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Oxidation and
Reduction
Section 9.1
Electron Transfer Theory
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According to modern theory
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the gain of electrons is called reduction.
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Ag  e  Ag
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the loss of electrons is called oxidation.
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Ag  Ag  e
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Redox
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Theory suggests:
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the total number of e- gained in a reaction must
equal the total number of e- lost
Reduction and oxidation are separate processes
that occur simultaneously
The chemical equations of the oxidation,
reduction and the overall equation must be
balanced.
Writing and Balancing HalfReactions
Write a balanced net equation for the reaction of
copper with aqueous silver nitrate.
Step 1: Write each half reaction (the metal reacting
with the solution will undergo an oxidation –
forms a + ion thus a loss of electrons, while the
one in solution will undergo a reduction)
Cu( s )  Cu(2aq )  2e 
Ag(aq)  e   Ag( s )
Example Continued…
Step 2: Remember the number of electrons lost
must equal the number of electrons gained.
So we need to multiply the second reaction
by two so that we will have 2e- in each
equation
Cu( s )  Cu(2aq )  2e 
2  [ Ag(aq)  e   Ag( s ) ]
2 Ag(aq)  2e   2 Ag( s )
Example Cont…
Step 3: Now add the two half-reaction equations
2 Ag(aq)  2e  Cu( s )  2 Ag( s )  Cu(2aq )  2e
Step 4: Cancel out like terms
Step 5: Write net equation
2 Ag(aq)  Cu( s )  2 Ag( s )  Cu(2aq )
Oxidation States
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An oxidation number is a positive or
negative number corresponding to the
oxidation state of an atom
Using the following rules you can determine
the oxidation number of other atoms in
compounds (pg. 658)
Examples
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What is the oxidation number of sodium in
sodium hydride (NaH)?
- hydrogen in a hydride has an oxidation # of -1
- b/c there is only 1 Na, it must be +1
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What is the oxidation number of sulphur in
dihydrogen sulphide (H2S)?
- hydrogen is always +1
- b/c there are 2 of them the overall is 2(+1) = +2
- b/c only 1 S, it must be -2
Examples Cont…
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What is the oxidation number of carbon in the
carbonate ion (CO22-)?
- O has an oxidation number of -2
- there is an overall charge of 2- on the ion so we
can set up a mini equation:
x  2(2)  2
- solve for x
x  2(2)  2
x  (4)  2
x  4  2
x  2  4
x  2
Examples Cont…

What is the oxidation number of sulphur in
calcium sulphate ( CaSO4 )
 We can assumed that Ca is in ionic form  its oxidation
number is +2
 So we can set up our equation and solve
(2)  x  4(2)  0
(2)  x  (8)  0
6 x  0
x  6
Oxidation Numbers and Redox
Reactions
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An oxidation is an increase in oxidation
number
A reduction is a decrease in oxidation
number
In redox reactions, oxidation numbers
change.
Example:
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When methane burns, carbon dioxide and water
form. Identify the oxidation and reduction in this
reaction:
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First write the reaction:
CH 4( g )  2O2( g )  CO2( g )  2 H 2O( g )
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Now insert the oxidation numbers and arrows:
Seatwork:
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Pg. 653 #2
Pg. 656 #8-9
Pg. 659 #12, 13
Pg. 662 #18, 19, 20