Wording in the Original) 14 H (Aq) + Cr2o7 (Aq) + 6 I (Aq) → 2 Cr (Aq) + I2(S) + 7 H2O(L) 8 H+(Aq) + Mno –(Aq) + 5 Fe2+(Aq) → Mn2+(Aq) + 5 Fe3+(Aq) + 4 H O(L) 8

Chemistry 1010 Loader (Fall 2003) Page 1 of 3 Chemistry 1010

Handout 4: Reactions, equations and Redox 1. Write molecular and net-ionic (where applicable) equations (p) aluminum sulfate solution is mixed with barium hydroxide for the reactions that take place in the following experiments. solution → (a) Sodium hydroxide solution reacts with zinc nitrate Al2(SO4)3(aq) + 3 Ba(OH)2(aq) 3 BaSO4(s) + 2 Al(OH)3(s) 3+ 2– 2+ – → solution to give a precipitate of zinc hydroxide. 2 Al (aq) + 3 SO4 + 3 Ba + 6 OH (aq) 3 BaSO4(s) + 2 Al(OH)3(s)

NaOH(aq) + Zn(NO3)2(aq) → Zn(OH)2(s) + NaNO3 (aq) (q) lithium oxide(s) is added to water – 2+ → 2 OH (aq) + Zn (aq) → Zn(OH)2(s) Li2O(s) + H2O(l) 2 LiOH(aq) → + – (b) Copper(II) oxide dissolves in dilute hydrochloric acid. Li2O(s) + H2O(l) 2 Li (aq) + 2 OH (aq)

CuO(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l) (r) sodium bromide solution is added to an acidified (H2SO4) + 2+ CuO(s) + 2 H (aq) → Cu (aq) + H2O(l) potassium chromate solution → (c) Potassium chromate reacts with lead(II) nitrate solution 8 H2SO4(aq) + 6 NaBr(aq) + 2 K2CrO4(aq) 3 Br2(aq) + Cr2(SO4)3(aq) to give a yellow precipitate. + 8 H2O(l) + 2 K2SO4(aq) + 3 Na2SO4(aq) + – 2– 3+ 16 H (aq) + 6 Br (aq) + 2 CrO4 (aq) → 3 Br2(aq) + 2 Cr (aq) + 8 H2O(l) K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s) + 2 KNO3 (aq) 2– 2+ CrO4 (aq) + Pb (aq) → PbCrO4(s) 2. What is the oxidation number of the UNDERLINED atom? 2– (d) Magnesium ribbon reacts with dilute hydrochloric acid. (a) HCl (b) HOCl (c) S2O3 (d) As2O3 (e) 2– Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Na3AlCl6 (f) MnO4 (g) NH3 (h) Mg2P2O7 + → 2+ Mg(s) + 2 H (aq) Mg (aq) + H2(g) (a) -1 (b) +1 (c) +2 (d) +3 (e) +3 (f)+6 (g) -3 (h) +5 (e) Solid calcium carbonate reacts with dilute nitric acid to 3. Write the formula for each of the following compounds: give carbon dioxide gas. (a) potassium sulfide; K2S CaCO3(s) + 2 HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l) + 2+ (b) barium chloride dihydrate; BaCl2.2H2O CaCO3(s) + 2 H (aq) → Ca (aq) + CO2(g) + H2O(l) (f) Potassium metal reacts with cold water to give hydrogen (c) antimony(III) sulfide; Sb2S3 gas. (d) calcium hydride; CaH2

2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g) (e) potassium chlorate; KClO3 + – 2 K(s) + 2 H2O(l) → 2 K (aq)+ 2 OH (aq) + H2(g) (f) sulfurous acid; H2SO3(aq) (g) Sodium hydrogen carbonate is heated strongly in a (g) calcium sulfite dihydrate; CaSO3.2H2O crucible leaving a residue of sodium carbonate. (h) chromium(III) oxide; Cr2O3 → 2 NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) 4. Which of the following reactions is an oxidation-reduction (h) Potassium chlorate is heated strongly with a catalyst to (REDOX) reaction. Write the NET-IONIC equation for each produce oxygen gas. reaction. Indicate where applicable, (i) the species being

2 KClO3(s) → 2 KCl(s) + 3 O2(g) oxidized and the species being reduced; and (ii) the oxidant (oxidizing agent) and the reductant (reducing agent). (i) P4O10(s) react and dissolves in water to give a solution that turns blue litmus paper red. If REDOX reaction : Cu being reduced/oxidant S being oxidized/reductant P4O10(s) + 6 H2O(l) → 4 H3PO4(aq) → (j) Strontium oxide dissolves in water to give a solution that (a) 2 Cu2O + Cu2S 6 Cu + SO2 → turns red litmus paper blue. 2 Cu2O + Cu2S 6 Cu + SO2 (k) Aqueous barium nitrate reacts with aqueous lithium (b) AgNO3(aq) + NaCl(aq) → AgCl(aq) + NaNO3(aq) sulfate to give a precipitate of solid barium sulfate and Not redox. Double replacement reaction

