Solved Balancing Redox Reactions using half-reaction method ... - Free Printable

Let's solve each redox reaction using the half-reaction method, balancing them in the specified conditions (neutral, acidic, or basic), and identify which species is oxidized and which is reduced.

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a. $ \text{Cu}^{2+} + \text{I}^- \rightarrow \text{Cu} + \text{I}_3^- $ (neutral)

#### Step 1: Assign oxidation states
- Cu²⁺ → Cu: +2 → 0 → reduction
- I⁻ → I₃⁻: I is -1 in both, but in I₃⁻, average oxidation state is -1/3? Wait — actually, each I atom is still -1, but we must consider the formation of triiodide ion.

Wait: I⁻ → I₃⁻ involves oxidation?

Actually:
- In I⁻: oxidation state = -1
- In I₃⁻: central I is +1, terminal are -1 → but average is -1/3

But more accurately, I⁻ is oxidized to I₃⁻ because two I⁻ ions lose electrons to form I₃⁻:

$$
3\text{I}^- \rightarrow \text{I}_3^- + 2e^-
$$

So:
- I⁻ is oxidized
- Cu²⁺ is reduced

#### Half-reactions:

Reduction:
$ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} $

Oxidation:
$ 3\text{I}^- \rightarrow \text{I}_3^- + 2e^- $

Balance electrons: both have 2e⁻, so combine:

$$
\text{Cu}^{2+} + 3\text{I}^- \rightarrow \text{Cu} + \text{I}_3^-
$$

✔ Balanced.

- Oxidized: I⁻
- Reduced: Cu²⁺

---

b. $ \text{MnO}_4^- + \text{Ag} \rightarrow \text{Mn}^{2+} + \text{Ag}^+ $ (acidic)

#### Oxidation states:
- Mn in MnO₄⁻: +7 → Mn²⁺: +2 → reduction
- Ag: 0 → Ag⁺: +1 → oxidation

#### Half-reactions:

Reduction: MnO₄⁻ → Mn²⁺ (in acidic)

1. MnO₄⁻ → Mn²⁺
2. Balance O: add 4 H₂O to right
3. Balance H: add 8H⁺ to left
4. Balance charge: add 5e⁻ to left

$$
\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}
$$

Oxidation: Ag → Ag⁺

$$
\text{Ag} \rightarrow \text{Ag}^+ + e^-
$$

Multiply oxidation by 5:

$$
5\text{Ag} \rightarrow 5\text{Ag}^+ + 5e^-
$$

Add:

$$
\text{MnO}_4^- + 8\text{H}^+ + 5\text{Ag} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Ag}^+
$$

✔ Balanced.

- Oxidized: Ag
- Reduced: MnO₄⁻

---

c. $ \text{Hg} + \text{NO}_3^- + \text{Cl}^- \rightarrow \text{HgCl}_4^{2-} + \text{NO}_2 $ (acidic)

#### Oxidation states:
- Hg: 0 → Hg in HgCl₄²⁻: +2 → oxidation
- N in NO₃⁻: +5 → NO₂: +4 → reduction

#### Half-reactions:

Oxidation: Hg → HgCl₄²⁻

1. Hg → HgCl₄²⁻
2. Add 4 Cl⁻ to left
3. Balance charge: add 2e⁻ to right

$$
\text{Hg} + 4\text{Cl}^- \rightarrow \text{HgCl}_4^{2-} + 2e^-
$$

Reduction: NO₃⁻ → NO₂ (acidic)

1. NO₃⁻ → NO₂
2. Balance O: add H₂O to right
3. Balance H: add 2H⁺ to left
4. Balance charge: add 1e⁻ to left

$$
\text{NO}_3^- + 2\text{H}^+ + e^- \rightarrow \text{NO}_2 + \text{H}_2\text{O}
$$

Multiply reduction by 2:

$$
2\text{NO}_3^- + 4\text{H}^+ + 2e^- \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O}
$$

Add to oxidation:

$$
\text{Hg} + 4\text{Cl}^- + 2\text{NO}_3^- + 4\text{H}^+ \rightarrow \text{HgCl}_4^{2-} + 2\text{NO}_2 + 2\text{H}_2\text{O}
$$

✔ Balanced.

