Oxidation and Reduction

Chemistry UNIT 5 OXIDATION AND REDUCTION CHAPTER 11 and 12 OXIDATION: REDUCTION: OXIDATION NUMBER: ALSO CALLED OXIDATION STATE -how many electrons an atom in a compound would gain or lose 1) 2) 3) 4) 5) The sum of oxidation numbers in a neutral compound is 0 An element by itself is 0 The oxidation number of a monatomic ion is the charge on that ion itself O is always -2 except in peroxides where it is -1 H is always +1 except in hydrides where it is -1 SEE ADDITIONAL RULES ON p. 484. WH AT IS TH E OXIDATION STATE OF: a) Cr in K2Cr2O7 b) Cr in K2CrO4 c) S in H2SO4 d) Mn in KMnO4 e) P in PO4 -3 f) N in NH3 g) S in SCN - Worksheet 5.1: Oxidation Numbers Assign the correct oxidation number to each atom in the following substances/ions. - K FeO CO IO4 RbCl N2 Na2S2O3 SCN Na2O N2O S8 HSO4 H2O2 NO Na2SO3 PO4-3 MgBr2 N2O4 MnSO4 HCO3 CaS NO2 KMnO4 Hg2+2 K2Cr2O7 N2O3 OH H3PO4 N2O5 CO3-2 (NH4)3PO4 Ag NO2 (NH4)2S HCH3COO CrO4-2 FeCl2 CaCO3 Cr2O7-2 FeCl3 CO2 IO3 - - - ClO - - - - ClO3 - S2O3-2 REDOX REACTIONS Consider the following reaction Copper (II) chloride, when dissolved in water it produces an acid solution containing copper (II) ions and chloride ions. When aluminum foil is added to the solution, an immediate reaction occurs, producing bubbles, gas, and heat. The aluminum foil changes to a reddish brown. The solution changes from turquoise to colourless. Two reactions have occurred here: The bubbles are formed when the aluminum reacts with the hydrogen ions in solution to from hydrogen gas bubbles. In the second reaction: Cu+2 (aq) + Al (s) à Cu (s) + Al+3 (aq) The copper ions are changing to copper atoms, and the aluminum atoms are changing to aluminum ions. The reddish brown forming on the aluminum foil is the copper atoms coming out of solution. The aluminum ions are colourless and are replacing the copper (II) ions that were in the blueish solution. Consider each metal separately: Al(s) à oxidation: Al+3 (aq) + 3e Al loses electrons Cu+2 (aq) reduction: + 2e à Cu(s) Cu gains electrons These two processes must occur together (electrons will not exist alone - when one species loses electrons, these electrons must go somewhere). These are known as oxidation - reduction reactions or redox reactions. Each of the reactions (the oxidation and reduction ) are called half reactions of the redox reaction. The full redox reaction is: 3Cu+2 (aq) + 2Al(s) à 3Cu(s) + 2Al+3 (aq) BALANCING REDOX REACTIONS We must always balance the redox reaction to accurately show the gain and loss of electrons. Balanced redox reactions do not show the electrons! Always balance the charges and then balance the number of atoms. Note that in the above redox reaction, Al brings about the reduction of Cu+2. Thus Al is called the reducing agent. (Cu+2 gains electrons, and these electrons came from Al) Cu+2 is called the oxidizing agent. (If the Cu+2 was not present in solution, the Al would not give up its electrons, and therefore Al would not oxidize) Reducing agent: Oxidizing agent: causes reduction gives up electrons (i.e. is oxidized) AGENT: does it to someone else causes oxidation takes up electrons (i.e. is reduced) BOOKLET PAGE 8 • List of oxidizing and reducing agents • The equations listed are presented as the reduction half reactions • Any oxidizing agent on the left will react with a reducing agent on the right that is LOWER on the list. examples: 1) Will I2 be able to oxidize zinc metal to Zn +2 ions? (If iodine is the oxidizing agent, it is causing the oxidation of Zinc; i.e. I2 is gaining electrons and Zn is losing electrons) 2) Name an oxidizing agent that can oxidize Br- to Br2 but cannot oxidize