Balancing Redox Reactions Worksheet for 10th - Higher Ed | Lesson ... - Free Printable

It looks like you've uploaded a worksheet titled "Balancing Redox Reactions Worksheet" with two sections: Acid Solutions and Basic Solutions. However, I don't see the actual image content in this chat — only the text of the worksheet.

To help you properly, I’ll assume you want me to balance the redox reactions listed in the worksheet. Below is a step-by-step explanation for balancing each reaction using the half-reaction method, which is standard for redox reactions in acidic or basic solutions.

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🔧 General Steps for Balancing Redox Reactions:

1. Assign oxidation states to identify what’s oxidized and reduced.
2. Split into half-reactions: oxidation and reduction.
3. Balance atoms (except O and H).
4. Balance O by adding H₂O.
5. Balance H by adding H⁺ (acidic) or OH⁻ (basic).
6. Balance charge by adding electrons (e⁻).
7. Equalize electrons in both half-reactions.
8. Add half-reactions together.
9. Simplify and cancel species that appear on both sides.
10. For basic solution, add OH⁻ to both sides to neutralize H⁺.

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Let’s go through each reaction one by one.

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## ✔ Acid Solutions

1. MnO₄⁻ + BrO₃⁻ → MnO₄²⁻ + Br₂

Step 1: Assign oxidation states
- Mn in MnO₄⁻: +7
- Mn in MnO₄²⁻: +6 → Reduced
- Br in BrO₃⁻: +5
- Br in Br₂: 0 → Oxidized

So:
- Reduction: MnO₄⁻ → MnO₄²⁻
- Oxidation: BrO₃⁻ → Br₂

Reduction Half-Reaction:
MnO₄⁻ → MnO₄²⁻
Already balanced for Mn.
Charge: -1 → -2 → Add 1 e⁻ to left:

→ MnO₄⁻ + e⁻ → MnO₄²⁻

Oxidation Half-Reaction:
BrO₃⁻ → Br₂
Balance Br: 2BrO₃⁻ → Br₂
Balance O: Add 6H₂O to right
Balance H: Add 12H⁺ to left
Balance charge: Add 10e⁻ to right

→ 2BrO₃⁻ + 12H⁺ + 10e⁻ → Br₂ + 6H₂O

Now balance electrons:
- Reduction: ×10 → 10MnO₄⁻ + 10e⁻ → 10MnO₄²⁻
- Oxidation: 2BrO₃⁻ + 12H⁺ + 10e⁻ → Br₂ + 6H₂O

Add:
10MnO₄⁻ + 2BrO₃⁻ + 12H⁺ → 10MnO₄²⁻ + Br₂ + 6H₂O

✔ Balanced in acid.

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2. MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻

Oxidation States:
- Mn: +7 → +4 (reduction)
- S: +4 → +6 (oxidation)

Reduction: MnO₄⁻ → MnO₂
Balance Mn: OK
Balance O: Add 2H₂O to right
Balance H: Add 4H⁺ to left
Balance charge: Add 3e⁻ to left

→ MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

Oxidation: SO₃²⁻ → SO₄²⁻
Add H₂O to left, H⁺ to right, e⁻ to right

SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Now equalize electrons:
- Reduction: ×2 → 2MnO₄⁻ + 8H⁺ + 6e⁻ → 2MnO₂ + 4H₂O
- Oxidation: ×3 → 3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻

Add:
2MnO₄⁻ + 3SO₃²⁻ + 2H⁺ + H₂O → 2MnO₂ + 3SO₄²⁻ + 4H₂O

Wait — simplify:
Cancel H₂O: 2H₂O on right, 3H₂O on left → net 1H₂O on left

Final:
2MnO₄⁻ + 3SO₃²⁻ + 2H⁺ → 2MnO₂ + 3SO₄²⁻ + H₂O

✔ Balanced in acid.

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3. Cr₂O₇²⁻ + Cl⁻ → Cr³⁺ + Cl₂

Cr: +6 → +3 (reduction)
Cl: -1 → 0 (oxidation)

Reduction: Cr₂O₇²⁻ → 2Cr³⁺
Balance O: Add 7H₂O to right
Balance H: Add 14H⁺ to left
Balance charge: Add 6e⁻ to left

→ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation: 2Cl⁻ → Cl₂ + 2e⁻

Multiply oxidation by 3: 6Cl⁻ → 3Cl₂ + 6e⁻

Add:
Cr₂O₇²⁻ + 14H⁺ + 6Cl⁻ → 2Cr³⁺ + 7H₂O + 3Cl₂

✔ Balanced.

