Additional notes to the packet that were given on Friday 5/9/2014
or Monday 5/12/14:
We are learning RedOx to:
- practice balancing equations
-review basic terminology: ion, cation, anion, reactant,
product, transition metal, alkali metal, alkali earth metals,
halogens, noble (inert) gas
- understand batteries better
-review activity series of metals ( and explore reactivity of
other compounds)
- etc. etc. etc
Thermite reaction (notes at end of this file) and magnesium
HCl battery will be done as demos
Balancing a chemical equation gives you lots of information. Some equations are more difficult to balance. The technique of
half reactions will allow you to balance some equations that you would find difficult otherwise.
Oxidation numbers help us keep track of the movement of electrons (what chemistry is really all about)
We need to learn how to assign oxidation numbers
RedOx notes:
Remember electrons are Negative (losing electron makes you
positive)
You MUST remember the basic definitions, which are often summarized as
Oil Rig
Oxidation is Loss of electrons; Reduction is Gain (of electrons)
Or
LEO GR
Loss of Electrons is Oxidation, Gain is Reduction
 In Oxidation half reactions, the oxidation number (value) of the
element increases (like a big Ox)
Na  Na+ + 1eIn Reduction half reactions, the oxidation number (value) gets smaller
(this is what reduction means)
2 e- + Cu+2  Cu
Oxidizing Agents allow oxidation to proceed (by being reduced)
Reducing agents allow reduction to proceed (by being oxidized)
Steps to take in solving the Redox problems
(how partial credit will be determined on the test)
1)
Assign oxidation numbers (elements have just as many
protons as electrons, Ions have a net charge).
2)
Identify the two half reactions: what is being oxidized and
what is being reduced.
3)
Balance each half reaction (keep the subscripts in mind now
so you don’t get confused later).
4)
Balance the two half reactions so that just as many electrons
are produced as are needed.
5)
Add the two half reactions and transfer these values to the
overall reaction.
6)
Take into account the spectators (adjust the numbers in step 5
as necessary).
OXIDATION NUMBERS; for every rule there is an exception; you must get the feel of
assigning oxidation numbers Please look at several sources (the example on the desk; this one;
the one in the book, others on-line, etcetera, etcetera, etcetera)
These are pretty good rules for assigning oxidation numbers
1. The oxidation number for an atom in its elemental form is always zero.
o A substance is elemental if both of the following are true:
 only one kind of atom is present
 charge = 0
o Examples:
 S8: The oxidation number of S = 0
 Fe: The oxidation number of Fe = 0
2. The oxidation number of a monoatomic ion = charge of the monatomic ion.
o Examples:
 Oxidation number of S2- is -2.
 Oxidation number of Al3+ is +3.
3. The oxidation number of all Group 1A metals = +1 (unless elemental).
4. The oxidation number of all Group 2A metals = +2 (unless elemental).
5. Hydrogen (H) has two possible oxidation numbers:
o +1 when bonded to a nonmetal
o -1 when bonded to a metal
6. Oxygen (O) has two possible oxidation numbers:
o -1 in peroxides (O22-)....pretty uncommon
o -2 in all other compounds...most common
7. The oxidation number of fluorine (F) is always -1.
8. The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.
9. The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the
polyatomic ion.
When assigning oxidation numbers to the elements in a substance, take a systematic approach.
Ask yourself the following questions:
1.
2.
3.
4.
5.
Is the substance elemental?
Is the substance ionic?
If the substance is ionic, are there any monoatomic ions present?
Which elements have specific rules?
Which element(s) do(es) not have rules?
Use rule 8 or 9 from above to calculate these.
Not great rules but get you the feel of assigning oxidation numbers:
0)
1)
elements have zero charge (or they would be ions); just like compounds
Poly atomics choose first (Know your poly-atomics if more than 2 elements)
(“The needs of the many outweigh the needs of the few”)
2)
Remember they are called oxidation numbers (oxygen chooses -2, usually (can be -1 in
peroxides); Fluorine is -1)
3)
Give a little, alkali metals choose +1 (by giving up one electron). Hydrogen is above it all
and is special (usually +1, but hydrogen can be -1 when in a metal hydride)
4)
Alkali earth metals choose +2
5)
With all this giving up of electrons there must be some taking, Halogens take one
electron to become -1
6)
Continue with elements picking their preferred charges (work from outside columns to
the inner “valley of confusion”) until there is only one left; if the element is last to choose it
must have the charge that makes everything else sum to zero.
