Download NAME UNIT 7: NOTES: REDOX (PART 1): OXIDATION #`S, An

NAME ______________________________ UNIT 7: NOTES: REDOX (PART 1): OXIDATION #'S,
An oxidation number is a positive or negative number assigned to a species. It is assigned to help understand
the number of electrons involved in bonding to a species of a different element, and to indicate the degree of
oxidation or reduction.
+1
-1
Given: Na + Cl2  2 NaCl
0
0
For instance, the sodium ion of NaCl is given an oxidation state of +1, compared to the original Na0. This +1
value indicates that 1 electron was involved in the bonding process to the Cl, (that's the "1") and that the electron
has been lost (that's the "+") to the chloride. Thus, that simple symbol of the +1 oxidation state, indicates two
pieces of important information.
The chloride conversely, is assigned a -1, indicating that the chloride species is a reduced species (compared to
Cl0 it is more negative), due to a gain of 1 electron. The “gained” electron is the same electron, as that lost by
the sodium. Thus, 1 electron was lost… 1 was gained… illustrating the Law of the Conservation of Charge.
***********************************************************************************************
I) Redox Reaction: Any reaction in which there is a change in oxidation numbers (LEO says GER)
REDuction / OXidation reaction: Electrons are * lost
of * oxidation states (or numbers)
and * gained
generally in a change
generally for two species. 
any chemical entity: a(n) molecule, atom or ion
A) For every oxidation there must be a reduction.
N
GER
LEO
G
B) Oxidation State (a.k.a: * Oxidation Number
): An arbitrarily assigned value which explains
or predicts the number of electrons of a species, involved in making a bond with a species of a
different element.
1) + or – values for oxidation states apply to species of compounds, or of ions in water.
0 is the oxidation state for pure elements. [thus, the oxidation state(s) of the oxygen species in a molecule
of O2 is 0, since the molecule is produced by species of the same (not different) element(s)]
1
2) The charge on an ion is only ONE category of oxidation states. Every encounter with an
oxidation state (number), is NOT necessarily an encounter with an ion
a) Oxidation states can be applied to species of a molecule as well. They are used to
describe the number *of shared electrons in a bond between the atoms of
a covalent (molecular) substance.
3) An oxidation number of a species may be * positive
more strongly to another nucleus
to be gained, relatively speaking.
or * negative
if its electrons are attracted
if the involved electrons tend
a) The metal ions of a compound will most likely be assigned a * positive
number
The nonmetals of a compound may be assigned a negative or a positive number.
i) The species with the positive oxidation # is often written first in a formula
OR RATHER: *The species with the lower electronegativity value is written
first in a formula
ii) There are some exceptions…For 1st year students, NH3 is the most important.

b) As written, there are 2 situations of serious importance when dealing with oxidation
states:
i) We use these + and – oxidation states, when dealing with species of
compounds or ions dissolved in water.
In light of the definition of oxidation state, why do pure elements get an oxidation
state of zero (0) assigned to them? * oxidation states account for the electrons
used to bond species of different elements. Thus, in pure elements, 0 electrons are
used to bond to species of different elements.
ii) When dealing with ions in water
 Ions in aqueous solution are also called *electrolytes or hydrated ions
2
4) For many species, the oxidation number is related to the ionic charge …. but first year
students must understand that virtually any single integer value may be appropriate.
Normally the values are whole numbers, but, they may be fractional.
a) it is vital to know the oxidation #s so we can track the electrons of a reaction...
0
+4 -2
0
2Mg(s) + SiO2(s) 
+2 -2
Si(s) + 2 MgO(s)
b) When only ONE value is written on the periodic table, then that is the one you will use
whenever the species is bonded in a compound.
Note that there are 2 val. electrons. Will Ca
tend to lose 2 e - or gain 6 e- to achieve a
stable octet?
Do you see from where the +2 comes?
+2
in a compound, every calcium will be a +2
Ca
2-8-8-2
e.g.) CaS,
Ca3(PO4)2
 c) For some species, the oxidation number will need to be calculated.
e.g.
[Ar] 3d ↑↓ ↑ ↑ ↑ ↑ 4s↑↓
FeO vs. Fe2O3
+2
+3
Fe
2-8-14-2
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II) SUMMARY: Assigning Oxidation Numbers:
A) When assigning, use the oxidation value listed on your periodic table, unless there is a special rule.
Rule … Description
Oxid. #
Examples
0 no electrons (0 electrons) used
1
Any element in its *elemental form or any *free element
0
Fe
N20 no electrons (0) used to bond N
to an atom of a different element
(*elemental form or free element refers to atoms or the diatomic molecules)
2
In ALL compounds, the Group 1 metal ions and the Ag ion
ALWAYS have a +1 (This applies to Ag in binary and ternary compounds)
+1
NaCl: sodium is Na+1
AgNO3: silver is Ag+1
3
In ALL compounds, the Group 2 metal ions and the Zn ion
ALWAYS have a +2 (This applies to Zn in binary and ternary compounds)
+2
CaS: calcium is Ca+2
ZnCl2: zinc is Zn+2
4
In ALL compounds, Al ion is ALWAYS +3
+3
Al2O3:each aluminum =Al+3
-1
NaF: fluoride is F-1
OF2: each F is F-1
(This applies to Al in binary and ternary compounds)
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In ALL compounds F (fluoride ion), is ALWAYS -1
(This applies to binary and to ternary compounds)
As a general rule, when the compound
is binary, pick the first oxid. # listed
In BINARY IONIC compounds:
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a) GROUP 15 ions = -3
(first one listed on the PT)
-3
Na3N: nitrogen is N-3
b) GROUP 16 ions = -2
(first one listed on the PT)
-2
CaS: sulfide is S-2
c) GROUP 17 ions = -1
(first one listed on the PT)
-1
NaCl: chloride is Cl-1
(Oxidation numbers for all of the above elements in ternary compounds must be
calculated arithmetically)
SOME RULE VARIATIONS EXIST
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Oxygen is usually a -2,
except when in compounds with fluorine
and in
compounds with the peroxide polyatomic ion (O2)-2
-2 usually NaOH: oxygen is –2
+1 or +2 OF2: oxygen is +2
-1
Na2O2 each oxygen is -1
8
Hydrogen is usually a +1,
except in metal hydrides (a binary ionic compound in which H is bonded
+1 usually H2O: each hydrogen is H+1
-1
NaH: hydrogen is H-1
to a metal but is the 2nd element…with the 1st being an oxidized metal)
9
In any compound, the sum of every species’s oxidation
number must add up to ZERO, because all compounds are
neutral, by their definition (no charges on compounds)
10 For any given simple ion in aqueous solution (monatomic
ion), the oxidation number = the charge of the ion
NaCl: Na+1 plus Cl-1 =
ZnCl2: Zn+2 plus Cl-1
plus Cl-1
0
=0
oxid # = in the ion Br-1 the
ion charge oxidation number = -1
11 In a polyatomic ion, the SUM of every oxidation number of
every particle must equal THE ASSIGNED CHARGE OF of
the polyatomic ion.
4
Practice: Assign the oxidation number to each species.
Please keep in mind that you need + values and – values.
RECAP of Rules
a)
2 AgI(s) + F2(g)  I2(s) + 2 AgF(s)
Pure elements =
Grp 1 and Ag ions in comps =
Grp 2 and Zn ions in comps =
b)
2 H2O(l) 
Al ions in comps are =
2 H2(s) + O2(s)
F ions (fluoride ions) in comps =
Grp 15 ions in binary ionic comps =
16 ions in binary ionic comps =
c)
F2(g) + 2NaI(aq)  2NaF(aq) +
17 ions in binary ionic comps =
I2(g)
Every O in a comp (usually) =
Every H in a comp (usually) =
d)
Mg(s) + 2 HCl(aq) 
e)
Zn(s) + 2HCl(aq)
f)
Hg(l) + Ag2S(s)  HgS(s) + 2 Ag(s)
can you guess at the oxidation state of mercury in this compound?

