Balancing Redox Reactions Worksheet for 10th - 12th Grade | Lesson ... - Free Printable

It looks like you've uploaded a worksheet titled "Balancing Redox Reactions" with six redox reactions that need to be balanced. Since I can't view images directly, I’ll solve each of the redox reactions listed in your text and explain the steps.

Let’s go through each reaction one by one, balancing them using the half-reaction method (for acidic or basic conditions as needed). Here are the reactions:

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1. SO₂(g) + HNO₃(aq) → H₂SO₄(aq) + NO(g)

#### Step 1: Assign oxidation states
- In SO₂: S = +4
- In H₂SO₄: S = +6 → S is oxidized
- In HNO₃: N = +5
- In NO: N = +2 → N is reduced

So:
- Oxidation: S⁴⁺ → S⁶⁺ + 2e⁻
- Reduction: N⁵⁺ + 3e⁻ → N²⁺

To balance electrons:
- Multiply oxidation by 3: 3S⁴⁺ → 3S⁶⁺ + 6e⁻
- Multiply reduction by 2: 2N⁵⁺ + 6e⁻ → 2N²⁺

Now balance atoms:
- 3SO₂ → 3H₂SO₄ (need H and O)
- 2HNO₃ → 2NO

But we need to balance H and O using H⁺ and H₂O (acidic medium).

#### Half-reactions:
Oxidation:
SO₂ → SO₄²⁻
Balance O: add 2H₂O to left
SO₂ + 2H₂O → SO₄²⁻
Balance H: add 4H⁺ to right
SO₂ + 2H₂O → SO₄²⁻ + 4H⁺
Balance charge: add 2e⁻ to right
SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻

Reduction:
NO₃⁻ → NO
Balance O: add 2H₂O to right
NO₃⁻ → NO + 2H₂O
Balance H: add 4H⁺ to left
NO₃⁻ + 4H⁺ → NO + 2H₂O
Balance charge: add 3e⁻ to left
NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O

Now make electrons equal:
Multiply oxidation by 3:
3SO₂ + 6H₂O → 3SO₄²⁻ + 12H⁺ + 6e⁻

Multiply reduction by 2:
2NO₃⁻ + 8H⁺ + 6e⁻ → 2NO + 4H₂O

Add together:
3SO₂ + 6H₂O + 2NO₃⁻ + 8H⁺ + 6e⁻ → 3SO₄²⁻ + 12H⁺ + 6e⁻ + 2NO + 4H₂O

Cancel:
- 6e⁻ cancels
- 6H₂O – 4H₂O = 2H₂O on left
- 8H⁺ – 12H⁺ = -4H⁺ → move to right as 4H⁺

Final ionic equation:
3SO₂ + 2NO₃⁻ + 2H₂O → 3SO₄²⁻ + 2NO + 4H⁺

Now add H⁺ and spectator ions (H⁺ from HNO₃ and H₂SO₄):
From HNO₃: 2HNO₃ → 2NO₃⁻ + 2H⁺
We have 2NO₃⁻ and need 4H⁺ → so use 4H⁺ total.

Add 2H⁺ from H₂SO₄? But H₂SO₄ produces 2H⁺ per molecule.

Better to write molecular form:

Try:
3SO₂ + 2HNO₃ + H₂O → 3H₂SO₄ + 2NO

Check atoms:
Left: S=3, O=6+6+1=13, H=2+2=4, N=2
Right: S=3, O=12+2=14, H=6, N=2 → not balanced

Wait — better approach:

From ionic:
3SO₂ + 2NO₃⁻ + 2H₂O → 3SO₄²⁻ + 2NO + 4H⁺

To make neutral, add 4H⁺ to both sides? No — instead, add 2H₂SO₄ to provide SO₄²⁻ and H⁺.