Mr P’s summary:
- Know the preferred charges (+1, +2, skip a few, +3,+/-4, -3, -2, -1, 0)
- Neutral atoms/compounds are neutral (zero)
-Know the polyatomics (“the needs of the many outweigh the needs of the few”)
- give a little (alkali metals and alkali earth metals tend to form cations (by giving up
electrons)
- take a little (halogens take electrons to become anions)
- they are called Oxidation numbers so oxygen is almost always the -2 anion; (except in
peroxides, H2O2)
- hydrogen can be both a cation or an anion (hence metal hydride batteries) SURPRISE
- the elements in the middle of the periodic table choose last so they have to take on a
charge to balance the rest
(you can’t always get what you wanted, but if you try real hard you (might just get) what
you(“We, the compound”) need)
Practice; YES I know some don’t make sense I put them in to see if you
are following the rules
Formula
Known oxidation
number
Inferred oxidation
number
ZnO
O = -2
Zn = +2
NaCl2
Na= +1
Cl =
MnCl2
Mn =
MnO2
Mn =
NiCl
Ni =
Sb2O3
Sb2 O5
MgO
Mg2N3
CaCl2
NH3
N2O5
SnO
BiCl3
CuCl
CCl4
CrCl3
P2O5
Compound
polyatomic ION
KNO3
NO3 =
K=
N=
O=
K2SO4
SO4 =
K=
S=
O=
Ca3(PO4)2
PO4 =
Ca =
P=
O=
NaClO3
ClO3 =
Na =
Cl =
O=
Na2B4O7
B4O7 =
Na =
B=
O=
L;
Practice; YES I know some don’t make sense I put them in to see if you
are following the rules ANSWERS
Compound
Formula
Known oxidation
number
Infered oxidation
number
ZnO
O = -2
Zn = +2
NaCl2
Na= +1
Cl = -1/2 Crazy
MnCl2
Cl =-1
Mn = +2
MnO2
O= -2
Mn = +4
NiCl
Cl= -1
Ni = +1
Sb2O3
O= -2
Sb = +3
Sb2 O5
O= -2
Sb= +5
MgO
O= -2
Mg= +2
Mg2N3
Mg= +2
N= -4/3 Crazy
CaCl2
Ca= +2
Cl= -1
NH3
H= +1
N= -3
N2O5
O= -2
N= +5
SnO
O= -2
Sn= +2
BiCl3
Cl= -1
Bi= +3
CuCl
Cl= -1
Cu= +1
CCl4
Cl= -1
C= +4
CrCl3
Cl= -1
Cr= +3
P2O5
O= -2
P= +5
polyatomic ION
KNO3
NO3 = -1
K = +1
N = +5
O = -2
K2SO4
SO4 = -2
K = +1
S = +6
O = -2
Ca3(PO4)2
PO4 = -3
Ca = +2
P = +5
O = -2
NaClO3
ClO3 = -1
Na = +1
Cl = +7 +5
O = -2
Na2B4O7
B4O7 = -2
Na = +1
B = +3
O = -2
Oxidation –Reduction Reactions Basic Skills
Oxidation Numbers
Using the guidelines from the Nomenclature packet, those described in the book, on-line or
elsewhere.
Identify the oxidation number for each of the elements in the following compounds (remember
the oxidation number is for the element NOT the gross charge on the group)
NaCl
H2O
CO2
FeBr2
NH4F
PbBr2
KNO3
CuS
CaCl2
H2SO4
AlF3
SbCl3
Now for some that may be trickier, remember that the oxidation state of the last remaining element is
determined by the electro negativity of the elements it combines with.
K3PO4
K2CO3
CaCO3
CO2
CO
K2CrO7
CaSO4
K2CrO4
Na2MoO4
FeO
Fe2O3
Na2B4O7
Oxidation –Reduction Reactions Basic Skills
Balancing equations
Identify as oxidation or reduction and fill in any missing numbers:
1)
2)
3)
4)
5)
Mg  Mg+2 + 2eFe  Fe+3 + __ e1 e- + Fe+3  Fe+2
Mg  Mg+2 + 2 e__ e- + Mn+5  Mn+2
Oxidation or Reduction
Oxidation or Reduction
Oxidation or Reduction
Oxidation or Reduction
Oxidation or Reduction
In the following examples identify the oxidation half-reaction and the reduction half
reaction.
