Chapter 7 Redox reaction7.1 What is redox reaction?There are two ways of finding out whether or not a substance has been oxidised or reduced during a chemical reaction:■■ electron transfer■■ changes in oxidation number.Oxidation is losing electrons. (OIL)Reduction is gaining electrons .(RIG)7.2 Redox and electron transferHalf-equationsExample: 2Na(s) + Cl2(g) →2NaCl(s)✓Na → Na++ e–Na – e– → Na+ Oxidation is losing electrons. (OIL)✓Cl2 + 2e– → 2Cl– Reduction is gaining electrons .(RIG)Balancing half-equationsExample : 1 Construct the balanced ionic equation for the reaction between nickel and iron(III) ions, Fe3+, from the half-equations:Ni(s) →Ni2+(aq) + 2e–Fe3+(aq) + e–→ Fe2+(aq)The balanced ionic equation is: ___________________________________________________Example : 2 Construct the balanced ionic equation for the reaction of iodide ions (I–) with manganate(VII) ions (MnO4–) inthe presence of hydrogen ions (H+). Use thefollowing two half-equations to help you:i 2I–(aq) →I2(aq) + 2e–ii MnO4–(aq) + 8H+(aq) + 5e–→ Mn2+(aq) + 4H2O(l)The balanced ionic equation is: ___________________________________________________Exercise : 2 a Write two half-equations for the following reactions. For each half-equation statewhether oxidation or reduction is occurring.i Cl2 + 2I– → I2 + 2Cl–ii 2Mg + O2→ 2MgOiii 4Fe + 3O2 → 2Fe2O3b Zinc metal reacts with IO3– ions in acidic solution.Construct a balanced ionic equation for thisreaction, using the two half-equations below:2IO3– + 12H+ + 10e– → I2 + 6H2OZn → Zn2+ + 2e–7.3 Oxidation numbersWhat are oxidation numbers(oxidation states)?Oxidising agent &reducing agentOxidation number rules1.单质化合价均为0.2.在化合物中，许多原子有固定化合价--第一主族（_________________________________________）元素为+1价--第二主族（_________________________________________）元素为+2价--氟元素（F）总为-1价--氢元素（H）为+1价（除非为金属氢化物，如NaH, H为-1 ）--氧元素（O）为-2 价（过氧化物peroxide中的氧为-1，F2O中氧为+2）3. 化合物中各元素化合价之和为0.4. 离子团的化合价之和等于离子的带电荷数。

（如NO3- : N元素与O元素的化合价之和为-1）5. 无论是离子团还是化合物中，电负性更强的元素带负电。

Exercise ：State the ox. no. of the bold atoms in these compounds or ions:a P2O5b S O42–c H2Sd Al2Cl6e N H3f Cl O2–g Ca C O37.4 Naming compoundOxides of nitrogen■■The ox. no. of N in N2O is +1. So this compound is nitrogen(I) oxide.■■The ox. no. of N in NO is +2. So this compound is nitrogen(II) oxide.■■The ox. no. of N in NO2 is +4. So this compound is nitrogen(IV) oxide.Nitrate ions■■ The ox. no. of N in the NO2– ion is +3. So NaNO2 is sodium nitrate(III).■■ The ox. no. of N in the NO3– ion is +5. So NaNO3 is sodium nitrate(V).7.5 From name to formulaExercise : Give the formulae of: a sodium chlorate(I) b iron(III) oxidec potassium nitrate(III)d phosphorus(III) chloride.7.6 Balancing chemical equations using oxidation numbersExample: Copper(II) oxide (CuO) reacts with ammonia (NH3) to form copper, nitrogen (N2) and water.Exercise: Use the oxidation number method to balance these equations.a H2SO4 + HI → S + I2 + H2Ob HBr + H2SO4→ Br2 + SO2 + H2Oc V3+ + I2 + H2O → VO2+ + I– + H+homeworkSection AQ2 A cheap carbon monoxide detector for a gas heater consists of a patch containing palladium chloride crystals. When carbon monoxide is present, the crystals turn from orange to black as the following reaction takes place.CO(g) + PdCl2(s) + H2O(l) → CO2(g) + Pd(s) + 2HCl(aq)orange blackWhich is the element whose oxidation number decreases in this reaction?A carbonB chlorineC hydrogenD palladiumQ4 The nickel-cadmium rechargeable battery is based upon the following overall reaction.Cd + 2NiOOH + 4H2O → Cd(OH)2 + 2Ni(OH)2.H2OWhat is the oxidation number of nickel