Oxidation-Reduction Reactions “Redox” LEO SAYS GER

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

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

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

Reducing Agents and Oxidizing Agents This slide refers to vocabulary that has been excluded from AP Chemistry by the College Board, and will not be tested. 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 Active nonmetals: Gain electrons easily Are easily reduced

Redox Reaction Prediction #1 Reduced in reaction 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 Oxidized in reaction 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

For any equation to be balanced: 1. The number of atoms of each type on the left side of the arrow must equal the number of atoms of each type to the right of the arrow. 2. The total charges of all the ions on the left side of the arrow must equal the total charges of all the ions to the right of the arrow.

In addition, for redox reactions: 3. The electrons lost (during oxidation) must equal the electrons gained (during reduction).

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 Oxygen, in a compound or ion, is -2

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 Use the combined ‘charges’ of the oxygens in each ion or compound to determine the oxidation number of Cr or C.

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 -14 -8 -4 Use the combined ‘charges’ of the oxygens in each ion or compound to determine the oxidation number of Cr or C.

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 -14 -8 -4 The sum of the oxidation numbers of the other element must add up to the charge on the ion or molecule.

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 +12 -14 +6 -8 +3 +4 -4 The sum of the oxidation numbers of the other element must add up to the charge on the ion or molecule.

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 +12 +6 +3 +4 Divide the sum of the charges by the number of atoms to get the oxidation number for chromium and carbon in the reactants.

Balancing Oxidation-Reduction Reactions: 1. Assign oxidation numbers to every atom in the reaction. Cr2O72- + C2O42-  Cr3+ + CO2 +12/2=+6 +6/2=+3 +3 +4 Divide the sum of the charges by the number of atoms to get the oxidation number for chromium and carbon in the reactants.

Balancing Oxidation-Reduction Reactions: 2. Write ‘bare bones’ half reactions. Include only the atom, ion or element that changes oxidation number. Cr+6 + 3e-  Cr+3 C+3  C+4 + 1e- Remember that each half reaction must also be balanced for charge. The total charges on the left must equal the total charges on the right.

Balancing Oxidation-Reduction Reactions: 3. Take into account any subscripts in the formulas of reactants and products, and multiply the half reactions accordingly. Cr2O72- + C2O42-  Cr3+ + CO2 2[Cr+6 + 3e-  Cr+3] 2[C+3  C+4 + 1e-]

Balancing Oxidation-Reduction Reactions: 3. Take into account any subscripts in the formulas of reactants and products, and multiply the half reactions accordingly. Cr2O72- + C2O42-  Cr3+ + CO2 2Cr+6 + 6e-  2 Cr+3 2C+3  2 C+4 + 2e-

Balancing Oxidation-Reduction Reactions: 4. Multiply each half reaction by the appropriate factor so that the number of electrons lost = number of electrons gained. 1[2Cr+6 + 6e-  2 Cr+3] 3[2C+3  2 C+4 + 2e-]

Balancing Oxidation-Reduction Reactions: 4. Multiply each half reaction by the appropriate factor so that the number of electrons lost = number of electrons gained. 2Cr+6 + 6e-  2 Cr+3 6C+3  6 C+4 + 6e-

Balancing Oxidation-Reduction Reactions: 4. Add the two half reactions together. 2Cr+6 + 6e-  2 Cr+3 6C+3  6 C+4 + 6e- 2Cr+6 + 6C+3  2 Cr+3 + 6 C+4

Balancing Oxidation-Reduction Reactions: 4. Add the two half reactions together. 2Cr+6 + 6e-  2 Cr+3 6C+3  6 C+4 + 6e- 2Cr+6 + 6C+3  2 Cr+3 + 6 C+4 At this point, the electrons lost = the electrons gained during the reaction.

Balancing Oxidation-Reduction Reactions: 5. You now have the number of each atom that undergoes oxidation or reduction in the balanced equation. Take any subscripts into account when inserting coefficients. 2Cr+6 + 6C+3  2 Cr+3 + 6 C+4 Cr2O72- + 3C2O42- 2Cr3+ + 6CO2

Balancing Oxidation-Reduction Reactions: 6. Balance the reaction for charge, using OH- (if in base) or H+ (if in acid). The equation below takes place in acid: Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 Charges: -2 + -6 = -8 (left)  +6 (right)

Balancing Oxidation-Reduction Reactions: Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 Charges: -2 + -6 = -8 (left)  +6 (right) Since the reaction takes place in acid, you need to add 14 H+ to the left side so that the charges become equal.

Balancing Oxidation-Reduction Reactions: 14 H+ + Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 Charges on left = +6 = Charges on right

Balancing Oxidation-Reduction Reactions: 7. Balance for H and O by adding water to the appropriate side of the reaction. 14 H+ + Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 + 7 H2O

Balancing Oxidation-Reduction Reactions: 8. Check the balance for all atoms in the reaction. 14 H+ + Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 + 7 H2O Left: 14 H Right: 14 H 2 Cr 2 Cr 19 O 19 O 6 C 6 C

Redox Stoichiometry Calculations involving concentrations and redox reactions are quite common. Many ores containing metals are analyzed using redox titrations. Since many compounds change color as they are oxidized or reduced, one of the reactants may serve as the indicator in the titration.

Redox Stoichiometry The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion. What is the concentration of iron(II) ion if 31.50 mL of 0.105M potassium bromate is required to completely react with 10.00 mL of the iron solution.

Redox Stoichiometry The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion. 1. Write the balanced chemical reaction. Fe2+(aq) + BrO31-(aq)  Fe3+(aq) + Br1-(aq)