Ch. 5 Oxidation and Reduction

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Transcript Ch. 5 Oxidation and Reduction

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
Oxygen
Lose Hydrogen
Gains hydrogen
Hydrogen
Lose Electrons
Gains electrons
e-
Leo the Lion!
• LEO the lion says GER
– 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
3. 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
Reduction: CuO is
the oxidizing agent
• CuO + H2  Cu + H2O
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! 
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)
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. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
A group of such cells is called a battery.
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 halfreactions, 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+ + 2eRed
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+ +
2eOxidizing 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.
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.
Ox
Zn ---> Zn2+
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+ + 2eRed
e- + 2 H+ + VO2+ ---> VO2+ +
H 2O
Step 4: Multiply by an appropriate factor.
Ox
Zn ---> Zn2+ + 2eRed 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!