Chapter 24 Oxidation-reduction Reactions
24-1 Oxidation states
First, let’s dispense with writing oxidation-reduction all the time. Like many
chemistry, let’s call these redox reactions
In Chapter 10 we looked at the Lewis formula of a couple of compounds to figure
out which atoms gained or lost electrons in a redox reaction. Having to write out
Lewis formulas for all compounds in a reaction is cumbersome and time
consuming, so we will now learn a set of rules that will help us to assign
oxidation states to help us locate which atoms gain or lose electrons.
As you learn how to assign oxidation state, please keep in mind that the idea of
an oxidation state is a simple booking trick that helps us to identify the material
getting oxidized or reduced. It does not necessarily equal an actual charge on at
atom.
The oxidation state for monoatomic ions is easy to do. The oxidation state of the
atom is equal to the charge on the ion.
The oxidation state in covalent compounds is found by arbitrarily assigning
electrons that are actually shared to one atom of a pair. When the electrons are
shared between two identical atoms, they are split evenly between the two
atoms. When they shared between two different atoms, the electrons are
assigned to the more electronegative atom.
A good example of that is water. Remember water is a covalent molecule. The
covalent bond between the Hydrogen and the oxygen comes from a sharing of
one electron from O and one electron from H. While these electrons are shared
between the two atoms (They orbit both atoms, not just one or the other), for the
oxidation state of this molecule we will assume that oxygen takes the electrons
from both of the hydrogens and keeps them to itself. Thus the O has a -2
oxidation state (Because it has 2 more electrons than it needs) while the
hydrogens have a +1 oxidation state (because each has 1 less electron than it
needs, and so has a +1 charge)
Key Concept:
Six rules for assigning oxidation states (a little simpler than rule from text)
1. Oxidation state for an atom in an element is 0 (examples Na(s), O2, O3, Hg)
2. Oxidation state of a monoatomic ion is the same as it’s charge ( Na+ +1, Cl- 1,)
3. Oxygen is assigned an oxidation state of -2 in covalent compounds. (H2O,
CO2) An exception occurs in peroxides O22-, where the oxidation state is -1
(Prime example H2O2)
4. In covalent compounds with nonmetals, H is given a +1 NH3, CH4
5. Fluorine, i.e., always -1
6. Sum of oxidation state must be zero for an electrically neutral species. For a
charged ion the sum of the oxidation state must be equal to the charge of
the ion
These rules aren’t perfect but they should apply for the problems you will get in
this class.
Example Problems:
NaCl
Ionic compound, Na+ and ClOxidation state of ionic ions is the same as the charge of the ion
Na = +1, Cl=-1
CO2
O is -2, two of them, for a net of -4, compound is neutral so C must
be +4 here
SF6
F is -1, six of them for -6, so S must be +6
NO3O is -2 times 3 = -6, net charge is -1 so N must be +5
CH4
H is +1, so C must be -4
Clicker questions: Following two examples from the lab
KMnO4
Break into ions; K+ & MnO4K+ so K=+1
MnO4-, O is -2, so Mn must be +7 to make net -1 charge
H2C2O4
Break into ions H+ and C2O4-2
H+ so H=+1
C2O4-2, O= -2, but there are 4 for a total of -8
Net charge on ion is -2 so there must be a net of +6 on the C
For +3 on each C
Note an interesting case of applying the rule to
Fe3O4 Magnetite (the iron ore that is magnetic)
The oxygen gives a -8, so the iron should be +8/3 state! This is not
a true oxidation state but occurs because of the way we apply the
rules. In this case there are 2 Fe 3+ and 1 Fe 2+ for a net effect of
8/3
24-2 Oxidation-Reduction Reactions
As we saw in chapter 10 redox reaction involve the transfer of electrons from one
atom to another, so they are sometimes called electron transfer reactions
Recall from the last chapter that
The material that loses electrons, gain + charge, is being oxidized, and is
considered the reducing agent in the reaction
The material that gains electron....loses + charge....is being reduced....is
considered to be the oxidizing reagent in the reaction.
Now that we have oxidation states it is much easier to come up with these
species
Example:
Fe2O3(s) + 3CO(g) 6 2Fe(s) + 3CO2(g)
Oxidation states
+3 -2
+2 0
0
+4 -2
So Fe goes from +3 to 0 so it is being reduced and Fe2O3(s) is the oxidizing agent and
C goes from +2 to +4 so it is being oxidized and CO(g) is a reducing agent
24-3 Half reactions
Key Concept:
All redox equations can be split into two half reactions, one that represents the
oxidation and the other that represents the reduction
Example:
Zn(s) + Cu2+(aq) 6 Zn2+(aq) + Cu(s)
has two ½ reactions
Zn(s) 6 Zn2+(aq) + 2eoxidation ½ reaction
Zn is being oxidized
Zn oxidation # 8
Electrons on right-hand side of equation
2+
Cu (aq) +2e- 6 Cu (s)
Reduction ½ reaction
Cu is being reduced
Cu oxidation # 9
Electrons on Left-hand side of equation
Break redox equation into ½ reactions is the key to balancing redox equations
24-4 Balancing Oxidation-Reductions reactions in Acidic and Neutral solutions
Balancing redox reactions is trickier than balancing non-redox reaction because
you have to balance the electrons in the ½ reactions. So let’s learn how to do
this.
Key Concept:
Rules for balancing redox reactions in Neutral or Acidic conditions
1. Write separate equations for reduction and oxidation half reactions
2. For Each half reactions
a. First Balance all elements except O and H
b. Add H2O to balance O
c. Add H+ to balance H
d. add electrons to balance net charge
3. If necessary multiply one or both ½ reaction equations by integer so the total
number of electrons used in one reaction is equal to the total number of
electrons furnished by the other.