Alternatively, use:
3SO₂ + 2HNO₃ + H₂O → 3H₂SO₄ + 2NO

But check H: left = 2 (from HNO₃) + 2 (from H₂O) = 4H
Right = 6H → no

Try:
3SO₂ + 2HNO₃ + 2H₂O → 3H₂SO₄ + 2NO

Left: H = 2 (HNO₃) + 4 (2H₂O) = 6H
Right: 6H → good
O: 6 (SO₂) + 6 (HNO₃) + 2 (H₂O) = 14
Right: 12 (H₂SO₄) + 2 (NO) = 14 → good
S: 3 = 3
N: 2 = 2

✔ Balanced!

Answer:
> 3SO₂(g) + 2HNO₃(aq) + 2H₂O(l) → 3H₂SO₄(aq) + 2NO(g)

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2. Al(s) + H₂SO₄(aq) → Al₂(SO₄)₃(aq) + H₂(g)

This is a single displacement.

Al → Al³⁺ + 3e⁻
2H⁺ + 2e⁻ → H₂

Balance electrons:
- Multiply Al by 2: 2Al → 2Al³⁺ + 6e⁻
- Multiply H⁺ by 3: 6H⁺ + 6e⁻ → 3H₂

So:
2Al + 6H⁺ → 2Al³⁺ + 3H₂

Now add SO₄²⁻ to both sides:
Al₂(SO₄)₃ has 3 SO₄²⁻, so need 3 SO₄²⁻ on both sides.

From H₂SO₄: provides 2H⁺ and SO₄²⁻ per molecule

So for 6H⁺ → need 3 H₂SO₄

Then:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Check:
Left: Al=2, H=6, S=3, O=12
Right: Al=2, S=3, O=12, H=6 → ✔

Answer:
> 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

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3. Au³⁺(aq) + I⁻(aq) → Au(s) + I₂(s)

Au³⁺ → Au: gains 3e⁻
I⁻ → I₂: loses 1e⁻ per I, so 2I⁻ → I₂ + 2e⁻

Balance electrons:
- Multiply Au³⁺ by 2: 2Au³⁺ + 6e⁻ → 2Au
- Multiply I⁻ by 3: 6I⁻ → 3I₂ + 6e⁻

So:
2Au³⁺ + 6I⁻ → 2Au + 3I₂

Answer:
> 2Au³⁺(aq) + 6I⁻(aq) → 2Au(s) + 3I₂(s)

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4. IO⁻(aq) + I⁻(aq) → I₂(s)

IO⁻ is iodate ion? Actually, IO⁻ is hypoiodite.

I in IO⁻: O is -2, so I = +1
I⁻: I = -1
I₂: I = 0

So:
- I⁻ → I₂: oxidation: 2I⁻ → I₂ + 2e⁻
- IO⁻ → I₂: reduction: IO⁻ + H⁺ + e⁻ → ½I₂ + H₂O (in acid)

Use acidic medium.

Reduction half:
IO⁻ → I₂
Balance I: 2IO⁻ → I₂
Balance O: add 2H₂O to right
2IO⁻ → I₂ + 2H₂O
Balance H: add 4H⁺ to left
2IO⁻ + 4H⁺ → I₂ + 2H₂O
Balance charge: left = 2(-1) + 4(+1) = +2, right = 0 → add 2e⁻ to left
2IO⁻ + 4H⁺ + 2e⁻ → I₂ + 2H₂O

Oxidation half:
2I⁻ → I₂ + 2e⁻

Add:
2IO⁻ + 4H⁺ + 2e⁻ + 2I⁻ → I₂ + 2H₂O + I₂ + 2e⁻

Cancel 2e⁻:
2IO⁻ + 4H⁺ + 2I⁻ → 2I₂ + 2H₂O

Divide by 2:
IO⁻ + 2H⁺ + I⁻ → I₂ + H₂O

But we don’t have H⁺ in original. Is this possible?

In neutral/basic conditions, it may involve OH⁻.

Alternatively, write in base.

But the problem doesn’t specify. Let's assume acidic.

But if no H⁺ given, maybe it's implied.

Alternatively, use water.

But standard way: the reaction is known as:
IO⁻ + I⁻ + H⁺ → I₂ + H₂O

But since H⁺ isn’t in original, perhaps we need to include it.