H2(g) + MgCl2(aq)
ZnCl2(aq) + H2(g)
There is no
rule that applies directly for assigning a value ...
g)
4 Al(s) + 3 O2(s)  2 Al2O3(s)
THINK: CATCH
h)
2 Li(g) + S(s)  Li2S
REACTANTS
i)
N2(g) + 6 Na(s) 
PRODUCTS
2 Na3N
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III) Calculating an oxidation number:
Very often, a species' oxidation number must be calculated. The rules can only go so far.
Elements with multiple oxidation number possibilities must have those oxidation numbers
calculated. You see, the oxidation number may be different from compound to compound.
A) The oxidation number reflects the number of electrons being used to bond, and thus, it makes sense
to imagine that the oxidation state of a particular species changes, as the species to which it is
bonded, changes.
1) This is especially true for most of the Transition metals and the all of the nonmetals.
These oxidation states should NOT be assumed, but rather, they should be algebraically
calculated
a) This is especially true for nonmetals in TERNARY INORGANIC compounds
b) PROCESS: Assign known oxidation numbers using the rules and assign the
unknown species, as "x". Then, algebraically solve for "x"
RECAP of Rules
Pure elements =
Fact: You cannot use the rules for Grp 15,16,17 elements in the following
compounds.
Grp 1 and Ag ions in comps =
Grp 2 and Zn ions in comps =
Al ions in comps are =
What is the oxidation number of S in Li2SO3?
F ions (fluoride ions) in comps =
Grp 15 ions in binary ionic comps =
16 ions in binary ionic comps =
17 ions in binary ionic comps =
Every O in a comp (usually) =
Every H in a comp (usually) =
What is the oxidation number of Fe in Fe2O3?
What is the oxidation number of Mn in Ca(MnO4)2
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PRACTICE: ASSIGNING OXIDATIONS NUMBERS IN TERNARY COMPOUNDS OR THOSE WITH TRANSITION METALS
PROCEDURE :
1) Assign "x" to the element you are investigating.
2) Use the rules and periodic table values to assign known oxidation numbers.
3) Multiply every oxidation number by the appropriate subscript to create an equation and set the
equation equal to zero (for compounds) and equal to the charge (for PAI).
4) Arithmetically solve for "x" by multiplying subscripts by oxidation # and set it all = to 0
Calculate and assign the oxidation number of :
a)
S in
Na2SO4
j)
b)
Cl in LiClO4
k)
c)
S in CaS2O3
d)
N in
e)
Mn in KMnO4
f)
KNO3
l)
N in Ca(NO3)2
S in Al2(SO3)3
m)
S in Al2(SO4)3
n)
Cu in CuO
o)
Cr in CrCl3
O in OF2 (exception)
g)
O in OF (exception)
h)
H in
i)
Fe in FeCl2
CaH2
(exception)
Cl in LiClO3
p)
Cr in Na2Cr2O7
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PRACTICE : Solve for the element in question. The sum, in this case is NOT equal to zero, but to the
charge of the polyatomic ion itself.
THIS IS AN APPLICATION OF RULE 11
-3
a) P in
(PO4)
b) N in
(NO3)
c) S in
(SO4)
e.g.:
x
-2
(PO4)-3
x – 8 = -3
x = +5
-1
(not 0, because it is an ION)
-2
d) Mn in (MnO4)
e) Mn in (MnO4)
-1
-2
Note: The oxidation number of a transition metal ion found in a polyatomic ion is a constant for that ion in that specific
polyatomic ion.
+1
Proof:
x
-2
Na2Cr2O7
-3 +1
CaCr2O7
Al2(Cr2O7)3
x
-2
(NH4)2Cr2O7
Make your life easier:
The N of (NO3)-1
is a * +5
The S of (SO4)-2
is a * +6
The P of (PO4)-3
is a * +5
As a side note for those who love biology:
The many faces of redox:
The Oxidized species of a rxn:
The Reduced species of a rxn:
becomes more positive (increases)
in oxidation number
becomes more negative
(decreases) in oxidation number
Na0 
Na+1
S0
+ 1e-
+ 2e- 
S-2
completely loses e- or experiences a
shift of e- away from it in a covalent
bond
completely gains e- or experiences a
shift of e- towards itself in a
covalent bond
loses hydrogen
gains hydrogen
gains oxygen (with a loss of e-)
e.g. alcohol to carboxylic acid
loses oxygen
Check Out:
http://science.jrank.org/pages/3776/Krebs-Cycle.html ...see also Tollens Test:
current / classic
inorganic
chemistry
Biology &/or
organic reactions
e.g. Krebs's cycle
http://www.youtube.com/watch?v=CMCVpE8p8yo&safety_mode=true&persist_safety_mode=1&safe=active
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NAME ______________________________ UNIT 7: NOTES: REDOX (PART 1 CONTINUED)
I-III) Oxidation Numbers
IV) How to identify a redox reaction on paper ….
A) A REDOX reaction is one in which there is a change in oxidation numbers when comparing
the oxidation numbers of the reactants to the oxidation numbers of the products
1) You should be able to identify a reaction as a redox (or as a non-redox reaction)
a) Essentially, your job is to assign an oxidation state to every species and to analyze
the oxidation states of the reactant species relatives to the oxidation states of the
product species.
B) One way to help identify the oxidized and reduced species
+3
+2
Fe + CuSO4  FeSO4 + Cu
+1
1) STRATEGY: Assign oxidation numbers to all species and compare
the reactant species to the species it becomes on the product side.
a) The answer must always come from the reactant side, when asked
for: the oxidized or reduced species
reducing agent or oxidizing agent...
0
-1
-2
-3
2)  We can identify the oxidized species because it is the reactant which
becomes * MORE POSITIVE (increases in oxidation #)
The reduced species is the reactant which becomes *MORE NEGATIVE (decreases in
oxidation number
0
a)
due to a gain of e0
2 Mg(s)
+ O2(g)

2+ 2-
2 MgO(s)
Mg0 was oxidized because it lost / gained negative charge (e-) and its oxidation
number became more * positive
O20 was reduced because it
lost / gained negative charge (e-) and its oxidation
number became more * negative
 the oxidized and reduced species are from the * reactant
side of the reaction equation!!!
Be sure to include the oxidation state with every answer. Some reactions may have “spectator ions”
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Guided Practice: Assign the oxidation numbers to each species and identify the oxidized and reduced species
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351.4 kJ
a) oxidation of aluminum:
0
0
+3 -2
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351.4 kJ
reduced species = * O20 or O0
because its oxidation state became * more negative
oxidized species = * Al0
because its oxidation state became * more positive
0
b) Making Ammonia (Haber Process):
reduced species = * N20 or N0
0
as a product
-3 +1
N2(g) + 3H2(g)  2NH3(g)
because its oxidation state became * more negative
oxidized species = * H20 or H0
because its oxidation state became * more positive
0
as a product
0
as a product
as a product
+1 -1
c) Making Table Salt: 2 Na(s) + Cl2(g) 
2 NaCl(s)
http://listverse.com/2008/03/04/top-10-amazing-chemical-reactions/
reduced species = * Cl20 or Cl0
oxidized species = * Na0
because its oxidation state became * more negative
because its oxidation state became * more positive
+6 -1
d) Purifying Uranium :
reduced species = * U+6
oxidized species = * F-1
0
as a product
as a product
0
UF6(s) U(s) + 3 F2(g)
notice there is only 1 reactant … but there
are 2 different species in that reactant.
because its oxidation state became * more negative
because its oxidation state became * more positive
as a product
as a product
(note: F6-1 is not correct, because the fluoride ions are not bonded to each other ...each is bonded via a covalent bond [surprisingly] to the uranium)
individually to the uranium ion)
0
0
reduced species: * O20 or O0
+3
f) Purifying Gold:
+1 -2
2 N2(g) + 2 O2(g)  2 N2O(g)
e) Making Laughing Gas:
-2
oxidized species: * N20 or N0
0
+1 -2
0
Au2S3(s) + 3 H2(g)  3 H2S(g) + 2 Au(s)
reduced species: *Au+3
oxidized species: * H20 or H0
10
0
g) Purifying Silicon:
+4 -2
0
+2 -2
Mg(s) + SiO2(s)  Si(s) + 2 MgO(s)
reduced species: *Si+4
oxidized species: *Mg0
0
h) Purifying Nickel:
+2 -1
+3 -1
0
2 Al(s) + 3 NiCl2(aq)  2 AlCl3(aq) + 3 Ni(s)
reduced species: *Ni+2
oxidized species: * Al0
+4 -2
0
+1 -2
+1 +5 -2
4 NO2 + O2 + 2 H2O  4 HNO3(aq)
i) Making Nitric Acid:
reduced species: * O20 or O0
oxidized species: * N+4
0
+2 +6 -2
0
+2 +6 -2
j) Corrosion of the Statue of Liberty: Fe(s) + CuSO4(aq)  Cu(s) + FeSO4(aq)
reduced species: * Cu+2
oxidized species: * Fe0
+3 -2
l) Ore Reduction:
0
0
+4 -2
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
reduced species: * Fe+3
oxidized species: * C0
N.B. Ore: a general term referring to a metal ion-containing mineral, that may be trapped in a larger mixture known as, a rock.
Iron ore: deemed valuable for its oxidized form of iron: Fe+2 and Fe+3, bonded in a compound with reduced oxygen (oxide).
The reduction of an ore refers to converting the metal CATION back into the metal ATOM by having the ion GAIN electrons back
History Buffs: The Hall Process and later the Bessemer Process are often seen as turning points of the Industrial Revolution in the West.
11
COOL CONCEPT!
(Purely Honors Chemistry)
+1 -1
m) Challenge: Teeth Whitening:
Disproportionation
Reaction: a redox reaction,
in which the * same reactant
species is both the oxidized
species and the reduced
species.
reduced species: * O-1
 2 H2O
2 H2O2
0
+ O2(g)
oxidized species: * O-1
+1 -2 +1
n) Challenge:
+1 -2
0