at the beginning and at the end of the reaction?Q5 In the treatment of domestic water supplies, chlorine is added to the water to form chloric(I) acid, HClO.Cl2(aq) + H2O(I) → H+(aq) + Cl–(aq) + HClO(aq)This reacts further to give the chlorate(I) ion.HClO(aq) + H2O(I) → H3O+(aq) + ClO–(aq)Both HClO and ClO– kill bacteria by oxidation.What is the change in oxidation number of chlorine in forming the chlorate(I) ion from the aqueous chlorine?A –1B 0C +1D +2Q6 Ammonium nitrate, NH4NO3, can decompose explosively when heated.NH4NO3 → N2O + 2H2OWhat are the changes in the oxidation numbers of the two nitrogen atoms in NH 4NO 3 when this reaction proceeds?A –2, –4B +2, +6C +4, –6D +4, –4 Q8 In which substance does nitrogen exhibit the highest oxidation state?A NOB N 2OC N 2O 4D NaNO 2 Q11 Chlorine dioxide is produced on a large scale as it is used for bleaching paper pulp. It is made by the following reaction.2Cl O 3(aq) + SO 2(g) → 2Cl O 2(g) + SO 4 (aq)How do the oxidation numbers of chlorine and sulphur change in this reaction?Q12 In some early paintings, lead(II) carbonate was used as a white pigment. In the 19th century hydrogen sulphide from burning coal reacted with this pigment to form black lead(II) sulphide, PbS. The original colour of the painting may be restored by carefully treating the area with dilute hydrogen peroxide, producing lead(II) sulphate which is also white. What is the role of the hydrogen peroxide?A catalystB oxidising agentC reducing agentD solvent Q13 In an experiment, 50.0 cm 3 of a 0.10 mol dm –3 solution of a metallic salt reacted exactly with 25.0 cm 3 of 0.10 mol dm –3 aqueous sodium sulphite.The half-equation for oxidation of sulphite ion is shown below.SO 3 (aq) +H 2O(I)→SO 4 (aq) +2H (aq) +2e If the original oxidation number of the metal in the salt was +3, what would be the new oxidation number of the metal? A+1 B+2 C+4 D+5 Q14 The amount of titanium dioxide in an ore can be determined by using the following reaction. 3TiO2 + 4BrF3 → 3TiF4 + 2Br2 + 3O2Which element increases in oxidation number in this reaction?A bromineB fluorineC oxygenD titaniumQ16 Sodium iodide reacts with concentrated sulfuric acid. The equation which represents one of the reactions that takes place is shown.8NaI + 9H2SO4 → 8NaHSO4 + 4I2 + H2S + 4H2OWhich species has been oxidised in this reaction?A H+B I–C Na+D SO42–Q17 In which reaction does an element undergo the largest change in oxidation state?Q19 Which conversion involves a reduction of chromium?Section BQ20 Many crude oils contain sulphur as H2S. During refining, by the Claus process, the H2S is converted into solid sulphur, which is then removed.reaction I 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)reaction II 2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)Which statements about the Claus process are correct?1 H2S is oxidised in reaction I.2 SO2 oxidises H2S in reaction II.3 SO2 behaves as a catalyst.Q21 Which reactions are redox reactions?1 CaBr2 + 2H2SO4→ CaSO4 + Br2 + SO2 + 2H2O2 CaBr2 + 2H3PO4 → Ca(H2PO4)2 + 2HBr3 CaBr2 + 2AgNO3 → Ca(NO3)2 + 2AgBrQ22 Zirconium, Zr, proton number 40, is a metal which is used in corrosion-resistant alloys. Zirconium metal is extracted from the oxide ZrO2 by the following sequence of reactions.reaction1 ZrO2+2Cl2+2C→ ZrCl4+2COreaction 2 ZrCl 4 + 2Mg→ Zr + 2MgCl 2Which statements about this extraction process are correct?1 Carbon in reaction 1 behaves as a reducing agent.2 Magnesium in reaction 2 behaves as a reducing agent.3 Chlorine in reaction 1 behaves as a reducing agent.Q2 Magnesium burns in nitrogen to give magnesium nitride, a yellow solid which has theformula Mg3N 2.Magnesium nitride reacts with water to give ammonia and magnesium hydroxide.