| Chem 109 |
General Chemistry |
Summer 2002 |
| Lecture Notes:: 17 June |
© R. Paselk 2002 |
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Aqueous Solutions, cont.
Oxidation/Reduction (Redox) Reactions
In these reactions we see a transfer of electrons from one
atom or molecule to another. First let's look at some terms.
- Oxidation refers to taking electrons away from a substance.
So to oxidize means to behave like oxygen normally does and "steal"
electrons.
- Reduction refers to recieving electrons - it is the opposite
of oxidation.
- Note that oxidation and reduction always go together. When
oxygen oxidizes it is itself reduced (it gains electrons).
- Consider the example of burning natural gas:
CH4 + 2 O2 Æ
CO2 + 2 H2O
Notice that the methane is oxidized by the oxygen. We say
that the carbon and hydrogen are both oxidized to give the new
covalent products, water and carbon dioxide, because the electrons
are not evenly shared, they are pulled toward the oxygens in
each bond.
Examples:
- A piece of clean copper is placed in a solution of silver
nitrate. Assuming the copper goes to Cu(II) write a balanced
equation. Notice that in this case you must balance the charge
in order to get the final equation.
Ag+ + NO3- + Cu0
Æ Ag0 + NO3-
+ Cu2+
Balancing: 2Ag+ + Cu0 Æ
2Ag0 + Cu2+
- Consider the reaction of calcium metal with hydrochloric
acid. This is similar to the reaction of sodium and water, and
hydrogen gas is given off. Write a net ionic equation for this
reaction:
H+ + Cl- + Ca0 Æ H2 (g) + Ca2+
+ Cl-
Balancing: 2 H+ + Ca0 Æ
H2 (g) + Ca2+
- A piece of clean iron is placed in a solution of copper(II)
sulfate. Assuming iron goes to Fe(III) write a balanced equation.
Again the problem is to balance the charge.
Cu2+ + SO42- + Fe0
Æ Cu0 + Fe3+
+ SO42-
Balancing: 3 Cu2+ + 2 Fe0 Æ 3 Cu0 + 2 Fe3+
Acid-Base Reactions
Neutralization: When we combine equal numbers of moles
of hydrogen ion and hydroxide ion a neutralization occurs. That
is, there is no reactive component left, all of the acid has been
consumed by all of the base, and water has been synthesized.
- Consider the reaction of 50.0 mL of 0.25 M hydrochloric acid
with 25.0 mL of 0.50 M sodium hydroxide.
- Write a net ionic equation for this reaction:
H+ + Cl- + Na+ + OH-
Æ H2O + Cl-
+ Na+
Giving: H+ + OH- Æ
H2O
- How much acid will be left over? Base? How much water was
made (synthesized)?
- First find out the moles of each:
- acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10-2moles;
- base = (25.0 mL)(1 L/1000 mL)(0.50 mole/L) = 1.25 x 10-2moles.
- Since the amounts are identical they will completely neutralize
each other. That is they will react completely and both acid
and base will be consumed with none left over.
- Since the ratio in the equation is 1:1:1 acid:base:water,
1.25 x 10-2moles of water are produced.
Oxidation Numbers
For simple elemental ions it is easy to determine the charge
on an atom, but in many other circumstances this is not the case.
In order to name compounds and understand reactions we frequently
need this information which is obtained from oxidation numbers.
Oxidation numbers are in essence an electronic accounting
method in which electrons are assigned to a particular
atom in a bond or interaction. As such they give an approximate
picture of where electrons actually reside in compounds. We will
find this information very useful later when we look at particular
types of chemical reactions. Oxidation numbers are essential for
nomenclature.
Oxidation numbers are most readily assigned using a simple
set of rules:
- In the formula for any substance the sum of the oxidation
numbers of all the atoms in the formula is equal to the charge
shown. Thus:
- For elements, such as Ar, O2, S8, etc.
in the uncombined state the oxidation number for each atom must
be 0, since no charge is shown and the atoms are equal to each
other.
- For monoatomic ions the oxidation number equals the charge.
- For a compound the sum of the oxidation numbers of the atoms
equals 0.
- For a polyatomic ion the sum of the oxidation numbers of
the atoms equals the charge on the ion.
- In compounds fluorine is always assigned an oxidation
number of -1.
- Alkali metals in compounds will always (for our class) be
assigned an oxidation number of +1.
- Alkaline-earth metals in compounds will always (for our class)
be assigned an oxidation number of +2
- In compounds oxygen is usually assigned an
oxidation number of -2.
- Exception 1: in peroxides it is -1 while in superoxides
it is -1/2. These will generally be obvious due to other rules
(or the names).
- Exception 2: in combination with fluorine oxygen can
be positive due to Rule 2 above, thus for OF2 oxygen
is assigned an oxidation number of +2.
- In compounds hydrogen is usually assigned an
oxidation number of +1
- Exception: in metallic hydrides hydrogen is assigned
an oxidation number of -1. These exceptions will be fairly obvious:
NaH, CaH2, etc.
- Aluminum will always (for our class) be assigned an oxidation
number of +3, other elements in this Group will usually be assigned
an oxidation number of +3.
Let's try these rules on some examples:
- OH-: Rule 1 & 5 Æ
-2 + H = -1, H = -1 - (-2) = +1
- MgH2: Rule 1 & 3 Æ
+2 + 2H = 0, 2H = -2, H = -1
- C2H3O2-:
Rule 1 & 5 & 6 Æ 2C
+ 3(+1) + 2(-2) = -1, 2C = -1 - (+3) - (-4) = 0, C = 0
- MnO2: Rule 1 & 5 Æ
Mn + 2(-2) = 0, Mn = - (-4) = +4
Additional practice examples:
- NH4+
- CO32-
- NO3-
- NO2-
- PO43-
- SO32-
- SO42-
- C2O42-
- HCO3-
- MnO4-
- HClO4
- HClO3
- HClO2
- HClO
Click here to go to the key.
Finally, note that in writing formulae, the element with the
more positive oxidation number comes first. There are, of course,
a few exceptions, the most well known being ammonia: NH3
(by the rules it should be H3N).
Balancing Redox Equations
There are two common methods for balancing redox reactions:
the oxidation number method and the half-reaction method. The
half-reaction method works very well for ionic reactions, it is
relatively easy to give partial credit, and it is the only method
I will use in this class. If you know how to do the other method
you are welcome to do so, but be careful to make sure you show
your work or I won't be able to give partial credit!
The Half-Reaction Method. In the half-reaction method
what we do is first break an equation into two parts and then
balance the parts individually. Presented stepwise:
Acid Solution:
- Separate the reaction into two half-reactions.
- Balance each half-reaction separately:
- Balance atoms other than O & H by inspection.
- Balance O by adding H2O to the opposite side.
- Balance H by addding H+ as appropriate.
- Balance the charge by adding electrons (e-) -
add to same side as excess of positive charge, or opposite side
if excess negative charge.
- Balance the charges of the two half-reactions by multiplying
appropriately.
- Add two equations together
- Cancel items appearing on both sides.
© R A Paselk
Last modified 17 June 2002