2 NaOH + Cl2
reduced species: *Cl20 or Cl0
+1 -1
NaCl
+1 +1 -2
+
+1 -2
NaClO + H2O
oxidized species: *Cl20 or Cl0
Individual Practice: You must assign each oxidation number and then analyzed the change so that you can
identify the oxidized and reduced species of each reaction. You must include the oxidation state of the species
with each answer.
1)
2 O2 (g) +
CH4(g)

CO2(g) + 2 H2O(l)
reduced species: *O20 or O0
2)
oxidized species: * C-4
Fe(s) + 2 Fe(NO3)3(aq) 
reduced species: *Fe+3 to the Fe+2
3)
Mg (s) +
4)
4 BCl3 (s) +
Cl2(g)
MgSO4(aq) + H2(g)
P4(s) + 6 H2(g) 
+ 2 NaI(s)
reduced species: *Cl20 or Cl0
(interesting n’est pas?)
oxidized species: * Mg0
reduced species: *P40 or P0
5)
oxidized species: *Fe0 to the Fe+2
H2(SO4)(aq) 
reduced species: * H+1
3 Fe(NO3)2(aq)
4 BP (s) + 12 HCl(g)
oxidized species: *H20 or H0

2 NaCl (s)
+
I2(s)
oxidized species: *I-1
12
6)
2 KClO3(s) 
2 KCl(s)
reduced species: *Cl+5
7)
2 SO2(g)
+
O2(g) 
+
oxidized species: *O-2
SO3(g)
reduced species: *O20 or O0
8)
2 HgO(s) 
2 Hg(l)
reduced species: *Hg+2
3 O2(g)
oxidized species: *S+4 to the S+6
+
O2(g)
oxidized species: *O-2
C) The agents: The agents are the opposite of their names. In short, the "agent" is that species whose
presence enables the activity for which it is named. ... HUH????
e.g.) The presence of the oxidizing agent allows oxidation to proceed, hence the
oxidizing agent is the reduced species.
1) Oxidizing Agent: * The reduced species
(a.k.a. an oxidizer or oxidant)
a) a strong oxidizing agent is reduced, easily (readily).
b) a weak oxidizing agent is reduced, but more slowly when compared to a stronger one.
2) Reducing Agent: * The oxidized species
(a.k.a. an antioxidant….Wait a minute … I have heard this term before …. Hey! )
3) Spectator ions: Not always present … but these are ions which do not change in oxidation
state.
TIME FOR A METAPHOR ... MAKE SOME CONNECTIONS!!
Think about a sports agent: Does the sports agent play the sport?
Does the travel agent take the arranged trip?
Does the insurance agent purchase the prepared policy?
So, is the oxidizing agent the oxidized species ????
13
Interpret: The N+5 of NaNO3 is a strong oxidizing agent (oxidizer)
*N+5 is easily / readily reduced (oxidizing agents or oxidizers are reduced species)
Interpret: The Mn+7 of KMnO4 is a strong oxidizing agent (oxidizer)
* Mn+7 is easily / readily reduced
Interpret: Sodium hypochlorite is a stronger oxidizing agent than 3% hydrogen peroxide
* Sodium hypochlorite is more easily reduced than hydrogen peroxide
D) A “real world” application: CLOROX bleach / Pool Chlorine
Recall that an oxidizing agent is any substance which causes another substance to * lose
electrons.
one or more
The decolorizing action of bleaches is due in part to their ability to remove these electrons which are activated
by visible light to produce the various colors. The hypochlorite ion [(ClO)-1], found in many commercial
preparations, is reduced to chloride ion and hydroxide ion forming a basic solution as it accepts electrons from
the colored material as shown below.
+1 -2
-1
(ClO)-1 + 2e- + H2O  Cl + 2(OH)-1
Robert Asato, Ph.D, Leeward Community College
PRACTICE
1) Given: F2(g) + 2KCl(aq)  2KF(aq) + Cl2(g)
a) Assign oxidation numbers to each species
b) What species is oxidized? ______
(be sure to record the oxidation number of your answer, even when it is 0)
c) What species is reduced? _______
(be sure to record the oxidation number of your answer, even when it is 0)
d) What species is the oxidizing agent? ______
e) What species is the reducing agent? ______
(be sure to record the oxidation number of your answer, even when it is 0)
(be sure to record the oxidation number of your answer, even when it is 0)
f) How are the species of a redox reaction related to the “agents”?
14
2) Given: 2 Al(s) + 3 NiCl2(aq)  2 AlCl3(aq) + 3 Ni(s)
a) Assign oxidation numbers to each species
b) What species is oxidized? ______
(be sure to record the oxidation number of your answer, even when it is 0)
c) What species is reduced? _______
(be sure to record the oxidation number of your answer, even when it is 0)
d) What species is the oxidizing agent? ______
e) What species is the reducing agent? ______
(be sure to record the oxidation number of your answer, even when it is 0)
(be sure to record the oxidation number of your answer, even when it is 0
3) Given: 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
a) What species is oxidized? ______ How do you know? ...The oxidation state ________________
b) What species is reduced? ______
4) Given: Mg(s) + SiO2(s)  Si(s) + 2 MgO(s)
a) What species is the oxidizing agent? ______ How do you know? ________________
b) What species is the reducing agent? ______
___5) Given: 2 N2(g) + 3 O2(g)  2 N2O3(g)
a)
N20
b) O20
c) N-3
0
6) Given:
Which species is the oxidizing agent?
d) O-2
+4
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
From a chemical (or redox) point of view, what happens to the three moles of carbon as they react with
the rust? The 3 mol of C0 are _______________ Defense: __________________________________
_______ 7) What is the oxidation number of silicon in the compound H2SiF6(s)?
___8) In which of the following compounds, does oxygen have the most positive oxidation number?
a) NaOH
b) Ag2O
c) NaClO
d) OF2
15
9) What is meant, in chemical terms, by the phrase: “Fluorine gas (F2(g)) is a strong oxidizing agent.”?
__________________________________________________________________________
For question 10 a –d, use your knowledge of chemistry and the following passage, which is an adaptation from
McQuarrie & Rock Descriptive Chemistry 1985 p. 153) Numbers for the lines of print have been provided.
1
3
6
9
Copper is slightly less abundant than nickel and is found in many different ores. An
ore is a general term referring to the rocks that are really mixtures of a number of
valuable minerals (a.k.a. elements or compounds).
Copper generally occurs as various sulfides, although in some ores copper is present in
the form of sulfates, carbonates and other oxygen containing compounds. Deposits of
the free metal are very rare, being found only in Michigan. Most copper-containing
ores have a copper content of less than 1 percent, but some richer ores have up to
4 percent copper.
Copper ores contain other metals and metalloids such as selenium, and tellurium,
which are important by-products when the copper ore is reduced to copper metal.
Some important copper minerals (a.k.a. compounds of copper) are chalcocite, (Cu2S),
chalcopyrite (CuFeS2), and malachite (CuCO3Cu(OH)2), which is a crystalline
mixture of two compounds.
10) a) What is the oxidation number of copper in the mineral (compound) chalcocite (Cu2S) ?
________
b) In lines 5-6 you read: “Deposits of the free metal are very rare…” What does the author mean by the
term “free metal”? _____________________________ What is the oxidation state associated with
the atoms of the free metal copper? _____
c) In line 9 a reference is made to copper ore being reduced to copper metal. Chemically speaking,
what must happen to the metallic ions of an ore, in order to be reduced to the metal? _____________
____________________________________________________________________________________
d) What elements, other than metals and nonmetals may be obtained from refining copper ores?
16
___11) Given the chemical equation:
KI + Pb(NO3)2  PbI2 + KNO3
you can state that
a) no changes in oxidation states occur
b) lead changes in oxidation state from Pb+2 to Pb0
c) iodide changes in oxidation state from I0
to I-1
d) potassium changes in oxidation state from K+1 to K0
___12) Given the chemical equation:
2 KCl + F2  2 KF + Cl2
The oxidation number of the fluorine atoms in F2 changes from:
a) -1
to -2
c) -1 to 0
b) -1
to +1
d) 0 to -1
____13) Which category is composed of elements that have both positive and negative oxidation states ?
a) the noble gases
b) the transition metals
c) the halogens
d) the alkaline-earth metals
___14) Which of the following reactions is classified as a redox reaction?
a) PCl5 + 4 H2O  H3PO4 + 5 HCl
b) Ca + 2 HNO3  Ca(NO3)2 + H2
c) H2SO4  SO3 + H2O
d) 3BaBr2 +
Al2(S2O3)3  3BaS2O3 + 2AlBr3
___15) Which of the following reactions is classified as a redox reaction?
a) 3 O2  2 O3
b) CaO + SiO2  CaSiO3
c) C2H5OH + 3 O2  2 CO2 + 3 H2O
d) Fe3(PO4)2 + 6 NaOH 
2 Na3PO4 + 3 Fe(OH)2
17
___16) In which of these compounds does Cl have the highest (most positive) oxidation number?
a) NaClO
c) NaClO3
b) NaClO2
d) NaClO4
____17) What is the oxidation state of nitrogen in the polyatomic ion (NH4)+1?
___18) Given the equation:
4 BCl3 + P4 + 6 H2  4 BP + 12 HCl
a) No redox reaction occurs
c) Phosphorous (P4) is reduced
b) The boron in BCl3 is oxidized
d) Hydrogen is a spectator ion
For 19 – 22 place a checkmark next to the reactions classified as a redox reaction. There may be more
than 1 checked answer
___19)
Mg (s) + H2(SO4)(aq)  MgSO4(aq) + H2(g)
___20)
2 KrF2 (g) + 2 H2O(g)  2 Kr (g) + O2(g) + 4 HF(g)
___21)
4 BCl3 (s) + P4(s) + 6 H2(g)  4 BP (s) + 12 HCl(g)
___22)
2 NH4Cl(s) + Ba(OH)2(aq)  BaCl2(s) + 2 NH3(g) + 2 H2O(l)
A very interesting reaction, n’est pas?
…one of the few reactions with a noble gas
ANSWERS
1) Given: F2(g) + 2KCl(aq)  2KF(aq) + Cl2(g)
a) Assign oxidation numbers to each species
b) What species is oxidized? Cl1c) What species is reduced? F20
d) What species is the oxidizing agent? F20
e) What species is the reducing agent? Cl1f) Answers will vary … think about how you identify an oxidized species …vs. an oxidizing agent … We’ll share answers in class….
2) Given: 2 Al(s) + 3 NiCl2(aq)  2 AlCl3(aq) + 3 Ni(s)
a) Assign oxidation numbers to each species
b) What species is oxidized? Al0
c) What species is reduced? Ni2+
d) What species is the oxidizing agent? Ni2+
e) What species is the reducing agent? Al0
18
19
3) Given: 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
a) What species is oxidized? Al0
b) What species is reduced? H1+
4) Given: Mg(s) + SiO2(s)  Si(s) + 2 MgO(s)
a) What species is the oxidizing agent? Si4+
b) What species is the reducing agent? Mg0
5) b) O20
6) From a chemical (or redox) point of view, what happens to the three moles of carbon as they react with
the rust? They are oxidized. Defense: The oxidation number of the C changes from 0 to +4
7) +4
8) d) OF2
9) It is easily / readily reduced
10) a) +1
b) The pure element
Zero (0)
c) They gain electrons
d) metalloids such as Se and Te
11) a) no changes in oxidation states occur
12) d) 0 to -1
13) c) the halogens
14) b) Ca + 2 HNO3  Ca(NO3)2 + H2
15) c)
C2H5OH + 3 O2  2 CO2 + 3 H2O
16) d) NaClO4
17) N = -3
18) c) tetra- phosphorous or phosphorous atoms (P4) are reduced
__19)
Mg (s) + H2(SO4)(aq)  MgSO4(aq) + H2(g)
__20)
2 KrF2 (g) + 2 H2O(g)  2 Kr (g) + O2(g) + 4 HF(g)
__21)
4 BCl3 (s) + P4(s) + 6 H2(g)  4 BP (s) + 12 HCl(g)
20
NAME ____________________________________ UNIT 7: REDOX NOMENCLATURE OF
INORGANIC COMPOUNDS
I-IV) Oxidation numbers through Oxidizing/Reducing Agents
V) NOMENCLATURE: (Latin: nomen = name)
