Ch. 19 Oxidation and Reduction REDOX

3 ways of looking at oxidation and reduction 1. oxidation is a gain of oxygen atoms, reduction is a loss of oxygen atoms 2. oxidation is a loss of hydrogen atoms, reduction is a gain of hydrogen atoms 3. oxidation is a loss of electrons, reduction is a gain of electrons Most fundamental explanation, what we will be dealing with the most

Oxidation Reduction Gains oxygen Lose Oxygen Lose Hydrogen Gains hydrogen Lose Electrons Gains electrons Oxygen Hydrogen e-

Leo the Lion! LEO the lion says GER OIL RIG Loss of Electrons is Oxidation Gain of Electrons is Reduction OIL RIG Oxidation Is the Lost of electrons Reduction Is the Gain of electrons

Examples Is the reactant oxidized or reduced? Pb  PbO3 SnO2  SnO KClO3 KCl C2H6O  C2H4O C2H2  C2H6

Pertaining to LEO… Mg + S  MgS Mg + S  Mg2+ + S2- Magnesium is oxidized Said to be the reducing agent Substance in the reaction that loses electrons Sulfide sulfur atom is reduced Said to be the oxidizing agent Substance in the reaction that gains electrons

Oxidation Numbers A count of the electrons transferred or shared in the formation or breaking of chemical bonds You must assign each element in the reaction an oxidation number Follow a set of rules…

Oxidation Number Rules 1. The total of the oxidation numbers of all the atoms in a neutral molecule, an isolated atom, or a formula unit is 0 2. In their compounds, the Group 1A metals all have an oxidation number of +1, and the Group 2A metals have an oxidation number of 2+

Rules Con’t In its compounds, hydrogen has an oxidation number of +1 (except in metal hydrides such as NaH, where it is -1) 4. In its compounds, oxygen has an oxidation number of -2 (except in peroxides such as H2O2, where it is -1) 5. In their binary compounds with metals, Group7A elements have an oxidation number of -1. Group 6A elements have an oxidation number of -2, and Groups 5A elements have an oxidation number of -3.

Problems What is the oxidation number of each element? I2 Cr2O3 AlCl3 Na2SO4 CaH2

Identifying Redox Reactions 0 +3 -2 0 +3 -2 2 Al + Fe2O3  2 Fe + Al2O3 Al increases from 0 to +3, it is ______ Oxidized! Fe decreases from +3 to 0, it is _______ Reduced!

Oxidizing and Reducing Agents Now the confusing part… CuO + H2  Cu + H2O Cu goes from +2 to 0 Cu is reduced, therefore it is called an oxidizing agent because it causes some other substance to be oxidized H goes from 0 to +1 H is oxidized, therefore it is called a reducing agent because it causes some other substance to be reduced.

Identifying Agents in an Equation CuO + H2  Cu + H2O Reduction: CuO is the oxidizing agent Oxidation: H2 is the reducing agent

Electrochemical Cells An apparatus that allows a redox reaction to occur by transferring electrons through an external connector. Product favored reaction ---> voltaic or galvanic cell ----> electric current Reactant favored reaction ---> electrolytic cell ---> electric current used to cause chemical change. Batteries are voltaic cells

Basic Concepts of Electrochemical Cells Anode Cathode

CHEMICAL CHANGE ---> ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Zn is oxidized and is the reducing agent Zn(s) ---> Zn2+(aq) + 2e- Cu2+ is reduced and is the oxidizing agent Cu2+(aq) + 2e- ---> Cu(s)

GOOD AFTERNOON!  Please turn in your redox problems to the bin. PRE-AP: Please turn in your redox problems to the bin. Grab the sheets from the corner. Homework DUE: FRIDAY REGULAR Please grab a worksheet from the corner Take out the homework from last night – on desk REMINDERS: Project Description and Outline DUE MAY 2nd. Project DUE May 23rd (or by May 9th for +5 pts)

•Electrons travel thru external wire. Zn --> Zn2+ + 2e- Cu2+ + 2e- --> Cu Oxidation Anode Negative Reduction Cathode Positive <--Anions Cations--> RED CAT •Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments.

Standard reduction Potential E° = standard potential of the cell. Represents the voltage of the cell when the electrons create a current by passing through the wire. E°= E (reduction) + E (oxidation)

Electrolysis of Water Electrolysis of water is the breaking apart of water from H20 into its ions by running an electrical charge through it. Example of the first use and type of a electrochemical cell

CHEMICAL CHANGE ---> ELECTRIC CURRENT To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. This is accomplished in a GALVANIC or VOLTAIC cell. A group of such cells is called a battery. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

Terms Used for Voltaic Cells

Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques. MnO4- + 5 Fe2+ + 8 H+ ---> Mn2+ + 5 Fe3+ + 4 H2O Mn = +7 Fe = +2 Mn = +2 Fe = +3

Balancing Equations Consider the reduction of Ag+ ions with copper metal. Cu + Ag+ --give--> Cu2+ + Ag

Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. Ox Cu ---> Cu2+ Red Ag+ ---> Ag Step 2: Balance each element for mass. Already done in this case. Step 3: Balance each half-reaction for charge by adding electrons. Ox Cu ---> Cu2+ + 2e- Red Ag+ + e- ---> Ag

Step 4: Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu ---> Cu2+ + 2e- Oxidizing agent 2 Ag+ + 2 e- ---> 2 Ag Step 5: Add half-reactions to give the overall equation. Cu + 2 Ag+ ---> Cu2+ + 2Ag The equation is now balanced for both charge and mass.

Add H2O on O-deficient side and add H+ on other side for H-balance. Balancing Equations Balance the following in acid solution— VO2+ + Zn ---> VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn ---> Zn2+ Red VO2+ ---> VO2+ Step 2: Balance each half-reaction for mass. Red 2 H+ + VO2+ ---> VO2+ + H2O Add H2O on O-deficient side and add H+ on other side for H-balance.

Balancing Equations Step 3: Balance half-reactions for charge. Ox Zn ---> Zn2+ + 2e- Red e- + 2 H+ + VO2+ ---> VO2+ + H2O Step 4: Multiply by an appropriate factor. Red 2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O Step 5: Add balanced half-reactions Zn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O

Tips on Balancing Equations Never add O2, O atoms, or O2- to balance oxygen. Never add H2 or H atoms to balance hydrogen. Be sure to write the correct charges on all the ions. Check your work at the end to make sure mass and charge are balanced. PRACTICE!