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4. P + Cl₂ → POCl₃

This is not redox? Wait — P: 0 → +5 in POCl₃
Cl: 0 → -1

But it's written as P + Cl₂ → POCl₃ — missing oxygen.

Actually, this likely requires water or O₂. But let’s assume it's in acid and we need to balance.

Wait — probably typo. Standard reaction is:
P₄ + 6Cl₂ + 6H₂O → 4H₃PO₃ + 12HCl — but not matching.

Alternatively, perhaps:
P + O₂ + Cl₂ → POCl₃ — but not given.

Wait — maybe it's meant to be P₄ + Cl₂ → PCl₃, but here it's POCl₃.

Assume: P + O₂ + Cl₂ → POCl₃ — but not balanced.

Better: In acidic conditions, phosphorus can oxidize.

Try:
P + 3/2 O₂ + 3/2 Cl₂ → POCl₃ — but not helpful.

Wait — better to write:
P + 3Cl₂ + 1/2 O₂ → POCl₃ — still messy.

But since it's not a redox reaction in the usual sense unless O comes from H₂O, perhaps it's incomplete.

But if we assume P is oxidized, and Cl₂ is reduced, and O comes from H₂O, then:

Oxidation: P → POCl₃
But POCl₃ has P=+5, so:

P → POCl₃
Balance O: add H₂O to left
Balance Cl: add Cl⁻ to left
But no ions.

Alternative: Write as molecular equation.

But it's ambiguous. Perhaps it's meant to be:

P₄ + 6Cl₂ → 4PCl₃ — but that’s not POCl₃.

Given the context, this might be a typo. Let's skip for now.

Wait — another possibility:
P + 3Cl₂ + 1/2 O₂ → POCl₃ — but not balanced.

Or:
P + 3Cl₂ + H₂O → POCl₃ + 2HCl — check:

Left: P, 6Cl, 2H, 1O
Right: P, 3O, 3Cl, 2H, 1Cl → total 4Cl → no.

Try:
P + 3Cl₂ + 2H₂O → POCl₃ + 2HCl
Left: P, 6Cl, 4H, 2O
Right: P, 3O, 3Cl, 2H, 1Cl → 4Cl, 2H, 3O — mismatch.

Better:
P + 3Cl₂ + 3H₂O → POCl₃ + 6HCl — too many H.

No. Actually, standard way is:
P₄ + 6Cl₂ → 4PCl₃
Then PCl₃ + Cl₂ → PCl₅
But not POCl₃.

POCl₃ is made from P₄ + 6Cl₂ + 3O₂ → 4POCl₃

So:
P₄ + 6Cl₂ + 3O₂ → 4POCl₃

But original says: P + Cl₂ → POCl₃

So scale:
P + 3/2 Cl₂ + 3/4 O₂ → POCl₃ — not nice.

Perhaps in acid, with H₂O?

But without more info, this reaction is incomplete. Skip for now.

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5. PbO₂ + I⁻ → PbI₂ + IO₃⁻

Pb: +4 → +2 (reduction)
I: -1 → +5 (oxidation)

Reduction: PbO₂ → PbI₂
But PbI₂ is solid, so write as Pb²⁺?

Better: PbO₂ → Pb²⁺
Balance O: add 2H₂O to right
Balance H: add 4H⁺ to left
Balance charge: add 2e⁻ to left

→ PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O

Oxidation: I⁻ → IO₃⁻
Balance I: OK
Balance O: add 3H₂O to left
Balance H: add 6H⁺ to right
Balance charge: add 6e⁻ to right

→ I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻

Equalize electrons:
Reduction ×3: 3PbO₂ + 12H⁺ + 6e⁻ → 3Pb²⁺ + 6H₂O
Oxidation: I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻

Add:
3PbO₂ + I⁻ + 12H⁺ + 3H₂O → 3Pb²⁺ + 6H₂O + IO₃⁻ + 6H⁺

Simplify:
3PbO₂ + I⁻ + 6H⁺ → 3Pb²⁺ + IO₃⁻ + 3H₂O

Now, Pb²⁺ + 2I⁻ → PbI₂(s), but I⁻ is already used.

