Redox

Redox rules for Acidic reactions.
Sb2(SO4)3 + KMnO4 + H2O ® H3SbO4 + K2SO4 + MnSO4 + H2SO4
Sb+3 + SO4-2 + K+ + MnO4- +H2O ® H+ + SbO4-3 + K+ + SO4-2 + Mn+2 + SO4- + H+ + SO4-2 (cross out specter ions – Shown)
Net Ionic equation:
Sb+3 + MnO4- + H2O ® H+ + SbO4-3 + Mn+2 + H+ / 1.  Write a net ionic equation. Use your solubility rules. If there are “+” or “-“ in the equation, you can skip this step.
Sb+3 + Mn+7O4-2 + H2+O-2 ® H+ + Sb+5O4-2 + Mn+2 + H+ / 2.  Find the charge of each element. See page 2 of the Redox instructions handout. Underline those that changed their charge. Elements in a space alone (no charge noted) have a zero charge.
a.  Sb+3® Sb+5O4-3
b.  Sb+3® Sb+5O4-3 + 2e-
c.  Sb is balanced.
d.  4 H2O + Sb+3® Sb+5O4-3 + 2e-
e.  4 H2O + Sb+3® Sb+5O4-3 + 2e- + 8 H+
/ a.  Write the two half reactions.
b.  Pick one element that changed its charge.
c.  Balance all atoms except hydrogen or oxygen. If the element that changed its charge has to be balanced at this point the number of electrons will be multiplied by the number of atoms.
d.  Add electrons to the side with the largest oxidation number. Note: zero is larger than -1.
e.  Balance the oxygen atoms by adding water.
f.  Balance the hydrogen atoms by adding H+
a.  Mn+7O4-2 ® Mn+2
b.  5 e- + Mn+7O4-2 ® Mn+2
c.  Mn is balanced
d.  5 e- + Mn+7O4-2 ® Mn+2 + 4 H2O
e.  8 H+ + 5 e- + Mn+7O4-2 ® Mn+2 + 4 H2O / 3.  Repeat step 3 for the next element.
5( 4 H2O + Sb+3® Sb+5O4-3 + 2e- + 8 H+
2( 8 H+ + 5 e- + Mn+7O4-2 ® Mn+2 + 4 H2O
20 H2O + 5 Sb+3® 5Sb+5O4-3 + 10e- + 40 H+
16 H+ + 10 e- + 2 Mn+7O4-2 ® 2 Mn+2 + 8 H2O / 4.  The number of electrons given up by the Sb must equal the number of electrons needed by the Mn. To fix this you should find the common denominator and multiply the entire equation by that number. Once this is done mark out the equations so you don’t become confused.
12 H2O + 5 Sb+3 + 2 Mn+7O4-2® 5Sb+5O4-3 + 24 H+ + 2 Mn+2 / 5.  Add the equations algebraically. Reduce when necessary and the electrons will cancel.

Redox

Redox rules for Basic reactions. All the rules are the same except for number 3 c and d.
Br2 + Ca(OH)2 à CaBr2 + Ca(BrO3)2 + H2O
Br2 + Ca+2 + OH- ® Ca+2 + Br - + Ca+2 + BrO3- + H2O / 1.  Write a net ionic equation. Use your solubility rules. If there are “+” or “-“ in the equation, you can skip this step.
Br20 + Ca+2 + O-2H+1- ® Ca+2 + Br - + Ca+2 + Br+5O-2 3 - + H+12O-2 / 2.  Find the charge of each element. See page 2 of the Redox instructions handout. Underline those that changed their charge. Elements in a space alone (no charge noted) have a zero charge. Note: in this example the bromine is both oxidized and reduced.
a.  Br20 ® Br –
b.  e- + Br20 ® Br –
c.  2 e- + Br20 ® 2 Br – / 3.  Write the two half reactions.
a. Pick one element that changed its charge.
b.  Balance all atoms except hydrogen or oxygen. If the element that changed its charge has to be balanced at this point the number of electrons will be multiplied by the number of atoms.
c.  Add electrons to the side with the largest oxidation number. Note: zero is larger than -1.
d.  Balance the oxygen atoms by adding twice as many OH- as you think you need. (unless you are balancing oxygen atoms in hydroxide.)
e. Balance the hydrogen atoms by adding water.
a.  e- + Br20 ® Br+5O3–
b.  Br20 ® Br+5O3– + 5 e-
c.  Br20 ® 2 Br+5O3– + 10 e-
d.  12 OH- + Br20 ® 2 Br+5O3– + 10 e-
e.  12 OH- + Br20 ® 2 Br+5O3– + 10 e- + 6 H2O / 4.  Repeat step 3 for the next element.
5( 2 e- + Br20 ® 2 Br –
12 OH- + Br20 ® 2 Br+5O3– + 10 e- + 6 H2O
10 e- + 5 Br20 ® 10 Br – / 5.  The number of electrons given up by the Sb must equal the number of electrons needed by the Mn. To fix this you should find the common denominator and multiply the entire equation by that number. Once this is done mark out the equations so you don’t become confused.
12 OH- + 6 Br20 ® 2 BrO3– + 10 Br - + 6 H2O / 6.  Add the equations algebraically. Reduce when necessary and the electrons should cancel. In this case since Br2 is in both equations you have to add these as well.


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?

Compound method Polyatomic Ion method

Na = +1 (see Rule #6) Na + Cl + 3(O) = 0 ClO3- is the polyatomic ion
of the compound

Cl = ? +1+ Cl + 3(-2) = 0 Cl + 3(O) = -1

O = -2 (always) +1 + Cl -6 = 0 Cl + 3(-2) = -1

Cl - 5 = 0 Cl – 6 = -1

Cl = +5 Cl =-1+6 = +5

Al2(SO4)3 SO4-2 is the polyatomic ion

2A1 + 3S + 12 O = 0 S + 4(O) = -2

2(+3) + 3S + 12(-2) = 0 S + 4(-2) = -2

+6 +3S - 24 = 0 S - 8 = -2

3S - l8 = 0 S = -2 + 8

3S = 0 + 18 S = +6

3 3

S = +6

Notice the answer is the same which ever 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 form 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 (Strong written apart/Weak written together)

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.)