Cl- to Cl2. YOU CAN AND SHOULD NOW READ THE TEXTBOOK PAGES 474-493 AND COMPLETE THE SECTION END QUESTIONS. A CLOSER LOOK AT REDOX HALF REACTIONS Let's take a look at a substance that is oxidized when it reacts with oxygen: OXIDATION: loss of electrons, GAIN OF OXYGEN! Fe + O2 à Fe2O3 (rust) C + O2 à CO2 OXIDATION involves a chemical change in which an element changes its oxidation number. LOOKING at examples of OXIDATION and REDUCTION SO32- à SO42- Oxidation is gain of oxygen CaO à Ca Reduction is loss of oxygen 2I- à I2 Oxidation is loss of electrons IO3- à I2 Reduction is loss of oxygen!! CrO4-2 à Cr2O7-2 Oxidation is gain of oxygen!! BALANCING REDOX EQUATIONS: ASSIGN OXIDATION NUMBER IDENTIFY REDOX PAIR BALANCE REDOX PAIR BALANCE OTHER SPECIES - conserve atoms ; conserve charge Balance O's by adding H2O's Balance H's by adding H+'s Balance charge by adding electrons MAJOR HYDROXIDE: (MAJOR OH-) 1. 2. 3. 4. Balance the MAJOR species Balance the O atoms Balance the H atoms Balance the charge using electrons (- ) TH ESE ARE NOT REDOX REACTIONS: SO3 + H2O à H2SO4 P4H10 + 6H2O à 4H3PO4 DO YOU KNOW WHY? BALANCING REDOX EQUATIONS…Examples ASSIGN OXIDATION NUMBER IDENTIFY REDOX PAIR BALANCE REDOX PAIR BALANCE OTHER SPECIES - conserve atoms -conserve charge Balance O's by adding H2O's Balance H's by adding H+'s Balance charge by adding electrons 1) Mg + O2 à MgO 2) H+ + Zn + VO3- à VO+2 + Zn2+ + 3) H3AsO3 + NaBrO3 à NaBr + H3AsO4 H2O 4) NF3 + AlCl3 à N2 5) Zn + HNO3 à Zn(NO3)2 6) NaI + H2SO4 à H2S + I2 + 7) SO3-2 + IO3- H+ à + + Cl2 + AlF3 + NH4NO3 + H2O Na2SO4 + H2O SO4-2 + I2 + H2O 8) 9) C2O4-2 MnO4- + + H2S Cr2O7-2 + H+ à S + Mn2+ HALF REACTIONS 10) à S2O8-2 ⇄ HSO4- CO2 + + H2O Cr+3 11) Balance the HALF REACTION which occurs in acidic solution: Cr2O7 -2 ⇄ Cr+3 1. Balance the MAJOR species 2. Balance the O atoms 3. Balance the H atoms 4. Balance the charge using electrons (-) 12) Balance the HALF REACTION which occurs in basic solution: Pb ⇄ HPbO 2 1. Balance the MAJOR species 2. Balance the O atoms 3. Balance the H atoms 4. Balance the charge using electrons (-) NOW CONVERT THIS EQUATION TO BASIC CONDITIONS BY ADDING THE WATER EQUILIBRIUM EQUATION IN SUCH A WAY AS TO CANCEL ALL THE H+ IN THE HALF REACTION: Complications to keep in mind: -subscript (eg. Cr2O7) -reaction occurs in acidic or basic conditions -disproportionation – where the same element is both oxidized AND reduced NOW COMPLETE WORKSHEET 5.2 and 5.3!!!!!!!!! Worksheet 5.2: Balancing Exercise 1 - + Fe+2 + H+ ó H2O + Fe+3 + Mn+2 - + Cl - + H2O ó OH - + NO + Cl2 3- SO3-2 + IO3 - + H+ ó SO4-2 + I2 + H2O 4- ClO - + OH - + CrO2 ó CrO4-2 + H2O + Cl 5- AsO3-3 + Br2 + OH ó AsO4-3 + Br - + H2O 6- H2O2 + F - + H+ ó H2O + F2 7- C2O4-2 + Cr2O7-2 + H+ ó CO2 + H2O + Cr+3 8- H2O + Al + OH 9- H2S + MnO4 ó H2O + OH 10- BrO - ó Br - + BrO3 11- NO3 - + NO + 12- SO2 + Cr2O7-2 + + H2O 1- MnO4 2- NO3 - - - - ó Al(OH)4- + H2 + Mn+2 + S8 H2O ó OH - + NO2 H+ ó Cr+3 + SO4-2 - - - The exercise contains two examples of disproportionation (reactions in which the same element is both oxidized and reduced). Which are they? Worksheet 5.3: Balancing Exercise 2 1- S2O8-2 ó HSO4- (acidic) 2- N2H4 ó N2 (basic) 3- HXeO4 4- C2H3O2 5- Cr(OH)3 ó 6- - ó HXeO6-3 (basic) - ó C2H5OH (acidic) CrO4-2 (basic) (acidic) PO3-3 ó P2O7-4 + 7- C4H6 + N2 ó CO3-2 + NH4 8- U+4 + Cr2O7-2 ó UO2+2 + Cr+3 (acidic) Na2O2 ó Na2O + O2 (basic) 9- (acidic) 10- AsO3-3 + I2 ó AsO4-3 + I - (basic) 11- MnO4 - + H2O2 ó Mn+2 + O2 (acidic) 12- C2H2 + ClO4 - ó CO2 + ClO3 - (acidic) 13- Zn + As2O3 ó AsH3 + Zn+2 (basic) ó CH3NH2 + Ti+4 (acidic) Mn+2 NO3 1415- CH3NO2 + CN - + Ti+3 MnO4 - ó + - + CO3-2 (acidic)