So instead, use PbI₂ directly.

So write:
3PbO₂ + I⁻ + 6H⁺ → 3Pb²⁺ + IO₃⁻ + 3H₂O
Then: 3Pb²⁺ + 6I⁻ → 3PbI₂(s)

Total I⁻: 1 + 6 = 7 I⁻

So final:
3PbO₂ + 7I⁻ + 6H⁺ → 3PbI₂(s) + IO₃⁻ + 3H₂O

✔ Balanced.

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6. NO₂ → NO + NO₃⁻

NO₂ → NO and NO₃⁻ — disproportionation.

N in NO₂: +4
In NO: +2
In NO₃⁻: +5

So some N reduced, some oxidized.

Let’s say:
a NO₂ → b NO + c NO₃⁻

Balance N: a = b + c
O: 2a = b + 3c

From N: b = a - c
Plug into O: 2a = (a - c) + 3c = a + 2c
→ 2a = a + 2c → a = 2c

So c = a/2, b = a - a/2 = a/2

So ratio: a : b : c = 2 : 1 : 1

So: 2NO₂ → NO + NO₃⁻

But charge: left neutral, right NO₃⁻ has -1

So add H⁺ and H₂O.

In acid:
2NO₂ → NO + NO₃⁻

Balance O: left 4, right 1 + 3 = 4 → OK
H: none → add H⁺ and H₂O

But no H — so must involve H⁺.

Write half-reactions.

Reduction: NO₂ → NO
NO₂ → NO
Add H₂O to right: NO₂ → NO + H₂O
Add H⁺ to left: NO₂ + 2H⁺ → NO + H₂O
Balance charge: add e⁻ to left: NO₂ + 2H⁺ + e⁻ → NO + H₂O

Oxidation: NO₂ → NO₃⁻
NO₂ → NO₃⁻
Add H₂O to left: H₂O + NO₂ → NO₃⁻
Add H⁺ to right: H₂O + NO₂ → NO₃⁻ + 2H⁺
Balance charge: add e⁻ to right: H₂O + NO₂ → NO₃⁻ + 2H⁺ + e⁻

Now add:
Reduction: NO₂ + 2H⁺ + e⁻ → NO + H₂O
Oxidation: H₂O + NO₂ → NO₃⁻ + 2H⁺ + e⁻

Add:
2NO₂ → NO + NO₃⁻ + H₂O

Wait — H₂O and H⁺ cancel?

Add:
Left: NO₂ + 2H⁺ + e⁻ + H₂O + NO₂
Right: NO + H₂O + NO₃⁻ + 2H⁺ + e⁻

Cancel: H₂O, H⁺, e⁻

→ 2NO₂ → NO + NO₃⁻

But charge: left 0, right -1 → need H⁺

So add H⁺ to left? But no.

Wait — in acid, we can have:

From above, after cancellation:
2NO₂ → NO + NO₃⁻

But charge imbalance.

Wait — actually, from oxidation: produces 2H⁺
Reduction consumes 2H⁺

So they cancel.

But NO₃⁻ has -1 charge, so overall right side: -1

Left: neutral → not balanced.

Wait — mistake.

In oxidation: H₂O + NO₂ → NO₃⁻ + 2H⁺ + e⁻ → charge: 0 → -1 + 2 + (-1) = 0 → OK

Reduction: NO₂ + 2H⁺ + e⁻ → NO + H₂O → charge: 0 + 2 -1 = +1 → right: 0 → not balanced!

Wait: NO₂ (neutral) + 2H⁺ (+2) + e⁻ (-1) → total +1
Right: NO (0) + H₂O (0) → 0 → not balanced.

Error.

