Solved Unit 5, Activity 6 Balancing Redox Reactions | 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 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
- That uses 3 Mg
- Remaining 3 Mg → form MgO

Each MgO needs 1 Mg and 1 O
- B₂O₃ provides 3 O → can make 3 MgO

So:
- B₂O₃ + 6 Mg → 3 MgO + Mg₃B₂

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

✔ Balanced.

Electron transfer:
- B reduced: +3 → -3 (6 e⁻ per B), 2 B → 12 e⁻ gained
- Mg oxidized: 0 → +2 (2 e⁻ per Mg), 6 Mg → 12 e⁻ lost

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

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

Assign oxidation numbers:
- Cr₂O₇²⁻: O = -2, total charge = -2
- 7×(-2) = -14 → 2 Cr must be +12 → each Cr = +6
- Fe²⁺ = +2
- Cr³⁺ = +3
- Fe³⁺ = +3

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 to balance electrons:
- 6 e⁻ gained → need 6 Fe²⁺

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

Balance charge:
- Left: -2 + 6×(+2) = -2 + 12 = +10
- Right: 2×(+3) + 6×(+3) = 6 + 18 = +24 → not balanced

Need to add H⁺ and H₂O since acidic.

We know Cr₂O₇²⁻ reduces to Cr³⁺ in acid:
- Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Fe²⁺ → Fe³⁺ + e⁻ → multiply by 6: 6Fe²⁺ → 6Fe³⁺ + 6e⁻

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

Now check:
- Atoms: Cr: 2=2, O: 7=7, H: 14=14, Fe: 6=6 ✔️
- Charge: Left: -2 + 14 + 6×(+2) = -2+14+12 = +24
Right: 2×(+3) + 6×(+3) = 6+18 = +24 ✔️

✔ 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, so N = +5)
- IO₃⁻: O = -2, total = -1 → 3×(-2)= -6 → I = +5
- NO₂: O = -2, total neutral → N = +4

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

But I₂ has two I atoms → each I₂ loses 10 e⁻ (2×5)

Each NO₃⁻ gains 1 e⁻ → need 10 NO₃⁻

So:
- I₂ + 10 NO₃⁻ → 2 IO₃⁻ + 10 NO₂

Now balance O and H.

Right: 2 IO₃⁻ → 6 O, 10 NO₂ → 20 O → total 26 O
Left: 10 NO₃⁻ → 30 O → too many on left

Wait — we need to add H⁺ and H₂O.

Use half-reactions.

Oxidation: I₂ → IO₃⁻

I₂ → 2IO₃⁻
Balance I: already good
Balance O: add 6 H₂O to right? No — need to add water to left?

Standard way:
I₂ → 2IO₃⁻
Add 6 H₂O to left to supply O: I₂ + 6H₂O → 2IO₃⁻
Now balance H: 12 H on left → add 12 H⁺ to right
Charge: left = 0, right = 2×(-1) + 12×(+1) = -2 + 12 = +10 → add 10 e⁻ to right

So:
I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻

Reduction: NO₃⁻ → NO₂

NO₃⁻ → NO₂
Add H⁺: NO₃⁻ + 2H⁺ → NO₂ + H₂O
Balance charge: left: -1 + 2 = +1, right: 0 → add 1 e⁻ to left

So:
NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O

Multiply reduction by 10:
10NO₃⁻ + 20H⁺ + 10e⁻ → 10NO₂ + 10H₂O

Now add to oxidation:
I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻
+
10NO₃⁻ + 20H⁺ + 10e⁻ → 10NO₂ + 10H₂O
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I₂ + 10NO₃⁻ + 6H₂O + 20H⁺ → 2IO₃⁻ + 10NO₂ + 10H₂O + 12H⁺

Simplify:
Cancel 6H₂O from both sides → 4H₂O on right
Cancel H⁺: 20H⁺ - 12H⁺ = 8H⁺ on left

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

Check atoms:
- I: 2 = 2 ✔️
- N: 10 = 10 ✔️
- O: 30 + 8 = 38; right: 6 (from IO₃⁻) + 20 (NO₂) + 4 (H₂O) = 30 → wait! Mismatch

Wait: left: 10 NO₃⁻ → 30 O, 8 H⁺ → no O
Total O: 30
Right: 2 IO₃⁻ → 6 O, 10 NO₂ → 20 O, 4 H₂O → 4 O → total 30 ✔️
H: 8 H⁺ → 8 H; 4 H₂O → 8 H ✔️
Charge: left: 0 + 10×(-1) + 8×(+1) = -10 + 8 = -2
Right: 2×(-1) + 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(II) to Pb(0) and Pb(IV).

