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Chapter 4B: Balancing Redox Reactions LEO SAYS GER West Valley High School AP Chemistry Mr. Mata.

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Presentation on theme: "Chapter 4B: Balancing Redox Reactions LEO SAYS GER West Valley High School AP Chemistry Mr. Mata."— Presentation transcript:

1 Chapter 4B: Balancing Redox Reactions LEO SAYS GER West Valley High School AP Chemistry Mr. Mata

2 Oxidation and Reduction (Redox)  Electrons are transferred  Spontaneous redox rxns can transfer energy  Electrons (electricity)  Heat  Non-spontaneous redox rxns can be made to happen with electricity

3 Oxidation Reduction Reactions (Redox) Each sodium atom loses one electron: Each chlorine atom gains one electron:

4 LEO says GER : LEO says GER : Lose Electrons = Oxidation Sodium is oxidized Gain Electrons = Reduction Chlorine is reduced

5 Not All Reactions are Redox Reactions Reactions in which there has been no change in oxidation number are not redox rxns. Examples:

6 Rules for Assigning Oxidation Numbers Rules 1 & 2 1.The oxidation number of any uncombined element is zero 2. The oxidation number of a monatomic ion equals its charge

7 Rules for Assigning Oxidation Numbers Rules 3 & 4 3. The oxidation number of oxygen in compounds is The oxidation number of hydrogen in compounds is +1

8 Rules for Assigning Oxidation Number Rule 5 5. The sum of the oxidation numbers in the formula of a compound is 0 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H

9 Rules for Assigning Oxidation Numbers Rule 6 6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge X + 3(-2) = -1 N O  X = +5  X = +6 X + 4(-2) = -2 S O

10 The Oxidation Number Rules - SIMPLIFIED 1. The sum of the oxidation numbers in ANYTHING is equal to its charge 2. Hydrogen in compounds is Oxygen in compounds is -2

11 Reducing Agents and Oxidizing Agents  The substance reduced is the oxidizing agent  The substance oxidized is the reducing agent Sodium is oxidized – it is the reducing agent Chlorine is reduced – it is the oxidizing agent

12 Trends in Oxidation and Reduction Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

13 Redox Reaction Prediction #1

14 Redox Reaction Prediction #2

15 Oxidation Reduction Oxidation 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.

16 Agents Oxidizing agent gets reduced. Gains electrons. More negative oxidation state. Reducing agent gets oxidized. Loses electrons. More positive oxidation state.

17 Half-Reactions All redox reactions can be thought of as happening in two halves. One produces e-’s = Oxidation half. One requires e-’s = Reduction half.

18 Balancing 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 reactions  For each half reaction balance all reactants except H and O  Balance O using H 2 O

19 Acidic Solution  Balance H using H +  Balance charge using e -  Multiply equations to make electrons equal  Add equations and cancel identical species  Check that charges and elements are balanced.

20 Balancing Redox Reactions __ Al + __ Cu 2+ --> __ Cu + __ Al 3+ Start by writing half reactions (Oxidation and reduction) (Electrons go on the more positive side) Oxidation: Al --> Al e - Reduction: 2e - + Cu 2+ --> Cu

21 Balancing 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 so Oxidation: 2 x (Al --> Al e - ) Reduction: 3 x ( 2e - + Cu 2+ --> Cu)

22 Balancing Redox Reactions Oxidation: 2 x (Al --> Al e - ) Reduction: 3 x ( 2e - + Cu 2+ --> Cu) _________________________________ Recombine: 6e - +2 Al + 3Cu 2+ -->2Al Cu + 6e - The electrons must cancel. 2 Al + 3 Cu 2+ --> 2 Al Cu Atoms and charges must be conserved.

23 Balancing Redox Reactions (Acidic Conditions) MnO I - --> I 2 + Mn 2+ (acidic) Step 1 Half Reactions: MnO > Mn 2+ I - --> I 2

24 Lets balance the reduction one first: for every Oxygen add a water on the other side: MnO > Mn H 2 O For every hydrogen add a H + to the other side: 8H + + MnO > Mn H 2 O Balance the imbalance of charge with electrons (+7 vs. +2): 5e - + 8H + + MnO > Mn H 2 O

25 Now for the oxidation I - --> I 2 Balance the atoms: 2I - --> I 2 Balance the imbalance of charge with electrons (-2 vs. 0): 2I - --> I 2 + 2e -

26 Balancing Redox Reactions Balance the electrons by finding the common multiple and multiply the half reactions accordingly. Common Multiple here is 10.