Wait — actually, in many cases, this reaction occurs in neutral or slightly acidic conditions.

But to balance, we must include H⁺.

So final balanced equation:
IO⁻(aq) + I⁻(aq) + 2H⁺(aq) → I₂(s) + H₂O(l)

But wait — we had 2IO⁻ above.

Let’s recheck.

From earlier:
2IO⁻ + 4H⁺ + 2I⁻ → 2I₂ + 2H₂O

So divide by 2:
IO⁻ + 2H⁺ + I⁻ → I₂ + H₂O

Yes.

But if H⁺ is not provided, then perhaps the reaction requires acidic conditions.

So unless specified otherwise, we include H⁺.

Answer:
> IO⁻(aq) + I⁻(aq) + 2H⁺(aq) → I₂(s) + H₂O(l)

If the problem expects only the species listed, it might be incomplete.

But chemically, this is correct.

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5. MnO₄⁻(aq) + CH₃OH(l) → CO₂(g) + Mn²⁺(aq)

This is a classic redox reaction in acidic solution.

MnO₄⁻ → Mn²⁺: Mn goes from +7 to +2 → gains 5e⁻
CH₃OH → CO₂: C goes from -2 to +4 → loses 6e⁻

Find LCM of 5 and 6 = 30

So:
- 6MnO₄⁻ → 6Mn²⁺ + 30e⁻
- 5CH₃OH → 5CO₂ + 30e⁻

Now balance each half-reaction.

Reduction: MnO₄⁻ → Mn²⁺
Add 4H₂O to right: MnO₄⁻ → Mn²⁺ + 4H₂O
Add 8H⁺ to left: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Add 5e⁻ to left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation: CH₃OH → CO₂
Balance C: already
Balance O: add H₂O to left? Wait, CH₃OH has 1O, CO₂ has 2O → add H₂O to left? No.

Better: CH₃OH → CO₂
Add H₂O to right? No.

Balance O: add 1H₂O to left? Not helpful.

Standard: CH₃OH → CO₂
Add H₂O to right to balance O? No.

Actually: CH₃OH → CO₂
C: 1
O: 1 vs 2 → add H₂O to left? No.

Better: CH₃OH → CO₂
Balance O: add 1H₂O to left? Still messy.

Use: CH₃OH → CO₂
Add H₂O to left to supply oxygen? No.

Instead:
CH₃OH → CO₂
Balance O: add 1H₂O to left? No.

Correct way:
CH₃OH → CO₂
Add H₂O to right? No.

Balance atoms:
- C: balanced
- O: left=1, right=2 → add 1H₂O to left? No.

Actually, CH₃OH → CO₂ + 4H⁺ + 4e⁻? But carbon oxidation state change.

C in CH₃OH: H is +1, O is -2, so C = -2
In CO₂: C = +4 → change of 6e⁻ per C

So:
CH₃OH → CO₂
Balance O: add H₂O to right? No.

Standard half-reaction:
CH₃OH → CO₂
Add H₂O to left to provide oxygen? No.

Better:
CH₃OH → CO₂
Balance O: add 1H₂O to left? No.

Actually:
CH₃OH → CO₂
Add H₂O to right? No.

Correct:
CH₃OH → CO₂
Balance O: left=1, right=2 → add 1H₂O to left? No.

Use:
CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻

Yes! That's standard.

Because:
- C: 1
- O: 1 (CH₃OH) + 1 (H₂O) = 2 → CO₂ has 2 → good
- H: 4 (CH₃OH) + 2 (H₂O) = 6 → 6H⁺ → good
- Charge: left=0, right=6+ → add 6e⁻ to right

So:
CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻

Now multiply:
- Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O ×6 → 6MnO₄⁻ + 48H⁺ + 30e⁻ → 6Mn²⁺ + 24H₂O
- Oxidation: CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻ ×5 → 5CH₃OH + 5H₂O → 5CO₂ + 30H⁺ + 30e⁻

