Redox problems
If the original problem has + or – written in the problem skip
this step.
MnO4- + H+ + Br-  Mn+2 + Br20 + H2O
Mn+7O4-2 + H+ + Br-  Mn+2 + Br20 + H2+O-2
Mn+7O4-2 + H+ + Br-  Mn+2 + Br20 + H2+O-2
First half reaction:
a.) Mn+7O4-2  Mn+2
b.) 5e- + Mn+7O4-2  Mn+2
c.) 5e- + Mn+7O4-2  Mn+2 + 4 H2O
d.) 8 H+ + 5e- + Mn+7O4-2  Mn+2 + 4 H2O
a.) 2 Br-  Br20
b.) 2 Br-  Br20 + 2 e2 (8 H+ + 5 e- + Mn+7O4-2  Mn+2 + 4 H2O)
5 (2 Br-  Br2 + 2 e-)
+
16 H + 10 e- + 2Mn+7O4-2  2Mn+2 + 8 H2O
10 Br-  5 Br20 + 10 e16 H+ + 10 e- + 2Mn+7O4-2  2Mn+2 + 8 H2O
10 Br-  5 Br20 + 10 e16 H+ + + 2Mn+7O4-2 + 10 Br-  2Mn+2 + 8 H2O + 5 Br2
Acidic redox (All are Acidic unless
otherwise noted)
1. Write a net ionic equation using your
solubility rules. This step is not done until
you get to the last problems of this
assignment.
2. Assign oxidation numbers to all elements.
Use your handout page 2 to refresh your
memory on how to get the charge of Mn
in the MnO4- particle.
3. Underline those things that change their
charge from one side to the other of the
equation.
4. Half reactions:
a. Balance all elements except hydrogen
and oxygen.
b.Balance the charges by adding electrons
to the side with the largest oxidation
number.
c. Balance oxygen by adding water.
d.Balance the hydrogen by adding H+.
Note: YOU will do all of this in one
equation. Don’t write it separate like I did.
Repeat for 2nd ½ reaction.
a. Balance all atoms except oxygen and
hydrogen.
b. Note on this step you have to add 2
electrons because you have two atoms
of Br
c. no oxygen or hydrogen to balance
5. Balance the electrons on each side of the
equations. Find the common denominator
and multiply each equation. Cross out the
equations after you finish the
multiplication.
6. Simplify and add the two equations. Mark
out the electrons. You will sometimes
mark out some water or H+ on each side.
7. Write what is left.
Redox problems
Basic
If the original problem has + or – written in the problem skip
this step. Use page 4 to determine if the compound ionizes
or should be written together.
NO3- + NH3  NO2
N+5O3-2 + N-3H3+1  N+4O2-2
N+5O3-2 + N-3H3+1  N+4O2-2
Note: in this case nitrogen is in both half reactions. This
happens occasionally.
N+5O3-2
 N+4O2-2
Done.
e- + N+5O3-2  N+4O2-2
e- + N+5O3-2  N+4O2-2 + 2 OHH2O + e- + N+5O3-2  N+4O2-2 + 2 OHNow the Oxygen and Hydrogen is balanced.
1. Write a net ionic equation using your
solubility rules. This step is not done until
you get to the last problems of this
assignment.
2. Assign oxidation numbers to all elements.
Use your handout page 3 to refresh your
memory on how to get the charge of Mn
in the MnO4- particle.
3. Underline those things that change their
charge from one side to the other of the
equation.
4. Half reactions:
a. Balance all elements except hydrogen
and oxygen.
b.Balance the charges by adding electrons
to the side with the largest oxidation
number.
c. Balance oxygen by adding twice as
many OH- as you think you need.
Unless you are balancing OH-‘s
d.Balance the hydrogen by adding H2O.
Note: YOU will do all of this in one
equation. Don’t write it separate like I did.
Repeat for 2nd ½ reaction.
N H3  N O2
a.) Balance all atoms except oxygen and
hydrogen.
b.
Balance
charges by adding electrons. It
N-3H3+1  N+4O2-2 + 7 etakes 7 electrons to move from -3 to +4
on the number line.
c. Balance the oxygen by adding twice as
4 OH- + N-3H3+1  N+4O2-2 + 7 emany OH- as you think
d. Balance the hydrogen by adding H2O.
