Q.Balance the following reactions by oxidation number method
a. Cr2O7²⁻ (aq) + SO3²⁻ (aq) → Cr³⁺ (aq) + SO4²⁻ (aq) (acidic)
b. MnO4⁻ (aq) + Br⁻ (aq) → MnO2 (s) + BrO3⁻ (aq) (basic)
c. H2SO4 (aq) + C (s) → CO2

a. Cr2O7²⁻ + SO3²⁻ → Cr³⁺ + SO4²⁻ (acidic). Cr falls +6→+3 (gains 3e⁻ per Cr, x2 = 6e⁻ per Cr2O7²⁻); S rises +4→+6 (loses 2e⁻ per S). Equalising 6e⁻ needs 3 SO3²⁻ per Cr2O7²⁻: Cr2O7²⁻ + 3SO3²⁻ → 2Cr³⁺ + 3SO4²⁻. Balancing O (16 on the left vs 12 on the right) and H (acidic medium, using H⁺) gives the final form: Cr2O7²⁻(aq) + 3SO3²⁻(aq) + 8H⁺(aq) → 2Cr³⁺(aq) + 3SO4²⁻(aq) + 4H2O(l). Check: charge both sides = 0.

b. MnO4⁻ + Br⁻ → MnO2 + BrO3⁻ (basic). Mn falls +7→+4 (gains 3e⁻); Br rises -1→+5 (loses 6e⁻). Equalising needs 2 MnO4⁻ per Br⁻: 2MnO4⁻ + Br⁻ → 2MnO2 + BrO3⁻. Balancing O and H with H2O/H⁺ then converting to basic medium by adding OH⁻ gives: 2MnO4⁻(aq) + Br⁻(aq) + H2O(l) → 2MnO2(s) + BrO3⁻(aq) + 2OH⁻(aq). Check: O both sides = 9; charge both sides = -3.

c. H2SO4 + C → CO2 + SO2 + H2O (acidic). C rises 0→+4 (loses 4e⁻); S falls +6→+4 (gains 2e⁻ per S). Equalising needs 2 H2SO4 per C: C(s) + 2H2SO4(aq) → CO2(g) + 2SO2(g) + 2H2O(l). Check: C=1, H=4, S=2, O=8, all balanced on both sides already, with no extra H⁺/H2O needed. …

Oxidation number method (Section 6.3.1): equalise total oxidation-number increase and decrease first, then balance O with H …

• Forgetting the final OH⁻-conversion step for the two basic-medium equations (b, d), leaving H⁺ ions in a basic solution, which is chemically inconsistent.
• In (a), only multiplying SO3²⁻'s coefficient without also checking that the FULL equation (O and H) still balances after the electron-equalising step -- electron balance and atom balance are two separate checks, both re …