(i)Construct an equation for the reaction of magnesium nitride with water. ................................................................................................................................................. (ii)Does a redox reaction occur when magnesium nitride reacts with water?Use the oxidation numbers of nitrogen to explain your answer. ....................................................................................................................................................... ....................................................................................................................................................... ....................................................................................................................................................... ................................................................................................................................. (June 2009) Acidified potassium dichromate(VI) can oxidise ethanedioic acid, H2C2O4.The relevant half-equations are shown.Cr2O72– + 14H+ + 6e–→ 2Cr3+ + 7H2OH2C2O4→ 2CO2 + 2H+ + 2e–(a) State the overall equation for the reaction between acidified dichromate(VI) ions and ethanedioic acid. (2)(b) In an experiment a 0.242 g sample of hydrated ethanedioic acid, H2C2O4.x H2O, was reacted with a 0.0200 mol dm–3 solution of acidified potassium dichromate(VI).32.0 cm3 of the acidified potassium dichromate(VI) solution was required for complete oxidation of the ethanedioic acid.(i) Calculate the amount, in moles, of dichromate(VI) ions used to react with the sample of ethanedioic acid.amount = ............................. mol [1] (ii) Calculate the amount, in moles, of ethanedioic acid in the sample.amount = ............................. mol [1] (iii) Calculate the relative molecular mass, M r, of the hydrated ethanedioic acid.M r = (1)(iv) Calculate the value of x in H2C2O4.x H2O.x = (1)[Total: 6] 3. A 6.30 g sample of hydrated ethanedioic acid, H2C2O4.x H2O, was dissolved in water andthesolution made up to 250 cm3.A 25.0 cm3 sample of this solution was acidified and titrated with 0.100 mol dm–3 potassium manganate(VII) solution. 20.0 cm3 of this potassium manganate(VII) solution was required to react fully with the ethanedioate ions, C2O42–, present in the sample.(a) The MnO4– ions in the potassium manganate(VII) oxidise the ethanedioate ions.(i) Explain, in terms of electron transfer, the meaning of the term oxidise in the sentence above. ............................................................................................................................................. (1)(ii) Complete and balance the ionic equation for the reaction between the manganate(VII) ions and the ethanedioate ions.2MnO4–(aq) + 5C2O42–(aq) + .........H+(aq) .........(aq) + 10CO2(aq) + .........H2O(l)[3] (b) (i) Calculate the number of moles of manganate(VII) used in the titration.[1] (ii) Use the equation in (a)(ii) and your answer to (b)(i) to calculate the number of moles ofC2O42– present in the 25.0 cm3 sample of solution used.[1] (iii) Calculate the number of moles of H2C2O4.x H2O in 6.30 g of the compound.[1] (iv) Calculate the relative formula mass of H2C2O4.x H2O.[1] (v) The relative formula mass of anhydrous ethanedioic acid, H2C2O4, is 90.Calculate the value of x in H2C2O4.x H2O.[1] [Total: 9]。