4 OH- + N-3H3+1  N+4O2-2 + 7 e- + 2 H2O
Unusual but happens:
-3
+1
+4
-2
* Double all to balance hydrogen
* 8 OH + 2N H3 2 N O2 + 14 e + 4H2O + 3H2O
**Repeat step 4 c&d
**6OH- + 8 OH- + 2NH3 2NO2 + 14 e- + 4H2O + 3H2O
-3
+1
+4
-2
***Simplify
*** 14 OH + 2N H3 2N O2 + 14 e + 7H2O
-3
+1
+4
-2
14 OH + 2NH3 2NO2 + 14 e + 10H2O
14 (H2O + e- + NO3-  NO2 + 2 OH-)
-
-
14 OH- + 2NH3 2NO2 + 14 e- + 10H2O
4 14 H2O + 14 e- + 14 NO3-  14 NO2 + 28 14 OH4 H2O + 14 NO3- + 2NH3  16 NO2 + 14 OH-
5. Balance the electrons on each side of the
equations. Find the common denominator
and multiply each equation. Cross out the
equations after you finish the
multiplication.
6. Simplify and add the two equations. Mark
out the electrons. You will sometimes
mark out some water or H+ on each side.
7. Write what is left.
RULES FOR ASSIGNING OXIDATION NUMBERS
1. The oxidation number of any free element is 0. A free element is one that is written alone in a
“space” on the equation. Example: Cu + O2  (both are free and have an oxidation number of zero)
2. The oxidation number of a monatomic ion (F-, I- Na+, Ca+2 or Ca2+, S-2 or S2-) is equal to the charge
of the ion.
3. The oxidation number of each hydrogen atom in most compounds is 1+.
4. The oxidation number of each oxygen atom in most compounds is 2-. Exception: Oxygen in peroxide
is assigned 1- (H2O2).
s. The sum of the oxidation numbers in a particle must equal the apparent charge of that particle. If
there is no apparent charge written then the apparent charge is zero. All compounds have an
apparent charge of zero. The apparent charge of polyatomic ion is equal to its charge. (Chlorate ClO3- The chloride ion here has a charge of 5+ and oxygen is always has a 2- charge. Cl + 3(O) = -1
Cl + 3(-2) = -1
Cl + (-6) = -1
Cl = -1 + 6
Cl = +5
6. In compounds, the elements of Group IA, Group IIA, and Aluminum have positive oxidation
numbers +1, +2, and +3 respectively.
7. In binary compounds, the group VII elements have a -1 charge when they are the second part of a
binary compound. Group VI have a -2 charge.
When you have a compound you can work out the oxidation numbers in two different ways. You can
see these demonstrated below.
NaClO3 What is the oxidation number of Cl?
Na = +1 (see Rule #6)
Cl = ?
O = -2 (always)
Compound method
Na + Cl + 3(O) = 0
+1+ Cl + 3(-2) = 0
+1 + Cl -6 = 0
Cl - 5 = 0
Cl = +5
Al2(SO4)3
2A1 + 3S + 12 O = 0
2(+3) + 3S + 12(-2) = 0
+6 +3S - 24 = 0
3S - l8 = 0
3S = 0 + 18
3
3
S = +6
Polyatomic Ion method
ClO3- is the polyatomic ion
of the compound
Cl + 3(O) = -1
Cl + 3(-2) = -1
Cl – 6 = -1
Cl =-1+6 = +5
SO4-2 is the polyatomic ion
S + 4(O) = -2
S + 4(-2) = -2
S - 8 = -2
S = -2 + 8
S = +6
Notice the answer is the same whichever way you choose but the polyatomic method is
shorter and can be done in your head.
Ionic Equations Rules
When you write a compound in ionic form you write the two parts separate. When you are
told to write a compound in molecular form, you write it like you always have. Strong acids
and bases are written in ionic form. Weak acids and bases are written in molecular form.
(Ionic form means that ions are formed in solution. Molecular from means no ions form in
solution.) (Example: HCl (aq) is written in ionic form H+ + Cl-. On the other hand HCl(g) is
written in molecular form as HCl because it is a gas.)
Solubility Rules
1. All common salts (any we may use) of Group 1 and ammonium are soluble.
2. All common acetates and nitrates are soluble.
3. All binary compounds of Group 17 elements (other than F) with metals are soluble except
those of silver, mercury (I), and lead.
4. All sulfates are soluble except: barium, strontium, lead, calcium, silver, and mercury (I).
5. Except for those in Rule 1, carbonates, silicates, and phosphates are insoluble.
6. All salts of sulfides are insoluble except for those of Group I and II elements and of
ammonium.
7. Gases: Are always written in molecular form (not soluble).
8. Oxides: Are always written in molecular form.
Strong Acids and Bases are soluble
9. All group 1 and 2 all make strong bases with hydroxide except beryllium otherwise they
are weak.
10.Binary acids: HCl, HBr, & HI are strong acids and written as ions (HCl is written as H+
and Cl- in a net ionic equation.), all other acids are written in molecular form.
11.Ternary acids: If the number of oxygen atoms is TWO more than the number of
hydrogen atoms it is strong. (H2SO4, HNO3, HClO3 are examples of strong acids. HP3O4,
H2SO3 are examples of weak acids.)