27 Balancing Redox Reactions 2( 5e - + 8H + + MnO > Mn H 2 O ) 5( 2I - --> I 2 + 2e - ) Step 3 Check electrons, atoms and charge. Clean it up. 10e H + + 2MnO I - -->5I 2 + 2Mn H 2 O + 10e - 16H + +2MnO I - ->5I 2 +2Mn H 2 O

28 Basic Solution Do everything you would with acid, but add one more step. Add enough OH - to both sides to neutralize the H + Makes water

29 Basic Solution Cr(OH) 3 +ClO 3 - ->CrO Cl - (basic) Step 1 Half Reactions: Lets balance the reduction one first: ClO > Cl - for every Oxygen add a water on the other side: ClO > Cl - + 3H 2 O For every hydrogen add a H + to the other side: 6H + + ClO > Cl - + 3H 2 O

30 Basic Solution Each H + will react with an OH - on both sides: 6 OH - + 6H + +ClO 3 - -> Cl - +3H 2 O + 6OH - H + and OH - make water: 6H 2 O + ClO > Cl - + 3H 2 O + 6OH - cancel the waters: 3H 2 O + ClO > Cl - + 6OH - Balance the imbalance of charge with e-’s (-1 vs. -7): 6e - + 3H 2 O + ClO > Cl - + 6OH -

31 Basic Solution Now for the oxidation: Cr(OH) 3 --> CrO 4 2- for every O, add a H 2 O on other side: H 2 O + Cr(OH) 3 --> CrO 4 2- For every H, add a H + to the other side: H 2 O + Cr(OH) 3 --> CrO H + Each H + will react with OH - on both sides: 5 OH - +H 2 O+Cr(OH) 3 ->CrO H + +5OH -

32 Basic Solution H + and OH - make water: 5 OH - +H 2 O + Cr(OH) 3 -> CrO H 2 O cancel the waters: 5 OH - + Cr(OH) 3 --> CrO H 2 O Balance the imbalance of charge with electrons (-5 vs.-2): 5 OH - + Cr(OH) 3 -> CrO H 2 O + 3e -

33 Basic Solution Balance the electrons by finding the common multiple and multiply the half reactions accordingly. Common Multiple here is 6: 1(6e - + 3H 2 O + ClO 3 - -> Cl - + 6OH - ) 2(5 OH - +Cr(OH) 3 -> CrO H 2 O + 3e - )

34 Basic Solution Step 3: Check electrons, atoms and charge then clean it up: 6e - + 3H 2 O + ClO OH - + 2Cr(OH) 3 -->Cl - + 6OH - + 2CrO H 2 O + 6e - ClO OH - + 2Cr(OH) 3 -->Cl - + 2CrO H 2 O

35 Practice Redox Reaction #1 Mn 2+ + BiO3 - => MnO4 - + Bi 3+ 2 (4 H2O + Mn 2+ => MnO H + 5 e) 5 ( 2 e + 6 H + + BiO3 - => Bi H2O) 14 H Mn BiO3 - => 2 MnO Bi H2O

36 Practice Redox Reaction #2 ClO3 - + Cl - => Cl2 + ClO2 2 (e + 2 H + + ClO3 - => ClO2 + H2O) 2Cl - => Cl2 + 2 e) 4 H ClO Cl - => 2 ClO2 + 2 H2O + Cl2

37 Practice Redox Reaction #3 P + Cu 2+ => Cu + H2PO4 - 2 (4 H2O + P => H2PO H e) 5 (2 e + Cu 2+ => Cu) 8 H2O + 2 P + 5 Cu 2+ => 2 H2PO H Cu)

38 Practice Redox Reaction #4 MnO4 - + C2O4 2- => MnO2 + CO2 2 (3 e + 4 H + + MnO4 - => MnO2 + 2 H2O) 3 (C2O4 2- => 2 CO2 + 2 e) 4 H2O + 2 MnO C2O4 2- => 2 MnO2 + 8 OH CO2

39 Practice Redox Reaction #5 ClO2 => ClO2 - + ClO3 - e + ClO2 => ClO2 - H2O + ClO2 => ClO H + + e 2 OH ClO2 => ClO2 - + ClO3 - + H2O

40 Practice Redox Reaction #6 H 2 O + Zn + NO3 - => Zn(OH) NH3 8 e + 9 H + + NO3 - => NH3 + 3 H2O 4 (4 H2O + Zn => Zn(OH) H e 6 H2O + NO OH Zn => NH3 + 4 Zn(OH)4 2-


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