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

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

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

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

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

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

Rules for Assigning Oxidation Numbers Rules 3 & 4 3. The oxidation number of oxygen in compounds is -2 4. The oxidation number of hydrogen in compounds is +1

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

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 + 4(-2) = -2 S O X + 3(-2) = -1 N O  X = +5  X = +6

The Oxidation Number Rules - SIMPLIFIED 1. The sum of the oxidation numbers in ANYTHING is equal to its charge 2. Hydrogen in compounds is +1 3. Oxygen in compounds is -2

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

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

Redox Reaction Prediction #1 Important Oxidizers Formed in reaction MnO4- (acid solution) MnO4- (basic solution) MnO2 (acid solution) Cr2O72- (acid) CrO42- HNO3, concentrated HNO3, dilute H2SO4, hot conc Metallic Ions Free Halogens HClO4 Na2O2 H2O2 Mn(II) MnO2 Cr(III) NO2 NO SO2 Metallous Ions Halide ions Cl- OH- O2

Redox Reaction Prediction #2 Important Reducers Formed in reaction Halide Ions Free Metals Metalous Ions Nitrite Ions Sulfite Ions Free Halogens (dil, basic sol) Free Halogens (conc, basic sol) C2O42- Halogens Metal Ions Metallic ions Nitrate Ions SO42- Hypohalite ions Halate ions CO2

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.

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

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.

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 H2O

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.

Balancing Redox Reactions __ Al + __ Cu2+ --> __ Cu + __ Al3+ Start by writing half reactions (Oxidation and reduction) (Electrons go on the more positive side) Oxidation: Al --> Al 3+ + 3e- Reduction: 2e- + Cu2+ --> Cu

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 3+ + 3e-) Reduction: 3 x ( 2e- + Cu2+ --> Cu)

Balancing Redox Reactions Oxidation: 2 x (Al --> Al 3+ + 3e-) 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 3+ + 3Cu Atoms and charges must be conserved.

Balancing Redox Reactions (Acidic Conditions) MnO4- + I- --> I2 + Mn2+ (acidic) Step 1 Half Reactions: MnO4- --> Mn2+ I- --> I2

Lets balance the reduction one first: for every Oxygen add a water on the other side: MnO4- --> Mn2+ + 4H2O For every hydrogen add a H+ to the other side: 8H+ + MnO4- --> Mn2+ + 4H2O Balance the imbalance of charge with electrons (+7 vs. +2): 5e- + 8H+ + MnO4- --> Mn2+ + 4H2O

Now for the oxidation I- --> I2 Balance the atoms: 2I- --> I2 Balance the imbalance of charge with electrons (-2 vs. 0): 2I- --> I2 + 2e-

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

Balancing Redox Reactions 2( 5e- + 8H+ + MnO4- --> Mn2+ + 4H2O ) 5( 2I- --> I2 + 2e- ) Step 3 Check electrons, atoms and charge. Clean it up. 10e- + 16H+ + 2MnO4- + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 16H+ +2MnO4- +10I-->5I2 +2Mn2+ + 8H2O

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

Basic 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- + 3H2O For every hydrogen add a H+ to the other side: 6H+ + ClO3- --> Cl- + 3H2O

Basic 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-

Basic 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 --> CrO42- + 5H+ Each H+ will react with OH- on both sides: 5 OH-+H2O+Cr(OH)3 ->CrO42-+5H++5OH-

Basic Solution H+ and OH- make water: 5 OH- +H2O + Cr(OH)3 -> CrO42- + 5H2O cancel the waters: 5 OH- + Cr(OH)3 --> CrO42- + 4H2O Balance the imbalance of charge with electrons (-5 vs.-2): 5 OH- + Cr(OH)3 -> CrO42- + 4H2O + 3e-

Basic Solution Balance 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-)

Basic Solution Step 3: Check electrons, atoms and charge then clean it up: 6e- + 3H2O + ClO3- + 10 OH- + 2Cr(OH)3 -->Cl- + 6OH- + 2CrO42- + 8H2O + 6e- ClO3- + 4 OH- + 2Cr(OH)3 -->Cl- + 2CrO42- + 5H2O

Practice Redox Reaction #1 Mn 2+ + BiO3 - => MnO4 - + Bi 3+   2 (4 H2O + Mn 2+ => MnO4 - + 8 H + 5 e) 5 ( 2 e + 6 H + + BiO3 - => Bi 3+ + 3 H2O) 14 H + + 2 Mn 2+ + 5 BiO3 - => 2 MnO4 - + 5 Bi 3+ + 7 H2O

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

Practice Redox Reaction #3 P + Cu 2+ => Cu + H2PO4 -   2 (4 H2O + P => H2PO4 - + 6 H + + 5 e) 5 (2 e + Cu 2+ => Cu) 8 H2O + 2 P + 5 Cu 2+ => 2 H2PO4 - + 12 H + + 5 Cu)

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 MnO4 - + 3 C2O4 2- => 2 MnO2 + 8 OH - + 6 CO2

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

Practice Redox Reaction #6 H2O + Zn + NO3 - => Zn(OH)4 2- + NH3   8 e + 9 H + + NO3 - => NH3 + 3 H2O 4 (4 H2O + Zn => Zn(OH)4 2- + 4 H + + 2 e 6 H2O + NO3 - + 7 OH - + 4 Zn => NH3 + 4 Zn(OH)4 2-