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

Let’s solve each redox reaction step by step using the half-reaction method. We’ll balance atoms, then charges, and finally identify what’s oxidized and reduced.

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a. Cu²⁺ + I⁻ → Cu + I₃¹⁻ (neutral)

Step 1: Split into half-reactions.

Reduction: Cu²⁺ → Cu
Oxidation: I⁻ → I₃¹⁻

Step 2: Balance atoms other than H and O.

Reduction: Already balanced — Cu²⁺ → Cu
Oxidation: 3I⁻ → I₃¹⁻ (need 3 iodides to make one triiodide)

Step 3: Balance charge with electrons.

Reduction: Cu²⁺ + 2e⁻ → Cu
Oxidation: 3I⁻ → I₃¹⁻ + 2e⁻ (left side: -3, right side: -1 → add 2e⁻ to right)

Step 4: Add half-reactions.

Cu²⁺ + 2e⁻ + 3I⁻ → Cu + I₃¹⁻ + 2e⁻
Cancel electrons: Cu²⁺ + 3I⁻ → Cu + I₃¹⁻

Check atom and charge balance:
Left: Cu=1, I=3; charge = +2 + (-3) = -1
Right: Cu=1, I=3; charge = 0 + (-1) = -1 ✔

Oxidized: I⁻ (lost electrons)
Reduced: Cu²⁺ (gained electrons)

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b. MnO₄¹⁻ + Ag → Mn²⁺ + Ag¹⁺ (acidic)

Step 1: Half-reactions.

Reduction: MnO₄¹⁻ → Mn²⁺
Oxidation: Ag → Ag¹⁺

Step 2: Balance atoms.

Reduction: Mn is balanced. Add water for oxygen:
MnO₄¹⁻ → Mn²⁺ + 4H₂O
Add H⁺ for hydrogen:
MnO₄¹⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Oxidation: Ag → Ag¹⁺ (already balanced)

Step 3: Balance charge with electrons.

Reduction: Left: -1 + 8 = +7; Right: +2 → add 5e⁻ to left:
MnO₄¹⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation: Ag → Ag¹⁺ + e⁻

Step 4: Equalize electrons. Multiply oxidation by 5:

5Ag → 5Ag¹⁺ + 5e⁻

Step 5: Add together:

MnO₄¹⁻ + 8H⁺ + 5e⁻ + 5Ag → Mn²⁺ + 4H₂O + 5Ag¹⁺ + 5e⁻
Cancel electrons: MnO₄¹⁻ + 8H⁺ + 5Ag → Mn²⁺ + 4H₂O + 5Ag¹⁺

Check: Atoms and charge balanced?
Left: Mn=1, O=4, H=8, Ag=5; charge = -1 +8 +0 = +7
Right: Mn=1, O=4, H=8, Ag=5; charge = +2 +0 +5 = +7 ✔

Oxidized: Ag
Reduced: MnO₄¹⁻

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c. Hg + NO₃¹⁻ + Cl → HgCl²⁻ + NO₂ (acidic)

Step 1: Half-reactions.

Oxidation: Hg → HgCl₄²⁻
Reduction: NO₃¹⁻ → NO₂

Step 2: Balance atoms.

Oxidation: Hg + 4Cl⁻ → HgCl₄²⁻ (add 4 chlorides)
Reduction: NO₃¹⁻ → NO₂
Balance O: add H₂O to right → NO₃¹⁻ → NO₂ + H₂O
Balance H: add 2H⁺ to left → NO₃¹⁻ + 2H⁺ → NO₂ + H₂O

Step 3: Balance charge.