PbSO₄ contains Pb²⁺ and SO₄²⁻

Products: Pb (0), PbO₂ (Pb⁴⁺), SO₄²⁻

So:
- Pb²⁺ → Pb⁰ → reduction (gain 2 e⁻)
- Pb²⁺ → Pb⁴⁺ → oxidation (lose 2 e⁻)

So one Pb²⁺ reduced, one oxidized → 1:1 ratio

But each PbSO₄ has one Pb and one SO₄²⁻

So need 2 PbSO₄: one gives Pb, one gives PbO₂

So:
2 PbSO₄ → Pb + PbO₂ + 2 SO₄²⁻

But now check O and H.

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

Also, PbO₂ has 2 O, but where do they come from?

In acidic solution, we may need H⁺ and H₂O.

But PbSO₄ is solid, and products are Pb(s), PbO₂(s), SO₄²⁻(aq)

Let’s assign oxidation states:

- Pb in PbSO₄: +2
- Pb in Pb: 0
- Pb in PbO₂: +4
- SO₄²⁻ unchanged: S = +6

So:
- One Pb²⁺ → Pb⁰ → gain 2 e⁻
- One Pb²⁺ → Pb⁴⁺ → lose 2 e⁻

So equal electrons.

So:
PbSO₄ → Pb + PbO₂ + SO₄²⁻ — but only one Pb on left

Need two PbSO₄

Try:
2 PbSO₄ → Pb + PbO₂ + 2 SO₄²⁻

Atoms:
- Pb: 2 = 1 + 1 ✔️
- S: 2 = 2 ✔️
- O: 8 = ? PbO₂ has 2 O, 2 SO₄²⁻ have 8 O → total 10 O → too many

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

So need to remove O — use H⁺ to form H₂O

Add 2 H⁺ to right to react with extra O?

Better: PbO₂ is formed from Pb²⁺ + 2 H₂O → PbO₂ + 4 H⁺ + 2 e⁻ (oxidation)

And Pb²⁺ + 2 e⁻ → Pb (reduction)

So combine:

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

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

So: 2 Pb²⁺ + 2H₂O → Pb + PbO₂ + 4H⁺

Now add SO₄²⁻ as spectator:

From 2 PbSO₄ → 2 Pb²⁺ + 2 SO₄²⁻

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

Check atoms:
- Pb: 2 = 1 + 1 ✔️
- S: 2 = 2 ✔️
- O: 8 (from SO₄) + 2 (H₂O) = 10; right: PbO₂ (2) + 2 SO₄²⁻ (8) = 10 ✔️
- H: 4 = 4 ✔️

Charge:
- Left: 0 (solids) + 0 (H₂O) = 0
- Right: 2×(-2) + 4×(+1) = -4 + 4 = 0 ✔️

✔ Balanced.

Oxidizing agent: Pb²⁺ (in PbSO₄) — it gets reduced
Reducing agent: Pb²⁺ — it gets oxidized
→ So Pb²⁺ is both oxidizing and reducing agent (disproportionation)

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

First, assign oxidation numbers:

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

So:
- Cl: -1 → +1 → oxidation, loses 2 e⁻
- Cr: +6 → +3 → reduction, gains 3 e⁻

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

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

Now balance O and H in basic solution.

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

Also, need to balance H.

Use half-reactions.

