RI 2025 Tut(Ans) Redox Reactions (Ans) for students

-1- Raffles Institution Year 5 H2 Chemistry 2025 Tutorial 1 – Redox Reactions (Suggested Answers) Practice Questions 5 (a) +5 –2 +3 0 2KNO3(s) → 2KNO2(s) + O2(g) This is a redox reaction where the oxidation number of N decreases from +5 in KNO3 to +3 in KNO2, and the oxidation number of O increases from -2 in KNO3 to 0 in O2. (b) +6 +6 2CrO42−(aq) + 2H+(aq) → Cr2O72−(aq) + H2O(l) This is not a redox reaction since there is no change in initial and final oxidation states for all elements. (c) +1 0 +2 Cu2O(s) + H2SO4(aq) → Cu(s) + CuSO4(aq) + H2O(l) This is a redox reaction where Cu is both reduced and oxidised. (d) -4 0 +2 -2 2CH4(g) + 2NH3(g) + 3O2(g) → 2HCN(g) + 6H2O(g) This is a redox reaction where the oxidation number of O decreases from 0 in O2 to -2 in H2O, and the oxidation number of C increases from -4 in CH4 to +2 in HCN. 6 (a) Oxidation: 2I− → I2 + 2e– Reduction: H2SO4 + 8H+ + 8e– → H2S + 4H2O Overall: 8I−(aq) + H2SO4(aq) + 8H+(aq) → 4I2(g) + H2S(g) + 4H2O(l) (b) Oxidation: MnO42− → MnO4– + e− Reduction: MnO42− + 4H+ + 2e– → MnO2 + 2H2O Overall: 3MnO42−(aq) + 4H+(aq) → MnO2(s) + 2MnO4−(aq) + 2H2O(l) This is another example of a disproportionation reaction. (c) Note: Both the cation and the anion in FeC2O4 undergo oxidation. Fe2+(aq) → Fe3+(aq) + e− C 2O42–(aq) → 2CO2(g) + 2e− Oxidation: FeC2O4 → Fe3+ + 2CO2 + 3e− Reduction: Ce3+(aq) + e− → Ce2+(aq) Overall: FeC2O4(aq) + 3Ce3+(aq) → Fe3+(aq) + 2CO2(g) + 3Ce2+(aq)

-2- 7 (a) Oxidation: Mn(OH)2 + 2OH− → MnO2 + 2H2O + 2e− Reduction: ClO− + H2O + 2e− → Cl− + 2OH− Overall: C lO−(aq) + Mn(OH)2(s) → Cl−(aq) + MnO2(s) + H2O(l) (b) Oxidation: C2O42− + 4OH− → 2CO32– + 2H2O + 2e− Reduction: MnO4− + 2H2O + 3e− → MnO2 + 4OH− Overall: 2MnO4−(aq) + 3C2O42−(aq) + 4OH−(aq) → 2MnO2(s) + 6CO32– (aq) + 2H2O(l) (c) Oxidation: ClO− + 4OH– → ClO3− + 2H2O + 4e− Reduction: ClO– + H2O + 2e− → Cl− + 2OH− Overall: 3C lO−(aq) → 2Cl−(aq) + ClO3−(aq) This is another example of a disproportionation reaction. 8 (a) 3SO2 + Cr2O72− + 2H+ → 2Cr3+ + 3SO42− + H2O [Oxidation: SO2 + 2H2O → SO42− + 4H+ + 2e– & Reduction: Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O] (b) 3SO2 + Cr2O72− + 2H+ → 3SO42− + 2Cr3+ + H2O Na2S2O5 + 2HCl → 2NaCl + 2SO2 + H2O Ratio of Cr2O72− : SO2 : Na2S2O5 = 1:3: 3 2 Amount of + Cr2O72− reacted = 14.20 × 10−3 × 0.0100 = 1.420 × 10−4 mol Amount Na2S2O5 reacted = 1.420 × 10−3 × 3 2 = 2.130 × 10−4 mol −= × × × +× +× =4 22 5Mass of Na S O present 2.130 10 (2 23.0 2 32.1 5 16.0) 0.04051 g 6 22 5 0