On balancing the given redox reaction,
aCr₂O₇²⁻ + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + (

Q: What is NEET?

NEET is a competitive examination conducted in India. This question is from the 2023 NEET Chemistry paper.

Q: How can I prepare for NEET Chemistry?

To prepare for NEET Chemistry, practice previous year questions (PYQs), understand concepts thoroughly, and solve mock tests regularly. This 2023 question covers Redox Reactions topic.

Q: What is the difficulty level of this question?

This NEET 2023 Chemistry question has a medium difficulty level and is marked as an important question.

Q: Where can I find more NEET previous year questions?

You can find more NEET previous year questions on AspirantMitraa. Browse by subject, topic, year, and difficulty level to practice effectively.

Question

On balancing the given redox reaction,
aCr₂O₇²⁻ + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + (c/2)H₂O(l)
the coefficients a, b and c are found to be, respectively-

Accepted Answer

Correct Answer: A — a = 1, b = 3, c = 8Step-by-step Solution (Ion-electron/Half-reaction method):Step 1 — Identify oxidation state changes:Cr in Cr₂O₇²⁻: +6. Cr in Cr³⁺: +3. Each Cr gains 3e⁻. Two Cr per dichromate → gain of 6e⁻ per Cr₂O₇²⁻.S in SO₃²⁻: +4. S in SO₄²⁻: +6. Each S loses 2e⁻ per SO₃²⁻.Step 2 — Balance electrons transferred:Oxidation (SO₃²⁻ → SO₄²⁻): 1 mol SO₃²⁻ loses 2e⁻.Reduction (Cr₂O₇²⁻ → 2Cr³⁺): 1 mol Cr₂O₇²⁻ gains 6e⁻.To balance electrons: 1 Cr₂O₇²⁻ requires 3 SO₃²⁻ (3 × 2e⁻ = 6e⁻). So a = 1, b = 3.Step 3 — Write half-reactions and balance fully:Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂OOxidation: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻. Multiply by 3: 3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻Step 4 — Add half-reactions:Cr₂O₇²⁻ + 14H⁺ + 3SO₃²⁻ + 3H₂O → 2Cr³⁺ + 7H₂O + 3SO₄²⁻ + 6H⁺Simplify: Cr₂O₇²⁻ + 8H⁺ + 3SO₃²⁻ → 2Cr³⁺ + 4H₂O + 3SO₄²⁻Therefore: a = 1, b = 3, c = 8. ✓Verification: Charge balance: LHS = (–2) + 3(–2) + 8(+1) = –2 – 6 + 8 = 0. RHS = 2(+3) + 3(–2) = 6 – 6 = 0. ✓Key Concepts:Ion-electron (half-reaction) method: Separately balance oxidation and reduction half-reactions, then combine after equalising electrons.In acidic medium: Use H⁺ and H₂O to balance O and H. In basic medium: use OH⁻ and H₂O.Cr₂O₇²⁻ is a powerful oxidising agent in acidic solution (orange → green as Cr³⁺ is formed).Oxidation states: S in SO₃²⁻ is +4; in SO₄²⁻ is +6. Cr in Cr₂O₇²⁻ is +6; in Cr³⁺ is +3.

On balancing the given redox reaction, aCr₂O₇²⁻ + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + (c/2)H₂O(l) the coefficients a, b and c are found to be, respectively-

Master NEET with Our Test Series

More Questions from Redox Reactions

Frequently Asked Questions

On balancing the given redox reaction, aCr₂O₇²⁻ + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + (c/2)H₂O(l) the coefficients a, b and c are found to be, respectively-

Master NEET with Our Test Series

More Questions from Redox Reactions

Frequently Asked Questions