Practice worksheet for balancing redox reactions, including oxidation number assignment and half-reaction balancing.

It looks like you've uploaded a "Balancing REDOX Reactions Practice Worksheet" with solutions already filled in. The worksheet includes five redox reactions, each with:

- Assigning oxidation numbers
- Identifying the oxidation and reduction half-reactions
- Balancing the redox reaction using the half-reaction method
- Circling the oxidizing agent and underlining the reducing agent

Let me walk through each problem step by step to explain the solution, verify correctness, and clarify any concepts.

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🔬 General Approach to Balancing Redox Reactions (in Acidic Solution):

1. Assign oxidation numbers
2. Identify what is oxidized and reduced
3. Write half-reactions: oxidation and reduction
4. Balance each half-reaction:
- Atoms first (except O and H)
- Balance O with H₂O
- Balance H with H⁺
- Balance charge with electrons (e⁻)
5. Equalize electrons lost and gained
6. Add half-reactions together
7. Simplify and cancel species appearing on both sides
8. Check atom and charge balance

Now let's go through each question.

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✔ Problem 1:

Reaction:
`H₂SO₄ + HI → I₂ + SO₂ + H₂O`

#### Step 1: Assign Oxidation Numbers
- H₂SO₄: S = +6 (since O is -2, H is +1)
- HI: I = -1
- I₂: I = 0
- SO₂: S = +4
- H₂O: H = +1, O = -2

#### Step 2: Identify Changes
- I⁻ → I₂: Oxidation (from -1 to 0) → Loss of electrons
- S⁶⁺ → S⁴⁺: Reduction (from +6 to +4) → Gain of electrons

#### Half-Reactions:
- Oxidation: `2I⁻ → I₂ + 2e⁻`
- Reduction: `SO₄²⁻ → SO₂`

Balance reduction half-reaction:
- SO₄²⁻ → SO₂
- Add 2 H₂O to right: SO₄²⁻ → SO₂ + 2H₂O
- Add 4H⁺ to left: SO₄²⁻ + 4H⁺ → SO₂ + 2H₂O
- Add 2e⁻ to left: SO₄²⁻ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O

Now match electrons:
- Oxidation: `2I⁻ → I₂ + 2e⁻`
- Reduction: `SO₄²⁻ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O`

Add them:
`2I⁻ + SO₄²⁻ + 4H⁺ → I₂ + SO₂ + 2H₂O`

Now use molecular forms:
- H₂SO₄ provides SO₄²⁻ and 2H⁺ → need 2 H₂SO₄ for 4H⁺
- But we only need one SO₄²⁻ → so use 1 H₂SO₄ and add extra H⁺ from acid? Wait — actually, in this case, H₂SO₄ acts as both source of sulfate and acid.

But note: In the answer key, it says:
> `1 H₂SO₄ + 2 HI → 1 I₂ + 1 SO₂ + 1 H₂O`

Wait — that can't be right because H atoms don’t balance:
- Left: 2H (from H₂SO₄) + 2H (from 2HI) = 4H
- Right: 2H (from H₂O)

So clearly, the given answer is incorrect.

Let’s fix it.

Correct balanced equation:
From earlier:
`SO₄²⁻ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O`
`2I⁻ → I₂ + 2e⁻`

So total:
`SO₄²⁻ + 4H⁺ + 2I⁻ → SO₂ + 2H₂O + I₂`

Now convert to molecules:
- Use H₂SO₄ as source of SO₄²⁻ and 2H⁺
- Need 2 more H⁺ → so need additional acid, or assume excess H⁺

But in this reaction, HI provides H⁺ and I⁻, and H₂SO₄ provides SO₄²⁻ and H⁺

Let’s try:
- 1 H₂SO₄ → SO₄²⁻ + 2H⁺
- 2 HI → 2I⁻ + 2H⁺
- Total H⁺: 4H⁺ → perfect!

So:
`H₂SO₄ + 2HI → I₂ + SO₂ + 2H₂O`

✔ This balances:
- Atoms: H: 2+2=4 → 4H in 2H₂O
- O: 4 from H₂SO₄ → 2 in SO₂ + 2 in 2H₂O = 4
- S: 1 → 1
- I: 2 → 2

Charge: neutral on both sides.

So correct balanced equation is:
> H₂SO₄ + 2HI → I₂ + SO₂ + 2H₂O

But the worksheet says:
> `1 H₂SO₄ + 2 HI → 1 I₂ + 1 SO₂ + 1 H₂O` ✘

This has only 1 H₂O, but should be 2 H₂O.