Add:
6MnO₄⁻ + 48H⁺ + 30e⁻ + 5CH₃OH + 5H₂O → 6Mn²⁺ + 24H₂O + 5CO₂ + 30H⁺ + 30e⁻

Cancel:
- 30e⁻ cancel
- H⁺: 48 - 30 = 18H⁺ on left
- H₂O: 5 on left, 24 on right → net 19H₂O on right

So:
6MnO₄⁻ + 5CH₃OH + 18H⁺ → 6Mn²⁺ + 5CO₂ + 19H₂O

Check atoms:
Left: Mn=6, O=24 + 5 + 18 = 47? Wait:
- MnO₄⁻: 6×4=24 O
- CH₃OH: 5×1=5 O
- H⁺: no O
Total O: 29
H: 5×4=20 (CH₃OH) + 18 = 38 H
Right: H₂O: 19×1=19 H? No — 19H₂O has 38H and 19O
CO₂: 5×2=10 O
Mn²⁺: 6
Total O: 19 + 10 = 29 → good
H: 38 → good
C: 5 → good
Mn: 6 → good

✔ Balanced.

Answer:
> 6MnO₄⁻(aq) + 5CH₃OH(l) + 18H⁺(aq) → 6Mn²⁺(aq) + 5CO₂(g) + 19H₂O(l)

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6. Br₂(l) + OH⁻(aq) → Br⁻(aq) + BrO₃⁻(aq)

This is a disproportionation reaction.

Br₂ → Br⁻ (reduction) and BrO₃⁻ (oxidation)

In basic solution.

Reduction: Br₂ → Br⁻
Br₂ + 2e⁻ → 2Br⁻

Oxidation: Br₂ → BrO₃⁻
Br₂ → 2BrO₃⁻
Balance O: add 6H₂O to right
Br₂ → 2BrO₃⁻ + 6H₂O
Balance H: add 12H⁺ to left
Br₂ + 12H⁺ → 2BrO₃⁻ + 6H₂O
Balance charge: left=12+, right=0 → add 10e⁻ to left
Br₂ + 12H⁺ + 10e⁻ → 2BrO₃⁻ + 6H₂O

But we’re in basic solution, so add 12OH⁻ to both sides:
Br₂ + 12H⁺ + 12OH⁻ + 10e⁻ → 2BrO₃⁻ + 6H₂O + 12OH⁻
→ Br₂ + 12H₂O + 10e⁻ → 2BrO₃⁻ + 6H₂O + 12OH⁻
Simplify: subtract 6H₂O
Br₂ + 6H₂O + 10e⁻ → 2BrO₃⁻ + 12OH⁻

Now reduction: Br₂ + 2e⁻ → 2Br⁻

Make electrons equal:
- Multiply reduction by 5: 5Br₂ + 10e⁻ → 10Br⁻
- Oxidation: Br₂ + 6H₂O + 10e⁻ → 2BrO₃⁻ + 12OH⁻

Add:
5Br₂ + 10e⁻ + Br₂ + 6H₂O + 10e⁻ → 10Br⁻ + 2BrO₃⁻ + 12OH⁻

Wait — left has 6Br₂? No:

Add:
5Br₂ + 10e⁻ → 10Br⁻
+ Br₂ + 6H₂O + 10e⁻ → 2BrO₃⁻ + 12OH⁻
Total: 6Br₂ + 6H₂O + 20e⁻ → 10Br⁻ + 2BrO₃⁻ + 12OH⁻ + 10e⁻

No — we added two half-reactions with 10e⁻ each, but they are separate.

Actually, we want same number of electrons.