- Oxidized: Hg
- Reduced: NO₃⁻

---

d. $ \text{AsH}_3 + \text{Zn}^{2+} \rightarrow \text{H}_3\text{AsO}_4 + \text{Zn} $ (acidic)

#### Oxidation states:
- As in AsH₃: -3 → H₃AsO₄: +5 → oxidation
- Zn²⁺: +2 → Zn: 0 → reduction

#### Half-reactions:

Oxidation: AsH₃ → H₃AsO₄

1. AsH₃ → H₃AsO₄
2. Balance O: add 4 H₂O to left
3. Balance H: add 8H⁺ to right
4. Balance charge: add 8e⁻ to right

$$
\text{AsH}_3 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{AsO}_4 + 8\text{H}^+ + 8e^-
$$

Reduction: Zn²⁺ → Zn

$$
\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}
$$

Multiply reduction by 4:

$$
4\text{Zn}^{2+} + 8e^- \rightarrow 4\text{Zn}
$$

Add:

$$
\text{AsH}_3 + 4\text{H}_2\text{O} + 4\text{Zn}^{2+} \rightarrow \text{H}_3\text{AsO}_4 + 8\text{H}^+ + 4\text{Zn}
$$

✔ Balanced.

- Oxidized: AsH₃
- Reduced: Zn²⁺

---

e. $ \text{CN}^- + \text{MnO}_4^- \rightarrow \text{CNO}^- + \text{MnO}_2 $ (basic)

#### Oxidation states:
- C in CN⁻: +2 (N is -3, so C is +2) → in CNO⁻: C is +4 (O is -2, N is -3, so C = +4) → oxidation
- Mn in MnO₄⁻: +7 → MnO₂: +4 → reduction

#### Half-reactions (basic):

Oxidation: CN⁻ → CNO⁻

1. CN⁻ → CNO⁻
2. Balance O: add H₂O to left
3. Balance H: add H⁺ to right → but in basic, use OH⁻

Better way:

CN⁻ → CNO⁻

Add water to balance oxygen:

CN⁻ + H₂O → CNO⁻ + 2H⁺ + 2e⁻

But in basic, add 2OH⁻ to both sides:

CN⁻ + H₂O + 2OH⁻ → CNO⁻ + 2H⁺ + 2OH⁻ + 2e⁻
→ CN⁻ + H₂O + 2OH⁻ → CNO⁻ + 2H₂O + 2e⁻

Simplify:

$$
\text{CN}^- + 2\text{OH}^- \rightarrow \text{CNO}^- + \text{H}_2\text{O} + 2e^-
$$

Reduction: MnO₄⁻ → MnO₂ (basic)

1. MnO₄⁻ → MnO₂
2. Balance O: add 2H₂O to right
3. Balance H: add 4H⁺ to left → then add 4OH⁻ to both sides

MnO₄⁻ → MnO₂

Add 2H₂O to right: MnO₄⁻ → MnO₂ + 2H₂O
Add 4H⁺ to left: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O
Add 4OH⁻ to both sides:
MnO₄⁻ + 4H⁺ + 4OH⁻ → MnO₂ + 2H₂O + 4OH⁻
→ MnO₄⁻ + 4H₂O → MnO₂ + 2H₂O + 4OH⁻
→ MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻

Now balance charge: add 3e⁻ to left

$$
\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-
$$

Now balance electrons:

Oxidation: CN⁻ + 2OH⁻ → CNO⁻ + H₂O + 2e⁻
Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

LCM of 2 and 3 is 6.

Multiply oxidation by 3:
3CN⁻ + 6OH⁻ → 3CNO⁻ + 3H₂O + 6e⁻

Multiply reduction by 2:
2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻

Add:

3CN⁻ + 6OH⁻ + 2MnO₄⁻ + 4H₂O + 6e⁻ → 3CNO⁻ + 3H₂O + 6e⁻ + 2MnO₂ + 8OH⁻

Cancel:

- 6e⁻ cancels
- 6OH⁻ on left, 8OH⁻ on right → move 2OH⁻ to right
- 4H₂O left, 3H₂O right → net 1H₂O left

So:

3CN⁻ + 2MnO₄⁻ + H₂O → 3CNO⁻ + 2MnO₂ + 2OH⁻

✔ Balanced.

