Q4. Balance MnO4− + SO2 → Mn2+ + HSO4− in acidic solution (NCERT Exercise 7.18 (b)).
Reduction: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (× 2). Oxidation: SO2 + 2H2O → HSO4− + 3H+ + 2e− (× 5). Adding: 2MnO4− + 16H+ + 5SO2 + 10H2O → 2Mn2+ + 8H2O + 5HSO4− + 15H+. Cancelling 8H2O and 15H+: 2MnO4− + 5SO2 + 2H2O + H+ → 2Mn2+ + 5HSO4−. Charge: −1 on each side.
Q5. Balance Cr2O72− + SO2 → Cr3+ + SO42− in acidic solution (NCERT Exercise 7.18 (d)).
Cr falls 6 per Cr2O72− (+6 → +3, two atoms); S rises 2 (+4 → +6), so three SO2. Cr2O72− + 3SO2 → 2Cr3+ + 3SO42− has charge −2 on the left and 0 on the right, so add 2H+ on the left; one H2O on the right then balances hydrogen: Cr2O72− + 3SO2 + 2H+ → 2Cr3+ + 3SO42− + H2O. Charge: 0 on each side.
Q6. Mn3+ is unstable in solution and disproportionates to Mn2+, MnO2 and H+. Write the balanced ionic equation (NCERT Exercise 7.21).
One Mn3+ goes down to +2 (gains 1e−) and one goes up to +4 (loses 1e−): 2Mn3+ + 2H2O → Mn2+ + MnO2 + 4H+. Charge: +6 on each side.
Q7. Excess chlorine in drinking water is removed with sulphur dioxide. Write the balanced equation for the reaction in water (NCERT Exercise 7.23).
Cl2 + SO2 + 2H2O → 2Cl− + SO42− + 4H+, or Cl2 + SO2 + 2H2O → 2HCl + H2SO4. Chlorine is reduced (0 → −1) and sulphur oxidised (+4 → +6).
Q8. Thiosulphate gives tetrathionate with iodine but sulphate with bromine. Explain using oxidation numbers, and balance the bromine reaction (NCERT Exercise 7.14).
Iodine is a mild oxidant: it takes sulphur only from +2 (average) in S2O32− to +2.5 in S4O62−, 2S2O32− + I2 → S4O62− + 2I−. Bromine is stronger and takes sulphur all the way to +6. Each S2O32− then loses 8 electrons, so it needs 4Br2: S2O32− + 4Br2 + 5H2O → 2SO42− + 8Br− + 10H+. NCERT prints 2Br2 and 4Br−, which leaves the charges unbalanced (−2 on the left, +2 on the right).
Q9. Sulphur dioxide and hydrogen peroxide can act as both oxidising and reducing agents, but ozone and nitric acid act only as oxidants. Why (NCERT Exercise 7.8)?
In SO2, sulphur is +4, between its lowest (−2) and highest (+6) states, and in H2O2 oxygen is −1, between −2 and 0, so each can go either way. In HNO3, nitrogen is already at its highest state, +5, so it can only be reduced. In O3, oxygen is at 0 and can in practice only go down (only fluorine could make it positive), so ozone acts as an oxidant.
Q10. In Ostwald’s process, 4NH3 + 5O2 → 4NO + 6H2O. What is the maximum mass of NO from 10.00 g of NH3 and 20.00 g of O2 (NCERT Exercise 7.25)? (N = 14.0, H = 1.0, O = 16.0)
NH3: 10.00/17.0 = 0.588 mol, which would need 0.588 × 5/4 = 0.735 mol O2. Only 20.00/32.0 = 0.625 mol O2 is present, so oxygen is the limiting reagent. NO = 0.625 × 4/5 = 0.500 mol, that is 0.500 × 30.0 = 15.0 g.
Q11. 20.0 mL of a copper(II) solution is treated with excess KI, and the iodine freed needs 25.0 mL of 0.100 M sodium thiosulphate. Find the concentration of Cu2+.
2Cu2+ + 4I− → Cu2I2 + I2 and I2 + 2S2O32− → 2I− + S4O62−, so Cu2+ : S2O32− = 1 : 1. Thiosulphate = 0.100 mol L−1 × 0.0250 L = 2.50 × 10−3 mol = moles of Cu2+; [Cu2+] = 2.50 × 10−3/0.0200 = 0.125 M.
Q12. Balance P4 + OH− → PH3 + H2PO2− in basic solution and name the oxidising and reducing agents (from NCERT Exercise 7.19 (a)).
P goes 0 → −3 in PH3 and 0 → +1 in H2PO2−, so for one PH3 there are three H2PO2−. P4 → PH3 + 3H2PO2− has charge 0 on the left and −3 on the right, so add 3OH− on the left. Hydrogen is then 3 on the left and 9 on the right, so add 3H2O on the left: P4 + 3OH− + 3H2O → PH3 + 3H2PO2−. Phosphorus is both the oxidising and the reducing agent (disproportionation). NCERT’s exercise prints the product as HPO2−; the hypophosphite ion is H2PO2−, as in NCERT’s own reaction 7.46.