Oxidation: Left: 0 + 4(-1) = -4; Right: -2 → add 2e⁻ to right:
Hg + 4Cl⁻ → HgCl₄²⁻ + 2e⁻

Reduction: Left: -1 + 2 = +1; Right: 0 → add 1e⁻ to left:
NO₃¹⁻ + 2H⁺ + e⁻ → NO₂ + H₂O

Step 4: Equalize electrons. Multiply reduction by 2:

2NO₃¹⁻ + 4H⁺ + 2e⁻ → 2NO₂ + 2H₂O

Step 5: Add:

Hg + 4Cl⁻ + 2NO₃¹⁻ + 4H⁺ + 2e⁻ → HgCl₄²⁻ + 2e⁻ + 2NO₂ + 2H₂O
Cancel electrons: Hg + 4Cl⁻ + 2NO₃¹⁻ + 4H⁺ → HgCl₄²⁻ + 2NO₂ + 2H₂O

Check: Atoms and charge?
Left: Hg=1, Cl=4, N=2, O=6, H=4; charge = 0 -4 -2 +4 = -2
Right: Hg=1, Cl=4, N=2, O=4+2=6, H=4; charge = -2 +0 +0 = -2 ✔

Oxidized: Hg
Reduced: NO₃¹⁻

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d. AsH₃ + Zn²⁺ → H₃AsO₄ + Zn (acidic)

Step 1: Half-reactions.

Oxidation: AsH₃ → H₃AsO₄
Reduction: Zn²⁺ → Zn

Step 2: Balance atoms.

Oxidation: As is balanced. Add water for oxygen:
AsH₃ + 4H₂O → H₃AsO
Balance H: left has 3+8=11H, right has 3H → add 8H⁺ to right:
AsH₃ + 4H₂O → H₃AsO₄ + 8H⁺

Reduction: Zn²⁺ → Zn (balanced)

Step 3: Balance charge.

Oxidation: Left: 0; Right: 0 + 8 = +8 → add 8e⁻ to right:
AsH₃ + 4H₂O → H₃AsO + 8H⁺ + 8e⁻

Reduction: Zn²⁺ + 2e⁻ → Zn

Step 4: Equalize electrons. Multiply reduction by 4:

4Zn²⁺ + 8e⁻ → 4Zn

Step 5: Add:

AsH₃ + 4H₂O + 4Zn²⁺ + 8e⁻ → H₃AsO₄ + 8H⁺ + 8e⁻ + 4Zn
Cancel electrons: AsH₃ + 4H₂O + 4Zn²⁺ → H₃AsO₄ + 8H⁺ + 4Zn

Check: Atoms and charge?
Left: As=1, H=3+8=11, O=4, Zn=4; charge = 0+0+8 = +8
Right: As=1, H=3+8=11, O=4, Zn=4; charge = 0+8+0 = +8 ✔

Oxidized: AsH₃
Reduced: Zn²⁺

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e. CN¹⁻ + MnO₄¹⁻ → CNO¹⁻ + MnO₂ (basic)

Step 1: Half-reactions.

Oxidation: CN¹⁻ → CNO¹⁻
Reduction: MnO₄¹⁻ → MnO₂

Step 2: Balance atoms (in basic, we’ll adjust later).

Oxidation: CN¹⁻ → CNO¹⁻
Add H₂O to left for oxygen: CN¹⁻ + H₂O → CNO¹⁻
Balance H: add 2H⁺ to right → CN¹⁻ + H₂O → CNO¹⁻ + 2H⁺

Reduction: MnO₄¹⁻ → MnO₂
Add 2H₂O to right for oxygen: MnO₄¹⁻ → MnO₂ + 2H₂O
Add 4H⁺ to left: MnO₄¹⁻ + 4H⁺ → MnO₂ + 2H₂O

Step 3: Balance charge.

Oxidation: Left: -1; Right: -1 + 2 = +1 → add 2e⁻ to right:
CN¹⁻ + H₂O → CNO¹⁻ + 2H⁺ + 2e⁻

Reduction: Left: -1 + 4 = +3; Right: 0 → add 3e⁻ to left:
MnO₄¹⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

Step 4: Equalize electrons. LCM of 2 and 3 is 6.

