6.1.3 REDOX REACTIONS
• Oxidation numbers identify and indicate which element
is oxidized and which is reduced. Here's an example the reaction between sodium metal and chlorine gas:
2 Na
+
Cl2
→
2 NaCl
• It is often useful to write the oxidation number for every
element above the element in the equation. Thus for our
reaction we have:
0
2 Na
0
+
Cl2
+1 -1
→
2 NaCl
• balancing coefficients in the equation do not affect the
value of the oxidation numbers.
elemen
t
initial
ox no
Na
0
Cl
0
final
ox no
change in
electrons (e-)
oxidized or
reduced
→
+1
lost 1 e-
oxidized
→
-1
gain 1 e
reduced
EXAMPLE
2 Mg
+
O2
→
2 MgO
Add in the oxidation numbers
• which are oxidized and which are reduced?
An increase in oxidation number indicates oxidation
A decrease in oxidation number indicates reduction
Reducing agent
the substance that is oxidized.
It allows another element to be
reduced.
Oxidizing agent
the substance that is reduced.
It allows another element to be
oxidized.
EXAMPLE
N2
element
initial
ox no
N
0
H
0
+
2H2
→
2 NH3
final
ox no
e-
oxidized or
reduced
Agent
→
-3
gain 3e-
reduced
oxidizing agent: N2
→
+1
lose 1e-
oxidized
reducing agent: H2
ASSIGNMENTS
• Practice 6.1.3
• Assignment 6.1.3
BALANCING REDOX REACTIONS
• Many redox reactions cannot easily be balanced
just by counting atoms. Consider the following net
ionic equation
Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)
• If you simply count atoms, the equation is balanced
but the charges aren’t balanced! Charges
represent gain or loss of electrons, and, like atoms,
electrons are conserved during chemical reactions.
1. BALANCING EQUATIONS USING
OXIDATION NUMBERS
• Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s
element
initial
ox no
final
ox no
Cu
0
→
+2
lost 2
×
1
=
2
Ag
+1
→
0
gain 1
×
2
=
2
change in
e-
balance for electrons
• 1 Cu(s) + 2Ag+(aq) → 1 Cu2+(aq) + 2Ag(s)
EXAMPLE
• MnO41- + Fe2+ + H1+ → Mn2+ + Fe3+ + H2O
element
initial
ox no
final
ox no
Mn
+7
→
+2
5
×
1
=
5
Fe
+2
→
+3
1
×
5
=
5
change in
e-
balance for electrons
• 1 MnO41- + 5 Fe2+ + H1+ → 1 Mn2++ 5 Fe3+ + H2O
• H and O are not balanced
• 1 MnO41- + 5 Fe2+ + 8 H1+ → 1 Mn2+ + 5 Fe3+ + 4 H2O
EXAMPLE
• NH3 + O2 → NO2 + H2O
element
initial
ox no
N
-3
O
0
final
ox no
change in
e-
→
+4
7
→
-2
2
balance for electrons
• Before using a multiplier to get the electrons to match, notice
the subscript with oxygen - O2. In our summary chart we base
our oxidation number changes on a single atom, but our
formula tells us that we must have at least two oxygen. You will
save some time and frustration if we take this into
account now. So in our summary table we will add some
columns to change our minimum number of atoms and
electrons involved. Then we complete the chart:
el
e
m
e
nt
initial
ox no
N
-3
→
+4
7
x 1 =
7
×
4
=
2
8
O
0
→
-2
2
×
4
×
7
=
2
8
final
ox no
No.
change in atom
es
2 =
No.
e-
balance for
electrons
• Since we were counting oxygen atoms in
the O2 molecule on the reactant side of the
equation, that's where we'll use the "7". Since
nitrogen's oxidation number also changed we will
use nitrogen's balancing coefficient there
4 NH3 + 7 O2 → 4 NO2 + H2O
• The last step is to balance for hydrogen atoms (and
finishing oxygen), which will mean placing a 6 in
front of H2O:
4 NH3 + 7 O2 → 4 NO2 + 6 H2O
• K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O +
Na2SO4 + K2SO4
• Your first concern is to make sure you correctly
determine all oxidation numbers. Since the only
place you see sulfur in this reaction is in SO42-, sulfur's
oxidation number is not going to change.
