RI 01 Redox Reactions Tutorial (Ans)

-1- Raffles Institution Year 5 H2 Chemistry 2024 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. A disproportionation reaction is a redox reaction in which the same element in a substance is simultaneously oxidised and reduced. If the question asks to state the type of reaction, a better answer would be disproportionation rather than redox. (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− C2O42–(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: ClO−(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: 3ClO−(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 2 3 4 27Amount of Cr O reacted 14.20 10 0.0100 1.420 1 0 mol− − − =   =  44 2Amount of SO produced 3 1.420 10 4.260 10 mol−−=   =  Na2S2O5 + 2HCl → 2NaCl + 2SO2 + H2O 44 2 2 5Amount of Na S O present 4.260 10 2 2.130 10 mo l−−=   =  −=    +  +  =4 2 2 5Mass of Na S O present 2.130 10 (2 23.0 2 32.1 5 1 6.0) 0.04051 g 6 2 2 5 0.04051Concentration of Na S O in ppm 10 405 ppm100=  = Mass of Na2S2O5 / g : Mass of preserved meat / g 0.04051 : 100 0.04051 100 : 1 60.04051 10 405 100 = : 106 9 Since Oxidation: H2O2 → O2 + 2H+ + 2e– or 4H2O2 → 4O2 + 8H+ + 8e– 27Cl On : en − : 22HOn 1 : : 4 1 : 8 : 4 +7 Reduction (unbalanced): Cl2O7 + 8e– → ?  Each Cl atom accepted 4 e–.  Oxidation number of Cl decreases from +7 to +3. –1 +1 +3 +5 A Cl– B ClO– C ClO2– D ClO3– Answer: C

-3- 10 2NaNOn : 4 2 4()NH SOn +3 ? –3 Since NO2– Final Product NH4+ Accept x mol of e– Lose x mol of e–  New oxidation number of N = 0 0.010 : 0.005 2NOn − : en − : 4NHn + 0.010 : : 0.010 1 : x : 1 0 +2 +1 –1 A N2 B NO C N2O D NH2OH Answer: A 11 (a) 2NH2OH → N2 + 2H2O + 2H+ + 2e− 2NH2OH → N2O + H2O + 4H+ + 4e− (b) 3 3 3Amount of Fe reduced 10 10 0.50 5.00 10 mol+ − − =   =  33 2Amount of NH OH reacted 50 10 0.050 2.50 10 mo l−−=   =  2NH OHn : en − : 3Fen + Since Reduction: Fe3+ + e– → Fe2+ or 2Fe3+ + 2e– → 2Fe2+ Oxidation: NH2OH → product + 2e– 2.50 x 10–3 : : 5.00 x 10–3 1 : : 2 1 : 2 : 2 Using NH2OH → product + 2e–, oxidation number of N in the reactant = oxidation number of N in product + (−2) Oxidation number of N in product = −1 − (−2) = +1 NH2OH is oxidised to N2O. The relevant oxidation half-equation is therefore 2NH2OH → N2O + H2O + 4H+ + 4e− in which 1 mole of NH2OH loses 2 mol of electrons. NH2OH is oxidised to N2O. Overall equation: 2NH2OH + 4Fe3+ → N2O + 4Fe2+ + H2O + 4H+ 12 (a) 2 2 3 2 3 2 23 12.1[Na S O .5H O] in 2 23.0 2 32.1 3 16.0 5(2 1.0 16.0) 0.04875 mol dm [S O ] in − − =  +  +  +  + = = FA1 FA1 I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O62–(aq) 2 3 3 23Amount of S O used 24.40 10 0.04875 1.190 10 m ol− − − =   =  I 34 2Amount of liberated 1.190 10 2 5.948 10 mol−−=   =  Concentration of XO3− ions in FA2 = 7.95 x 10–3 mol dm–3

-4- II 33 32 4 Amount of XO which reacted with to form 25.0 10 7.95 10 1.988 10 mol − − − − − =    = 3XOn − : 2 nI 1.988 x 10–4 : 5.948 x 10–4 1 : 3 Hence, 3 mol of I2 is liberated by 1 mol of XO3−. (Note: It must be an integer.) (b) Since Oxidation: 2I− → I2 + 2e– or 6I− → 3I2 + 6e– 2 nI : en − : 3XOn − 3 : : 1 3 : 6 : 1 +5 Reduction (unbalanced): XO3− + 6e– → ?  Oxidation number of X decreases from +5 to –1. Hence, the product obtained is X− where the oxidation number of X is −1. (c) Oxidation: 2I− → I2 + 2e− Reduction: XO3− + 6H+ + 6e− → X− + 3H2O Overall: XO3−(aq) + 6I−(aq) + 6H+(aq) → X−(aq) + 3H2O(l) + 3I2(aq) 13 Amount of thiosulfate ions = 0.0500  24.80 1000 = 1.24  10−3 mol Mole ratio of Cu2+ : I2 : S2O32− = 2 : 1 : 2 Amount of Cu2+ in 25.0 cm3 of diluted solution = amount of S2O32− = 1.24  10−3 mol [Cu2+] in diluted solution = 1.24×10-3 25.0×10-3 = 0.0496 mol dm⁻3 [Cu2+] in original undiluted solution = 0.0496 x 250 41 = 0.302 mol dm⁻3

-5- Colour changes as titration progresses Just before titration begins. Addition of starch indicator during titration. End point of titration. Colour of solution changes from dark blue to colourless, but pale cream ppt remains. 14 9 (a) H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l) 33Amount of NaOH used 14.75 10 0.100 1.475 10 m ol−−=   =  34 2 2 4Amount of H C O present 1.475 10 2 7.375 10 mol−−=   =    4 3 2 2 4 3 7.375 10H C O in solution 0.0295 mol dm25.0 10 − − − ==  9 (b) (i) oxidation: C2O42ˉ(aq) → 2CO2(g) + 2eˉ reduction: MnO4ˉ(aq) + 8H+(aq) + 5eˉ → Mn2+(aq) + 4H2O(l) Overall: 2MnO4−(aq) + 5C2O42−(aq) + 16H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) 9 (ii) 34 4Amount of MnO used 32.00 10 0.0205 6.560 10 m ol− − − =   =  4 23 24 6.560 10Total amount of C O present 5 1.64 10 mol2 − −− =  =  (c) 24 2 4 2 2 4Amount of C O from H C O 7.375 10 mol−− = 2 3 4 4 2 4 2 2 4 2 2 4 Amount of C O from Na C O 1.640 10 7.375 10 9.02 5 10 mol Amount of Na C O − − − − =  −  =  = Hence,   4 3 2 2 4 3 9.025 10Na C O in solution 0.0361 mol dm25.0 10 − − − == 