👉 Mistake in worksheet answer.

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✔ Problem 2:

Reaction:
`KNO₃ + HCl → NO + KCl + H₂O`

Oxidation numbers:
- N in KNO₃: +5 (K⁺, O⁻² → x + 3(-2) = -1 → x = +5)
- N in NO: +2 → reduced
- Cl in HCl: -1 → stays -1 (no change)
- So Cl not involved?

Wait — but KCl is formed → no change in Cl oxidation state.

But what about H? No.

Wait — is this possible?

Let’s see: KNO₃ → NO → N goes from +5 to +2 → reduction

What gets oxidized?

Only possibility: Cl⁻ could be oxidized to Cl₂?

But product is KCl, not Cl₂.

So unless there's an error...

Wait — perhaps HCl is oxidized? But Cl⁻ remains Cl⁻.

Unless it's a different reaction.

Actually, this reaction is not balanced as written.

Let’s look at the answer key:
> `2 KNO₃ + 4 HCl → 2 NO + 2 KCl + 2 H₂O`

But check oxidation states:
- N: +5 → +2 → gains 3e⁻ per N → 2N gain 6e⁻
- Cl: -1 → still -1 → no change

So where do electrons come from?

Ah! No species is oxidized! That can't happen.

So something is wrong.

Wait — maybe HCl is oxidized to Cl₂?

But product is KCl, not Cl₂.

Alternatively, perhaps H⁺ is involved in reduction, but H⁺ doesn't change.

Wait — maybe the intended reaction is:
> KNO₃ + HCl → NO + KCl + Cl₂ + H₂O

But that's not what’s written.

Looking back at the worksheet: it says:
> `2 KNO₃ + 4 HCl → 2 NO + 2 KCl + 2 H₂O`

But that implies:
- 2 N: +5 → +2 → gain of 6e⁻
- 4 Cl: -1 → -1 → no loss

No electron donor → impossible.

So likely, this reaction is missing Cl₂ as a product.

In reality, nitrate can oxidize HCl to Cl₂:
> 2NO₃⁻ + 6H⁺ + 2Cl⁻ → 2NO + Cl₂ + 3H₂O

Yes! That makes sense.

So correct version:
> 2KNO₃ + 4HCl → 2NO + Cl₂ + 2KCl + 2H₂O

But the worksheet shows only KCl and H₂O, no Cl₂.

So either:
- The worksheet is wrong, or
- There’s a typo in the reaction

But the answer key shows:
> Oxidation: Cl⁻ → Cl⁺ (but Cl⁺ isn't stable)

Wait — they wrote:
> Oxidation: Cl⁻ → Cl⁺ → but that's not standard.

They wrote:
> Oxidation: Cl⁻ → Cl⁺ (not possible)

And reduction: N⁵⁺ → N²⁺

But no species is oxidized if Cl⁻ stays Cl⁻.

So the answer key is incorrect here.

Correct half-reactions:
- Reduction: NO₃⁻ → NO
- NO₃⁻ → NO
- Add 2H₂O → NO + 2H₂O
- Add 4H⁺ → NO₃⁻ + 4H⁺ → NO + 2H₂O
- Add 3e⁻ → NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O

- Oxidation: Cl⁻ → Cl₂
- 2Cl⁻ → Cl₂ + 2e⁻

To balance electrons:
- Multiply reduction by 2: 2NO₃⁻ + 8H⁺ + 6e⁻ → 2NO + 4H₂O
- Multiply oxidation by 3: 6Cl⁻ → 3Cl₂ + 6e⁻

Total:
> 2NO₃⁻ + 8H⁺ + 6Cl⁻ → 2NO + 3Cl₂ + 4H₂O

With spectator ions:
> 2KNO₃ + 8HCl → 2NO + 3Cl₂ + 2KCl + 4H₂O

But worksheet says:
> `2 KNO₃ + 4 HCl → 2 NO + 2 KCl + 2 H₂O` ✘

This is wrong — unbalanced and no electron transfer.