- Oxidized: CN⁻
- Reduced: MnO₄⁻

---

f. $ \text{H}_2\text{O}_2 + \text{ClO}_2 \rightarrow \text{ClO}_2^- + \text{O}_2 $ (basic)

#### Oxidation states:
- O in H₂O₂: -1 → O₂: 0 → oxidation
- Cl in ClO₂: +4 → ClO₂⁻: +3 → reduction

#### Half-reactions (basic):

Oxidation: H₂O₂ → O₂

1. H₂O₂ → O₂
2. Balance O: already balanced
3. Balance H: add 2H⁺ to right → but basic → use OH⁻

H₂O₂ → O₂ + 2H⁺ + 2e⁻

Add 2OH⁻ to both sides:

H₂O₂ + 2OH⁻ → O₂ + 2H⁺ + 2OH⁻ + 2e⁻
→ H₂O₂ + 2OH⁻ → O₂ + 2H₂O + 2e⁻

So:

$$
\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^-
$$

Reduction: ClO₂ → ClO₂⁻

ClO₂ → ClO₂⁻

Add 1e⁻ to left:

$$
\text{ClO}_2 + e^- \rightarrow \text{ClO}_2^-
$$

Now balance electrons:

Oxidation produces 2e⁻, reduction uses 1e⁻ → multiply reduction by 2:

2ClO₂ + 2e⁻ → 2ClO₂⁻

Add:

H₂O₂ + 2OH⁻ + 2ClO₂ → O₂ + 2H₂O + 2ClO₂⁻

✔ Balanced.

- Oxidized: H₂O₂
- Reduced: ClO₂

---

g. $ \text{ClO}^- + \text{CrO}_2^- \rightarrow \text{CrO}_4^{2-} + \text{Cl}_2 $ (basic)

#### Oxidation states:
- Cl in ClO⁻: +1 → Cl₂: 0 → reduction
- Cr in CrO₂⁻: +3 → CrO₄²⁻: +6 → oxidation

Note: CrO₂⁻ is chromite ion; Cr is +3.

#### Half-reactions (basic):

Oxidation: CrO₂⁻ → CrO₄²⁻

1. CrO₂⁻ → CrO₄²⁻
2. Balance O: add 2H₂O to left
3. Balance H: add 4H⁺ to right → now add 4OH⁻ to both sides

CrO₂⁻ → CrO₄²⁻
Add 2H₂O to left: 2H₂O + CrO₂⁻ → CrO₄²⁻
Add 4H⁺ to right: 2H₂O + CrO₂⁻ → CrO₄²⁻ + 4H⁺
Add 4OH⁻ to both sides:
2H₂O + CrO₂⁻ + 4OH⁻ → CrO₄²⁻ + 4H⁺ + 4OH⁻
→ 2H₂O + CrO₂⁻ + 4OH⁻ → CrO₄²⁻ + 4H₂O
→ CrO₂⁻ + 4OH⁻ → CrO₄²⁻ + 2H₂O + 3e⁻? Wait — check charge.

Left: CrO₂⁻ (-1) + 4OH⁻ (-4) = -5
Right: CrO₄²⁻ (-2) + 2H₂O (0) = -2 → need 3e⁻ on right?

No — oxidation means loss of electrons → put on right.

But let’s balance atoms and charge properly.

CrO₂⁻ → CrO₄²⁻

Add 2H₂O to left: 2H₂O + CrO₂⁻ → CrO₄²⁻
Add 4H⁺ to right: 2H₂O + CrO₂⁻ → CrO₄²⁻ + 4H⁺
Now add 4OH⁻ to both sides:

Left: 2H₂O + CrO₂⁻ + 4OH⁻
Right: CrO₄²⁻ + 4H⁺ + 4OH⁻ → CrO₄²⁻ + 4H₂O

So: 2H₂O + CrO₂⁻ + 4OH⁻ → CrO₄²⁻ + 4H₂O
→ CrO₂⁻ + 4OH⁻ → CrO₄²⁻ + 2H₂O

Now balance charge:

Left: -1 + (-4) = -5
Right: -2 + 0 = -2 → difference of 3 → add 3e⁻ to right

So:

$$
\text{CrO}_2^- + 4\text{OH}^- \rightarrow \text{CrO}_4^{2-} + 2\text{H}_2\text{O} + 3e^-
$$