Multiply oxidation by 3:
3CN¹⁻ + 3H₂O → 3CNO¹⁻ + 6H⁺ + 6e⁻

Multiply reduction by 2:
2MnO₄¹⁻ + 8H⁺ + 6e⁻ → 2MnO₂ + 4H₂O

Step 5: Add:

3CN¹⁻ + 3H₂O + 2MnO₄¹⁻ + 8H⁺ + 6e⁻ → 3CNO¹⁻ + 6H⁺ + 6e⁻ + 2MnO₂ + 4H₂O

Cancel electrons and simplify H⁺ and H₂O:

H⁺: 8 on left, 6 on right → leave 2H⁺ on left
H₂O: 3 on left, 4 on right → move 1 H₂O to left → net: 1 H₂O on right

So: 3CN¹⁻ + 2MnO₄¹⁻ + 2H⁺ → 3CNO¹⁻ + 2MnO₂ + H₂O

But this is acidic! Convert to basic: add 2OH⁻ to both sides to neutralize 2H⁺:

3CN¹⁻ + 2MnO₄¹⁻ + 2H⁺ + 2OH⁻ → 3CNO¹⁻ + 2MnO₂ + H₂O + 2OH⁻
→ 3CN¹⁻ + 2MnO₄¹⁻ + 2H₂O → 3CNO¹⁻ + 2MnO₂ + H₂O + 2OH⁻

Simplify H₂O: subtract 1 H₂O from both sides:

3CN¹⁻ + 2MnO₄¹⁻ + H₂O → 3CNO¹⁻ + 2MnO₂ + 2OH⁻

Check: Atoms and charge?
Left: C=3, N=3, Mn=2, O=8+1=9, H=2; charge = -3 -2 +0 = -5
Right: C=3, N=3, Mn=2, O=3+4+2=9, H=2; charge = -3 +0 -2 = -5 ✔

Oxidized: CN¹⁻
Reduced: MnO₄¹⁻

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f. H₂O₂ + ClO₂ → ClO₂¹⁻ + O₂ (basic)

Step 1: Half-reactions.

Oxidation: H₂O₂ → O₂
Reduction: ClO₂ → ClO₂¹⁻

Step 2: Balance atoms.

Oxidation: H₂O₂ → O₂ + 2H⁺ (balance H)
Reduction: ClO₂ → ClO₂¹⁻ (Cl and O balanced)

Step 3: Balance charge.

Oxidation: Left: 0; Right: 0 + 2 = +2 → add 2e⁻ to right:
H₂O₂ → O₂ + 2H⁺ + 2e⁻

Reduction: Left: 0; Right: -1 → add 1e⁻ to left:
ClO₂ + e⁻ → ClO₂¹⁻

Step 4: Equalize electrons. Multiply reduction by 2:

2ClO₂ + 2e⁻ → 2ClO₂¹⁻

Step 5: Add:

H₂O₂ + 2ClO₂ + 2e⁻ → O₂ + 2H⁺ + 2e⁻ + 2ClO₂¹⁻
Cancel electrons: H₂O₂ + 2ClO₂ → O₂ + 2H⁺ + 2ClO₂¹⁻

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

H₂O₂ + 2ClO₂ + 2OH⁻ → O₂ + 2H⁺ + 2OH⁻ + 2ClO₂¹⁻
→ H₂O₂ + 2ClO₂ + 2OH⁻ → O₂ + 2H₂O + 2ClO₂¹⁻

Check: Atoms and charge?
Left: H=2+2=4, O=2+4+2=8, Cl=2; charge = 0+0-2 = -2
Right: H=4, O=2+2+4=8, Cl=2; charge = 0+0-2 = -2 ✔

Oxidized: H₂O₂
Reduced: ClO₂

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g. ClO¹⁻ + CrO₂¹⁻ → CrO₄²⁻ + Cl₂ (basic)

Step 1: Half-reactions.

Reduction: ClO¹⁻ → Cl₂
Oxidation: CrO₂¹⁻ → CrO₄²⁻

Step 2: Balance atoms.

Reduction: 2ClO¹⁻ → Cl₂ (balance Cl)
Add 2H₂O to right for oxygen: 2ClO¹⁻ → Cl₂ + 2H₂O
Add 4H⁺ to left: 2ClO¹⁻ + 4H⁺ → Cl₂ + 2H₂O

Oxidation: CrO₂¹⁻ → CrO₄²⁻
Add 2H₂O to left for oxygen: CrO₂¹⁻ + 2H₂O → CrO₄²⁻
Add 4H⁺ to right: CrO₂¹⁻ + 2H₂O → CrO₄²⁻ + 4H⁺

Step 3: Balance charge.