CONCEPT: Nomenclature is the process of naming a compound, when given its formula
A)
Refect: We can categorize the
INORGANIC COMPOUNDS
Molecular (or Covalent)
Compounds
Ionic Compounds
They have different types of bonds & properties
so it’s reasonable that they are named differently
IUPAC strategy
Use the
Prefix system
IUPAC strategy
Use
Stock’s Nomenclature System
So when given a formula of a compound, we must know how to get to a name.....
B) Before we go any further you must understand that the “endings” of many names can help you.
 Have you noticed that many binary compounds end in –IDE? The last syllable of the
nonmetal’s element name is changed to –IDE, when a nonmetal is the more electronegative
element of the compound.
NaCl is binary inorganic with Cl as the
more electronegative element
 Have you noticed that the names of the metals never change? Given:
electronegativity values:
NaCl
0.9 vs. 3.1
The name is sodium chloride
The cation’s name is the same … the anion’s name has changed to -ide
 Have you noticed that ternary inorganics (compounds with polyatomic ions)
use the name of the polyatomic ion with no change in ending?
Given: NaNO3
The name is sodium nitrate
Notice in a ternary inorganic, there are no changes in the names
21
Structured Overview:...Here is what you need to know for nomenclature strategies for inorganic compounds.
INORGANIC COMPOUNDS
Molecular (a.k.a. Covalent)
Compounds
Ionic Compounds
Use Stock’s Nomenclature system
when naming
Use the prefix system
when naming
We will name only
binary molecular
compounds using
prefixes
About endings
The more
electronegative
element’s ending
changes to -IDE
Prefixes to use are:
Mono = 1
Di
=2
Tri
=3
Tetra = 4
Pent(a) = 5
Hex(a) = 6
Hept(a) = 7
Oct(a) = 8
Non(a) = 9
Dec(a) = 10
We will name binary
ionic compounds
AND
ternary ionic compounds
with Stock’s system
When binary
When ternary
ENDINGS:
The more
electronegative
element’s ending
changes to -IDE
There are no
special endings.
When the metal cation has more
than 1 positive oxidation state
possibility as listed on the periodic
table, include a roman numeral
that matches / identifies the value
of the oxidation state. This
enhances communication and
eliminates confusion as to the
compound’s identity.
refers to
Fe2O3 = iron (III) oxide
the +
oxidation
state
and it is different
chemically from
FeO = iron (II) oxide
22
WHEN ASKED TO NAME AN INORGANIC COMPOUND, WHEN GIVEN A FORMULA
ASK: IS THE COMPOUND I NEED TO NAME….?
A BINARY MOLECULAR COMPOUND
ASK : IS IT…?
ASK: IS THE COMPOUND…?
A BINARY IONIC COMPOUND
A TERNARY IONIC COMPOUND
The overall rule is to use the appropriate
The overall rule is to AVOID prefixes and identify
prefix that matches with each subscript of the
the positive oxidation state with a roman numeral
compound’s formula:
(if warranted)
1 =mono (used mostly w/ oxides & for 2nd species)
I = +1
2= di
II = +2
3= tri
III= +3
4= tetra
IV = +4
5 =penta
V = +5
6= hex(a)
VI = +6
7= hept(a)
VII = +7
8= octa
9= nona
10 =deca
Process:
Process:
Process:
1) Name the 1st element using any applicable prefix
1) Name the metal species
1) Name the positive species
2) Name the 2nd element with any applicable prefix
2) Name the nonmetal species
2) Name the negative species (often
but end it in IDE
but end it with IDE
a polyatomic ion)
3) Stop
3) Reflect on the metal species .
Ask: Is there more than 1 positive state?
3) reflect on the metal species.
Ask: Is there more than 1 pos. state?
Examples:
N2O5 = dinitrogen pentoxide
If yes
assign oxidation numbers and include the positive
value in the name as a roman numeral & Stop
CCl4 = carbon tetrachloride
If yes
assign oxidation numbers &
include the positive value in
the name as a r.n. & Stop
If no: Stop… do not use a roman numeral
NO = nitrogen monoxide
Examples:
If no: Stop… leave it alone. Do not
include a roman numeral
Examples:
CO2 = carbon dioxide
H2S = dihydrogen sulfide or more commonly just
hydrogen sulfide (an exception to the rule)
MnO2 = manganese (IV) oxide
Mn3(PO4)2 = manganese (II) phosphate
Cu2S = copper (I) sulfide
CuS = copper (II) sulfide
Cu 2SO4 = copper (I) sulfate
CuSO4 = copper (II) sulfate
CaCl2 = calcium chloride
ZnSO 4 = zinc sulfate
23
Use the flow chart and name each of the following compounds. Ionic compounds and molecular compounds
are mixed freely … thus, you first analyze critically as to the category to which each compound belongs …then
name….
1) ZnS _____________________________________
(is this binary molecular / binary ionic / ternary ionic? … now, apply the rules)
2) Au2S3 ___________________________________
(is this binary molecular / binary ionic / ternary ionic? …. now, apply the rules)
3) OF2 _____________________________________
(is this binary molecular / binary ionic / ternary ionic? …. now, apply the rules)
4) Cr2O3 __________________________________
(is this binary molecular / binary ionic / ternary ionic? …, now apply the rules)
5) N2O ____________________________________
(is this binary molecular / binary ionic / ternary ionic?)
6) Ag2SO4__________________________________
(is this binary molecular / binary ionic / ternary ionic?)
7) SF4____________________________________
These answers are given in reverse
order …just in case you don’t want
to “peek”.
21) dihydrogen oxide
or dihydrogen monoxide
20) zinc phosphate (no roman numeral)
19) iron (III) phosphate
18) aluminum phosphate
(no roman numeral)
17) phosphorus pentachloride
8) MnO2 ____________________________________
9) N2O5 _____________________________________
16) iron (III) chloride
15) titantium (II) nitrate
14) calcium chloride (no r.n)
10) PCl3 ______________________________________
13) nitrogen dioxide
11) NO ______________________________________
12) lead (II) oxide
12) PbO _____________________________________
11) nitrogen oxide or
nitrogen monoxide
13) NO2 _____________________________________
10) phosphorus trichloride
9) dinitrogen pentoxide
14) CaCl2 ____________________________________
8) manganese (IV) oxide
15) Ti(NO3)2 _________________________________
7) sulfur tetrafluoride
16) FeCl3 ____________________________________
6) silver sulfate
17) PCl5 _____________________________________
18) AlPO4 ___________________________________
19) FePO4 ___________________________________
20) Zn3(PO4)2 _________________________________
21) H2O (technically& with a giggle)_______________________
(no roman numeral)
5) dinitrogen oxide
or dinitrogen monoxide
4) chromium (III) oxide
3) oxygen difluoride
2) gold (III) sulfide
1) zinc sulfide
(no roman numeral)
Now for a laugh… check out: http://www.dhmo.org/facts.html
24
Check your grasp of the strategies for inorganic compound nomenclature
1)
Fact1 : Mn in the compound MnCl4 has an oxidation state of +4
Fact2 : V in the compound VCl4 has an oxidation state of +4
Fact3 : The most correct name for MnCl4 is
Fact4 : The most correct name for VCl4 is
manganese (IV) chloride
vanadium (IV) chloride
Question: Why do both of the above compound names use a Roman numeral in their name?
* Both metal ions have multiple (more than one) positive oxidation state
____________________________________________________________________________________
2)
Fact1 : Na in the compound Na2SO4 has an oxidation state of +1
Fact2 : Ca in the compound CaSO4 has an oxidation state of +2
Fact3 : The most correct name for Na2SO4 is sodium sulfate
Fact4 : The most correct name for CaSO4 is calcium sulfate
Question: Why do neither of the above compound names contain a Roman numeral in their name?
* Each metal ion has only 1 possible positive oxidation number (neither has multiple positive oxidation states)
____________________________________________________________________________________
3)
Fact1 : Fe in the compound FeCl3 has an oxidation state of +3
Fact2 : Al in the compound AlCl3 has an oxidation state of +3
Fact3 : The most correct name for FeCl3 is iron (III) chloride
Fact4 : The most correct name for AlCl3 is just aluminum chloride
Question: Why does one of the above compound names include a Roman numeral while the name
of the other compound does not include a Roman numeral ?
* The iron has more than one possible positive oxidation state … and aluminum has only one
25
NOMENCLATURE: 1-25 are all ionic compounds. Practice the use of Stock’s system. Twenty six through
thirty one are each molecules, so, use the prefix system to name them. The answers are on the next page.
1) CaO _____________________________
(use PT: do you need to include a Roman numeral ?)
14) NiCO3 _____________________________
(use PT: do you need to include a Roman numeral ?)
2) NiCl3 _____________________________
(use PT: do you need to include a Roman numeral ?)
15) K3PO4 _____________________________
(use PT: do you need to include a Roman numeral ?)
3) CuSO4 _____________________________
(use PT: do you need to include a Roman numeral ?)
16) MnO _____________________________
17) FeO
4) MnO3 _____________________________
(use PT: do you need to include a Roman numeral ?)
5) NaClO _____________________________
_____________________________
18) Li2CO3 ____________________________
19) SnF2
_____________________________
(use PT: do you need to include a Roman numeral ?)
20) HgCl2
_____________________________
6) Sr3(PO4)2 _____________________________
(use PT: do you need to include a Roman numeral ?)
7) AgF _____________________________
(use PT: do you need to include a Roman numeral ?)
21) Ca(OH)2 _____________________________
22) Li2S2O3 ____________________________
23) CuO
_____________________________
8) Au2Te _____________________________
(use PT: do you need to include a Roman numeral ?)
24) Na2SO3 _____________________________
9) PbO2 _____________________________
(use PT: do you need to include a Roman numeral ?)
10) Pb(NO3)4 ____________________________
(use PT: do you need to include a Roman numeral ?)
11) Ca3(PO4)2 __________________________
(use PT: do you need to include a Roman numeral ?)
12) TiO2 _____________________________
(use PT: do you need to include a Roman numeral ?)
13) CrO _____________________________
(use PT: do you need to include a Roman numeral ?)
25) Ni2(Cr2O7)3 ________________________
Naming specific molecules using prefixes
26) NI3 ______________________________
27) SF6_____________________________
28) NO _______________________________
29) SO2 _______________________________
30) OF _______________________________
31) SiO2 ______________________________
26
ANSWERS :
1) calcium oxide (binary, and no Roman numeral is needed,
since calcium has only a single Roman numeral
2) nickel (III) chloride (binary, nickel has at least 2 oxidation
#'s, so you must use a roman numeral. If you assigned
oxidation #'s correctly, you predicted that Ni = +3 state)
3) copper (II) sulfate (ternary : if you assign oxidation #’s
like this:
Cux(SO4)-2
you'll see that X must equal +2. Copper has at
least 2 oxidation #’s so you must use the roman numeral
16)
17)
18)
19)
20)
21)
22)
23)
24)
25)
manganese (II) oxide
iron (II) oxide
lithium carbonate