Reduction: ClO⁻ → Cl₂

ClO⁻ → Cl₂

Balance Cl: 2ClO⁻ → Cl₂

Balance O: add 2H₂O to right

2ClO⁻ → Cl₂ + 2H₂O

Balance H: add 4H⁺ to left

2ClO⁻ + 4H⁺ → Cl₂ + 2H₂O

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

2ClO⁻ + 4H⁺ + 4OH⁻ → Cl₂ + 2H₂O + 4OH⁻
→ 2ClO⁻ + 4H₂O → Cl₂ + 2H₂O + 4OH⁻
→ 2ClO⁻ + 2H₂O → Cl₂ + 4OH⁻

Now balance charge:

Left: 2(-1) = -2
Right: 4(-1) = -4 → add 2e⁻ to left

$$
2\text{ClO}^- + 2\text{H}_2\text{O} + 2e^- \rightarrow \text{Cl}_2 + 4\text{OH}^-
$$

Now balance electrons:

Oxidation: CrO₂⁻ + 4OH⁻ → CrO₄²⁻ + 2H₂O + 3e⁻
Reduction: 2ClO⁻ + 2H₂O + 2e⁻ → Cl₂ + 4OH⁻

LCM of 3 and 2 is 6.

Multiply oxidation by 2:
2CrO₂⁻ + 8OH⁻ → 2CrO₄²⁻ + 4H₂O + 6e⁻

Multiply reduction by 3:
6ClO⁻ + 6H₂O + 6e⁻ → 3Cl₂ + 12OH⁻

Add:

2CrO₂⁻ + 8OH⁻ + 6ClO⁻ + 6H₂O + 6e⁻ → 2CrO₄²⁻ + 4H₂O + 6e⁻ + 3Cl₂ + 12OH⁻

Cancel:

- 6e⁻ cancels
- 6H₂O left, 4H₂O right → 2H₂O left
- 8OH⁻ left, 12OH⁻ right → 4OH⁻ on right

So:

2CrO₂⁻ + 6ClO⁻ + 2H₂O → 2CrO₄²⁻ + 3Cl₂ + 4OH⁻

Divide whole equation by 2:

$$
\text{CrO}_2^- + 3\text{ClO}^- + \text{H}_2\text{O} \rightarrow \text{CrO}_4^{2-} + \frac{3}{2}\text{Cl}_2 + 2\text{OH}^-
$$

To avoid fractions, keep as:

$$
2\text{CrO}_2^- + 6\text{ClO}^- + 2\text{H}_2\text{O} \rightarrow 2\text{CrO}_4^{2-} + 3\text{Cl}_2 + 4\text{OH}^-
$$

✔ Balanced.

- Oxidized: CrO₂⁻
- Reduced: ClO⁻

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✔ Final Answers Summary:

| Reaction | Balanced Equation | Oxidized | Reduced |
|--------|-------------------|---------|--------|
| a | $ \text{Cu}^{2+} + 3\text{I}^- \rightarrow \text{Cu} + \text{I}_3^- $ | I⁻ | Cu²⁺ |
| b | $ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Ag} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Ag}^+ $ | Ag | MnO₄⁻ |
| c | $ \text{Hg} + 4\text{Cl}^- + 2\text{NO}_3^- + 4\text{H}^+ \rightarrow \text{HgCl}_4^{2-} + 2\text{NO}_2 + 2\text{H}_2\text{O} $ | Hg | NO₃⁻ |
| d | $ \text{AsH}_3 + 4\text{Zn}^{2+} + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{AsO}_4 + 8\text{H}^+ + 4\text{Zn} $ | AsH₃ | Zn²⁺ |
| e | $ 3\text{CN}^- + 2\text{MnO}_4^- + \text{H}_2\text{O} \rightarrow 3\text{CNO}^- + 2\text{MnO}_2 + 2\text{OH}^- $ | CN⁻ | MnO₄⁻ |
| f | $ \text{H}_2\text{O}_2 + 2\text{ClO}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{ClO}_2^- + 2\text{H}_2\text{O} $ | H₂O₂ | ClO₂ |
| g | $ 2\text{CrO}_2^- + 6\text{ClO}^- + 2\text{H}_2\text{O} \rightarrow 2\text{CrO}_4^{2-} + 3\text{Cl}_2 + 4\text{OH}^- $ | CrO₂⁻ | ClO⁻ |

All reactions are balanced using the half-reaction method, with correct identification of oxidized and reduced species.