, since it's neutral and Mg is typically +2.
So: 3(+2) + 2(x) = 0 → 6 + 2x = 0 → x = -3 → B = -3

Changes:

- B: +3 → -3 → gain of 6 electrons per B atom → reduction
- Mg: 0 → +2 → loss of 2 electrons per Mg atom → oxidation

Now balance:

- Each B gains 6 e⁻ → 2 B atoms gain 12 e⁻
- Each Mg loses 2 e⁻ → need 6 Mg atoms to lose 12 e⁻

So:
- B₂O₃ provides 2 B atoms
- Need 6 Mg atoms

Products:
- 6 MgO (from 6 Mg)
- Mg₃B₂ (contains 3 Mg and 2 B)

But we have 6 Mg and 2 B from reactants.

So let’s write:

B₂O₃ + 6Mg → 6MgO + Mg₃B₂

Check atoms:
- B: 2 = 2 ✔
- O: 3 = 6? ✘ → 3 on left, 6 on right → not balanced

Wait — problem: MgO has one O, so 6 MgO → 6 O, but B₂O₃ only has 3 O.

So this can't work.

Wait! The product Mg₃B₂ is a boride (ionic compound), and MgO is also formed.

But where does the oxygen go?

Actually, all oxygen from B₂O₃ must go to MgO.

So: B₂O₃ has 3 O → needs 3 MgO → 3 Mg used

Then remaining Mg goes to Mg₃B₂.

So:

- B₂O₃ → 2 B and 3 O
- 3 O → 3 MgO → 3 Mg
- 2 B → Mg₃B₂ → requires 3 Mg
- Total Mg needed: 3 + 3 = 6 Mg

So:

B₂O₃ + 6Mg → 3MgO + Mg₃B₂

Now check atoms:
- B: 2 = 2 ✔
- O: 3 = 3 ✔
- Mg: 6 = 3 + 3 = 6 ✔

Now check oxidation changes:

- B: +3 → -3 → change of 6 e⁻ per B → 2 B → 12 e⁻ gained
- Mg: 0 → +2 → 6 Mg → lose 12 e⁻ → balanced

✔ Balanced.

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

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

Assign oxidation numbers:

- Cr₂O₇²⁻: O = -2, total O = -14 → 2Cr + (-14) = -2 → 2Cr = +12 → Cr = +6
- Fe²⁺: +2
- Cr³⁺: +3
- Fe³⁺: +3

Changes:

- Cr: +6 → +3 → gain of 3 e⁻ per Cr → 2 Cr → gain 6 e⁻
- Fe: +2 → +3 → lose 1 e⁻ per Fe → need 6 Fe to lose 6 e⁻

So:
- 1 Cr₂O₇²⁻ → 2 Cr³⁺
- 6 Fe²⁺ → 6 Fe³⁺

Now balance:

Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺

Now balance O and H — acidic solution.

Left: 7 O in Cr₂O₇²⁻
Right: no O → add 7 H₂O to right

→ But then H is unbalanced.

Add H⁺ to left.

Standard way: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
To balance O: add 7 H₂O to right
Then add 14 H⁺ to left to balance H

So:

Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O

Now add electrons: Cr₂O₇²⁻ → 2Cr³⁺ → gain 6e⁻

So:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Now Fe²⁺ → Fe³⁺ + e⁻ → multiply by 6:

6Fe²⁺ → 6Fe³⁺ + 6e⁻

Add both:

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

✔ 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
- NO₃⁻: N = +5 (O = -2 → 3×(-2) = -6; N + (-6) = -1 → N = +5)
- IO₃⁻: O = -2 → 3×(-2) = -6 → I + (-6) = -1 → I = +5
- NO₂: O = -2 → 2×(-2) = -4 → N = +4

Changes:

- I: 0 → +5 → lose 5 e⁻ per I atom → I₂ has 2 I → lose 10 e⁻
- N: +5 → +4 → gain 1 e⁻ per N → need 10 N atoms

So:

I₂ → 2IO₃⁻ → lose 10 e⁻
10 NO₃⁻ → 10 NO₂ → gain 10 e⁻

Now balance:

I₂ + 10NO₃⁻ → 2IO₃⁻ + 10NO₂

Now balance O and H (acidic).

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

Also, H needed.

Better to use half-reactions.

Oxidation half: I₂ → IO₃⁻

I₂ → 2IO₃⁻
Balance I: done
Balance O: add 6 H₂O to left? No — right has more O.

Add water to left, H⁺ to right.

I₂ → 2IO₃⁻
Add 6 H₂O to left → 6 O on left? No — I₂ has no O.

Better:

I₂ → 2IO₃⁻
Add 6 H₂O to right to provide O? No.