tin (II) fluoride (stannous fluoride)
mercury (II) chloride
calcium hydroxide
lithium thiosulfate
copper (II) oxide
sodium sulfite
nickel (III) dichromate
4) manganese (VI) oxide (binary : assign the
oxidation numbers as such :
MnXO3-2 Mn must equal +6
26) nitrogen triiodide
27) sulfur hexafluoride
28) nitrogen monoxide
29) sulfur dioxide
30) oxygen monofluoride
5) sodium hypochlorite (this is a ternary inorganic)
(ClO)-1 is called "hypochlorite".
31) silicon dioxide (in this case the metalloid, Si,
acts like a metal….and is the oxidized species)
6) strontium phosphate
7) silver fluoride
8) gold (I) telluride
9) lead (IV) oxide
10) lead (IV) nitrate
11) calcium phosphate
12) titantium (IV) oxide
13) chromium (II) oxide
14) nickel (II) carbonate
15) potassium phosphate
27
MORE: Nomenclature: This worksheet is designed to practice the naming of inorganic compounds. Complete
each and every question by providing the most correct answer. You must have your Periodic Table and your
rules for assigning oxidation states, ready.
Name the following using the most correct names.
4) TiO2 ________________________________
14) CrO ___________________________
5) CuNO3 ______________________________
15) CrO3 __________________________
6) Fe(ClO3)2 ___________________________
16) NiS2O3 ________________________
7) Li3PO4 _____________________________
17) Ni2(Cr2O7)3 ___________________
8) NiC2O4 _____________________________
18) Au(NO3)3 ______________________
9) AgNO3 ______________________________
19) ZnO
10) MnO2 ______________________________
20) HgO __________________________
11) MnO3 ______________________________
21) CaCO3 _________________________
12) Mn2O7 ______________________________
22) Mg(SCN)2 ______________________
__________________________
13) Cr2O3 ______________________________
Answers:
4) Titanium (IV) oxide
Ti+4 O-2 since Ti has at least 3 possible positive oxidation states, you must use a Roman
numeral to identify the exact nature of the metal ion, for the reader. This allows chemists to differentiate between this
oxide of titanium and the other possible formulations of oxides of titanium, TiO [titanium (II) oxide]
and Ti2O3 [titanium (III) oxide]
5) Copper (I) nitrate
Cu+1 (NO3)-1 since copper has at least two possible positive oxidation states, you must use the
Roman numeral, (I), to differentiate CuNO3 from the other nitrate of copper, which is, copper (II) nitrate [(Cu(NO3)2]
28
6) iron (II) chlorate
Fe+2 (ClO3)-1 since Fe has at least 2 possible positive oxidation states, you must use a Roman
numeral . This allows chemists to differentiate between Fe(ClO3)2 and the other chlorate of iron...... Fe(ClO3)3 , in
which the Fe is in a +3 oxidation state.
Li1+ (PO4)3- Lithium has only a single oxidation value ... so no other formulations of
7) lithium phosphate
phosphates of lithium may exist – so you don’t need to use the Roman numeral – in fact, if you put a Roman
numeral into the name the name is considered to be incorrect.
8) nickel (II) oxalate
9) silver nitrate (no Roman numeral is used)
10) manganese (IV) oxide (must have a roman numeral …. Mn has more than one positive oxidation state)
11) manganese (VI) oxide (must have a roman numeral)
12) manganese (VII) oxide (must have a roman numeral)
13) chromium (III) oxide
(must have a roman numeral)
14) chromium (II) oxide
(must have a roman numeral)
15) chromium (VI) oxide
(must have a roman numeral)
16) nickel (II) thiosulfate (must have a roman numeral)
17) nickel (III) dichromate (must have a roman numeral)
18) gold (III) nitrate
(must have a roman numeral)
19) zinc oxide (No roman numeral is to be used ...it’s wrong if you use one!!!! Zn has only 1 oxid. state)
20) mercury (II) oxide (must have a roman numeral)
21) calcium carbonate (no Roman numeral)
22) magnesium thiocyanate (no Roman numeral)
29
NAME __________________________ NOTES: UNIT 7: REDOX: WRITING FORMULAE FOR INORGANICS
I-IV) Oxidation Numbers through Inorganic Nomenclature
V) Given The Name...Write The Empirical Formulae: This is the opposite of nomenclature … in this
section we will learn how to go from a name
to an empirical formula …without the stoichiometry 
A) Review: Empirical Formula: The chemical formula of a substance representing the * simplest
mole (molar) ratio between the elements (that is… the subscript ratio)
Let’s get a few things out in the open...
1) You need to know your polyatomic ions.
2) You need to analyze,.You must actively interpret, consistently
3) Please note that inorganic compounds are often categorized by the anion or more
electronegative element
e.g) oxides = binary ionic or molecular compounds w/ Oxygen
in the -2 oxidation state.
e.g. Fe2O3, ZnO, SO3, Al2O3, UO3, SiO2, NO, N2O3
halides = binary ionic or molecular compounds of Grp 17 elements
in the -1 oxidation state
e.g. CaI2, NaF, AgCl, AlBr3, NH4Cl
phosphates = any compound with the polyatomic ion (PO4)3e.g. Na3PO4, (NH4)3PO4, AlPO4
hydroxides=any inorganic compound w/ the polyatomic ion (OH)1e.g. NaOH, Ca(OH)2, Al(OH)3 ...operationally,
Formula
Name
Formula
Name
H3O1+
hydronium
CrO42-
chromate
Hg2
2+
dimercury I
Cr2O72-
dichromate
NH41+
ammonium
MnO41-
permanganate
C2H3O21-
acetate
NO21-
nitrite
cyanide
NO31-
nitrate
CO32-
carbonate
O22-
peroxide
HCO31-
hydrogen
carbonate
OH1-
hydroxide
(bicarbonate)
C2O42-
oxalate
PO43-
phosphate
ClO1-
hypochlorite
SCN1-
thiocyanate
ClO2
1-
chlorite
SO32-
sulfite
ClO31-
chlorate
SO42-
sulfate
ClO41-
perchlorate
HSO41-
hydrogen
sulfate
S2O32-
thiosulfate
CN
1-
metal hydroxides are classified as.
Arrhenius bases, when dissolved in water.
30
A) Here’s the challenge:
How can you get the correct subscripts for a compound’s formula? ...
1) For Molecules: * Turn the prefixes of the name into the subscripts.
prefix
value
mono
1
di
2
tri
3
tetra
4
penta
5
hexa
6
hepta
7
octa
8
Follow the prefixes as your guide ... there’s no need to play around with the oxidation
states
Note that when naming binary molecular compounds, prefixes are used rather exclusively.
However, just to confuse issues, some texts will use prefixes also with ionic compounds. According to IUPAC, is is
acceptable to use prefixes (almost any time …we won’t, because NYS wants you to limit their use to molecules.)
Practice: When no prefix is used, you may assume a subscript of “1”.
carbon dioxide * CO2
dinitrogen pentoxide * N2O5
sulfur trioxide * SO3
nitrogen trichloride * NCl3
carbon monoxide * CO
dinitrogen monoxide * N2O
dichlorine heptoxide * Cl2O7
carbon tetrachloride * CCl4
a) N.B. When asked to write the formula of any of the diatomic elements,
always record the symbol with a subscript of 2.
H
17
N
7
O
F
Cl
e.g) hydrogen gas or dihydrogen *H2
oxygen gas or dioxygen *O2
Br
I
i) In some texts you might actually read the names of the diatomic elements
such as pure N2(g) as dinitrogen, or pure O2(g) as dioxygen. You’ll note
that I have called diatomics like H2(g) , hydrogen gas and dihydrogen, just
to get you use to this… It is a pretty new trend … but you’re the next
generation of scientists, so I want you exposed to it….
31
2) For Ionic Compounds:
a) Review: With few exceptions, for most inorganic compounds, the symbol of the
lower electronegative element is written * first in the formula
it’s the one with the * positive
(so
oxidation number. The second element is,
the element with the greater electronegativity value or the species with the
* negative
oxidation number.
3) Process: Use oxidation states and subscripts to write a formula in which all state add to 0
4) Process: The Criss-Cross Method.
Concept: The absolute values of the oxidation numbers become the subscripts for
the opposite species. These are then reduced to the simplest whole-number
ratio (the empirical formula)
a) Assign the oxidation state associated with each cation & anion species
b) Make that oxidation state of the first species into the subscript of the second species
etc. Reduce the subscripts to the simplest whole number ratio.
Note1 Instead of criss-crossing, you may arithmetically determine the relationship.
Note2 The sum of the positive charge and the sum of the negative charges MUST add up to zero.
Note3 You MAY NOT adjust the charges of the cations or anions to get a total charge of zero.
Note4 You MAY adjust the subscripts to get a total charge of zero.
3) e.g.) sodium sulfide
Na+1 S-2
calcium nitrate
Ca+2 (NO3)-1 = Ca(NO3)2
32
potassium chloride
calcium sulfate
aluminum oxide
aluminum phosphate
aluminum chloride
zinc carbonate
magnesium phosphate
calcium fluoride
4) Writing the formula by interpreting Stock’s system
this roman numeral tells you the  oxidation # to use
titanium (IV) oxide
this roman numeral identifies
the positive oxid. # to use
chromium (III) oxide
Ti+4 O-2
manganese (VI) sulfide
iron (III) oxide
Mn+6 S-2
Be sure to use iron
in the +2 state
iron (II) oxide
nickel (III) carbonate
copper (II) sulfate
gold (III) thiosulfate
33
Given a written name and asked for the chemical formula of a compound...
Confirm it is a molecule, made
from nonmetal atoms
Confirm the first element name
represents a metal, telling you that the
compound is an ionic compound
 correctly interpret the elements’
names and write down the correct
symbols
 correctly interpret each (or any)
prefix and convert it into a
subscript for the nonmetal species
to which the prefix applies
 correctly interpret the elements’
names and write down the correct
symbols
or
When the name has
a roman numeral be
sure to...
 correctly interpret the
roman numeral & assign it
to the METAL species, as
an oxidation state (it is not a
When there is no
roman numeral be
sure to...
 assign the correct positive
oxidation state to the metal
species, using the periodic
table
subscript)
 correctly assign the negative
oxidation state to the other
species, using the periodic
table or knowledge of a PAI
Summary:
compounds
ionic
molecular
X+# Y-#
use the
prefixes
 determine the relationship
between the oxidation states,
either arithmetically or by
the criss-cross method, so
that the total sum of the
oxidations states adds up to
be zero
 write the final formula as an
empirical formula
 draw parentheses around any
polyatomic ion, which has an
assigned subscript of 2 or
greater (be especially careful
when dealing with (OH)1and (CN)1-
 erase all oxidation states
from your final answer.
There should be no charge
 correctly assign the negative
oxidation state to the other
species, using the periodic
table or Table E
 determine the relationship
between the oxidation states,
either arithmetically or by
the criss-cross method, so
that the total sum of the
oxidations states adds up to
be zero
 write the final formula as an
empirical formula
 draw parentheses around any
polyatomic ion, which has an
assigned subscript of 2 or
greater (be especially careful
when dealing with (OH)1and (CN)1-
 erase all oxidation states
from your final answer.
There should be no charge
34
DIRECTIONS: Write the empirical formula for each compound, using either method from the previous pages.
Remember for the arithmetic method:


