Redox

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Electron Transfer Reactions

Electron transfer reactions are oxidation-reduction or
redox reactions.

Results in the generation of an electric current (electricity)
or be caused by imposing an electric current.

Therefore, this field of chemistry is often called
ELECTROCHEMISTRY.
Terminology for Redox Reactions

OXIDATION—loss of electron(s) by a species;
increase in oxidation number; increase in oxygen.

REDUCTION—gain of electron(s); decrease in
oxidation number; decrease in oxygen; increase in
hydrogen.

OXIDIZING AGENT—electron acceptor; species is
reduced. (an agent facilitates something; ex. Travel
agents don’t travel, they facilitate travel)

REDUCING AGENT—electron donor; species is
oxidized.
You can’t have one… without the other!
Reduction (gaining electrons) can’t happen without an
oxidation to provide the electrons.
 You can’t have 2 oxidations or 2 reductions in the same
equation. Reduction has to occur at the cost of oxidation

LEO the lion says GER!
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Another way to remember
OIL
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Oxidation Numbers
OBJECTIVES
Determine the oxidation
number of an atom of any
element in a pure substance.
Oxidation Numbers
OBJECTIVES
Define oxidation and
reduction in terms of a
change in oxidation number,
and identify atoms being
oxidized or reduced in redox
reactions.
Assigning Oxidation Numbers
• An “oxidation number” is a positive or
negative number assigned to an atom
to indicate its degree of oxidation or
reduction.
• Generally, a bonded atom’s oxidation
number is the charge it would have if
the electrons in the bond were
assigned to the atom of the more
electronegative element
Rules for Assigning Oxidation Numbers
1) The oxidation number of any
uncombined element is zero.
2) The oxidation number of a
monatomic ion equals its charge.
0
0
1
1
2 Na  Cl 2  2 Na Cl
Rules for Assigning Oxidation Numbers
3) The oxidation number of oxygen in
compounds is -2, except in peroxides,
such as H2O2 where it is -1.
4) The oxidation number of hydrogen in
compounds is +1, except in metal
hydrides, like NaH, where it is -1.
1
2
H2O
Rules for Assigning Oxidation Numbers
5) The sum of the oxidation numbers of the
atoms in the compound must equal 0.
1
2
H2O
2(+1) + (-2) = 0
H
O
2
2 1
Ca (O H ) 2
(+2) + 2(-2) + 2(+1) = 0
Ca
O
H
Rules for Assigning Oxidation Numbers
6) The sum of the oxidation numbers in
the formula of a polyatomic ion is equal
to its ionic charge.
? 2
N O3

? 2
S O4
X + 3(-2) = -1
N
O
X + 4(-2) = -2
S
O
thus X = +5
thus X = +6
2
Review of Oxidation numbers
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1.
Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2.
In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3.
The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1.
4.4
4.
The oxidation number of hydrogen is +1 except when it is bonded to metals in
binary compounds. In these cases, its oxidation number is –1.
5.
Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal
to the charge on the molecule or ion.
HCO3Oxidation numbers of all the atoms
in HCO3- ?
O = -2
H = +1
3x(-2) + 1 + ? = -1
C = +4
4.4
Oxidation and Reduction (Redox)
A process called “reduction” is the
opposite of oxidation, and originally
meant the loss of oxygen from a
compound
Oxidation and reduction always occur
simultaneously
The substance gaining oxygen (or
losing electrons) is oxidized, while the
substance losing oxygen (or gaining
electrons) is reduced.
Oxidation and Reduction (Redox)
Today, many of these reactions may
not even involve oxygen
Redox currently says that electrons
are transferred between reactants
Mg
+
S→
Mg2+
+
S2-
(MgS)
•The magnesium atom (which has zero charge) changes to a magnesium ion by
losing 2 electrons, and is oxidized to Mg2+
•The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2
electrons, and is reduced to S2-
Oxidation and Reduction (Redox)
0
1
0
1
2 Na  Cl 2  2 Na Cl
Each sodium atom loses one electron:
1
0
Na  Na  e