Standard:
I₂ → 2IO₃⁻
Balance O: add 6 H₂O to left
Balance H: add 12 H⁺ to right
Balance charge: add 10 e⁻ to right

I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻

Reduction half: NO₃⁻ → NO₂

NO₃⁻ → NO₂
Balance O: add H₂O to right? NO₃⁻ has 3 O, NO₂ has 2 → add 1 H₂O to right
Balance H: add 2 H⁺ to left
Balance charge: add 1 e⁻ to left

NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O

Multiply by 10:

10NO₃⁻ + 20H⁺ + 10e⁻ → 10NO₂ + 10H₂O

Now add:

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

Simplify:

Cancel H⁺: 20H⁺ - 12H⁺ = 8H⁺ on left
Cancel H₂O: 10H₂O - 6H₂O = 4H₂O on right

So:

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

✔ 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 PbSO₄.

Assign oxidation states:

- PbSO₄: SO₄²⁻ → S = +6, O = -2
Pb: let x → x + (-2) = 0? Wait, PbSO₄ is neutral → Pb + SO₄²⁻ → Pb²⁺ → so Pb = +2

- Pb (elemental): 0
- PbO₂: O = -2 → 2×(-2) = -4 → Pb = +4

So Pb²⁺ → Pb (0) → reduction
Pb²⁺ → Pb⁴⁺ → oxidation

So Pb²⁺ disproportionates.

Each Pb²⁺:

- One goes to Pb(0): gains 2 e⁻
- One goes to Pb⁴⁺: loses 2 e⁻

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

But PbSO₄ provides both.

So:

2PbSO₄ → Pb + PbO₂ + 2SO₄²⁻

But now check O and S.

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

Wait: PbO₂ has 2 O, 2SO₄²⁻ has 8 O → total 10 O
Left: 2 PbSO₄ → 8 O → imbalance

So we need to account for oxygen.

But PbSO₄ has SO₄²⁻ → sulfate ion.

In products: PbO₂ and SO₄²⁻

But PbO₂ is solid, SO₄²⁻ is aqueous.

But oxygen count:

Left: 2 × 4 = 8 O
Right: PbO₂ has 2 O, 2SO₄²⁻ has 8 O → 10 O → too many

So we must have water involved.

Since it's acidic, we can use H⁺ and H₂O.

But here, PbSO₄ is decomposing.

Alternative: think of it as:

2Pb²⁺ → Pb + Pb⁴⁺

But Pb⁴⁺ forms PbO₂ in acidic medium.

So:

Pb²⁺ → Pb (reduction)
Pb²⁺ → PbO₂ (oxidation)

For oxidation: Pb²⁺ → PbO₂
Balance O: add H₂O to left → Pb²⁺ + 2H₂O → PbO₂ + 4H⁺ + 2e⁻

For reduction: Pb²⁺ + 2e⁻ → Pb

Add:

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

Now include SO₄²⁻: since PbSO₄ dissociates, and SO₄²⁻ is spectator.

So:

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

Check atoms:

- Pb: 2 = 1 + 1 ✔
- S: 2 = 2 ✔
- O: left: 2×4 (from SO₄) + 2×1 (from H₂O) = 8 + 2 = 10
Right: PbO₂ → 2 O, 2SO₄²⁻ → 8 O → 10 ✔
- H: 4 H on right → 4 H on left → 2H₂O → 4 H ✔

Charge:
- Left: 2PbSO₄ → neutral, H₂O → neutral → 0
- Right: 2SO₄²⁻ → -4, 4H⁺ → +4 → net 0 ✔

✔ Balanced.

Oxidizing agent: Pb²⁺ (it oxidizes itself) → disproportionation
Reducing agent: Pb²⁺ (same) → self-oxidizing and self-reducing

So Pb²⁺ is both oxidizing and reducing agent.

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

First, correct formula: CrO₂⁻ is likely Cr(OH)₃ or CrO₂⁻?
But CrO₂⁻ is not standard. Probably meant Cr(OH)₃ or Cr₂O₃?

Wait — common in basic: CrO₄²⁻ → Cr(OH)₃ or Cr(OH)₄⁻?

Actually, Cr(VI) → Cr(III)

Let’s assign oxidation numbers:

- Cl⁻: -1
- CrO₄²⁻: Cr = +6
- ClO⁻: O = -2 → Cl + (-2) = -1 → Cl = +1
- CrO₂⁻: probably typo — should be Cr(OH)₃ or Cr₂O₃? But CrO₂⁻ suggests Cr = +4? That doesn’t make sense.

Wait — likely meant Cr(OH)₃ or Cr(OH)₄⁻?

Standard: CrO₄²⁻ → Cr(OH)₃ in basic medium.

But here it says CrO₂⁻ — maybe it's CrO₂⁻ as in chromite? But that's Cr(II).

