1. Defining Oxidation and Reduction
Lesson 1
Test 9.1

2. Defining Oxidation and Reduction
SO4-2 is considered the spectator ion which will become eliminated
Consider:
Overall Eq’n Zn(s) + CuSO4(aq)  Cu(s) + ZnSO4(aq)
Total Ionic Eq’n Zn(s) + Cu2+(aq) + SO42-(aq)  Cu(s) + Zn2+(aq) + SO42- (aq)
Net Ionic Eq’n Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
● reducing agent
● donates electrons
● undergoes oxidation
● oxidizing agent
● accepts electrons
● undergoes reduction

3. Defining Oxidation and Reduction
Notice what happens to the reactants in this equation.
The zinc atoms lose electrons to form zinc ions. (becoming more positive)
The copper ions gain electrons to form copper atoms. (becoming more negative)
The following chemical definitions describe these changes
OXIDATION - loss of electrons
REDUCTION - gain of electrons
If a reaction does not have an oxidation or reduction component then it is not a redox reaction

4. Half-Reactions
a balanced equation that shows the number of electrons involved in either oxidation or reduction
two half-reactions are needed to represent a complete REDOX reaction (one shows oxidation, the other shows reduction
atoms and charges must balance
coefficients must be in smallest whole-number ratio
Zn(s)  Zn2+(aq) + 2e-
Cu2+(aq) + 2e-  Cu(s)
Oxidation
Reduction

5. Example Problem Solid Magnesium reacts with aqueous aluminum sulfate.
Write a balanced net ionic equation for the reaction, including physical states.
Step 1: Write the balanced overall equation
3 Mg(s) Al2(SO4)3 (aq)  3 MgSO4 (aq) Al(s)
Step 2: Separate into the total ionic equation and cancel out the spectator ions
3 Mg(s) Al3+(aq) SO42-(aq)  2 Al(s) + 3 Mg2+(aq) + 3 SO42- (aq)
Step 3: Write the Net Ionic Equation from step 2
3 Mg(s) Al3+(aq)  2 Al(s) + 3 Mg2+(aq)
Identify the reactant oxidized and the reactant reduced.
Mg is oxidized and Al is reduced
Identify the oxidizing agent and the reducing agent.
Mg is the reducing agent and Al is the oxidizing agent

6. Example Problem
Write balanced half-reactions for each of the following:
 a) Sn(s) + PbCl2(aq)  SnCl2(aq) + Pb(s)
Step 1: Separate into the balanced total ionic equation and cancel out the spectator ions.
Sn(s) + Pb+2 (aq) + 2 Cl- (aq)  Sn+2 (aq) + 2 Cl- (aq) + Pb(s)
Step 2: Write the net ionic equation
Sn(s) + Pb+2 (aq)  Sn Pb(s)
Step 3: Separate the equation into 2 half-reactions one for each element
Sn(s)  Sn+2
Pb+2 (aq)  Pb(s)
Step 4: Add electrons to both equations so that the net charge on both sides of each equation is equal. If you add electrons on one side of the arrow in one equation you will have to add it to the other side on the different equation. Then indicate which reaction is oxidation and which is reduction.
Oxidation: Sn(s)  Sn e-
Reduction: Pb+2 (aq) + 2 e-  Pb(s) 

7. Example Problem
Write balanced half-reactions for each of the following:
 b) Au(NO3)3(aq) + 3Ag(s)  3AgNO3(aq) + Au(s)
Step 1: Separate into the balanced total ionic equation and cancel out the spectator ions.
3 Ag(s) + Au+3 (aq) + 3 NO3- (aq)  3 Ag+ (aq) + 3 NO3 - (aq) + Au(s)
Step 2: Write the net ionic equation
3 Ag(s) + Au+3 (aq)  3 Ag+ (aq) + Au(s)
Step 3: Separate the equation into 2 half-reactions one for each element. You will have to divide the one equation to ensure the coefficients are 1.
3 Ag(s)  3 Ag+ (aq) divide by 3 -> Ag(s)  Ag+ (aq)
Au+3 (aq)  Au(s)
Step 4: Add electrons to both equations so that the net charge on both sides of each equation is equal. If you add electrons on one side of the arrow in one equation you will have to add it to the other side on the different equation. Then indicate which reaction is oxidation and which is reduction.
Oxidation: Ag(s)  Ag+ (aq) + 1e-
Reduction: Au+3 (aq) + 3 e-  Au(s)

8. Answer Practice Problems Pg. 601 #1-4, Pg. 607 #1-2
Homework
Answer Practice Problems
Pg. 601 #1-4,
Pg. 607 #1-2