Download Balancing Redox Equations by the Half

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+ +
1. Write a net ionic equation. Use your
SO4-2 + Mn+2 + SO4- + H+ + SO4-2 (cross out specter ions –
solubility rules. If there are “+” or “-“ in the
Shown)
equation, you can skip this step.
Net Ionic equation:
Sb+3 + MnO4- + H2O  H+ + SbO4-3 + Mn+2 + H+
Sb+3 + Mn+7O4-2 + H2+O-2  H+ + Sb+5O4-2 + Mn+2 + H+
a. Sb+3 Sb+5O4-3
b. Sb+3 Sb+5O4-3 + 2ec. Sb is balanced.
d. 4 H2O + Sb+3 Sb+5O4-3 + 2ee. 4 H2O + Sb+3 Sb+5O4-3 + 2e- + 8 H+
a.
b.
c.
d.
e.
Mn+7O4-2  Mn+2
5 e- + Mn+7O4-2  Mn+2
Mn is balanced
5 e- + Mn+7O4-2  Mn+2 + 4 H2O
8 H+ + 5 e- + Mn+7O4-2  Mn+2 + 4 H2O
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
12 H2O + 5 Sb+3 + 2 Mn+7O4-2 5Sb+5O4-3 + 24 H+ + 2
Mn+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.
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+
3. Repeat step 3 for the next element.
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.
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  CaBr2 + Ca(BrO3)2 + H2O
Br2 + Ca+2 + OH-  Ca+2 + Br - + Ca+2 + BrO3- + H2O
Br20 + Ca+2 + O-2H+1-  Ca+2 + Br - + Ca+2 + Br+5O-2 3 - +
H+12O-2
a. Br20  Br –
b. e- + Br20  Br –
c. 2 e- + Br20  2 Br –
a.
b.
c.
d.
e.
e- + Br20  Br+5O3–
Br20  Br+5O3– + 5 eBr20  2 Br+5O3– + 10 e12 OH- + Br20  2 Br+5O3– + 10 e12 OH- + Br20  2 Br+5O3– + 10 e- + 6 H2O
5( 2 e- + Br20  2 Br –
12 OH- + Br20  2 Br+5O3– + 10 e- + 6 H2O
10 e- + 5 Br20  10 Br –
12 OH- + 6 Br20  2 BrO3– + 10 Br - + 6 H2O
1. Write a net ionic equation. Use your
solubility rules. If there are “+” or “-“ in the
equation, you can skip this step.
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.
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.
4. Repeat step 3 for the next element.
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.
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
S + 4(O) = -2
S + 4(-2) = -2
S - 8 = -2
S = -2 + 8
S = +6
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
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.)