Correct reduction:

NO₂ → NO
N: +4 → +2 → gain 2e⁻

NO₂ → NO
Add H₂O to right: NO₂ → NO + H₂O
Add H⁺ to left: NO₂ + 2H⁺ → NO + H₂O
Balance charge: left: +2, right: 0 → add 2e⁻ to left

→ NO₂ + 2H⁺ + 2e⁻ → NO + H₂O

Oxidation: NO₂ → NO₃⁻
N: +4 → +5 → lose 1e⁻

NO₂ → NO₃⁻
Add H₂O to left: H₂O + NO₂ → NO₃⁻
Add H⁺ to right: H₂O + NO₂ → NO₃⁻ + 2H⁺
Add e⁻ to right: H₂O + NO₂ → NO₃⁻ + 2H⁺ + e⁻

Now balance electrons.

Reduction: ×1 → NO₂ + 2H⁺ + 2e⁻ → NO + H₂O
Oxidation: ×2 → 2H₂O + 2NO₂ → 2NO₃⁻ + 4H⁺ + 2e⁻

Add:
NO₂ + 2H⁺ + 2e⁻ + 2H₂O + 2NO₂ → NO + H₂O + 2NO₃⁻ + 4H⁺ + 2e⁻

Simplify:
3NO₂ + 2H₂O + 2H⁺ → NO + 2NO₃⁻ + H₂O + 4H⁺

Cancel H₂O: 2H₂O - H₂O = H₂O on left
H⁺: 2H⁺ - 4H⁺ = -2H⁺ → move to right

So:
3NO₂ + H₂O → NO + 2NO₃⁻ + 2H⁺

Check atoms:
N: 3 = 1 + 2 → OK
O: 6 + 1 = 7; right: 1 + 6 + 0 = 7 → OK
H: 2 = 2 → OK
Charge: left 0, right: -2 + 2 = 0 → OK

✔ Balanced:
3NO₂ + H₂O → NO + 2NO₃⁻ + 2H⁺

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## ✔ Basic Solutions

1. MnO₄⁻ + C₂O₄²⁻ → MnO₂ + CO₂

Mn: +7 → +4 (reduction)
C: +3 → +4 (oxidation)

Reduction: MnO₄⁻ → MnO₂
In base:
MnO₄⁻ → MnO₂
Add H₂O to right: MnO₄⁻ → MnO₂ + 2H₂O
Add H⁺ to left? No — in base, use OH⁻.

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

Oxidation: C₂O₄²⁻ → 2CO₂
C₂O₄²⁻ → 2CO₂
Balance: OK
Add 2e⁻ to right

→ C₂O₄²⁻ → 2CO₂ + 2e⁻

Equalize electrons:
Reduction ×2: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
Oxidation ×3: 3C₂O₄²⁻ → 6CO₂ + 6e⁻

Add:
2MnO₄⁻ + 3C₂O₄²⁻ + 4H₂O → 2MnO₂ + 6CO₂ + 8OH⁻

✔ Balanced in base.

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2. Cu(OH)₂ + N₂O₄ → CuO + NO₂⁻ + Cu²⁺

Cu(OH)₂ → CuO: Cu²⁺ → Cu²⁺? Same oxidation state?

Wait: Cu(OH)₂ → CuO + H₂O — decomposition, not redox.

But also Cu²⁺ produced — so perhaps Cu is oxidized?

Wait: Cu(OH)₂ → CuO: Cu remains +2
But Cu²⁺ is also +2 — so no change.

N₂O₄: N is +4
NO₂⁻: N is +3 → reduced

So N₂O₄ is reduced, but what is oxidized?

Cu(OH)₂ → CuO + H₂O — not redox.

But there's Cu²⁺ — perhaps Cu is being oxidized?

Wait — maybe it's:
Cu(OH)₂ → Cu²⁺ + 2OH⁻ — but that’s just dissociation.

But Cu²⁺ and CuO are both present.

Possibly:
Some Cu(OH)₂ → CuO (decomposition)
Other Cu(OH)₂ → Cu²⁺ (dissolution)

But no redox.

But N₂O₄ → NO₂⁻: N from +4 to +3 — reduction → needs oxidation.

So Cu must be oxidized — but Cu is already +2.

Unless Cu(OH)₂ is oxidized to CuO₂ or something.

Wait — perhaps it's a typo.

Maybe:
Cu(OH)₂ → CuO + H₂O (no redox)
N₂O₄ → 2NO₂⁻ — but N₂O₄ + 2e⁻ → 2NO₂⁻ — possible.

