Presentation on theme: "Chapter 4B: Balancing Redox Reactions"— Presentation transcript:
1Chapter 4B: Balancing Redox Reactions LEO SAYS GERWest Valley High SchoolAP ChemistryMr. Mata
2Oxidation and Reduction (Redox) Electrons are transferredSpontaneous redox rxns can transfer energyElectrons (electricity)HeatNon-spontaneous redox rxns can be made to happen with electricity
3Oxidation Reduction Reactions (Redox) Each sodium atom loses one electron:Each chlorine atom gains one electron:
4LEO says GER :Lose Electrons = OxidationSodium is oxidizedGain Electrons = ReductionChlorine is reduced
5Not All Reactions are Redox Reactions Reactions in which there has been no change in oxidation number are not redox rxns.Examples:
6Rules for Assigning Oxidation Numbers Rules 1 & 2 The oxidation number of any uncombined element is zero2. The oxidation number of a monatomic ionequals its charge
7Rules for Assigning Oxidation Numbers Rules 3 & 4 3. The oxidation number of oxygen incompounds is -24. The oxidation number of hydrogen in compounds is +1
8Rules for Assigning Oxidation Number Rule 5 5. The sum of the oxidation numbers in the formula of a compound is 02(+1) + (-2) = 0H O(+2) + 2(-2) + 2(+1) = 0Ca O H
9Rules for Assigning Oxidation Numbers Rule 6 6. The sum of the oxidation numbers in theformula of a polyatomic ion is equal toits chargeX + 4(-2) = -2S OX + 3(-2) = -1N O X = +5 X = +6
10The Oxidation Number Rules - SIMPLIFIED 1. The sum of the oxidation numbers inANYTHING is equal to its charge2. Hydrogen in compounds is +13. Oxygen in compounds is -2
11Reducing Agents and Oxidizing Agents The substance reduced is the oxidizing agentThe substance oxidized is the reducing agentSodium is oxidized – it is the reducing agentChlorine is reduced – it is the oxidizing agent
12Trends in Oxidation and Reduction Active metals:Lose electrons easilyAre easily oxidizedAre strong reducing agentsActive nonmetals:Gain electrons easilyAre easily reducedAre strong oxidizing agents
13Redox Reaction Prediction #1 Important OxidizersFormed in reactionMnO4- (acid solution)MnO4- (basic solution)MnO2 (acid solution)Cr2O72- (acid)CrO42-HNO3, concentratedHNO3, diluteH2SO4, hot concMetallic IonsFree HalogensHClO4Na2O2H2O2Mn(II)MnO2Cr(III)NO2NOSO2Metallous IonsHalide ionsCl-OH-O2
15Oxidation ReductionOxidation means an increase in oxidation state - lose electrons.Reduction means a decrease in oxidation state - gain electrons.The substance that is oxidized is called the reducing agent.The substance that is reduced is called the oxidizing agent.
17Half-ReactionsAll redox reactions can be thought of as happening in two halves.One produces e-’s = Oxidation half.One requires e-’s = Reduction half.
18Balancing Redox Reactions In aqueous solutions the key is the number of electrons produced must be the same as those required.For reactions in acidic solution an 8 step procedure.Write separate half reactionsFor each half reaction balance all reactants except H and OBalance O using H2O
19Acidic Solution Balance H using H+ Balance charge using e- Multiply equations to make electrons equalAdd equations and cancel identical speciesCheck that charges and elements are balanced.
20Balancing Redox Reactions __ Al + __ Cu2+ --> __ Cu + __ Al3+Start by writing half reactions (Oxidation and reduction)(Electrons go on the more positive side)Oxidation: Al --> Al e-Reduction: 2e- + Cu2+ --> Cu
21Balancing Redox Reactions 2. Balance the electrons by finding the common multiple and multiply the half reactions accordingly.The common multiple of the electrons is 6 soOxidation: 2 x (Al --> Al e-)Reduction: 3 x ( 2e- + Cu2+ --> Cu)
22Balancing Redox Reactions Oxidation: 2 x (Al --> Al e-)Reduction: 3 x ( 2e- + Cu2+ --> Cu)_________________________________Recombine: 6e-+2 Al + 3Cu2+-->2Al 3++ 3Cu + 6e-The electrons must cancel.2 Al + 3 Cu2+--> 2 Al CuAtoms and charges must be conserved.
24Lets balance the reduction one first: for every Oxygen add a water on the other side:MnO4- --> Mn2+ + 4H2OFor every hydrogen add a H+ to the other side:8H+ + MnO4- --> Mn2+ + 4H2OBalance the imbalance of charge with electrons (+7 vs. +2):5e- + 8H+ + MnO4- --> Mn2+ + 4H2O
25Now for the oxidation I- --> I2 Balance the atoms: 2I- --> I2 Balance the imbalance of charge with electrons (-2 vs. 0):2I- --> I2 + 2e-
26Balancing Redox Reactions Balance the electrons by finding the common multiple and multiply the half reactions accordingly.Common Multiple here is 10.
28Basic SolutionDo everything you would with acid, but add one more step.Add enough OH- to both sides to neutralize the H+Makes water
29Basic Solution Cr(OH)3 +ClO3- ->CrO42- + Cl- (basic) Step 1 Half Reactions:Lets balance the reduction one first:ClO3- --> Cl-for every Oxygen add a water on the other side:ClO3- --> Cl- + 3H2OFor every hydrogen add a H+ to the other side:6H+ + ClO3- --> Cl- + 3H2O
30Basic Solution Each H+ will react with an OH- on both sides: 6 OH- + 6H++ClO3- -> Cl- +3H2O + 6OH-H+ and OH- make water:6H2O + ClO3- --> Cl- + 3H2O + 6OH-cancel the waters:3H2O + ClO3- --> Cl- + 6OH-Balance the imbalance of charge with e-’s (-1 vs. -7):6e- + 3H2O + ClO3- --> Cl- + 6OH-
31Basic Solution Now for the oxidation: Cr(OH)3 --> CrO42- for every O, add a H2O on other side:H2O + Cr(OH)3 --> CrO42-For every H, add a H+ to the other side:H2O + Cr(OH)3 --> CrO H+Each H+ will react with OH- on both sides:5 OH-+H2O+Cr(OH)3 ->CrO42-+5H++5OH-
32Basic Solution H+ and OH- make water: 5 OH- +H2O + Cr(OH)3 -> CrO H2Ocancel the waters:5 OH- + Cr(OH)3 --> CrO H2OBalance the imbalance of charge with electrons (-5 vs.-2):5 OH- + Cr(OH)3 -> CrO H2O + 3e-
33Basic SolutionBalance the electrons by finding the common multiple and multiply the half reactions accordingly.Common Multiple here is 6:1(6e- + 3H2O + ClO3- -> Cl- + 6OH-)2(5 OH-+Cr(OH)3 -> CrO42-+4H2O + 3e-)