Oxidation: Cl⁻ → ClO⁻

Cl⁻ → ClO⁻
Add H₂O to left: Cl⁻ + H₂O → ClO⁻
Balance O: done
Balance H: add 2H⁺ to right → Cl⁻ + H₂O → ClO⁻ + 2H⁺
But in basic, convert H⁺ to OH⁻: add 2OH⁻ to both sides:

Cl⁻ + H₂O + 2OH⁻ → ClO⁻ + 2H⁺ + 2OH⁻ → ClO⁻ + 2H₂O

So: Cl⁻ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻

Reduction: CrO₄²⁻ → CrO₂⁻

CrO₄²⁻ → CrO₂⁻
Balance O: add 2 H₂O to right → CrO₄²⁻ → CrO₂⁻ + 2H₂O
Balance H: add 4H⁺ to left → CrO₄²⁻ + 4H⁺ → CrO₂⁻ + 2H₂O
Balance charge: left: -2 + 4 = +2, right: -1 → add 3e⁻ to left

So: CrO₄²⁻ + 4H⁺ + 3e⁻ → CrO₂⁻ + 2H₂O

Now convert to basic: add 4OH⁻ to both sides:

CrO₄²⁻ + 4H⁺ + 4OH⁻ + 3e⁻ → CrO₂⁻ + 2H₂O + 4OH⁻
→ CrO₄²⁻ + 4H₂O + 3e⁻ → CrO₂⁻ + 2H₂O + 4OH⁻
Simplify: CrO₄²⁻ + 2H₂O + 3e⁻ → CrO₂⁻ + 4OH⁻

Now multiply oxidation by 3, reduction by 2:

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⁻ cancel
- H₂O: 4H₂O - 3H₂O = 1H₂O on left
- OH⁻: 6OH⁻ - 8OH⁻ = -2OH⁻ → move to right

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

Check atoms:
- Cl: 3 = 3 ✔️
- Cr: 2 = 2 ✔️
- O: left: 8 (CrO₄) + 1 (H₂O) = 9; right: 3 (ClO⁻) + 4 (CrO₂⁻) + 2 (OH⁻) = 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 states:

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

So:
- Ni: 0 → +2 → oxidation, loses 2 e⁻
- Mn: +7 → +4 → reduction, gains 3 e⁻

LCM: 6 → 3 Ni and 2 MnO₄⁻

So:
3Ni + 2MnO₄⁻ → 3NiO + 2MnO₂

Now balance O and H in basic.

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

Also, need H.

Half-reactions.

Oxidation: Ni → NiO

Ni → NiO
Add H₂O to right: Ni → NiO
Add H⁺ to left: Ni + H₂O → NiO + 2H⁺
Then in basic: add 2OH⁻ to both sides:

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

Reduction: MnO₄⁻ → MnO₂

MnO₄⁻ → MnO₂
Add 2H₂O to right: MnO₄⁻ → MnO₂ + 2H₂O
Add 4H⁺ to left: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O
Add 3e⁻ to left: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

Convert to basic: add 4OH⁻ to both sides:

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

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

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

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

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

Check:
- Ni: 3 = 3 ✔️
- Mn: 2 = 2 ✔️
- O: left: 8 (MnO₄) + 1 = 9; right: 3 + 4 + 2 = 9 ✔️
- H: 2 = 2 ✔️
- 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)

Oxidation states:

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

So:
- I: -1 → +5 → oxidation, loses 6 e⁻
- Ce: +4 → +3 → reduction, gains 1 e⁻

So need 1 I⁻ and 6 Ce⁴⁺

So:
I⁻ + 6Ce⁴⁺ → IO₃⁻ + 6Ce³⁺

Now balance O and H in basic.

Left: no O, no H
Right: IO₃⁻ has 3 O

So need to add H₂O and OH⁻.

Half-reactions.

Oxidation: I⁻ → IO₃⁻

I⁻ → IO₃⁻
Add 3H₂O to left: I⁻ + 3H₂O → IO₃⁻
Add 6H⁺ to right: I⁻ + 3H₂O → IO₃⁻ + 6H⁺
Add 6e⁻ to right: I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻

Convert to basic: add 6OH⁻ to both sides:

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

Reduction: Ce⁴⁺ → Ce³⁺

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:
- I: 1 = 1 ✔️
- Ce: 6 = 6 ✔️
- O: 6 (OH⁻) = 3 (IO₃⁻) + 3 (H₂O) = 6 ✔️
- H: 6 = 6 ✔️
- 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⁺ (in PbSO₄)
- Reducing agent: Pb²⁺ (in PbSO₄)
*(Disproportionation)*

#### e.
Balanced: 3Cl⁻ + 2CrO₄²⁻ + H₂O → 3ClO⁻ + 2CrO₂⁻ + 2OH⁻
- 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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Let me know if you'd like these written neatly in a table format!