Transcript (+1) + - Edublogs

Oxidation-Reduction Reactions
Oxidation Numbers
Oxidation Numbers and Nomenclature
Identifying Oxidation-Reduction Reactions
Writing Equations for Oxidation-Reduction
Reactions
Oxidation-Reduction Titrations
Oxidation by Oxygen
Oxidation Numbers
Definition:
An oxidation number is a number that
reflects the electrons gained, lost, or
shared when an element reacts.
Remember that when you lose electrons
you become more positive and when you
gain electrons you become more negative.
Rules for Assigning Oxidation
Numbers
1. The oxidation # of any uncombined element is zero.
2. The oxidation # of a monoatomic ion equals its’ charge.
3. The more electrongative element in a binary compound is
assigned the number equal to the charge it would have if it
were an ion.
4. The oxidation number of fluorine is always -1.
5. Oxygen has an oxidation number of -2.
6. Hydrogen has an oxidation number of +1.
7. In compounds, Group 1 and 2 elements and aluminum
have oxidation numbers of +1, +2, and +3 respectively.
8. The sum of the oxidation numbers of atoms in a
compound is zero.
9. The sum of the oxidation # in a polyattomic ion is equal to
the charge of the ion.
Oxidation number or state
When dealing with simple ions, this is easy
to determine. It is simply the charge on
the ion.
Examples






Group IA (1)
Group IIA (2)
Group VIIA (17)
+1
+2
-1
Oxygen
-2 usually
Hydrogen +1 if bonded to nonmetal
Hydrogen -1 if bonded to metal
Oxidation number
For elements in their elemental state, the
oxidation number is also pretty straightforward.
Since all of the atoms are the same, the
electrons are shared equally so the oxidation
number is zero.
Examples
The atoms in N2, Na, P4, H2 and O2 all
have oxidation numbers of zero.
Oxidation numbers
With polar covalent bonds electrons are
shared but not equally.
For electrons that are shared in these
compounds, we assign the shared
electrons to the most electronegative
element.
We are just acting as though the
electronegativity difference was large
enough for the transfer of electrons to
occur.
Example
Assign the oxidation states for all elements
in water.
The electronegativities are:
H = 2.2, O = 3.5
The electrons from both hydrogen are
assigned to the oxygen.
Oxidation numbers: O = -2
H = +1
Oxidation numbers
Many elements have more than one
possible oxidation number.
Often, it is possible to determine the
oxidation number of those elements in a
compound simply by looking at what you do
know.
Follow the previous rules and then assign
an oxidation number that insures that the
overall compound has no net charge.
Example
Find the oxidation state for all elements in:
HNO3
Hydrogen we know - it must be +1
Oxygen should be -2 in this case.
What about nitrogen?
Example
See what you do know and find the
difference.
We know that H
is assigned a
value of +1