aqueous lithium nitrate. (c) 2 Al(s) + 3 F2(g) → 2 AlF3(s) Ba(NO3)2(aq) + Li2SO4(aq) → BaSO4(s) + 2 LiNO3(aq) 2 Al(s) + 3 F2(g) → 2 AlF3(s) 2+ 2– Ba (aq) + SO4 (aq) → BaSO4(s) (d) 2 K2CrO4(aq) + H2SO4(aq) → K2Cr2O7(aq) + K2SO4(aq) (l) lead(II) nitrate solution is mixed with sodium sulfate Not redox solution (e) Cl2(aq) + 2 KI(aq) → I2(aq) + 2 KCl(aq) Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2 NaNO3(aq) 2– 2+ Cl2(aq) + 2 KI(aq) → I2(aq) + 2 KCl(aq) SO4 (aq) + Pb (aq) → PbSO4(s) → (m) sodium hydroxide solution is added to a solution of (f) Na2O(s) + H2O(l) 2 NaOH(aq) ammonium chloride (ammonia gas is evolved) Not redox

NaOH(aq) + NH4Cl(aq) → NaCl(aq) + NH3(g) + H2O(l) (g) 2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g) – + → OH (aq) + NH4 (aq) NH3(g) + H2O(l) 2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g) (n) calcium reacts with water 5. Calculate the oxidation number of the indicated element in

Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g) each of the following species. → 2+ – Ca(s) + 2 H2O(l) Ca (aq) + 2 OH (aq) + H2(g) (a) sulfur in thiosulfate ion; +2 (o) copper(II) sulfide(s) reacts with dilute sulfuric acid (b) nitrogen in ammonium chloride; -3 CuS(s) + H2SO4(aq) → CuSO4(s) + H2S(g) (c) zinc in Na2ZnO2 ; +2 + 2– 2 H (aq) + S (aq) → H2S(g) (d) uranium in UO2(NO3)2; +2 Chemistry 1010 Loader (Fall 2003) Page 2 of 3

(e) silicon in silicic acid (like carbonic acid); +4 1 –3 –4 (f) phosphorus in the fertilizer called triple superphosphate, mole NaOH = 2 x 1.6459 x 10 mol = 8.2293 x 10 mol –1 Ca(H2PO4)2; +5 concentration of sulfuric acid = 0.08229 mol.L 6. Write balanced half-reactions for each of the following partial (c) A 0.64753 g sample of the hydroxide of an unknown half reactions. metal (M) required 34.30 mL of 0.4244 hydrochloric acid – for complete reaction. The hydroxide is insoluble in water (a) ClO2 (aq) → Cl2(g) in ACID solution and may have the formula MOH, M(OH)2 or M(OH)3. Use 6 e– + 8 H+(aq) + 2 ClO –(aq) → Cl (g) + 4 H O(l) 2 2 2 the information from the titration to deduce the most 2– → 2– (b) S2O3 S4O6 in ACID solution likely formula and hence the molar mass of the metal. 2– 2– – 2 S2O3 → S4O6 + 2 e The possible reactions are 2– (c) Cr(OH)3 → CrO4 in BASIC solution MOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) – → 2– – 5 OH + Cr(OH)3 CrO4 + 3 e + 4 H2O(l) M(OH)2(aq) + 2 HCl(aq) → 2 NaCl(aq) + 2 H2O(l)

(d) N2H4 → N2 in BASIC solution M(OH)3(aq) + 3 HCl(aq) → 3 NaCl(aq) + 3 H2O(l) – – –3 –1 4 OH + N2H4 → N2 + 4 e + 4 H2O(l) mol HCl used in the reaction = 34.30 x 10 L x 0.4244 mol.L 7. Balance the following redox equations in acid solution. = 0.014557 mol For the first equation one mole of acid reacts with one mole of → – 2+ (a) MnO2(s) + PbO2(aq) MnO4 (aq) + Pb (aq) base so if MOH is the formula then the molar mass would be 2+ reduction PbO2(aq) → Pb (aq) 0.64753 g -1 – = 44.482 g mol giving about 44.48 - 17.00 = oxidation MnO2(s) → MnO4 (aq) 0.014557 mol -1 – 2+ 37.48 g mol for the molar mass of M 2 e + PbO2(aq) → Pb (aq) change in oxidation number = -2 -1 MnO (s) → MnO –(aq) + 3 e– change in oxidation number = +3 for M(OH)2(aq) the molar mass would be 2 x 44.482 g mol = 2 4 88.965 g mol-1 giving about 89.87 - 34.00 = 55.87 g mol-1 for 2 MnO (s) + 3 Pb–O (aq) → 2 MnO –(aq) + 3 Pb2+(aq) :balance change 2 2 4 the molar mass of M. in ON for M(OH) (aq) the molar mass would be 3 x 44.482 g mol-1 = + → – 2+ 3 4 H (aq) + 2 MnO2(s) + 3 PbO2(aq) 2 MnO4 (aq) + 3 Pb (aq) 133.45 g mol-1 giving about 133.45 - 51.00 = 82.45 g mol-1 for + :balance charges with H the molar mass of M. + → – 2+ 4 H (aq) + 2 MnO2(s) + 3 PbO2(aq) 2 MnO4 (aq) + 3 Pb (aq) + 2 The is no metal with a molar mass close to 37.48 or 82.45 g H2O(l) :balance H with H2O mol-1 with an insoluble hydroxide. However, for 55.87 g mol-1 + → – -1 BALANCED: 4 H (aq) + 2 MnO2(s) + 3 PbO2(aq) 2 MnO4 (aq) + 3 this is almost exactly the molar mass of iron (55.85 g mol ) so 2+ Pb (aq) + 2 H2O(l) the metal hydroxide is probably Fe(OH)2.