Be sure you have written the appropriate symbol for each species
Assign the oxidation numbers and multiply them by whole-number subscripts that have their total add u to 0.
Remember for the criss/cross method:
•
Read the name of each given compound and re-write the compound using the elements’ symbols.
•
Assign oxidation #’s to each cation and each anion (you can assign any PAI its overall charge)
•
“Criss Cross” the oxidation numbers, ignoring the charges, thus making the ox. numbers into subsripts
•
Reduce to the simplest whole-number ratio (the empirical formula)
•
N.B. DO NOT carry over the + and - signs when making subscripts
When the subscript is 1 it is not written (CaCl2 NOT Ca1Cl2)
When one of the ions is a PAI and its subscript is 2 or more, enclose it with ( )
Compound Name
1.
2.
titanium (II) oxide
(binary or ternary ?)
vanadium (IV) oxide
Symbols and Ox. #’s
Ti2+
O2-
+4
-2
V
O
(binary or ternary ?)
3.
copper (II) nitrate
Cu+2
(NO3)-1
Criss Cross For
Subscripts
Ti2O2
simplify to empirical formula
Empirical Formula
TiO
Make those oxidation states into
subscripts, using absolute values!!
V2O4
Cu1(NO3)2
Cu(NO3)2
(binary or ternary ?)
4.
cadmium oxide
(binary or ternary ?)
calcium hydroxide
5.
6.
(binary or ternary ?)
nickel (III) phosphate
(binary or ternary ?)
barium phosphate
7.
(binary or ternary ?)
gold (I) thiosulfate
8.
(binary or ternary ?)
chromium (II) sulfate
9.
(binary or ternary ?)
manganese (VI) oxide
10
(binary or ternary ?)
manganese (IV) oxide
11
(binary or ternary ?)
12 mercury (I) chloride
13 nickle (III) oxide
35
Symbols and Ox. #’s
Compound Name
Criss Cross For
Subscripts
Empirical Formula
14 potassium dichromate
15 silver nitrate
16 vanadium (V) oxide
17 lithium hydroxide
18 aluminum phosphate
19 tin (II) fluoride
20 chromium (III) sulfate
21 zinc chloride
22 sodium hypochlorite
Selected Answers:
2) VO2
3) Cu(NO3)2
10) MnO3
11) MnO2
18) AlPO4
19) SnF2
4) CdO
12) HgCl
5) Ca(OH)2
13) Ni2O3
20) Cr2(SO4)3
6) NiPO4
14) K2Cr2O7
21) ZnCl2
7) Ba3(PO4)2
15) AgNO3
8) Au2S2O3
16) V2O5
9) CrSO4
17) LiOH
22) NaClO
36
NAME _____________________________________ NOTES: UNIT 7: REDOX: HALF-REACTIONS
VI) Species of a redox reaction (REVIEW)
oxidized species
reduced species
loses electrons
its oxidation # becomes
more positive, due to the loss
of negative charge
gains electrons
so its oxidation # becomes
more negative, due to the
gain of negative charge
aka: reducing agent
aka: oxidizing agent
spectator ion
no change in oxidation number
A) When a redox reaction is investigated, it can be represented by the separate reduction reaction and
by the separate oxidation reaction. These “portions” are called half-reactions.
1) How to write a reduction half-reaction: * species + e-  new & more negative species
e.g) The reduction of phosphorus to P3-
The reduction of Cl7+ to Cl5+
2) How to write an oxidation half-reaction: * species  new (more positive) species + ee.g) The oxidation of copper to Cu2+
The oxidation of Fe2+ to Fe3+
3) Every oxidation half-reaction must be accompanied by a reduction half-reaction.
a) When the half reactions are balanced separately so that the laws of the conservation
of matter and the conservation of charge are obeyed, the * number of electrons gained
by the reduced species will be equal to the number of electrons lost by the oxidized
species.
37
4) Complete each reaction by providing the correct number of lost/gained electrons or species
and label each as an oxidation ½ reaction or as a reduction ½ reaction.
Half-Reaction
Type of Half-Reaction
a. Fe0 
Fe+3
b. Cs0 
*Cs+1 + 1e-
c.
+ Se0 
* 2e-
d. Pb0
+
 4e-
e. *3e- + N0 
* 3e-