• Similarly, hydrogen and oxygen are always in
compounds, so their oxidation numbers also won't
change during the reaction. That narrows down the
list of elements to check.
• K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O +
Na2SO4 + K2SO4
element
initial
ox no
final
ox no
No.
change
atoms
in e-
Cr
+6
→
+3
3
I
+1
→
0
1
No.
e-
balance for
electrons
• Next, check for any subscripts associated with either
of these two elements - we see that Cr always has a
subscript of "2" (in both K2Cr2O7 and Cr2(SO4)3), and I
has a subscript in I2. So we'll add that to our
summary chart to get a total number of electrons
transferred, and then balance
element
initial
ox no
final
No.
change in
ox no
atoms
e-
No.
e-
balance for
electrons
Cr
+6
→
+3
3
×
2 =
6
×
1 =6
I
+1
→
0
1
×
2 =
2
×
3 =6
• Our table now tells us to use a balancing coefficient of
"1" for Cr on both sides of the equation and "3" for iodine.
Since we counted the atoms in I2 (and not HI), the "3" will
go in front of I2:
1 K2Cr2O7 + NaI + H2SO4 → 1 Cr2(SO4)3 + 3 I2 + H2O +
Na2SO4 + K2SO4
• With these numbers in place, we now balance for atoms
in the remainder of the equation to get our final answer:
1 K2Cr2O7 + 6 NaI + 7 H2SO4 → 1 Cr2(SO4)3 + 3 I2 + 7
H2O + 3 Na2SO4 + 1 K2SO4
ONE MORE EXAMPLE
• One more tricky one. Balance
Zn + HNO3 → Zn(NO3)2 + NO2 + H2O
• Determine oxidation numbers and create your summary
chart:
element
initial
ox no
final
ox no
No.
change
atoms
in e-
Zn
0
→
+2
2
N
+5
→
+4
1
No.
e-
balance for
electrons
• The main thing to notice is that N appears in two separate products Zn(NO3)2 and NO2. Should we consider the subscript for nitrogen
from Zn(NO3)2? In this case no, because this compound also contains
Zn, the oxidized element. Also, the oxidation number for nitrogen
does not change from HNO3 to Zn(NO3)2 .
• We now get our balancing coefficients from our
summary table. A "1" will be placed in front of Zn, but
which N should we use for the "2"? If you put it in front of
both HNO3 and NO2 you'll find you cannot balance for
nitrogen atoms. Since the oxidation number for nitrogen
changed in becoming NO2, we will try it there first. Some
trial-and-error may be required:
1 Zn + HNO3 → 1 Zn(NO3)2 + 2 NO2 + H2O
• With the 2 in place in front of NO2, we can now balance
the rest of the equation for atoms. Doing so gives us the
final answer:
1 Zn + 4 HNO3 → 1 Zn(NO3)2 + 2 NO2 + 2 H2O
PRACTICE PROBLEMS
• Practice problems 6.1.5- question #1
6.1.5- HALF REACTIONS
• Another way to balance redox reactions is by the halfreaction method. This technique involves breaking an
equation into the oxidation reaction and the reduction
reaction.
The general technique involves the following:
• The overall equation is broken down into two half-reactions. If
there are any spectator ions, they are removed from the
equations.
• Each half-reaction is balanced separately -atoms and then
charge. Electrons are added to one side of the equation to
balance charge.
• Next the two equations are compared to make sure electrons
lost equal electrons gained. One will be an oxidation reaction,
the other will be a reduction reaction.
