Chapter 19

Oxidation-Reduction Reactions

Section 1: Oxidation and Reduction

• Standard 3.g.:– Students know how to identify reactions that

involve oxidation and reduction and how to balance oxidation-reduction reactions.

• Objective:–We will assign oxidation numbers to reactant

and product species and explain what an oxidation-reduction reaction is.

Oxidation States

• The oxidation number assigned to an element in a molecule is based on the distribution of electrons in that molecule.

• Rules for Assigning Oxidation Numbers are in Table 1 on page 631.

Rules• The oxidation number of any pure element is 0.• The oxidation number of a monatomic ion equals

that charge on the ion.• The more electronegative element in a binary

compound is assigned the number equal to the charge it would have if it were an ion.

• The oxidation number of fluorine in a compound is always -1.

• Oxygen has an oxidation number of -2 unless it is combined with F, in which it is +1 or +2, or it is in a peroxide, in which it is -1.

• Hydrogen is +1, unless combined with a metal, then it is -1.

• In compounds, Group 1 is +1, Group 2 is +2, and Aluminum is +3.

• The sum of the oxidation numbers of all atoms in a neutral compound is 0.

• The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion.

Oxidation States of Chromium• Orange – K2Cr2O7

Potassium Dichromate• Yellow – Na2CrO4

Sodium Chromate• Green – CrCl3

Chromium(III) Chloride• Violet – Cr(NO3)3

Chromium(III) Nitrate

Oxidation• The processes in which the atoms or ions of an

element experience an increase in oxidation state.2Na(s) + Cl2(g) 2NaCl(s)

• The formation of sodium ion illustrates an oxidation process because each sodium atom loses an electron to become a sodium ion.

Na Na+ + e-• Sodium is considered oxidized because it

increased in oxidation number.

Reduction• The processes in which the atoms or ions of an

element experience a decrease in oxidation state.

• The formation of a chloride ion illustrates a reduction process because each chlorine atom gains an electron to become a chloride ion.

Cl2 + 2e- 2Cl-

• Chlorine is considered reduced because it decreased in oxidation number.

Oxidation and Reduction as a Process

• Any chemical process in which elements undergo changes in oxidation number is an oxidation-reduction reaction or redox reaction for short.

• The part of the reaction involving oxidation or reduction alone can be written as a half-reaction.

“Leo the Lion says Ger”

• LEO = Loss of Electrons Oxidation• GER = Gain of Electrons Reduction

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e-

2. S + 2e- S2-

3. Mg Mg2+ + 2e-

4. 2F- F2 + 2e-

5. O2 + 4e- 2O2-

6. Mn2+ MnO4- + 5e-

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e- Oxidation2. S + 2e- S2-

3. Mg Mg2+ + 2e-

4. 2F- F2 + 2e-

5. O2 + 4e- 2O2-

6. Mn2+ MnO4- + 5e-

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e- Oxidation2. S + 2e- S2- Reduction3. Mg Mg2+ + 2e-

4. 2F- F2 + 2e-

5. O2 + 4e- 2O2-

6. Mn2+ MnO4- + 5e-

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e- Oxidation2. S + 2e- S2- Reduction3. Mg Mg2+ + 2e- Oxidation4. 2F- F2 + 2e-

5. O2 + 4e- 2O2-

6. Mn2+ MnO4- + 5e-

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e- Oxidation2. S + 2e- S2- Reduction3. Mg Mg2+ + 2e- Oxidation4. 2F- F2 + 2e- Oxidation

5. O2 + 4e- 2O2-

6. Mn2+ MnO4- + 5e-

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e- Oxidation2. S + 2e- S2- Reduction3. Mg Mg2+ + 2e- Oxidation4. 2F- F2 + 2e- Oxidation

5. O2 + 4e- 2O2- Reduction

6. Mn2+ MnO4- + 5e-

ExamplesIdentify the following as Oxidation or Reduction:1. K K+ + e- Oxidation2. S + 2e- S2- Reduction3. Mg Mg2+ + 2e- Oxidation4. 2F- F2 + 2e- Oxidation

5. O2 + 4e- 2O2- Reduction

6. Mn2+ MnO4- + 5e- Oxidation

Redox Reactions• In order for a reaction to be a redox reaction,

the atoms in the reaction must change oxidation state. If they don’t, it is not a redox reaction.

