Defining Oxidation and Reduction Lesson 1 Test 9.1

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

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

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

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) + 2 Al(s) Step 2: Separate into the total ionic equation and cancel out the spectator ions 3 Mg(s) + 2 Al3+(aq) + 3 SO42-(aq)  2 Al(s) + 3 Mg2+(aq) + 3 SO42- (aq) Step 3: Write the Net Ionic Equation from step 2 3 Mg(s) + 2 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

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+2 + 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+2 + 2 e- Reduction: Pb+2 (aq) + 2 e-  Pb(s) 

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)

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