👉 Error in worksheet

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✔ Problem 3:

Reaction:
`FeCl₃ + H₂ → FeCl₂ + HCl`

Oxidation numbers:
- Fe in FeCl₃: +3
- Fe in FeCl₂: +2 → reduced
- H in H₂: 0
- H in HCl: +1 → oxidized

So:
- Fe³⁺ → Fe²⁺: reduction (gains 1e⁻)
- H₂ → 2H⁺: oxidation (loses 2e⁻)

Half-reactions:
- Oxidation: H₂ → 2H⁺ + 2e⁻
- Reduction: Fe³⁺ + e⁻ → Fe²⁺

Multiply reduction by 2:
- 2Fe³⁺ + 2e⁻ → 2Fe²⁺

Add:
- H₂ + 2Fe³⁺ → 2H⁺ + 2Fe²⁺

Now add Cl⁻ as spectator:
- FeCl₃ → Fe³⁺ + 3Cl⁻
- So 2FeCl₃ → 2Fe³⁺ + 6Cl⁻
- H₂ + 2FeCl₃ → 2FeCl₂ + 2HCl

But wait: 2FeCl₂ uses 4Cl⁻, 2HCl uses 2Cl⁻ → total 6Cl⁻ → matches

So balanced equation:
> H₂ + 2FeCl₃ → 2FeCl₂ + 2HCl

Worksheet says:
> `1 H₂ + 2 FeCl₃ → 2 FeCl₂ + 2 HCl` ✔

Perfect.

Oxidation: H₂ → 2H⁺ → H₂ is oxidized → reducing agent
Reduction: Fe³⁺ → Fe²⁺ → Fe³⁺ is reduced → oxidizing agent

So:
- Circle: FeCl₃ (oxidizing agent)
- Underline: H₂ (reducing agent)

✔ Correct

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✔ Problem 4:

Reaction:
`HNO₃ + Se → H₂SeO₃ + NO`

Oxidation numbers:
- N in HNO₃: +5
- N in NO: +2 → reduced
- Se: 0 → in H₂SeO₃: +4 (since O is -2, H is +1 → 2(+1) + x + 3(-2) = 0 → x = +4)

So:
- Se: 0 → +4 → loses 4e⁻
- N: +5 → +2 → gains 3e⁻

Half-reactions:

Oxidation: Se → SeO₃²⁻
- Se → SeO₃²⁻
- Add 3H₂O → Se + 3H₂O → SeO₃²⁻
- Add 6H⁺ → Se + 3H₂O → SeO₃²⁻ + 6H⁺
- Add 4e⁻ → Se + 3H₂O → SeO₃²⁻ + 6H⁺ + 4e⁻

Reduction: NO₃⁻ → NO
- NO₃⁻ → NO
- Add 2H₂O → NO₃⁻ → NO + 2H₂O
- Add 4H⁺ → NO₃⁻ + 4H⁺ → NO + 2H₂O
- Add 3e⁻ → NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O

Now balance electrons:
- Oxidation: 4e⁻ lost
- Reduction: 3e⁻ gained

LCM = 12

Multiply oxidation by 3: 3Se + 9H₂O → 3SeO₃²⁻ + 18H⁺ + 12e⁻
Multiply reduction by 4: 4NO₃⁻ + 16H⁺ + 12e⁻ → 4NO + 8H₂O

Add:
- 3Se + 9H₂O + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 18H⁺ + 4NO + 8H₂O

Simplify:
- Cancel 9H₂O from both sides: 3Se + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 4NO + 9H⁺ + H₂O
- H⁺: 16H⁺ – 9H⁺ = 7H⁺ left → move to right? Wait

Wait — better:
Left: 3Se + 4NO₃⁻ + 16H⁺
Right: 3SeO₃²⁻ + 4NO + 8H₂O + 18H⁺

Bring all to one side:
- 3Se + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 4NO + 8H₂O + 18H⁺

Move 18H⁺ to left:
- 3Se + 4NO₃⁻ + 16H⁺ – 18H⁺ → ... → negative H⁺? Not good.

Instead, subtract 16H⁺ from both sides:
- 3Se + 4NO₃⁻ → 3SeO₃²⁻ + 4NO + 8H₂O + 2H⁺

Now add H⁺ to make HNO₃ and H₂SeO₃.

But H₂SeO₃ is weak acid, so write as molecule.

So:
- 3Se + 4HNO₃ → 3H₂SeO₃ + 4NO + 2H⁺ + H₂O? Not quite.

Better: use H⁺ and NO₃⁻ from HNO₃.

But in practice, we often write:

From above:
- 3Se + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 4NO + 8H₂O + 18H⁺

Wait — this gives net 2H⁺ on right.

So:
- 3Se + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 4NO + 8H₂O + 18H⁺

→ Net: 3Se + 4NO₃⁻ → 3SeO₃²⁻ + 4NO + 8H₂O + 2H⁺

Now, add 2H⁺ to left and 2H⁺ to right? No.

Instead, use HNO₃ as source.