0 0 1+ 1-

• Redox: 2Na + Cl2 2NaCl

4+ 2- 1+ 2- 1+ 4+ 2-

• Not Redox: SO2 + H2O H2SO3

Homework

Chapter 19.1 pg 635 #2-4 and Oxidation Worksheet

Section 2: Balancing Redox Equations

• Standard 3.g.:– Students know how to identify reactions that

involve oxidation and reduction and how to balance oxidation-reduction reactions.

• Objective:–We will balance redox equations by using the

half-reaction method

Balancing Redox Equations

• In a normal equation, when we balance, we are only conserving mass, we are not looking at charge.

• In a Redox Equation, when properly balanced, both mass and charge are conserved.

Half-Reaction MethodConsists of 7 steps:1. Write the formula equation, if not given, and

then write the net ionic equation.2. Assign oxidation numbers and delete

substances containing only elements that do not change oxidation state.

3. Write the half-reaction for oxidation, balance atoms, and balance charges.

4. Write the half-reaction for reduction, balance atoms, and balance charges.

5. Conserve charge by adjusting the coefficients in front of electrons so that the number lost equals the number gained.

6. Combine the half-reactions and cancel out anything common to both sides of the equation.

7. Combine ions to form compounds shown in the original formula equation. Check all ions are balanced.

• When doing ionic equations, do not break up the covalent compounds

• Covalent compound consists of 2 or more non-metals!!!!

Example

• Copper reacts with hot, concentrated sulfuric acid to form copper(II) sulfate, sulfur dioxide, and water. Write and balance the equation for this reaction.

Step 1: Write Formula and Ionic EquationsFormula: Cu + H2SO4 CuSO4 + SO2+ H2O

Ionic: Cu + 2H+ + SO42- Cu2+ + SO4

2- + SO2 + H2O

Step 2: Assign Oxidations Numbers and Delete 0 1+ 6+ 2- 2+ 6+ 2- 4+ 2- 1+ 2-

Cu + 2H+ + SO42- Cu2+ + SO4

2- + SO2 + H2ODelete things that don’t change oxidation state.

Cu + 2H+ + SO42- Cu2+ + SO4

2- + SO2 + H2O

Cu + SO42- Cu2+ + SO2

Step 3: Write Oxidation Half-ReactionCu Cu2+

Atoms are balanced, look at charges 0 2+Cu Cu2+ + 2e-

Step 4: Write Reduction Half-ReactionSO4

2- SO2

Need to balance Oxygen, so add H2O as needed

SO42- SO2+ 2H2O

Need to balance H now, add H+ as neededSO4

2- + 4H+ SO2+ 2H2O

Now balance charges, 6+ 4+ on SulfurSO4

2- + 4H+ + 2e- SO2+ 2H2O

Step 5: Conserve charge by adjusting the coefficients in front of electrons so that the number lost equals the number gained.

Electron charge are equal so SKIP!Step 6: Combine Half-Reactions and Cancel

Cu Cu2+ + 2e-

SO42- + 4H+ + 2e- SO2+ 2H2O

Cu + SO42- + 4H+ + 2e- Cu2+ + 2e- + SO2+ 2H2O

Step 7: Combine ions to form compoundsCu + SO4

2- + 4H+ Cu2+ + SO2+ 2H2O

Need a SO42- to bond with Cu2+, must add to both

sides, which uses up the 2 extra H+ Cu + 2H2 SO4 Cu SO4 + SO2+ 2H2O

Check all elements are balanced and thenYOU ARE DONE!!!

• If a redox reaction is happening with a base rather than an acid, sometimes you will need to add H2O and OH- rather than H2O and H+

• Let’s do another one:

K2Cr2O7 + HCl + C2H5OH

CrCl3 + CO2 + KCl + H2O

Homework

Chapter 19.2 pg 641 #1-3