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?

No — let’s clarify:

In I₃⁻, the central I is +1, terminal I are -1 → average -1/3 per I. But I⁻ to I₃⁻ is not oxidation — it's a disproportionation-like process?

Actually, I⁻ is oxidized to I₃⁻, because I goes from -1 to an average of -1/3 → oxidation.

But wait — that would mean I⁻ is oxidized, and Cu²⁺ is reduced.

Yes.

So:
- Cu²⁺ → Cu: reduction
- I⁻ → I₃⁻: oxidation

#### Half-reactions:

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

Balance charge: add 2e⁻
→ $ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} $

Oxidation:
$ \text{I}^- \rightarrow \text{I}_3^- $

Balance I atoms:
$ 3\text{I}^- \rightarrow \text{I}_3^- $

Balance charge: left has -3, right has -1 → add 2e⁻ to right? No — electrons go on product side for oxidation.

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

Now balance electrons:

- Reduction: 2e⁻ consumed
- Oxidation: 2e⁻ produced

So they match.

Add together:

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

✔ Balanced.

Oxidized: I⁻
Reduced: Cu²⁺

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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

So:
- MnO₄⁻ → Mn²⁺ (reduction)
- Ag → Ag⁺ (oxidation)

#### Half-reactions (acidic):

Reduction: $ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} $

1. Balance Mn: already balanced.
2. Balance O: add 4 H₂O to right
→ $ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} $
3. Balance H: add 8 H⁺ to left
→ $ \text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} $
4. Balance charge:
Left: -1 + 8 = +7
Right: +2
Add 5e⁻ to left
→ $ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} $

Oxidation: $ \text{Ag} \rightarrow \text{Ag}^+ $

Add 1e⁻ to right:
→ $ \text{Ag} \rightarrow \text{Ag}^+ + e^- $

Now balance electrons:

Multiply oxidation by 5:
$ 5\text{Ag} \rightarrow 5\text{Ag}^+ + 5e^- $

Add to reduction:

$ \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⁻

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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

So:
- Hg → HgCl₄²⁻ (oxidation)
- NO₃⁻ → NO₂ (reduction)

#### Half-reactions (acidic):

Oxidation: $ \text{Hg} \rightarrow \text{HgCl}_4^{2-} $

1. Balance Hg: done.
2. Add 4 Cl⁻ to left:
$ \text{Hg} + 4\text{Cl}^- \rightarrow \text{HgCl}_4^{2-} $
3. Charge: left: -4, right: -2 → add 2e⁻ to right
→ $ \text{Hg} + 4\text{Cl}^- \rightarrow \text{HgCl}_4^{2-} + 2e^- $

Reduction: $ \text{NO}_3^- \rightarrow \text{NO}_2 $

1. Balance N: done.
2. Balance O: add 1 H₂O to right
→ $ \text{NO}_3^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} $
3. Balance H: add 2 H⁺ to left
→ $ \text{NO}_3^- + 2\text{H}^+ \rightarrow \text{NO}_2 + \text{H}_2\text{O} $
4. Balance charge:
Left: -1 + 2 = +1
Right: 0
Add 1e⁻ to left
→ $ \text{NO}_3^- + 2\text{H}^+ + e^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} $

Now balance electrons:

Oxidation: 2e⁻ released
Reduction: 1e⁻ consumed → 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⁻

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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⁺ → Zn: +2 → 0 → reduction

So:
- AsH₃ → H₃AsO₄ (oxidation)
- Zn²⁺ → Zn (reduction)

#### Half-reactions (acidic):

Oxidation: $ \text{AsH}_3 \rightarrow \text{H}_3\text{AsO}_4 $

1. Balance As: done.
2. Balance O: add 4 H₂O to left
→ $ \text{AsH}_3 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{AsO}_4 $
3. Balance H: left has 3 + 8 = 11 H, right has 3 → add 8 H⁺ to right
→ $ \text{AsH}_3 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{AsO}_4 + 8\text{H}^+ $
4. Balance charge:
Left: 0
Right: +8
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: $ \text{Zn}^{2+} \rightarrow \text{Zn} $

Add 2e⁻ to left:
→ $ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} $

Now balance electrons:

Oxidation: 8e⁻
Reduction: 2e⁻ → 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⁺

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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 is +4) → oxidation
- Mn in MnO₄⁻: +7 → MnO₂: +4 → reduction

So:
- CN⁻ → CNO⁻ (oxidation)
- MnO₄⁻ → MnO₂ (reduction)

#### Half-reactions (basic):

Oxidation: $ \text{CN}^- \rightarrow \text{CNO}^- $

1. Balance atoms: done.
2. Balance O: add 1 H₂O to left
→ $ \text{CN}^- + \text{H}_2\text{O} \rightarrow \text{CNO}^- $
3. Balance H: add 2 H⁺ to right
→ $ \text{CN}^- + \text{H}_2\text{O} \rightarrow \text{CNO}^- + 2\text{H}^+ $
4. Balance charge:
Left: -1
Right: -1 + 2 = +1
Add 2e⁻ to right
→ $ \text{CN}^- + \text{H}_2\text{O} \rightarrow \text{CNO}^- + 2\text{H}^+ + 2e^- $

Now convert to basic:
Add 2 OH⁻ to both sides:
→ $ \text{CN}^- + \text{H}_2\text{O} + 2\text{OH}^- \rightarrow \text{CNO}^- + 2\text{H}^+ + 2\text{OH}^- + 2e^- $

Simplify: H⁺ + OH⁻ → H₂O → 2H₂O

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

Simplify water: subtract 1 H₂O from both sides:

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

✔ Oxidation half-reaction.

Reduction: $ \text{MnO}_4^- \rightarrow \text{MnO}_2 $

1. Balance Mn: done.
2. Balance O: add 2 H₂O to right
→ $ \text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} $
3. Balance H: add 4 H⁺ to left
→ $ \text{MnO}_4^- + 4\text{H}^+ \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} $
4. Balance charge:
Left: -1 + 4 = +3
Right: 0
Add 3e⁻ to left
→ $ \text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} $

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

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

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

Simplify: subtract 2 H₂O

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

✔ Reduction half-reaction.

Now balance electrons:

Oxidation: 2e⁻
Reduction: 3e⁻ → LCM = 6

Multiply oxidation by 3:
$ 3\text{CN}^- + 6\text{OH}^- \rightarrow 3\text{CNO}^- + 3\text{H}_2\text{O} + 6e^- $

Multiply reduction by 2:
$ 2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- $

Add:

Left: $ 3\text{CN}^- + 6\text{OH}^- + 2\text{MnO}_4^- + 4\text{H}_2\text{O} $
Right: $ 3\text{CNO}^- + 3\text{H}_2\text{O} + 2\text{MnO}_2 + 8\text{OH}^- $

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

Better: move all to one side:

Cancel 3 H₂O from both sides → 1 H₂O left
OH⁻: 6 on left, 8 on right → net 2 OH⁻ on right

So final:

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

✔ Balanced.

Oxidized: CN⁻
Reduced: MnO₄⁻

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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

So:
- H₂O₂ → O₂ (oxidation)
- ClO₂ → ClO₂⁻ (reduction)

#### Half-reactions (basic):

Oxidation: $ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 $

1. Balance O: done.
2. Balance H: add 2 H⁺ to right
→ $ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ $
3. Balance charge: add 2e⁻ to right
→ $ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^- $

Convert to basic: add 2 OH⁻ to both sides:

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

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

✔ Oxidation half.

Reduction: $ \text{ClO}_2 \rightarrow \text{ClO}_2^- $

1. Balance Cl: done.
2. Balance O: done.
3. Balance charge: left: 0, right: -1 → add 1e⁻ to left
→ $ \text{ClO}_2 + e^- \rightarrow \text{ClO}_2^- $

But is this correct? ClO₂ is neutral, ClO₂⁻ is -1, so yes.

But we need to check if it's balanced in basic medium.

It is already balanced: atoms and charge.

But often, ClO₂ disproportionates, but here it's reduction.

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

Now balance electrons:

Oxidation: 2e⁻
Reduction: 1e⁻ → multiply reduction by 2:

$ 2\text{ClO}_2 + 2e^- \rightarrow 2\text{ClO}_2^- $

Add to oxidation:

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

✔ Balanced.