Let’s go back to the answer key:
> `3 Se + 10 HNO₃ → 3 H₂SeO₃ + 10 NO + 5 H₂O`

Wait — their answer is:
> `3 Se + 10 HNO₃ → 3 H₂SeO₃ + 10 NO + 5 H₂O`

Let’s check:

Atoms:
- Se: 3 → 3
- H: 10 → 6 (from 3H₂SeO₃) + 10 (from 10NO)? No — NO has no H
- H: left: 10 H from HNO₃
- Right: 6 H from 3H₂SeO₃ + 10 H from 5H₂O = 16 H → too many

Wait — 5H₂O has 10H → total H: 6 + 10 = 16 → but left has only 10H → mismatch

So cannot be.

Try balancing properly.

From earlier:
We had:
- Oxidation: Se → SeO₃²⁻ + 6H⁺ + 4e⁻ (wait — earlier was Se + 3H₂O → SeO₃²⁻ + 6H⁺ + 4e⁻)
- Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O

Multiply:
- Oxidation ×3: 3Se + 9H₂O → 3SeO₃²⁻ + 18H⁺ + 12e⁻
- Reduction ×4: 4NO₃⁻ + 16H⁺ + 12e⁻ → 4NO + 8H₂O

Add:
- 3Se + 9H₂O + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 18H⁺ + 4NO + 8H₂O

Cancel:
- H₂O: 9–8 = 1 H₂O left
- H⁺: 16–18 = -2 → move 2H⁺ to right

So:
- 3Se + 4NO₃⁻ + 16H⁺ + H₂O → 3SeO₃²⁻ + 4NO + 18H⁺ + 8H₂O

Wait — better:
Final: 3Se + 4NO₃⁻ + 16H⁺ → 3SeO₃²⁻ + 4NO + 8H₂O + 2H⁺

Now move 2H⁺ to left:
- 3Se + 4NO₃⁻ + 14H⁺ → 3SeO₃²⁻ + 4NO + 8H₂O

Now add 14H⁺ and 4NO₃⁻ → need 14 HNO₃

But also need to form H₂SeO₃: 3SeO₃²⁻ + 6H⁺ → 3H₂SeO₃

So total H⁺ needed: 14 + 6 = 20H⁺

So use 20 HNO₃ → 20H⁺ + 20NO₃⁻

But we only need 4NO₃⁻ for reduction → so 16 extra NO₃⁻ → will form 16 HNO₃

Wait — messy.

Better approach: use HNO₃ as reactant.

Let’s assume:
- 3Se + a HNO₃ → 3H₂SeO₃ + b NO + c H₂O

From oxidation: Se → H₂SeO₃ → Se(0) → Se(+4) → lose 4e⁻ per Se → 3×4 = 12e⁻ lost
- N: +5 → +2 → gain 3e⁻ per N → need 4 N atoms to gain 12e⁻

So b = 4

So 4 NO produced → so 4 HNO₃ used for reduction

But also, 3 H₂SeO₃ needs 6H and 3SeO₃

Each H₂SeO₃ requires 2H and SeO₃

But Se comes from Se metal

Also, H comes from HNO₃

So total H needed: 6H for H₂SeO₃ + ? for water

Let’s write:

3Se + 4HNO₃ → 3H₂SeO₃ + 4NO + ?

H: left: 4H, right: 6H → need 2H → so need more HNO₃

Suppose we use 10 HNO₃

Then:
- 10 HNO₃ → 10H⁺, 10NO₃⁻

For reduction: 4 NO₃⁻ → 4NO → 4×3e⁻ = 12e⁻ gained

For oxidation: 3Se → 3SeO₃²⁻ → 3×4e⁻ = 12e⁻ lost → balanced

So:
- 3Se + 4NO₃⁻ + 12H⁺ → 3SeO₃²⁻ + 4NO + 6H₂O

But we have 10 NO₃⁻ → so 6 NO₃⁻ left → will form 6 HNO₃

But we want to produce only 4 NO

So total: 3Se + 10 HNO₃ → 3H₂SeO₃ + 4NO + 6H₂O + 6HNO₃

But then HNO₃ on both sides — not desired.

Instead, use fewer HNO₃.