4. Add the two ½ reaction equations together and cancel any common terms
5. Double check that all species and charge balance.
Let’s try this with the reaction
VO2+ + Sn2+ 6 V3+ + Sn4+
Step 1 break into ½ reactions
Sn2+ 6 Sn4+
Step 2 a,b,c Balance atoms
OK
Step 2 d Balance electrons
Sn2+ 6 Sn4+ + 2e-
VO2+6 V3+
VO2+6 V3+ + H2O
VO2+ + 2H+6 V3+ + H2O
VO2+ + 2H+ + e-6 V3+ + H2O
Step 3 Make # of electrons in ½ reaction match
Multiply V equation by 2
2VO2+ + 4H+ + 2e-6 2V3+ + 2H2O
Step 4 Add equation together
Sn2+ + 2VO2+ + 4H+6 Sn4+ + 2V3+ + 2H2O
Let’s try another one:
Ag + NO3- 6 Ag+ + NO
Separate
Ag6 Ag+
NO3- 6 NO
Balance all but H and O
(Ok as is)
Balance O with water
Ag6 Ag+
Balance H with H+
Ag6 Ag+
Balance charge with electrons
Ag6 Ag+ + e-
NO3- 6 NO+ 2H2O
NO3- +4H+6 NO+ 2H2O
NO3- +4H+6 NO+ 2H2O (+4,-1 = +3 on
left, 0 on right, need 3- on left
NO3- +4H++3e-6 NO+ 2H2O
Multiply so electrons are even in two half rxns
(Ag6 Ag+ + e-) x 3
NO3- +4H++3e-6 NO+ 2H2O
+
3Ag6 3Ag + 3e
NO3- +4H++3e-6 NO+ 2H2O
Sum 2 half rxns
3Ag + NO3- +4H+6 3Ag+ + NO+ 2H2O
Double check that it’s all balanced (it is)
Let’s try one from the lab
H2C2O4 + MnO4- 6CO2 + Mn+2
Separate:
H2C2O4 6 CO2
MnO4- 6 Mn+2
Balance all but H and O
H2C2O4 6 2 CO2
MnO4- 6 Mn+2
Balance O with H2O
H2C2O4 6 2 CO2
MnO4- 6 Mn+2 + 4H2O
Balance H with H+
H2C2O4 6 2 CO2 + 2 H+
8H+ + MnO4- 6 Mn+2 + 4H2O
Balance Charge with eH2C2O4 6 2 CO2 + 2 H+ + 2e(Reduction ½ rxn)
8H+ + MnO4- +5e- 6 Mn+2 + 4H2O
(Oxidation ½ rxn)
Multiply the oxidation by 5 and the reduction by 2
5H2C2O4 6 10 CO2 + 10 H+ + 10e16H+ + 2MnO4- +10e- 6 2Mn+2 + 8H2O
Add and check
5H2C2O4 + 16H+ + 2MnO4- +10e-6 10 CO2 + 10 H+ + 10e- + 2Mn+2 + 8H2O
Remove common terms
5H2C2O4 + 6H+ + 2MnO4-6 10 CO2 + 2Mn+2 + 8H2O
24-5 Balancing Oxidation-Reductions reactions in Basic solutions
Occasionally you will get a problem that specifies that the reaction takes place in
a basic solution. What does this mean (pH > 8, OH- predominates, H+
concentration is insignificant.)
What would this do to the procedure we just did?? Can’t use H+ as part of the
reaction because there isn’t any around!
How would we balance this?
If you try to add OH- directly to balance the equations it is easy to get it all
screwed up because you add oxygen and throw the oxygen balance off.
McQuarrie has a way around this that works, but I find a little counterintuitive.
Let’s try this method instead. The trick is to solve it just as you did before, but
just before the final answer, add OH- to both sides of the equation so the OH- is
equal to any H+ and they combine to form water.
Key Concept:
To balance redox equation under basic conditions add one last step
New step (insert between step 4 and 5)
Count up the number of H+ in your final equation
Add an equal number of OH- to BOTH side of the equation
The H+ and OH- become water on one side of the equation
The OH- remains on the other side of the equation
Double check to see if any of the new waters can be canceled out
Example:
Ag(s) + CN-(aq) + O2 (g) 6(Basic) Ag(CN)2½ reactions?
Ag(s) + CN-(aq) + 6 Ag(CN)2-
O2 (g) 6H2O
(Notice this is a bit tricky because water
was not in the original equation, but
remember how we balance oxygen by
adding water?
Mass balance
Ag + 2CN- + 6 Ag(CN)2-
O2 (g) 62H2O;
Charge balance
Ag + 2CN- + 6 Ag(CN)2- + e-
4H+ + O2 + 4e- 62H2O
Multiply by 4
4Ag + 8CN- + 6 4Ag(CN)2- + e-
4H+ + O2 + 4e- 62H2O
4H+ + O2 (g) 62H2O
Sum
4Ag + 8CN- + 4H+ + O2 + 6 4Ag(CN)2- + 2H2O
Now remove H+ by adding OH4Ag + 8CN- + 4H+ + O2 +4 OH- 6 4Ag(CN)2- + 2H2O + 4OH4Ag + 8CN- + O2 +4H2O 6 4Ag(CN)2- + 2H2O + 4OHNotice that at this point you have 2 moles of water as a reactant and product,
and these two moles could be removed
4Ag + 8CN- + O2 +2H2O 6 4Ag(CN)2-+ 4OH-
24-6 Oxidation-Reduction Reactions and Chemical Analysis
Skip for now
24-7 Corrosion
Now that you have the basics, I think I will skip these details. However if you are
curious about how rust forms or how galvanizing iron prevents rust, take a look at
section 24-7 It is a pretty quick read.