Each chlorine atom gains one electron:
0

1
Cl  e  Cl
LEO says GER :
Lose Electrons = Oxidation
1
0
Na  Na  e

Sodium is oxidized
Gain Electrons = Reduction
0

1
Cl  e  Cl
Chlorine is reduced
LEO says GER :
- Losing electrons is oxidation, and the
substance that loses the electrons is
called the reducing agent.
- Gaining electrons is reduction, and
the substance that gains the electrons is
called the oxidizing agent.
Mg is the
reducing
agent
Mg is oxidized: loses e-, becomes a Mg2+ ion
Mg(s) + S(s) → MgS(s)
S is the oxidizing agent
S is reduced: gains e- = S2- ion
Oxidation and Reduction (Redox)
It is easy to see the loss and gain of
electrons in ionic compounds, but what
about covalent compounds?
In water, we learned that oxygen is
highly electronegative, so:
the oxygen gains electrons (is
reduced and is the oxidizing agent),
and the hydrogen loses electrons (is
oxidized and is the reducing agent)
Not All Reactions are Redox Reactions
- Reactions in which there has been no
change in oxidation number are NOT
redox reactions.
Examples:
1 5 2
1
1
1
1
1 5 2
Ag N O 3 (aq )  Na Cl (aq )  Ag Cl ( s )  Na N O 3 (aq )
1 2 1
1
6 2
1
6 2
1
2
2 Na O H (aq )  H 2 S O 4 (aq )   Na 2 S O 4 (aq )  H 2 O(l )
Corrosion
•Damage done to metal is costly to
prevent and repair
•Iron, a common construction metal often
used in forming steel alloys, corrodes by
being oxidized to ions of iron by oxygen.
•This corrosion is even faster in the
presence of salts and acids, because
these materials make electrically
conductive solutions that make
electron transfer easy
Corrosion
•Luckily, not all metals corrode easily
•Gold and platinum are called noble
metals because they are resistant to
losing their electrons by corrosion
•Other metals may lose their electrons
easily, but are protected from corrosion by
the oxide coating on their surface, such as
aluminum
•Iron has an oxide coating, but it is not
tightly packed, so water and air can
penetrate it easily
Corrosion
•Serious problems can result if bridges,
storage tanks, or hulls of ships corrode
•Can be prevented by a coating of oil,
paint, plastic, or another metal
•If this surface is scratched or worn away,
the protection is lost
•Other methods of prevention involve the
“sacrifice” of one metal to save the second
•Magnesium, chromium, or even zinc
(called galvanized) coatings can be applied
Reducing Agents and Oxidizing Agents
•
• An increase in oxidation number = oxidation
• A decrease in oxidation number = reduction
1
0
Na  Na  e

