Transcript Chapter 20 Oxidation-Reduction Reactions

WarmUp: Oxidation Review
In one of Ms. Lyall’s favorite reactions, solid zinc combines with
solid sulfur to produce zinc sulfide. Here is the skeleton equation
Zn(s) +
S(s) 
ZnS(s)
a. Label the oxidation number of each species.
b. Label the species that is oxidized and which is is reduced.
c. Label the oxidizing and the reducing agents
d. Determine how many electrons were lost by each atom of
the oxidized species. Draw an arrow from the oxidized
reactant to the oxidized product and label the arrow with the
number of electrons lost each atom. Repeat for the reduced
species.
WarmUp: Oxidation Review
In one of Ms. Lyall’s favorite reactions, solid zinc combines with
solid sulfur to produce zinc sulfide. Here is the skeleton equation
Zn is the
reducing
agent
Zn is oxidized loses 2 e0
Zn(s) +
S is the oxidizing agent
0
S(s) 
+2 -2
ZnS(s)
S is reduced gains 2 e-
a.
the oxidation
number
of
each
d. Label
Determine
howofmany
electrons
were
lost reducing
by each atom
of
e.
The number
electrons
lost
byspecies.
the
agent
b.
Label
the species
that
is of
oxidized
which
is by
is reduced.
the
oxidized
species.
Draw
anelectrons
arrowand
from
the oxidized
must
equal
the number
gained
the
reactant
oxidized
product
and
thetoarrow
c.
Label to
thethe
oxidizing
theuse
reducing
agents
oxidizing
agent.
Youand
can
thislabel
ratio
helpwith
youthe
number of electrons lost each atom. Repeat for the reduced
balance a redox reaction equation.
species.
Try to balance this skeleton equation:
HNO3(aq) + H3AsO3(aq) --> NO(g) + H3AsO4(aq) + H2O(l)
It can be done.
But it’s not easy to do by trial and error.
Balancing Equations Using Oxidation Numbers
Oxidation numbers can help you!
As is oxidized loses 2 e- X 3 = 6 electrons lost
+1 +5 -2
+1 +3 -2
+2 -2
2 HNO3(aq) + 3 H3AsO3(aq) --> NO(g) +
N is reduced
+1 +5 -2
+1
-2
H3AsO4(aq) + H2O(l)
gains 3 e- X 2 = 6 electrons gained
1. Label the oxidation number of each species.
2. Label which species is oxidized and which is is reduced.
3. Determine how many electrons are “lost” by each
oxidized atom and how many are “gained” by each reduced
atom.
4. Determine the ratio needed to balance electrons lost with
electrons gained.
Use that ratio to write locked coefficient ratio.
Balancing Equations Using Oxidation Numbers
Now that you have locked the ratio between HNO3 and H3AsO3, the
equation is easy to balance.
2 HNO3(aq) + 3 H3AsO3(aq) --> 2NO(g) + 3H3AsO4(aq) + H2O(l)
11 H ______
11
______
______
2 N ______
2
______
15
15 O ______
______ As ______
3
3
Balancing Equations Using Oxidation Numbers
Oxidation numbers can help you!
Cu is oxidized loses 2 e- X 3 = 6 electrons lost
0
+1 +5 -2
+2 +5 -2
+2 -2
+1 -2
Cu(s) + HNO3(aq) --> 3Cu(NO3)2(aq) + 2NO(g) + H2O(l)
Some
N is reduced gains 3 e- X 2 = 6 electrons gained
1. Label the oxidation number of each species.
2. Label which species is oxidized and which is is reduced.
3. Determine how many electrons are “lost” by each
oxidized atom and how many are “gained” by each reduced
atom.
4. Determine the ratio needed to balance electrons lost with
electrons gained.
Use that ratio to write locked coefficient ratio.
Balancing Equations Using Oxidation Numbers
3 Cu(s) + 8HNO3(aq) --> 3 Cu(NO3)2(aq) + 2 NO(g) + 4H2O(l)
3
______
3 Cu ______
______
8 N ______
8
______
24 O ______
24
______ H ______
8
8
Balancing Equations Using Oxidation Numbers
Oxidation numbers can help you!
Some S is oxidized loses 2 e- X 5 = 10
electrons lost
+1 +7 -2
+1 +6-2
+1+1+4 -2
+2 +6-2
+1+6-2
+1 +6-2
+1 -2
2KMnO4+ H2SO4+5NaHSO3  MnSO4+ K2SO4 + Na2SO4 + H2O
Mn is reduced gains 5 e-
X 2 = 10 electrons gained
1. Label the oxidation number of each species.
2. Label which species is oxidized and which is is reduced.
3. Determine how many electrons are “lost” by each
oxidized atom and how many are “gained” by each reduced
atom.
4. Determine the ratio needed to balance electrons lost with
electrons gained.
Use that ratio to write locked coefficient ratio.
Balancing Equations Using Oxidation Numbers
2KMnO4+ H2SO4+5NaHSO3 2MnSO4+ K2SO4 + Na2SO4 + H2O
2 Mn ______
______
2
______
2 K ______
2
______
5 Na ______
______ S ______
______ H ______
______ O ______
Because of the
even subscript
for Na in the
product you
need to double
everything
Balancing Equations Using Oxidation Numbers
Na2SO4
is locked
4KMnO4+ H2SO4+10NaHSO3 4MnSO4+2K2SO4 +5Na2SO4 +6H2O
4 Mn ______
______
4
______
4 K ______
4
______
10 Na ______
10
______ S ______
13
13
______ H ______
16
12 O ______
______
50
50
All the S in
product is locked.
Because of the
even subscript
for Na in the
product you
need to double
everything
Balancing Equations Using Oxidation Numbers
Try this one on your own
K2Cr2O7 + 6NaI + 7H2SO4 
Cr2(SO4)3 + 3I2 + 7H2O + 3Na2SO4 + K2SO4