From earlier: 3Se + 4NO₃⁻ + 12H⁺ → 3SeO₃²⁻ + 4NO + 6H₂O

But 3SeO₃²⁻ + 6H⁺ → 3H₂SeO₃

So total H⁺ needed: 12 + 6 = 18H⁺

So need 18 HNO₃

Then:
- 3Se + 18 HNO₃ → 3H₂SeO₃ + 4NO + 6H₂O + 14 HNO₃? No

Wait — 18 HNO₃ provide 18H⁺ and 18NO₃⁻

Use 4 NO₃⁻ for reduction → produce 4NO

Remaining 14 NO₃⁻ → stay as HNO₃

But we have 18H⁺ used: 12 for reduction, 6 for forming H₂SeO₃

So:
- 3Se + 18 HNO₃ → 3H₂SeO₃ + 4NO + 6H₂O + 14 HNO₃

Cancel 14 HNO₃ from both sides:
- 3Se + 4 HNO₃ → 3H₂SeO₃ + 4NO + 6H₂O

Now check atoms:
- Se: 3 → 3
- N: 4 → 4
- O: 12 from 4HNO₃ → right: 9 (3H₂SeO₃) + 4 (4NO) + 6 (6H₂O) = 19 → too many
- H: 4 → 6 (from H₂SeO₃) + 12 (from 6H₂O) = 18 → too many

So not balanced.

After research, the correct balanced equation is:
> 3Se + 4HNO₃ → 3H₂SeO₃ + 4NO + H₂O

Check:
- Se: 3 → 3
- N: 4 → 4
- O: 12 → 9 (H₂SeO₃) + 4 (NO) + 1 (H₂O) = 14 → no
- H: 4 → 6 + 2 = 8 → no

Still off.

Standard balanced equation:
> 3Se + 4HNO₃ + H₂O → 3H₂SeO₃ + 4NO

Check:
- H: 4 + 2 = 6 → right: 6 → ok
- O: 12 + 1 = 13 → right: 9 + 4 = 13 → ok
- Se: 3 → 3
- N: 4 → 4

Yes!

So:
> 3Se + 4HNO₃ + H₂O → 3H₂SeO₃ + 4NO

But worksheet says:
> `3 Se + 10 HNO₃ → 3 H₂SeO₃ + 10 NO + 5 H₂O`

Which is way off.

So worksheet answer is wrong.

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✔ Problem 5:

Reaction:
`H₂SO₄ + Fe → FeSO₄ + H₂`

Oxidation numbers:
- Fe: 0 → +2 → oxidized
- H: +1 → 0 → reduced

So:
- Fe → Fe²⁺ + 2e⁻
- 2H⁺ + 2e⁻ → H₂

So:
- Fe + 2H⁺ → Fe²⁺ + H₂

With sulfate:
- H₂SO₄ → 2H⁺ + SO₄²⁻
- FeSO₄ → Fe²⁺ + SO₄²⁻

So:
> Fe + H₂SO₄ → FeSO₄ + H₂

Balanced.

Worksheet says:
> `1 H₂SO₄ + 1 Fe → 1 FeSO₄ + 1 H₂` ✔

Oxidation: Fe → Fe²⁺ → Fe is oxidized → reducing agent
Reduction: H⁺ → H₂ → H⁺ is reduced → oxidizing agent

So:
- Circle: H₂SO₄ (oxidizing agent)
- Underline: Fe (reducing agent)

✔ Correct

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## 🟩 Summary of Errors in Worksheet:

| Problem | Error |
|--------|-------|
| 1 | Incorrect H₂O count — should be 2 H₂O, not 1 |
| 2 | Reaction not balanced; missing Cl₂ as product; no oxidation shown |
| 3 | ✔ Correct |
| 4 | Wrong coefficients — actual balanced equation is complex; worksheet answer is invalid |
| 5 | ✔ Correct |

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## ✔ Final Answer:

The worksheet contains errors in problems 1, 2, and 4. Only problems 3 and 5 are correctly solved.

Corrected Answers:

1. H₂SO₄ + 2HI → I₂ + SO₂ + 2H₂O
- Oxidation: I⁻ → I₂
- Reduction: SO₄²⁻ → SO₂
- Oxidizing agent: H₂SO₄
- Reducing agent: HI

2. 2KNO₃ + 8HCl → 2NO + 3Cl₂ + 2KCl + 4H₂O
- Oxidation: Cl⁻ → Cl₂
- Reduction: NO₃⁻ → NO
- Oxidizing agent: KNO₃
- Reducing agent: HCl

3. ✔ Correct

4. 3Se + 4HNO₃ + H₂O → 3H₂SeO₃ + 4NO
- Oxidation: Se → H₂SeO₃
- Reduction: HNO₃ → NO
- Oxidizing agent: HNO₃
- Reducing agent: Se

5. ✔ Correct

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