Sodium is oxidized – it is the reducing agent
0


1
Cl  e  Cl
Chlorine is reduced – it is the oxidizing agent
Trends in Oxidation and Reduction
Active metals:
Lose electrons easily
Are easily oxidized
Are strong reducing agents
Active nonmetals:
Gain electrons easily
Are easily reduced
Are strong oxidizing agents
Balancing Redox Equations
OBJECTIVES
Describe how oxidation
numbers are used to identify
redox reactions.
Balancing Redox Equations
OBJECTIVES
Balance a redox equation
using the oxidation-numberchange method.
Balancing Redox Equations
OBJECTIVES
Balance a redox equation
by breaking the equation into
oxidation and reduction halfreactions, and then using the
half-reaction method.
Identifying Redox Equations
In general, all chemical reactions can
be assigned to one of two classes:
1) oxidation-reduction, in which
electrons are transferred:
• Single-replacement, combination,
decomposition, and combustion
2) this second class has no electron
transfer, and includes all others:
• Double-replacement and acidbase reactions
Identifying Redox Equations
In an electrical storm, nitrogen and
oxygen react to form nitrogen monoxide:
N2(g) + O2(g) → 2NO(g)
•Is this a redox reaction?
YES!
•If the oxidation number of an element
in a reacting species changes, then
that element has undergone either
oxidation or reduction; therefore, the
reaction as a whole must be a redox.
Balancing Redox Equations
It is essential to write a correctly
balanced equation that represents
what happens in a chemical reaction
• Fortunately, two systematic methods
are available, and are based on the
fact that the total electrons gained in
reduction equals the total lost in
oxidation. The two methods:
1) Use oxidation-number changes
2) Use half-reactions
Using Oxidation-Number Changes
Sort of like chemical bookkeeping, you compare the increases and
decreases in oxidation numbers.
•start with the skeleton equation
•Step 1: assign oxidation numbers to all atoms; write above their
symbols
•Step 2: identify which are oxidized/reduced
•Step 3: use bracket lines to connect them
•Step 4: use coefficients to equalize
•Step 5: make sure they are balanced for both atoms and charge
Using half-reactions
A half-reaction is an equation showing just
the oxidation or just the reduction that takes
place
they are then balanced separately, and
finally combined
Step 1: write unbalanced equation in ionic
form
Step 2: write separate half-reaction
equations for oxidation and reduction
Step 3: balance the atoms in the halfreactions
Using half-reactions
continued
•Step 4: add enough electrons to one side
of each half-reaction to balance the charges
•Step 5: multiply each half-reaction by a
number to make the electrons equal in both
•Step 6: add the balanced half-reactions to
show an overall equation
•Step 7: add the spectator ions and balance
the equation
Choosing a Balancing Method
1) The oxidation number change
method works well if the oxidized
and reduced species appear only
once on each side of the equation,
and there are no acids or bases.
2) The half-reaction method works
best for reactions taking place in
acidic or alkaline solution.
Balancing Redox Equations
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
1.
Write the unbalanced equation for the reaction ion ionic form.
Fe2+ + Cr2O72-
2.
Fe3+ + Cr3+
Separate the equation into two half-reactions.
+2
Fe2+
Oxidation:
+6
Reduction:
3.
Cr2O7
2-
Fe3+
Cr3+
+3
+3
Balance the atoms other than O and H in each half-reaction.
Cr2O72-
2Cr3+
19.1
Balancing Redox Equations
4.
For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O72-
2Cr3+ + 7H2O
14H+ + Cr2O725.
Add electrons to one side of each half-reaction to balance the charges on the halfreaction.
Fe2+
Fe3+ + 1e-
6e- + 14H+ + Cr2O726.
2Cr3+ + 7H2O
2Cr3+ + 7H2O
If necessary, equalize the number of electrons in the two half-reactions by multiplying the
half-reactions by appropriate coefficients.
6Fe2+
6Fe3+ + 6e-
6e- + 14H+ + Cr2O72-
2Cr3+ + 7H2O
19.1
Balancing Redox Equations
7.
Add the two half-reactions together and balance the final equation by inspection. The
number of electrons on both sides must cancel.You should also cancel like
species.
Oxidation:
Reduction:
14H+ + Cr2O72- + 6Fe2+
8.
6Fe2+
6Fe3+ + 6e-
6e- + 14H+ + Cr2O72-
2Cr3+ + 7H2O
6Fe3+ + 2Cr3+ + 7H2O
Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
9.
For reactions in basic solutions, add OH- to both sides of the equation for every H+ that
appears in the final equation. You should combine H+ and OH- to make H2O.
19.1
CHEMICAL CHANGE --->
ELECTRIC CURRENT
To
obtain a useful current, we
separate the oxidizing and
reducing agents so that
electron transfer occurs thru
an external wire.
This is accomplished in a GALVANIC or VOLTAIC
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
cell.
A group of such cells is called a battery.
Galvanic Cells
anode
oxidation
cathode
reduction
-
+
spontaneous
redox reaction
19.2
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