(b) Br2(aq) + SO2(g) → H2SO4(aq) + HBr(aq) 10. A 50.00 mL sample of a solution containing iron(II) ions, 2+ –1 2 H2O(l) + Br2(aq) + SO2(g) → H2SO4(aq) + 2 HBr(aq) Fe (aq), required 20.63 mL of a 0.0216 mol.L solution of 2– – 3+ KMnO4 for complete reaction according to the equation: (d) Cr2O7 (aq) + I (aq) → Cr (aq) + I2(s) (acid solution) + 2– – 3+ (incorrect wording in the original) 14 H (aq) + Cr2O7 (aq) + 6 I (aq) → 2 Cr (aq) + I2(s) + 7 H2O(l) 8 H+(aq) + MnO –(aq) + 5 Fe2+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H O(l) 8. Concentrated sulfuric acid has a density of 1.84 g.cm–3 (1.00 4 2 3 (a) Calculate the concentration of iron(II) ions in the solution. mL = 1.00 cm ) and contains 96% of H2SO4 by weight. 0.04456 mol L-1 (a) Calculate the concentration of the concentrated sulfuric –1 (b) What volume of a 0.0150 mol.L solution of KMnO4 acid in mol.L–1. Ans: 18.0 mol L–1 solution would be required to react completely with the (b) What volume of the concentrated sulfuric acid would be iron(II) sulfate solution formed when 0.23245 g of iron –1 required to make 2.50 litres of a 3.0 mol.L solution of wire reacted with sulfuric acid? the bench acid? Ans: 0.417 L Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) (c) What is the mass of the volume of the acid calculated in 0.23245 g mol iron = = 4.1620 x 10-3 mol (b)? Ans: 0.767 kg 55.85gmol-1 9. (a) Calculate the concentration in mol.L–1 of a sodium 1 -3 mol KMnO4 required = 5 x 4.1620 x 10 mol hydroxide solution 33.45 mL of which neutralizes 25.00 1 x 4.1620 x10-3 mol –1 5 -–3 mL of 0.1500 mol.L of nitric acid. so volume required = -1 = 5.549 x 10 L → 0.150 mol L NaOH(aq) + HNO3(aq) NaNO3(aq) + H2O(l) Ans: 5.549 mL –3 –1 mol HNO3 neutralized = 25.00 x 10 L x 0.1500 mol.L = 11. Balance the following equations in acid or base as indicated. –3 3.750 x 10 mol Identify the oxidant and the reductant. Since one mole of nitric acid reacts with one mole of sodium – – → – hydroxide.. (a) ClO3 + MnO4 MnO2 + ClO4 (acidic) –3 + – – → – mole NaOH = 3.750 x 10 mol 2 H + 3 ClO3 + 2 MnO4 2 MnO2 + 3 ClO4 + H2O –1 – concentration of NaOH = 0.1121 mol.L (b) NO3 + S → NO + H2SO3 (acidic) + – (b) A 10.00 mL aliquot of dilute sulfuric acid required 16.12 4 H + 4 NO3 + 3 S + H2O → 4 NO + 3 H2SO3 –1 mL of 0.1021 mol.L sodium hydroxide solution for (c) Cr O 2– + I → IO – + Cr3+ (acidic) neutralization. What was the concentration of the sulfuric 2 7 2 3 + 2– → – 3+ acid solution? 34 H + 5 Cr2O7 + 3 I2 6 IO3 + 10 Cr + 17 H2O – 2– (d) Zn + NO3 → ZnO2 + NH3 (basic) 2 NaOH(aq) + H2SO4(aq) → Na2SO4 (aq) + 2 H2O(l) – – → 2– NaOH used = 16.12 x 10–3 L x 0.1021 mol.L–1 = 1.6459 x 10–3 mol 7 OH + 4 Zn + NO3 4 ZnO2 + NH3 + 2 H2O 2– – – Since one mole of sulfuric acid reacts with two moles of (e) S + ClO3 → Cl + S (basic) sodium hydroxide. Chemistry 1010 Loader (Fall 2003) Page 3 of 3

2– – – – 3 H2O + 3 S + ClO3 → Cl + 3 S + 6 OH