___________________
Se-2
___________________
+ * Pb+4
___________________
N-3
___________________
f. Mn+6 + * 3e-  Mn+3
g. Cl+5
___________________
2e- + * Cl+7
___________________
___________________
Notice, that the blanks are no longer provided. You must now figure out, by analyzing the charges of the
reactants and the products as to what must be "filled in" to complete the half-reaction.
Half-Reaction
Type of Half-Reaction
h.
Si0
 Si-4
__________________
i.
Mn0
 4e-
__________________
j
k
O0
Answers : a = oxidation, 3ee = reduction, 3e- f=reduction, 3ei= oxid., Mn+4 right side of arrow
k= reduction, 2e- left side of arrow
 Ca+2 + 2e-
__________________
 O-2
__________________
b = oxidation, Cs+1
c = reduction, 2ed = oxidation, Pb+4
+7
g= oxidation, Cl
h = reduction, 4e , left side of arrow
0
j= oxidation, Ca left side of arrow
38
5) The special difficulties of diatomic elements and writing half-reactions:

a) The reduction of F2 to F1-__________________________________________
stop
Before you go on,
analyze the species ...
What issue do you
notice with respect
to the law of the
conservation of
matter & writing
half-reactions for
the diatomics?
b) The reduction of Cl2 to Cl1-_________________________________________
c) The reduction of Br2 to Br1-____________________________________________
d) The reduction of I2 to I1-____________________________________________
e) The reduction of O2 to O2- * O2 + 4e- 
2 O-2
f) The reduction of N2 to N3-_____________________________________________
g) The oxidation of H2 to H1+_____________________________________________
h) The oxidation of O2- to O2___________________________________________
i) The oxidation of Cl1- to Cl2____________________________________________
j) The reduction of H1+ to H2 ____________________________________________
VII) Balancing Simple Redox Reactions via Half-reactions
 Determine which species were reduced and oxidized (always on the reactant side)
 Write each half reaction
 Balance by mass (balance according to the number of ALL particles: really used for diatomics)
 Balance by charge (obey Law the Conservation of Charge: multiply to get = # electrons)
 Rewrite the two half-reactions into 1 reaction, now using coefficients
A) Balance the given reaction :
___Cu2+ + ___Al0  ___Cu0
+ ___ Al3+
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED: __________________________________________________________________
The 2 half reactions are recombined, with the correct coefficients BUT excluding the electrons (for they have been "cancelled out") To recombine, you could just put
the coefficients on the spaces of the original equation.
39
B) Balance the given reaction :
___H1+
+