6)
2 Mg + O2  2 MgO
Oxidation:
Reduction:
2 Fe + O2  2 FeO
7)
Oxidation:
Reduction:
Mg + S  MgS
8)
Oxidation:
Reduction:
__ H2 + __ O2  __ H2O
9)
Oxidation:
Reduction:
20 tough (and not so tough) RedOx practice problems
(Select 10 from this sheet of 20 practice problems, show all work as described in class
1.)
__Cu + __HNO3  Cu(NO3)2 + __ NO2 + __ H2O
2.)
__ H2S + __HNO3  __ NO2 + __S + __ H2O
3.)
__ HCl + __MnO2  __ MnCl2
4.)
__Cu + __HNO3  __Cu(NO3)2 + __ NO + __ H2O
5.)
__HNO3 + __ H2SO4 + __FeSO4  __Fe2(SO4)3 + __ H2O + __ NO
6.)
__Br2 + __SO2 + __H2O  __ H2SO4 + __HBr
7.)
__Na2CrO4 + __ SnCl2 + __HCl  __ NaCl + __ SnCl4 + __ H2O + __ CrCl3
8.)
__K2Cr2O7 + __ H2SO4 + __FeSO4  __K2SO4 + __ Cr2(SO4)3 +__ Fe2(SO4)3 + __ H2O
9.)
__KClO3 + __ HCl  __KCl + __ H2O + __ Cl2
10.)
__S + __ HNO3  __H2SO4 __ NO
11.)
__Bi2S3+ __HNO3  __ Bi(NO3)3 + __S + __NO + __ H2O
12.)
__FeCl3 + __ H2SO3 + __ H2O  __FeCl2 + __ H2SO4 + __ HCl
13.)
__K2Cr2O7 + __HCl  KCl + __ CrCl3 + __ Cl2 + __ H2O
14.)
__Cu + __H2SO4  CuSO4 + __ SO2 + __ H2O
15.)
__H2SO4 + __HI  __ H2S + __ I2 + __ H2O
16.)
__HNO2 + __ H2SO4 + __KMnO4  __ K2SO4 + __ MnSO4 + __ HNO3 + __ H2O
17.)
__PH3 + __ KMnO4 + __H2SO4  __ K2SO4 + __ MnSO4 + __ H3PO4 + __ H2O
18.)
__Zn + __HNO3  Zn(NO3)2 + __ NH4NO3 + __ H2O
19.)
__K2Cr2O7 + __ NaNO2 + __H2SO4  __K2SO4 + __ Cr2(SO4)3 +__ NaNO3 + __ H2O
20.)
__Sb2S3 + __ HNO3 + __H2O  __H3SbO4 + __ H2SO4 +__ NO
(Can do all 20 for extra credit)
+ __ H2O + __Cl2
RedOx practice sheet for the test on: do at least 10 problems use your own notebook to practice
(select from the sheet of 20 practice problems)
In the following problems identify:
a) The oxidation numbers of all reactants and products.
b) The oxidation half-reaction.
c) The reduction half-reaction
d) Balance the half-reactions.
e) Balance the complete reaction (by transferring the coefficients and balancing the spectator ions).
__)
Oxidation:
Reduction:
Substance being oxidized _______
Substance being reduced ________
(if it’s an ion make sure you include the charge) Reducing
agent _______
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Reducing agent _______
Substance being reduced ________
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Reducing agent _______
Substance being reduced ________
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Substance being reduced ________
(if it’s an ion make sure you include the charge) Reducing
agent _______
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Reducing agent _______
Substance being reduced ________
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Reducing agent _______
Substance being reduced ________
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Substance being reduced ________
Reducing agent _______
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Substance being reduced ________
(if it’s an ion make sure you include the charge) Reducing
agent _______
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Reducing agent _______
Substance being reduced ________
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Reducing agent _______
Substance being reduced ________
oxidizing agent _______
__)
Oxidation:
Reduction:
Substance being oxidized _______
Substance being reduced ________
Reducing agent _______
Thermite is a memorable demonstration of an exothermic reaction
It can be described by the following reaction:
Fe2O3 + 2Al → 2Fe + Al2O3 + heat (lots of it)
Can you use your knowledge of stoichiometery to tell me
exactly how many grams of aluminum to mix with 20 grams
of rust (Fe2O3)
Do NOT try this at home; you do not understand all the
precautions that I am taking.