Reduction: Left: 2(-1) + 4 = +2; Right: 0 → add 2e⁻ to left:
2ClO¹⁻ + 4H⁺ + 2e⁻ → Cl₂ + 2H₂O

Oxidation: Left: -1; Right: -2 + 4 = +2 → add 3e⁻ to right:
CrO₂¹⁻ + 2H₂O → CrO₄²⁻ + 4H⁺ + 3e⁻

Step 4: Equalize electrons. LCM of 2 and 3 is 6.

Multiply reduction by 3:
6ClO¹⁻ + 12H⁺ + 6e⁻ → 3Cl₂ + 6H₂O

Multiply oxidation by 2:
2CrO₂¹⁻ + 4H₂O → 2CrO₄²⁻ + 8H⁺ + 6e⁻

Step 5: Add:

6ClO¹⁻ + 12H⁺ + 6e⁻ + 2CrO₂¹⁻ + 4H₂O → 3Cl₂ + 6H₂O + 2CrO₄²⁻ + 8H⁺ + 6e⁻

Cancel electrons, H⁺, H₂O:

H⁺: 12 - 8 = 4H⁺ left
H₂O: 4 - 6 = -2 → so 2H₂O on right

So: 6ClO¹⁻ + 2CrO₂¹⁻ + 4H⁺ → 3Cl₂ + 2CrO₄²⁻ + 2H₂O

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

6ClO¹⁻ + 2CrO₂¹⁻ + 4H⁺ + 4OH⁻ → 3Cl₂ + 2CrO₄²⁻ + 2H₂O + 4OH⁻
→ 6ClO¹⁻ + 2CrO₂¹⁻ + 4H₂O → 3Cl₂ + 2CrO₄²⁻ + 2H₂O + 4OH⁻

Simplify H₂O: subtract 2H₂O from both sides:

6ClO¹⁻ + 2CrO₂¹⁻ + 2H₂O → 3Cl₂ + 2CrO₄²⁻ + 4OH⁻

Check: Atoms and charge?
Left: Cl=6, O=6+4+2=12, Cr=2, H=4; charge = -6 -2 +0 = -8
Right: Cl=6, O=8+4=12, Cr=2, H=4; charge = 0 -4 -4 = -8 ✔

Oxidized: CrO₂¹⁻
Reduced: ClO¹⁻

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Final Answer:

a. Balanced: Cu²⁺ + 3I⁻ → Cu + I₃¹⁻
Oxidized: I⁻, Reduced: Cu²⁺

b. Balanced: MnO₄¹⁻ + 8H⁺ + 5Ag → Mn²⁺ + 4H₂O + 5Ag¹⁺
Oxidized: Ag, Reduced: MnO₄¹⁻

c. Balanced: Hg + 4Cl⁻ + 2NO₃¹⁻ + 4H⁺ → HgCl₄²⁻ + 2NO₂ + 2H₂O
Oxidized: Hg, Reduced: NO₃¹⁻

d. Balanced: AsH₃ + 4H₂O + 4Zn²⁺ → H₃AsO₄ + 8H⁺ + 4Zn
Oxidized: AsH₃, Reduced: Zn²⁺

e. Balanced: 3CN¹⁻ + 2MnO₄¹⁻ + H₂O → 3CNO¹⁻ + 2MnO₂ + 2OH⁻
Oxidized: CN¹⁻, Reduced: MnO₄¹⁻

f. Balanced: H₂O₂ + 2ClO₂ + 2OH⁻ → O₂ + 2H₂O + 2ClO₂¹⁻
Oxidized: H₂O₂, Reduced: ClO₂

g. Balanced: 6ClO¹⁻ + 2CrO₂¹⁻ + 2H₂O → 3Cl₂ + 2CrO₄²⁻ + 4OH⁻
Oxidized: CrO₂¹⁻, Reduced: ClO¹⁻