___Fe0
___Fe2+
+ ___H20
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED : ___________________________________________________________________
C) Balance the given reaction :
___Pb2+ + ___Fe0

___ Pb0
+ ___Fe3+
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED : ___________________________________________________________________
D) Balance the given reaction :
___Al0

+ ___Cr+3
___Cr0
+ ___Al3+
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED : ___________________________________________________________________
E) Balance the given reaction: ___Au+3 + ___ H20

___ H+1
+ ___ Au0
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION: ________________________________________________________
RECOMBINED: ___________________________________________________________________
40
Au+3
F) Balance the given reaction:
+ K0
* Au+3 + 3 e- 
Au0
3(K0  K+1 + 1e-)
1Au+3 + 3K0  3K+1
* 3(Cu+ + 1e-  Cu0 )
Cr0  Cr3+ + 3eCr0 + 3Cu+  Cu0
Al+3
H) Balance the given reaction:
I) Balance:
*
Ag+
+
+ Cr3+
+
Ba0

Al0
+ Ba+2
multiplying to get 6 e-
2Al0
+ 3Ba+2
Cu0 
Cu+2
+ Ag0
2(Ag+ + 1 e-  Ag0)
Cu0  Cu+2 + 2e2Ag+
J) Balance:
+ 1Au0

3Ba0
+ Au0
+ 3Cr3+
* 2(Al+3 + 3e- 
Al0 )
3(Ba0  2e- + Ba+2 )
2 Al+3 +
3K+1
+ Cu+  Cu0
Cr0
G) Balance the given reaction:

Al0
+
+
H1+
Cu0 

Al3+
Cu+2 + 2 Ag0
+
H20
* 3 (2 H1+ 2 e-  H20 )
2 (Al0

Al3+ + 3 e- )
2 Al0
+ 6 H1+  2 Al3+ + 3 H20
41
K) When correctly balanced using the smallest whole-number ratios, what is the coefficient of H1+?
Sn4+
a) 1
+
H20

b) 2
Sn0 + H1+
c) 3
d) 4
L) When correctly balanced using the smallest whole-number ratios, what is the coefficient of Hg0?
Br20
a) 1
+
b) 2
Hg0

Br1- + Hg2+
c) 3
d) 4
M) Determining the moles of electrons in a redox reaction.
+3 -2
1) In the reaction: 4 Al + 3 O2  2 Al2O3
how many moles of electrons are lost/gained?
Just multiply the moles a product ion by the oxidation number of the ion : (4)(3+) = 12
* Think about how you were taught to write a half-reaction…
* 6 mol
2) In the reaction: 2 Al + 3 I2  2 AlI3(s) how many moles of electrons are lost/gained?
*16 mol
3) Assuming the reaction goes to completion: 8 Cu + S8  8 CuS how many moles of
electrons are gained by the sulfur? (which of course, must be the same number of mols of e- lost by the Cu ...due
to the Law of the Conservation of Charge)
* 4 mol
*2 mol
4) Assuming the reaction goes to completion: 2 Mg + O2  2 MgO
electrons are lost by magnesium?
how many moles of
5) Assuming the reaction goes to completion: Fe + CuSO4  Cu + FeSO4 how many moles of
electrons are exchanged between the reactants?
42
VIII) Types of Inorganic Reaction:
A) Single Replacement (all are redox)
1)
element + compound → new element + new compound
F2(g) + 2KCl(aq) → 2KF(aq) + Cl2(g)
B) Synthesis
2) smaller compound or element + smaller compound or element → single larger product
2 NH3(g)
+ H2SO4(aq)  (NH4)2SO4 (aq)
C) Decomposition
1) single larger reactant → smaller products
2 NaNO3(s)  2 NaNO2(s) + O2(g)
D) Double Replacement (none are redox)
1) compound + compound → new compound + new compound
Pb(NO3)2(aq) + 2 KI(aq) 
PbI2(s) + 2 KNO3(aq)
Practice:
1) Ca(s) + 2HCl(aq)  CaCl2(aq) + H2(g)
2) 2 Al(OH)3(aq) + 3 Sr(s)  3 Sr(OH)2(aq) + 2 Al(s)
3) Ni(NO3)2(aq) + Na2CO3(aq)  NiCO3(s) + 2 NaNO3(aq)
4) 8 Cu(s) + S8(s) 
8CuS(s)
5) 2 NO + O2  2 NO2
6) 3Na2Cr2O7(aq)
+ 2 AlCl3(aq) 
6NaCl(aq) + Al2(Cr2O7)3(aq)
7) 2 Hg(NO3)2(s)  2 HgO(s) + 4 NO2(g) + O2(g)
43
8) U(s) + 3 F2(g)  UF6(s)
9) RhO3(s)  RhO(s) + O2(g)
10) 2 Al(s) + 3 Cu(NO3)2(aq)  2 Al(NO3)3(aq) + 3 Cu(s)
11) Mg(s) + 2 HF(aq) 
MgF2(aq) + H2(g)
12) 2 NaNO3(s)  2 NaNO2(s) + O2(g)
13) Pb(NO3)2(aq) + 2 KI(aq) 
PbI2(s) + 2 KNO3(aq)
14) SO3(g) + H2O(l)  H2SO4(aq)
15) Pb(NO3)2(aq) + H3AsO4(aq)  PbHAsO4(s) + 2 HNO3(aq)
16) 2 Al(s) + 3 Fe(NO3)2(aq)  2 Al(NO3)3(aq) + 3 Fe(s)
Answers;
1) Single Replacement
2) Single Replacement
3) Double Replacement
4) Synthesis
5) Synthesis
6) Double Replacement
7) Decomposition
8) Synthesis
9) Decomposition
10) Single Replacement
11) Single Replacement
12) Decomposition
13) Double Replacement
14) Synthesis
15) Double Replacement
16) Single Replacement
44
PRACTICE:
1)
7)
2)
8)
3)
4)
9)
5)
6)
Answers:
1) 4
2) 1
3) 1
4) 4
5) 1
6) 2
7) 4
8) 1
9) 2
45