You can watch a video of some engligh guys melting a car at
http://www.youtube.com/watch?v=rdCsbZf1_Ng

The chemical reaction that is occurring in this demonstration is as follows:
Fe2O 3 (s) + 2 Al (s)  Al2O 3 (s) + 2 Fe (s)
The enthalpy change for this reaction is -849 kJ/mol of iron(III) oxide. To give some idea of what this
released heat is doing, keep in mind that iron melts at 1,530 oC. The amount of themite used in this
reaction is suitable for a large lecture hall. The amount may be scaled down for smaller rooms while still
providing a spectacular demonstration. There are numerous methods listed in the literature for setting off
this reaction, magnesium ribbon, sulfuric acid and potassium chlorate and sugar, potassium
permanganate and glycerin.
In this reaction the iron(III) oxide (rust) is reduced to pure iron
and the pure aluminum is oxidized into aluminum oxide (this is
the new stuff we are learning; who gets the electrons and who gives
them up)
Fe2O3 + 2Al → 2Fe + Al2O3 + heat (lots of it)
Review you stoichiometery: (convert to moles, compare moles, and convert to desired quantity)
20 g Fe2O3 (____________) (____________) (__________) =
(
) (
) (
)
Why don’t you want to mix in extra aluminum?
Would it work the same is you used aluminumoxide?
g Al
Good general guidelines for assigning oxidation numbers (ON)
1. The Elements: The oxidation number of a pure element is zero
e.g. Fluorine in F2 has oxidation number of 0
2. Simple Ion Charge = Oxidation Number
e.g. Na1+ has oxidation number of +1, Mg2+ has oxidation number of +2
3. Hydrogen: Hydrogen has oxidation number of +1 in all compounds except in metal hydrides where it is
-1.
e.g. Water: H is bonded to O by a polar covalent bond. The e- pair
is closer
+
to O -> H is partially positive (δ ) and O is partially negative (δ ). H
has
oxidation number of +1.
4. Oxygen
O usually has oxidation number of -2, except in peroxides where it
assigned -1, and in OF2 where it is assigned a +2 due to the higher
negativity of F.
-In calcium oxide, CaO, oxide ion has a 2- charge. Its oxidation number is -2
-In H2O2, Each O is assigned the oxidation number of -1
is
electro
5. Covalent Compounds: In compounds that do not contain hydrogen or oxygen, the more electronegative
element is assigned the oxidation number it would have in an ionic compound. (Provide electro
negativity table)
e.g. In SCl2, chlorine has a higher electro negativity so it is assigned -1 first and sulfur is assigned +2.
6. Compounds: The sum of the oxidation numbers in a compound is zero.
e.g. The sum of oxidation numbers in NaCl is (+1) + (-1) = 0
7. Polyatomic Ions
The sum of the oxidation numbers of the elements in a polyatomic ion must equal the ion charge. Consider
these examples. If there are two poly atomic ions in a compound deal with them first.
e.g. Carbonate ion,
. O has oxidation number of -2. There are 3O’s in the formula so the total
negative charge is 6-. Since the carbonate ion has a charge of 2-, the oxidation number of C must be
+4.
Or to say it another way
(have your periodic table out and know the families, alkali, alkaline earth, halogens and noble gases)
Not great rules but get you the feel of assigning oxidation numbers:
0)
elements have zero charge (or they would be ions); just like compounds
1)
Poly atomics choose first (Know your poly-atomics if more than 2 elements)
(“The needs of the many outweigh the needs of the few”)
2)
Remember they are called oxidation numbers (oxygen chooses -2, usually (can be -1 in
peroxides); Fluorine is -1)
3)
Give a little, alkali metals choose +1 (by giving up one electron). Hydrogen is above it all
and is special (usually +1, but hydrogen can be -1 when in a metal hydride)
4)
Alkali earth metals choose +2
5)
With all this giving up of electrons there must be some taking, Halogens take one
electron to become -1
6)
Continue with elements picking their preferred charges (work from outside columns to
the inner “valley of confusion”) until there is only one left; if the element is last to choose it
must have the charge that makes everything else sum to zero.
If you don’t choose first you might not get your first choice
(you can’t always get what you wanted, but if you try real hard you (might just get) what
you (“We, the compound”) need)