So oxidation: Br₂ → 2BrO₃⁻ takes 10e⁻
Reduction: Br₂ → 2Br⁻ takes 2e⁻ per Br₂ → so for 10e⁻, need 5Br₂ → 10Br⁻

So total:
Oxidation: Br₂ + 6H₂O → 2BrO₃⁻ + 12OH⁻ + 10e⁻
Reduction: 5Br₂ + 10e⁻ → 10Br⁻

Add:
Br₂ + 6H₂O + 5Br₂ + 10e⁻ → 2BrO₃⁻ + 12OH⁻ + 10e⁻ + 10Br⁻

Cancel 10e⁻:
6Br₂ + 6H₂O → 2BrO₃⁻ + 10Br⁻ + 12OH⁻

Divide by 2:
3Br₂ + 3H₂O → BrO₃⁻ + 5Br⁻ + 6OH⁻

Check:
Left: Br=6, H=6, O=3
Right: Br=1+5=6, O=3+6=9? Wait: BrO₃⁻ has 3O, OH⁻ has 6O → total 9O
Left: 3H₂O has 3O → no

Wait: 3H₂O has 3O and 6H
Right: BrO₃⁻ has 3O, 6OH⁻ has 6O → total 9O → mismatch

Mistake.

Earlier: oxidation half:
Br₂ + 6H₂O → 2BrO₃⁻ + 12OH⁻

That’s 6H₂O → 12OH⁻? But 6H₂O has 12H, 12OH⁻ has 12H and 12O → but 6H₂O has only 6O → impossible.

Error.

Let’s fix oxidation half-reaction.

Oxidation: Br₂ → BrO₃⁻
Balance Br: Br₂ → 2BrO₃⁻
Balance O: add 6H₂O to right? No — add to left.

Standard: Br₂ → 2BrO₃⁻
Add 6H₂O to left to provide oxygen? No.

Better: Br₂ → 2BrO₃⁻
Add 6H₂O to right? No.

Actually:
Br₂ → 2BrO₃⁻
Balance O: add 6H₂O to left
Br₂ + 6H₂O → 2BrO₃⁻
Balance H: add 12H⁺ to right
Br₂ + 6H₂O → 2BrO₃⁻ + 12H⁺
Balance charge: left=0, right=12+ → add 10e⁻ to right
Br₂ + 6H₂O → 2BrO₃⁻ + 12H⁺ + 10e⁻

Now in basic solution, add 12OH⁻ to both sides:
Br₂ + 6H₂O + 12OH⁻ → 2BrO₃⁻ + 12H⁺ + 12OH⁻ + 10e⁻
→ Br₂ + 6H₂O + 12OH⁻ → 2BrO₃⁻ + 12H₂O + 10e⁻
Subtract 6H₂O:
Br₂ + 12OH⁻ → 2BrO₃⁻ + 6H₂O + 10e⁻

Now reduction: Br₂ + 2e⁻ → 2Br⁻

Multiply reduction by 5: 5Br₂ + 10e⁻ → 10Br⁻

Add to oxidation:
Br₂ + 12OH⁻ + 5Br₂ + 10e⁻ → 2BrO₃⁻ + 6H₂O + 10e⁻ + 10Br⁻

Cancel 10e⁻:
6Br₂ + 12OH⁻ → 2BrO₃⁻ + 6H₂O + 10Br⁻

Divide by 2:
3Br₂ + 6OH⁻ → BrO₃⁻ + 3H₂O + 5Br⁻

Check:
Left: Br=6, O=6, H=6
Right: Br=1+5=6, O=3 (BrO₃⁻) + 3 (H₂O) = 6, H=6 → ✔

Answer:
> 3Br₂(l) + 6OH⁻(aq) → BrO₃⁻(aq) + 5Br⁻(aq) + 3H₂O(l)

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✔ Final Answers:

1. 3SO₂(g) + 2HNO₃(aq) + 2H₂O(l) → 3H₂SO₄(aq) + 2NO(g)
2. 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)
3. 2Au³⁺(aq) + 6I⁻(aq) → 2Au(s) + 3I₂(s)
4. IO⁻(aq) + I⁻(aq) + 2H⁺(aq) → I₂(s) + H₂O(l)
5. 6MnO₄⁻(aq) + 5CH₃OH(l) + 18H⁺(aq) → 6Mn²⁺(aq) + 5CO₂(g) + 19H₂O(l)
6. 3Br₂(l) + 6OH⁻(aq) → BrO₃⁻(aq) + 5Br⁻(aq) + 3H₂O(l)

Let me know if you'd like these explained further or written in a different format!