HNO3
Oxygen is -2
and we have 3
of them = -6
OK, so what's left over??
(+1) + (-6) + charge on nitrogen = 0
Nitrogen must have a value of +5
Determining Oxidation Numbers
SO2
NO3-
PCl5
SO42-
H2O
NH4+
H2SO3
MnO4-
B
Li Be
+1
+2
H
+3
He
C
N
+4 +5 +4
-2 +3 +2
-4 +1 -3
O
F
-1
-2
-1
S
Cl
Ne
+1
Al
Na Mg
+1
+3
+2
K Ca Sc
+1
+2
Rb Sr
+1
+2
+2
Y
Mo Tc Ru Rh Pd
Nb
Zr
+6
+7 +8 +6 +4
+4 Ag Cd
+3
+3
+2
+3
+4
+5
+4
Hf
Ta
+4
+5
+4
-4
P
+5 +6 +4 +7 +5
+3
+2 +3 +1
-3
-2
-1
Cr Mn Fe Co
As Se Br
Cu
Ge
Ni
Zn Ga
+6 +7 +6
5+
6+
+5
3+
Fr Ra Lr
+1
V
+4 +5 +4
+3
+3
+3
+3 +4 +3
+2
+2
+2
+2
+2
Cs Ba Lu
+1
Ti
Si
+4
+3
+6
+4
+3
+2
+2
+2
+1
In
+4
-4
3+
3-
+3
+4
+2
+5
+3
-3
+2
W Re Os
Ir
Pt Au Hg
Tl
Pb
Bi
+6
+4
+4
+3
+4
+2
+3
+1
+4
+2
+5
+3
+3
+1
+2
+1
4+
2-
+1
-1
Sn Sb Te
+3
+2
+8
+6
+2
+3
+4
+3
+7
+6
+4
+1
+2
I
Ar
Kr
+4
+2
Xe
+6 +7 +5 +6
+4
+1
+4
-2
-1
+2
Po
+2
At Rn
Common oxidation numbers
-1
Oxidation numbers and
the periodic table
Some observed trends in compounds.
Metals have positive oxidation numbers.
Transition metals typically have more than
one oxidation number.
Nonmetals and semimetals have both
positive and negative oxidation numbers.
No element exists in a compound with an
oxidation number greater than +8.
Oxidation number and nomenclature
Stock system
For metals with several possible oxidation
numbers, use Roman numeral in the name.
FeSO4
Fe2(SO4)3
Cu2O
CuO
PbCl2
PbCl4
iron(II) sulfate
iron (III) sulfate
copper(I) oxide
copper(II) oxide
lead(II) chloride
lead(IV) chloride
Identifying oxidation-reduction
reactions.
Oxidation-Reduction - REDOX
A chemical reaction where there is a net
change in the oxidation number of one or
more species.
Both an oxidation and a reduction must
occur during the reaction.
Mg (s) + Cl2 (g)
MgCl2 (s)
Here the oxidation number of Mg has changed from
zero to +2. Cl has changed from zero to -1.
REDOX reactions
Oxidation
An increase in oxidation number.
Reduction
A decrease in oxidation number.
If the oxidation number of any element
changes in the course of a reaction, the
reaction is oxidation-reduction.
Example.
2 Fe(NO3)3 (aq) + Zn(s)
Zn(NO3)2 (aq)
2 Fe(NO3)2 (aq) +
Redox Reaction
Fe(s) + O2(g)  Fe2O3(s)
Iron is oxidized (0 to +3)
and O2 is reduced (0 to -2)
Half Reactions
Fe  Fe3+ + 3e- (oxidation)
O2 + 4e-  2 O2-
(reduction)
Half reactions
Example.
Half-reactions can be of the ‘net ionic’
form. Balance the follow
Fe3+ + Zn (s)
Fe2+ + Zn2+
2 ( Fe3+ + e-
Zn(s)
Fe2+)
(reduction)
Zn2+ + 2e2Fe3+ + Zn (s)
(oxidation)
2Fe2+ + Zn2+
Example
+3
0
2Fe(NO3)3 (aq) + Zn(s)
+2
+2
2Fe(NO3)2 (aq) Zn(NO3)2 (aq)
Fe3+ is reduced to Fe2+
Zn is oxidized to Zn2+
NO3- is a spectator ion.
Zinc - Copper Voltaic Cell
Zn (s) Zn
2
(aq)
2
(aq)
Cu
Cu(s)
A Zinc-Copper Voltaic Cell
Zn electrode
Cu electrode
Anode (-)
Cathode (+)
Oxidation
Reduction
Half Reaction:
Half Reaction:
Zn  Zn2+ + 2 e-
Cu2+ + 2 e-  Cu
Voltaic Cell cont…
In a voltaic cell the redox reaction is
spontaneous and produces an electric
current. This current can be measured
using a voltmeter.
Standard Electrode Potentials
• A standard electrode potential, Eo, is based on
the tendency for reduction to occur at the
electrode.
• The cell voltage, called the standard cell
potential (Eocell), is the difference between the
standard potential of the cathode and that of the
anode.
Eocell = Eo (cathode) – Eo (anode)
Standard Electrode Potentials
Balancing REDOX equations
Many REDOX equations can be balanced
by inspection.
H2S (g) + H2O2 (aq)
S (s) + 2 H2O (l)
However, others are more difficult.
2KMnO4 (aq) + H2O2 (l) + 3H2SO4 (aq)
2MnSO4 (aq) + K2SO4 (aq) + 3O2 (g) + 4H2O (l)
Balancing REDOX equations