Oxidized: H₂O₂
Reduced: ClO₂

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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

So:
- ClO⁻ → Cl₂ (reduction)
- CrO₂⁻ → CrO₄²⁻ (oxidation)

#### Half-reactions (basic):

Oxidation: $ \text{CrO}_2^- \rightarrow \text{CrO}_4^{2-} $

1. Balance Cr: done.
2. Balance O: add 1 H₂O to left
→ $ \text{CrO}_2^- + \text{H}_2\text{O} \rightarrow \text{CrO}_4^{2-} $
3. Balance H: add 2 H⁺ to right
→ $ \text{CrO}_2^- + \text{H}_2\text{O} \rightarrow \text{CrO}_4^{2-} + 2\text{H}^+ $
4. Balance charge:
Left: -1
Right: -2 + 2 = 0
Add 3e⁻ to right
→ $ \text{CrO}_2^- + \text{H}_2\text{O} \rightarrow \text{CrO}_4^{2-} + 2\text{H}^+ + 3e^- $

Convert to basic: add 2 OH⁻ to both sides:

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

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

Simplify: subtract H₂O

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

✔ Oxidation.

Reduction: $ \text{ClO}^- \rightarrow \text{Cl}_2 $

1. Balance Cl: 2 ClO⁻ → Cl₂
→ $ 2\text{ClO}^- \rightarrow \text{Cl}_2 $
2. Balance O: add 2 H₂O to right
→ $ 2\text{ClO}^- \rightarrow \text{Cl}_2 + 2\text{H}_2\text{O} $
3. Balance H: add 4 H⁺ to left
→ $ 2\text{ClO}^- + 4\text{H}^+ \rightarrow \text{Cl}_2 + 2\text{H}_2\text{O} $
4. Balance charge:
Left: -2 + 4 = +2
Right: 0
Add 2e⁻ to left
→ $ 2\text{ClO}^- + 4\text{H}^+ + 2e^- \rightarrow \text{Cl}_2 + 2\text{H}_2\text{O} $

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

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

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

Simplify: subtract 2 H₂O

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

✔ Reduction.

Now balance electrons:

Oxidation: 3e⁻
Reduction: 2e⁻ → LCM = 6

Multiply oxidation by 2:
$ 2\text{CrO}_2^- + 4\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} + 2\text{H}_2\text{O} + 6e^- $

Multiply reduction by 3:
$ 6\text{ClO}^- + 6\text{H}_2\text{O} + 6e^- \rightarrow 3\text{Cl}_2 + 12\text{OH}^- $

Add:

Left: $ 2\text{CrO}_2^- + 4\text{OH}^- + 6\text{ClO}^- + 6\text{H}_2\text{O} $
Right: $ 2\text{CrO}_4^{2-} + 2\text{H}_2\text{O} + 3\text{Cl}_2 + 12\text{OH}^- $

Simplify:
- H₂O: 6 – 2 = 4 H₂O on left
- OH⁻: 4 on left, 12 on right → net 8 OH⁻ on right

So:

$ 2\text{CrO}_2^- + 6\text{ClO}^- + 4\text{H}_2\text{O} \rightarrow 2\text{CrO}_4^{2-} + 3\text{Cl}_2 + 8\text{OH}^- $

Divide entire equation by 2:

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

To avoid fractions, keep as:

$ 2\text{CrO}_2^- + 6\text{ClO}^- + 4\text{H}_2\text{O} \rightarrow 2\text{CrO}_4^{2-} + 3\text{Cl}_2 + 8\text{OH}^- $

✔ Balanced.

Oxidized: CrO₂⁻
Reduced: ClO⁻

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✔ Final Answers:

| 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{H}_2\text{O} + 4\text{Zn}^{2+} \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{OH}^- + 2\text{ClO}_2 \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2\text{ClO}_2^- $ | H₂O₂ | ClO₂ |
| g | $ 2\text{CrO}_2^- + 6\text{ClO}^- + 4\text{H}_2\text{O} \rightarrow 2\text{CrO}_4^{2-} + 3\text{Cl}_2 + 8\text{OH}^- $ | CrO₂⁻ | ClO⁻ |

All equations are balanced using the half-reaction method in the specified conditions.