RI 01. Redox Reactions Tutorial (Suggested Answers to Self-Check Questions)

-1- Raffles Institution Year 5 H2 Chemistry 2025 Tutorial 1 – Redox Reactions Answers to Self-Check Questions 1 Let the oxidation number of the underlined atoms in the following species be x. (a) Note: Ba2+ & SO42– 2 4( 2) 0 6 x x ++ + − = =+ (b) Note: Na+ & VO43– 3( 1) 4( 2) 0 5 x x +++−= =+ (c) ( 2) 2 4 x x +− = + =+ (d) 2 4( 2) 2 3 x x +−= − =+ (e) 2( 1) 2 0 1 x x ++ = =− (f) 2 4( 2) 0 4 x x +−= =+ (g) Note: F is more electronegative than Br. ( 1) 0 1 x x +− = =+ (h) Note: F is more electronegative than I. 7( 1) 0 7 x x +−= =+ 2 (a) –3 –1 –3 –1 NH3(aq) + HCl(aq) → NH4Cl(aq) This is not a redox reaction since there is no change in initial and final oxidation states for all elements. (b) 0 +1 +2 0 3Mg(s) + 2NH3(g) → Mg3N2(s) + 3H2(g) This is a redox reaction where the oxidation number of H decreases from +1 in NH3 to 0 in H2, and the oxidation number of Mg increases from 0 in Mg to +2 in Mg3N2. 3 (a) 2Na2S2O3 + I2 → Na2S4O6 + 2NaI 33 223Amount of Na S O required 15.0 10 0.100 1.50 10 mol−−=×× =× Amount of I2 produced = 1.50 x 10ˉ3 ÷ 2 = 7.50 x 10ˉ4 mol (b) O3 + 2KI + H2O → I2 + O2 + 2KOH 4 3Amount of O in the mixture 7.50 10 mol−= × 4 33 3Volume of O in the mixture 7.50 10 22.7 0.01703 dm 17.03 cm−=××= = 3 17.03% of O in the mixture 100% 3.41%500=×= 4 (a) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 33Amount of HC reacted 22.50 10 0.100 2.250 10 mol−−= ×× = ×l 33Amount of NaOH in 25.0cm of solution 2.2 50 10 mol−= ×H 3 33 2.250 10Amount of NaOH in 100cm of solution 100 9 .00 10 mol25.0 − −×= ×= ×H

-2- (b) 2MnO4−(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 5O2(g) + 8H2O(l) 34 4Amount of MnO reacted 10.0 10 0.0200 2.00 10 mol−− − =×× =× 4 34 22 2.00 10Amount of H O in 25.0cm of solution 5 5.00 10 mol2 − −×= ×= ×H 4 33 22 5.00 10Amount of H O in 100cm of solution 100 2.0 0 10 mol25.0 − −×= ×= ×H (c) Na2O2 + 2H2O → 2NaOH + H2O2 Amount of Na2O2 = Amount of H2O2 produced = 2.00 × 10−3 mol Amount of NaOH from Na2O2 = 2 × 2.00 × 10−3 = 4.00 × 10−3 mol Amount of NaOH from Na2O = 9.00 × 10−3 − 4.00 × 10−3 = 5.00 × 10−3 mol Na2O + H2O → 2NaOH Amount of Na2O = 5.00 × 10−3 ÷ 2 = 2.50 × 10−3 mol