Half-Reaction method.
With this approach, the reaction is
broken into two parts.
Oxidation half-reaction. The portion of
the reaction where electrons are lost.
A
An+ + ne-
Reduction half-reaction. The portion of
the reaction where electrons are gained.
Balancing REDOX equations
The goal is then to make sure that the
same number of electrons are being
produced and consumed.
(m) (
A
(n) (me- + B
nB + mA
An+ + ne- )
Bm-
)
mAn+ + nBm+
When properly balanced, the electrons
Half reactions
Another Example
Determine the balanced equation for
the reaction of Fe2+ with Cr2O72- in an
acidic solution.
Fe2+ + Cr2O72-
H+
Fe3+ + Cr3+
The two half-reactions would be:
Fe2+
Fe3+
Half reactions
First, balance each half-reaction for all
elements except hydrogen and oxygen.
Fe2+
Fe3+
Cr2O72-
2Cr3+
Next, balance each half-reaction with
respect to oxygen by adding an
appropriate number of H2O.
Fe2+
Fe3+
Half reactions
Remember that this reaction occurs in an
acid solution so we can add H+ as
needed.
Fe2+
Fe3+
14H+ + Cr2O72-
2Cr3+ + 7H2O
Now we need to know how many
electrons are produced or consumed and
place them in our half-reactions.
Half-reactions
Fe2+
Fe3+ + e-
6e- + 14 H+ + Cr2O72-
2Cr3+ + 7H2O
We need the same number of electrons
produced and consumed so:
6Fe2+
6Fe3+ + 6e6e- + 14H+ + Cr2O72-
2Cr3+ + 7H2O
As our final step, we need to combine
Half-reactions
6Fe2+ + 14H+ + Cr2O726Fe3+ + 2Cr3+ + 7H2O
In this reaction, Fe2+ is oxidized and the
dichromate ion is reduced.
This reaction is used for the determination
of iron by titration.
Disproportionation reactions
In some reactions, the same species is
both oxidized and reduced.
Examples
2H2O2 (l)
2H2O (l) + O2 (g)
3Br2 (aq) +6OH- (aq)
+3H2O(l)
BrO3-(aq) +5Br-(aq)
For this to occur, the species must be
Oxidation-reduction titrations
REDOX reactions can also serve as the
basis for titrations.
For example, we can determine the
amount of iron in an ore by titration.
Initially, we must dissolve the sample.
This results in both iron(II) and iron(III)
being produced in solution.
Sample preparation
One option is to use a reductor. You
slowly wash your sample through the
column with water.
Jones Reductor
Zn(Hg)
Zn2+ + Hg(l) + 2e
An amalgam is used to prevent
Zn + 2H+reductorZn2+ + H2(g)
For iron, we get
Titrants
Now that all of our iron is in a single
oxidation state, we’re ready to do a
titration.
We need an oxidizing agent to convert
all the iron from Fe2+ to Fe3+.
Primary standard
A material that is available in pure form.
Common titrants
Oxidizing titrants
Dichromate - Cr2O72



The potassium salt is a primary standard
material.
Very stable solutions. If air is kept out, it can
last for years.
It is a very strong oxidizing agent.
Need an indicator such as diphenylamine
sulfonic acid.
Common titrants
Oxidizing titrants
Permanganate - MnO4



The potassium salt is the most commonly
used. It is not a primary standard.
Solutions must be standardized - typically use
Na2C2O4 ( a primary standard material.)
Reagent slowly degrades and MnO2 must be
removed
No indicator is needed - excess reagent
produces a pink solution.
Common titrants
The standardization of MnO42- with
oxalate involves the following
reaction.
2MnO42-(aq) + 5C2O42-(aq) + 8H+(aq)
2Mn2+(aq) + 10CO2 (g) + 8H2O (l)
Since MnO42- in an intense
Oxidation by oxygen
Oxygen is not the strongest of oxidizing
agents but it is about 19% of our
atmosphere.
It is able to react with all other elements
except:
noble gases
halogens
noble metals like gold
Combustion
Examples
CH4(g)
2H2O(g)
+ 2O2(g)
CO2(g) +
S(s)
+ O2(g)
SO2(g)
N2(g) + O2(g)
2NO(g)
+ O2 (g)
2 NO(g)
2 NO2(g)