No — perhaps it's a typo. Let's assume Cr(OH)₃ or Cr(OH)₄⁻?

But the formula is written as CrO₂⁻ — that would imply Cr = +4 → but Cr(VI) to Cr(IV)? Unlikely.

Alternatively, could be Cr₂O₃?

Wait — better look at common reactions.

Actually, in basic solution, CrO₄²⁻ is reduced to Cr(OH)₃ (Cr³⁺), and Cl⁻ is oxidized to ClO⁻.

So likely: CrO₂⁻ is a typo → should be Cr(OH)₃ or Cr(OH)₄⁻?

But let's suppose Cr(OH)₃ is the product.

But formula given is CrO₂⁻ — perhaps it's CrO₂⁻ as in Cr(IV)? But that's rare.

Wait — another possibility: CrO₂⁻ might be Cr(OH)₄⁻ written poorly?

No.

Alternatively, in some notations, CrO₂⁻ may represent [Cr(OH)₄]⁻, but oxidation state of Cr is still +3.

Let’s assume the intended product is Cr(OH)₃ or Cr(OH)₄⁻.

But to proceed, let's assign:

- Cl⁻ → ClO⁻: Cl from -1 to +1 → lose 2 e⁻
- CrO₄²⁻ → Cr³⁺: Cr from +6 to +3 → gain 3 e⁻

So least common multiple: 6 e⁻

So:

3Cl⁻ → 3ClO⁻ → lose 6 e⁻
2CrO₄²⁻ → 2Cr³⁺ → gain 6 e⁻

Now balance in basic solution.

Oxidation: Cl⁻ → ClO⁻

Cl⁻ → ClO⁻
Add H₂O to right? No — add H₂O to left, OH⁻ to right.

Cl⁻ + H₂O → ClO⁻ + 2H⁺ + 2e⁻ → but basic

So convert to basic:

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

Cl⁻ + H₂O + 2OH⁻ → ClO⁻ + 2H⁺ + 2OH⁻ + 2e⁻
→ Cl⁻ + H₂O + 2OH⁻ → ClO⁻ + 2H₂O + 2e⁻
Simplify: Cl⁻ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻

Reduction: CrO₄²⁻ → Cr(OH)₃

CrO₄²⁻ → Cr(OH)₃
Balance Cr: done
Balance O: add H₂O to right? CrO₄²⁻ has 4 O, Cr(OH)₃ has 3 O → add 1 H₂O to right? No.

Better:

CrO₄²⁻ → Cr(OH)₃
Add 3 H₂O to left → 3 H₂O
Add 3 H⁺ to right? But basic.

Standard:

CrO₄²⁻ + 3e⁻ + 4H₂O → Cr(OH)₃ + 5OH⁻

Yes — known half-reaction.

So:

CrO₄²⁻ + 3e⁻ + 4H₂O → Cr(OH)₃ + 5OH⁻

Now multiply:

Oxidation: 3Cl⁻ + 6OH⁻ → 3ClO⁻ + 3H₂O + 6e⁻
Reduction: 2CrO₄²⁻ + 6e⁻ + 8H₂O → 2Cr(OH)₃ + 10OH⁻

Add:

3Cl⁻ + 6OH⁻ + 2CrO₄²⁻ + 8H₂O → 3ClO⁻ + 3H₂O + 6e⁻ + 2Cr(OH)₃ + 10OH⁻ + 6e⁻

Cancel:

- H₂O: 8 on left, 3 on right → 5 H₂O on left
- OH⁻: 6 on left, 10 on right → move 4 OH⁻ to right

So:

3Cl⁻ + 2CrO₄²⁻ + 5H₂O → 3ClO⁻ + 2Cr(OH)₃ + 4OH⁻

Now check atoms:

- Cl: 3 = 3 ✔
- Cr: 2 = 2 ✔
- O: left: 2×4 (CrO₄) + 5 = 8 + 5 = 13
Right: 3ClO⁻ → 3 O, 2Cr(OH)₃ → 6 O, 4OH⁻ → 4 O → total 13 ✔
- H: left: 10 H
Right: 2Cr(OH)₃ → 6 H, 4OH⁻ → 4 H → 10 H ✔

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

✔ Balanced.

But the problem says CrO₂⁻ — if that means Cr(OH)₃, okay.

If CrO₂⁻ is meant to be something else, but likely a typo.

Assume Cr(OH)₃ is intended.

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

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f. Ni + MnO₄⁻ → NiO + MnO₂ (basic)

Assign oxidation numbers:

- Ni: 0
- MnO₄⁻: Mn = +7
- NiO: O = -2 → Ni = +2
- MnO₂: Mn = +4

Changes:

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

LCM: 6 e⁻

So:

3Ni → 3NiO → lose 6 e⁻
2MnO₄⁻ → 2MnO₂ → gain 6 e⁻

Now balance in basic solution.

