Download Balancing Redox Equations

Electrochemistry : Oxidation and Reduction
Electrochemical Reaction - Chemical reaction that involves the flow of electrons.
Redox Reaction (oxidation-reduction reaction) - A reaction in which at
least one atom changes in oxidation state.
Reduction - Any process in which the oxidation number of an atom
decreases (becomes more negative).
Oxidation - Any process in which the oxidation number of an atom
increases (becomes more positive).
Oxidation Number - The charge that an atom would have if the compound
in which it were found were ionic. (Next page is a refresher on “How to”.)
To help remember oxidation and reduction, remember the following:
OILRIG: Oxidation Is Loss Reduction Is Gain
Types of Redox Reactions
Corrosion - A type of redox reaction in which a metal is destroyed.
4 Fe(s) + 3 O2(g) 2 Fe2O3 • 3 H2O
Metathesis Reaction - A reaction in which atoms are interchanged
and there is no change in oxidation number.
Disproportionation Reaction - A reaction in which a single reactant
undergoes both oxidation and reduction.
Disproportionation
Don't confuse this with the following definitions:
Oxidizing Agent - Causes oxidation; undergoes reduction (gains electrons).
Reducing Agent - Causes reduction; undergoes oxidation (loses electrons).
**In spontaneous redox reactions, stronger oxidizing and reducing agents are converted into weaker oxidizing and reducing agents.
Good Oxidizing Agents
Atoms, ions, and molecules with large electron affinities.
e.g. F2, Cl2
Compounds with large oxidation states.
WHY? - The electronegativity increases as oxidation state increases
Electronegativity - The tendency of an atom to draw electrons toward itself.
e.g. MnO4-, CrO42-
Good Reducing Agents
Active metals
e.g. Na, Mg, Al, Zn
Metal hydrides
e.g. NaH, CaH2
H2 can act as either:
Oxidizing agent when it combines with metals.
Reducing agent when it combines with nonmetals.
Before we balance a Redox equation lets first refresh our memory on how to calculate oxidation
numbers.
Oxidation Number - The charge that an atom would have if the compound in which it were found were ionic.
The rules:
1) The sum of the oxidation numbers of the atoms in a molecule must be equal to the overall charge on the
molecule.
2) To assign a number to a transition metal ion (not listed in the table below) start with the overall charge, add
the total number of negative charges for oxygen (if there were four as in the case of MnO4- then you would
add 8 for a total of +7 for Mn), continue until all other species listed in the table below are considered (subtract
if it is a positive value.) The result is the oxidation number of the transition metal ion.
3)The most electronegative element will have a negative oxidation number.
Assigning Oxidation Numbers
Category
1) Neutral substances containing only a single element
2) Monatomic ions
Oxidation #
Example
0
N2, He
same as the charge
Na+ = +1
3) Hydrogen combined with a nonmetal
+1
HBr, CH4, OH-
4) Hydrogen combined with a metal
-1
NaH, CaH2
5) Metals in Group IA
+1
Li3N, Na2S
6) Metals in Group IIA
+2
Mg3N2
7) Oxygen
-2
H2O, NO
(Exceptions: H2O2, O22-)
8) Halogens
-1
-1
AlF3, HCl
Balancing Redox equations using the Oxidation number method
(Basic solution is demonstrated)
Al(s) +
OH-(aq)
H2O
Al(OH)-4(aq) + H2(g)
1. What are the reduction and oxidation pairs?
a) Al(s) and Al(OH)-4(aq) (oxidized)
b) ? and H2(g) (reduced)
Hint: 1. The reaction is taking place in a basic, aqueous media.
2. Look for a reduction potential for H2(g) in a table.
2. Calculate the Oxidation numbers and transfer to the redox partner
-3e-
0
2H2O + Al(s) +
-3
OH-(aq)
H2O
2Al(OH)-4(aq) + 3H2(g)
2 x -1e-
2 x +1
2x0
3. Mass Balance
4. Charge Balance
6 H2O + 2Al(s) +
OH-(aq)
H2O
2Al(OH)-4(aq) + 3H2(g)
Balancing Redox equations using the Half Reaction method
(Basic solution is demonstrated)
Pb(OH)-3(aq) + OCl-(aq)
OH-
PbO2(s) + Cl-(aq)
1. What are the reduction and oxidation pairs?
2. Mass Balance both equations (add H2O to balance extra oxygens
then add extra H+ to balance extra hydrogens from the added H2O)
3. Charge balance both equations (add extra e-)
4. Cancel any common terms.
5. Is the reaction taking place in a basic solution? Are there are any H+ left?
add OH- to both sides. H+ and OH- will make H2O on one side.
6. Add the two half reactions and cancel any extra water.
Pb(OH)-3(aq)
- ++22e
OH- +H
2e2O
H-+++ OCl-(aq)
H2O + Pb(OH)-3(aq) + OCl-(aq)
OH-
OH-
OH-
PbO2(s)+ H2O + H+ + 2e-
Cl-(aq) + H2O + OHPbO2(s) + Cl-(aq)+ 2H2O + OH-