Q.Balance the following redox reaction by ion-electron method. Cr2O7^-2(aq) + SO2(g) -> Cr^+3(aq) + SO4^-2(aq) (in acidic medium)
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Start your 14-day free trial to unlock the full solution →Splitting into reduction (Cr₂O₇²⁻ → Cr³⁺) and oxidation (SO₂ → SO₄²⁻) half-reactions, balancing atoms/charge with H⁺ and H₂O, equalising electrons, and adding gives Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.
Step 1 — Write the two half-reactions (unbalanced, skeletal):
Reduction: Cr₂O₇²⁻ → Cr³⁺
Oxidation: SO₂ → SO₄²⁻
Step 2 — Balance atoms other than O and H:
Reduction: Cr₂O₇²⁻ → 2Cr³⁺ (Cr balanced: 2 = 2)
Oxidation: SO₂ → SO₄²⁻ (S balanced: 1 = 1)
Step 3 — Balance O by adding H₂O, then balance H by adding H⁺ (acidic medium):
Reduction: Cr₂O₇²⁻ has 7 O, Cr³⁺ side has none, so add 7H₂O to the right:
Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
Now balance H (14 H on right) by adding 14H⁺ to the left:
Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
Oxidation: SO₄²⁻ has 4 O, SO₂ has 2 O, so add 2H₂O to the left:
SO₂ + 2H₂O → SO₄²⁻
Now balance H (4 H on left) by adding 4H⁺ to the right:
SO₂ + 2H₂O → SO₄²⁻ + 4H⁺
Step 4 — Balance charge by adding electrons:
Reduction: left charge = −2 + 14 = +12; right charge = 2(+3) = +6. Add 6e⁻ to the left to balance:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Oxidation: left charge = 0; right charge = −2 + 4 = +2. Add 2e⁻ to the right:
SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻
Step 5 — Equalise electrons (LCM of 6 and 2 = 6): …
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