Skip to content
Question of 78

Q.Balance the following redox reaction by Ion Electron method:
Cr2O7^2-(aq) + SO2(g) → Cr3+(aq) + SO4^2-(aq) (Acidic)

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2024Subjective· 2mImportance★★★★★
0% · 0/78 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Writing and balancing the reduction half (Cr₂O₇²⁻ → Cr³⁺) and the oxidation half (SO₂ → SO₄²⁻) separately, then equalising and cancelling the 6 electrons, gives the fully balanced equation: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.

Step 1 — Write the two half-reactions (skeletal):

Cr2O72−→2Cr3+Cr_2O_7^{2-} \rightarrow 2Cr^{3+} (reduction, Cr goes from +6 to +3)

SO2→SO42−SO_2 \rightarrow SO_4^{2-} (oxidation, S goes from +4 to +6)

Step 2 — Balance O by adding H₂O, then balance H by adding H⁺ (acidic medium):

Reduction: Cr2O72−+14H+→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O

Oxidation: SO2+2H2O→SO42−+4H+SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+

Step 3 — Balance charge by adding electrons:

Reduction: left charge = −2+14=+12-2+14=+12, right charge =+6=+6; add 6e⁻ to the left:

Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

Oxidation: left charge = 0, right charge =−2+4=+2=-2+4=+2; add 2e⁻ to the right:

SO2+2H2O→SO42−+4H++2e−SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-

Step 4 — Equalise electrons (multiply oxidation half by 3) and add the two halves:

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.