• Finally the two half-reactions are added together, and any
spectator ions that were removed are placed back into the
equation
• Mg(s) + Cl2 (g) → MgCl2 (s)
• In this reaction, Mg is oxidized and Cl is reduced
(Mg changes from 0 to +2; Cl changes from 0 to -1)
• Balance the two reactions for atoms.
Mg → Mg+2
Cl2 → 2 Cl-
• Next balance the equations for charge by adding
electrons
Mg → Mg+2 + 2 e-
Cl2 + 2 e- → 2 Cl-
oxidation
reduction
• Finally, add the two equations together:
Mg + Cl2 → Mg+2 + 2 Cl• and reform any compounds broken apart in the
earlier steps:
Mg + Cl2 → MgCl2
• We see that the original equation was already
balanced, not just for atoms but for electrons as
well
• Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag(s)
• Identify the elements undergoing oxidation (Cu) and
reduction (Ag). The nitrate group (NO3) is a spectator ion
which we will not include in our half-reactions.
Cu → Cu+2 + 2 e-
Ag+ + 1 e- → Ag
oxidation
reduction
• After balancing for atoms and for charge, we see that
the two equations do not have the same number of
electrons - there are 2 in the copper reaction but only
one in the silver reaction. Multiply everything in the silver
reaction by 2, then we will add the equations together:
Step 1
Step 2
Step 3
Write the balanced
half-reactions
Balance
electrons
Add half-reactions
Cu → Cu+2 + 2 eAg + 1 e- → Ag-
Cu → Cu+2 + 2 e×2
Add equations together
Reform compound/return spectator
ions
2 Ag+ + 2e- → 2 Ag
Cu + 2 Ag+ → Cu+2 + 2 Ag
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
• Here is a reaction occurring in an acid solution.
MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
• In this example, spectator ions have already been
removed. Even though hydrogen and oxygen do
not undergo changes in oxidation number they are
not spectators and we need to work with them in
our half-reactions.
• We determine that Mn undergoes reduction (+7 to
+2) while Fe undergoes oxidation (+2 to +3). The iron
half-reaction is straight forward, but the manganese
reaction is more complex - we must include
hydrogen and oxygen in its half-reaction:
Fe2+→ Fe +3 + 1e-
MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O
oxidation
reduction
• To balance the manganese half-reaction - first
balance for Mn and O atoms. Next balance the H
atoms, and finally add enough electrons to
balance the charge on both sides of the equation.
Step 1
Step 2
Write the balanced
half-reactions
Fe2+→ Fe +3 + 1eMnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Add equations together
Step 3
Add half-reactions
×
5
5Fe2+→ 5Fe +3 + 5eMnO4- + 8 H+ + 5e- → Mn2+ + 4 H2O
MnO4- + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ +
4H2O
• HNO3 + Cu + H+ → NO2 + Cu2+ + H2O
• Determine what is oxidized, what is reduced, and
write the two balanced half-reactions (Step 1)
• Balance for electrons lost = electrons gained (Step
2)
• Add equations together
Step 1
Step
2
Step 3
Write the balanced
half-reactions
Add half-reactions
Cu → Cu+2 + 2e-
Cu → Cu+2 + 2e-
HNO3 + H+ + 1 e- → NO2 +
H2O
×2
2HNO3 + 2H+ + 2e- → 2NO2 + 2H2O
2HNO3 + Cu + 2H+ → 2NO2 + Cu2+ +
Add equations together
2H2O
ASSIGNMENT
• Finish practice problems
• Assignment 6.1.5- hand-in
• When balancing redox reactions, either the oxidation
number method or the half-reaction method may be
used. Often you'll find that one method works best for
some equations, while the other method is more suited
for other reactions. Or you may find one method just
easier to use. The practice exercises and assignments tell
you which method to use for a reaction, but as you get
get more experience you'll be able to make your own
decision as to which method to use.
• Writing half-reactions, however, is a skill you will need for
our final topic in this course - Electrochemistry - so be
sure you can write balanced half-reactions.