But where does electron come from?

Perhaps Cu(OH)₂ provides H⁺ or something.

Wait — maybe it's:
Cu(OH)₂ → Cu²⁺ + 2OH⁻
Then Cu²⁺ + 2OH⁻ → Cu(OH)₂ — no.

Alternatively, perhaps it's a disproportionation of N₂O₄?

N₂O₄ ⇌ 2NO₂
But not helpful.

Perhaps:
N₂O₄ + 2OH⁻ → 2NO₂⁻ + H₂O + 1/2 O₂ — but not matching.

Alternatively, maybe Cu is oxidized to Cu²⁺, but Cu(OH)₂ already has Cu²⁺.

Wait — unless it's Cu⁺ in Cu(OH)₂ — but no.

This seems problematic.

Perhaps it's:
Cu(OH)₂ → CuO + H₂O
And N₂O₄ → 2NO₂⁻ + 2H⁺ — but in base, H⁺ not allowed.

In base:
N₂O₄ + 2OH⁻ → 2NO₂⁻ + H₂O + 1/2 O₂ — but O₂ not in products.

Alternatively:
N₂O₄ + 2e⁻ + 2H₂O → 2NO₂⁻ + 4H⁺ — but in base, add 4OH⁻ to both sides:

N₂O₄ + 2e⁻ + 2H₂O + 4OH⁻ → 2NO₂⁻ + 4H⁺ + 4OH⁻ → 2NO₂⁻ + 4H₂O

So: N₂O₄ + 2e⁻ + 4OH⁻ → 2NO₂⁻ + 2H₂O

Now, what is oxidized? If Cu(OH)₂ → CuO + H₂O, no redox.

But if Cu(OH)₂ → CuO₂⁻ or something — not possible.

Wait — perhaps it's:
Cu(OH)₂ → CuO + H₂O — no change
But Cu²⁺ is formed — so maybe it's not redox.

But the problem says "redox", so likely error.

Skip for now.

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3. Zn → Zn(OH)₄²⁻ + H₂

Zn → Zn(OH)₄²⁻: Zn⁰ → Zn²⁺ (oxidation)
H⁺ → H₂: reduction

But in base, H⁺ not present.

So use H₂O.

Oxidation: Zn → Zn(OH)₄²⁻
Zn → Zn(OH)₄²⁻
Add 4OH⁻ to left: Zn + 4OH⁻ → Zn(OH)₄²⁻
Balance charge: add 2e⁻ to right

→ Zn + 4OH⁻ → Zn(OH)₄²⁻ + 2e⁻

Reduction: H₂O → H₂
2H₂O + 2e⁻ → H₂ + 2OH⁻

Add:
Zn + 4OH⁻ + 2H₂O + 2e⁻ → Zn(OH)₄²⁻ + 2e⁻ + H₂ + 2OH⁻

Cancel: 4OH⁻ - 2OH⁻ = 2OH⁻ on left
2H₂O on left

Final:
Zn + 2H₂O + 2OH⁻ → Zn(OH)₄²⁻ + H₂

✔ Balanced in base.

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## 📝 Summary of Balanced Equations:

Acid Solutions:

1. 10MnO₄⁻ + 2BrO₃⁻ + 12H⁺ → 10MnO₄²⁻ + Br₂ + 6H₂O
2. 2MnO₄⁻ + 3SO₃²⁻ + 2H⁺ → 2MnO₂ + 3SO₄²⁻ + H₂O
3. Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ → 2Cr³⁺ + 3Cl₂ + 7H₂O
4. [Incomplete] — possibly typo
5. 3PbO₂ + 7I⁻ + 6H⁺ → 3PbI₂(s) + IO₃⁻ + 3H₂O
6. 3NO₂ + H₂O → NO + 2NO₃⁻ + 2H⁺

Basic Solutions:

1. 2MnO₄⁻ + 3C₂O₄²⁻ + 4H₂O → 2MnO₂ + 6CO₂ + 8OH⁻
2. [Unclear — may need clarification]
3. Zn + 2H₂O + 2OH⁻ → Zn(OH)₄²⁻ + H₂

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If you can upload the image again or clarify any specific reaction, I can give a more precise answer!