Oxidation: Ni → NiO

Ni → NiO
Add H₂O to left? Ni → NiO → add H₂O to right? No.

Ni → NiO
Add H₂O to left, OH⁻ to right.

Ni + H₂O → NiO + 2H⁺ + 2e⁻ → but basic

Convert:

Ni + H₂O → NiO + 2H⁺ + 2e⁻
Add 2OH⁻ to both sides:

Ni + H₂O + 2OH⁻ → NiO + 2H⁺ + 2OH⁻ + 2e⁻
→ Ni + H₂O + 2OH⁻ → NiO + 2H₂O + 2e⁻
→ Ni + 2OH⁻ → NiO + H₂O + 2e⁻

Reduction: MnO₄⁻ → MnO₂

MnO₄⁻ → MnO₂
Add 2H₂O to right? MnO₄⁻ has 4 O, MnO₂ has 2 → add 2H₂O to right? No.

Standard:

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Yes.

So:

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Now multiply:

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

Add:

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

Cancel:

- H₂O: 4 on left, 3 on right → 1 H₂O on left
- OH⁻: 6 on left, 8 on right → 2 OH⁻ on right

So:

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

Check atoms:

- Ni: 3 = 3 ✔
- Mn: 2 = 2 ✔
- O: left: 2×4 (MnO₄) + 1 = 8 + 1 = 9
Right: 3NiO → 3 O, 2MnO₂ → 4 O, 2OH⁻ → 2 O → 9 ✔
- H: left: 2 H, right: 2 H ✔

Charge:
- Left: 2(-1) = -2
- Right: 2(-1) = -2 ✔

✔ Balanced.

Oxidizing agent: MnO₄⁻
Reducing agent: Ni

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g. I⁻ + Ce⁴⁺ → IO₃⁻ + Ce³⁺ (basic)

Assign oxidation numbers:

- I⁻: -1
- Ce⁴⁺: +4
- IO₃⁻: I = +5
- Ce³⁺: +3

Changes:

- I: -1 → +5 → lose 6 e⁻
- Ce: +4 → +3 → gain 1 e⁻ → need 6 Ce

So:

I⁻ → IO₃⁻
6Ce⁴⁺ → 6Ce³⁺

Now balance in basic.

Oxidation: I⁻ → IO₃⁻

I⁻ → IO₃⁻
Add 3 H₂O to left → 3 H₂O
Add 6H⁺ to right? But basic.

Standard:

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

I⁻ + 3H₂O + 6OH⁻ → IO₃⁻ + 6H⁺ + 6OH⁻ + 6e⁻
→ I⁻ + 3H₂O + 6OH⁻ → IO₃⁻ + 6H₂O + 6e⁻
→ I⁻ + 6OH⁻ → IO₃⁻ + 3H₂O + 6e⁻

Reduction: Ce⁴⁺ → Ce³⁺

Ce⁴⁺ + e⁻ → Ce³⁺
Multiply by 6: 6Ce⁴⁺ + 6e⁻ → 6Ce³⁺

Add:

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

Check atoms:

- I: 1 = 1 ✔
- Ce: 6 = 6 ✔
- O: left: 6OH⁻ → 6 O, right: IO₃⁻ → 3 O, 3H₂O → 3 O → 6 ✔
- H: left: 6 H, right: 6 H ✔

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:

#### a.
Balanced: B₂O₃ + 6Mg → 3MgO + Mg₃B₂
- Oxidizing agent: B₂O₃
- Reducing agent: Mg

#### b.
Balanced: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
- Oxidizing agent: Cr₂O₇²⁻
- Reducing agent: Fe²⁺

#### c.
Balanced: I₂ + 10NO₃⁻ + 8H⁺ → 2IO₃⁻ + 10NO₂ + 4H₂O
- Oxidizing agent: NO₃⁻
- Reducing agent: I₂

#### d.
Balanced: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2SO₄²⁻ + 4H⁺
- Oxidizing agent: Pb²⁺ (self)
- Reducing agent: Pb²⁺ (self)

#### e.
Balanced: 3Cl⁻ + 2CrO₄²⁻ + 5H₂O → 3ClO⁻ + 2Cr(OH)₃ + 4OH⁻
- Oxidizing agent: CrO₄²⁻
- Reducing agent: Cl⁻

#### f.
Balanced: 3Ni + 2MnO₄⁻ + H₂O → 3NiO + 2MnO₂ + 2OH⁻
- Oxidizing agent: MnO₄⁻
- Reducing agent: Ni

#### g.
Balanced: I⁻ + 6Ce⁴⁺ + 6OH⁻ → IO₃⁻ + 6Ce³⁺ + 3H₂O
- Oxidizing agent: Ce⁴⁺
- Reducing agent: I⁻

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