Download Lecture 06 Slides

Oxidation - Reduction Reactions
Oxidation - Reduction
Chemistry
Oxidation - reduction (redox) reactions are chemical processes
that involve a transfer of electrons between substances
-- this can be a complete transfer to form ionic bonds
or a partial transfer to form covalent bonds
In all redox reactions:
• one substance loses electrons -- this substance is oxidized
• one substance gains electrons -- this substance is reduced
There are lots of processes in the natural world (and in the
laboratory) that involve redox reactions
• e.g., corrosion, batteries, photosynthesis/respiration, etc.
Reaction between zinc and sulfuric acid
Sulfuric acid (solution of H+ and SO42- ions)
Zn strip
Reaction between zinc and sulfuric acid
Overall equation (molecular equation)
H2 bubbles
Zn(s) + H2SO4(aq)
ZnSO4(aq) + H2(g)
Zinc loses electrons
Overall ionic equation
• zinc is oxidized
Zn2+(aq) + 2e-
Zn(s)
-- all dissolved ions are explicitly shown
Zn(s) + 2 H+(aq) + SO42–(aq)
Zn2+(aq) + SO42–(aq) + H2(g)
Hydrogen gains electrons
Net ionic equation
• hydrogen is reduced
2 H+(aq) + 2e-
H2(g)
-- includes only substances that undergo change
-- ions that are present but do not react (spectator ions) are not shown
Zn(s) + 2 H+(aq)
Electrons are transferred from zinc to hydrogen
Reaction between Cu and AgNO3
Cu(s)
Ag+(aq)
Ag(s)
initial
final
NO3–(aq)
Cu2+(aq)
NO3–(aq)
Oxidation of Cu: Cu(s) → Cu2+(aq) + 2eReduction of Ag+: 2 Ag+(aq) + 2e- → 2 Ag(s)
Zn2+(aq) + H2(g)
Oxidation - Reduction Reactions
Oxidation - reduction (redox) reactions are chemical processes
that involve a transfer of electrons between substances
• Oxidation occurs when a substance loses electrons
• Reduction occurs when a substance gains electrons
In a redox reaction, oxidation and reduction occur simultaneously
-- one cannot occur in the absence of the other
Electrons are transferred from Cu atoms to Ag+ ions in solution
Cu loses electrons (oxidation)
Overall Equation: Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Overall Ionic Equation:
Cu(s) + 2 Ag+(aq) + 2 NO3–(aq) → 2 Ag(s) + Cu2+(aq) + 2 NO3–(aq)
Net Ionic Equation: Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq)
initial
Cu(s)
Ag+(aq)
final
Ag(s)
Cu2+(aq)
Cu(s) + 2 Ag+(aq)
Cu2+(aq) + 2 Ag(s)
Ag+ gains electrons (reduction)
Oxidation number
Oxidation / reduction memory aid
The concept of oxidation numbers (also called oxidation states)
was devised as a bookkeeping system for keeping track of
electrons during redox reactions.
L ose
E lectrons
O xidation
-- oxidation of an element results in an increase in its
oxidation number
-- reduction of an element results in a decrease in its
oxidation number
G ain
E lectrons
R eduction
Oxidation numbers are assigned
according to a specific set of rules
Rules for assigning oxidation numbers
Rules for assigning oxidation numbers
1. An element in its elemental state is assigned an
oxidation number of zero
2. For any monoatomic ion, the oxidation number is equal
to the charge on the ion.
Ba
K
S
0
0
0
Au
0
barium
potassium
sulfur
gold
This includes elements that exist as a diatomic molecules
H2
Cl2
Cl-
Ba2+
+2
barium ion
(oxidation number: +2)
chloride ion
(oxidation number: –1)
O2
0
0
0
hydrogen
chlorine
oxygen
BaCl2
-1
+2
-1
In an ionic compound,
the ions retain their
oxidation number
ionic
compound
free ions
Rules for assigning oxidation numbers
Rules for assigning oxidation numbers
3. Nonmetals usually have negative oxidation numbers
(but sometime they can be positive)
3. Nonmetals usually have negative oxidation numbers
(but sometime they can be positive)
a. The oxidation number of oxygen is usually –2
b. The oxidation number of hydrogen is usually +1
when bonded to nonmetals and –1 when bonded
to metals.
Exceptions:
Peroxide ion (O22-): oxidation number = -1
Oxygen difluoride (OF2): oxidation number = +2
H 2O
H 2O
-2
water
MgO
-2
magnesium
oxide
H2O2
-1
hydrogen
peroxide
OF2
+2
oxygen
difluoride
+1
water
CH4
+1
methane
NaH
-1
sodium
hydride
Rules for assigning oxidation numbers
3. Nonmetals usually have negative oxidation numbers
(but sometime they can be positive)
c. The oxidation number of fluorine is –1 in all compounds.
Other halogens have an oxidation number of -1 in most
binary compounds.
Exceptions:
Rules for assigning oxidation numbers
4. The sum of the oxidation numbers of all atoms in a
neutral compound is zero.
5. The sum of the oxidation numbers of all atoms in a
polyatomic ion is equal to the charge of the ion.
Halogens (except fluorine) combined with oxygen have
positive oxidation numbers)
HF
NaCl
-1
-1
hydrogen
fluoride
sodium
chloride
MgBr2
-1
magnesium
bromide
ClO4-1
+7
perchlorate
ion
Many elements have multiple
oxidation numbers
Many elements have multiple
oxidation numbers
It depends on the types of compounds they form
It depends on the types of compounds they form
N oxidation
number
N2
N2O
NO
N2O3
NO2
N2O5
NO3–
0
+1
+2
+3
+4
+5
+5
Note: The oxidation number of oxygen is –2 in all of these compounds
Elemental copper (Cu)
Copper (II) sulfate
(CuSO4)
Cu oxidation state: 0
Cu oxidation state: +2
Rules for determining the oxidation number
of an element within a compound
Step 1: Write the oxidation number of each known atom
below the atom in the formula
Step 2: Multiply each oxidation number by the number of
atoms of that element in the compound
Step 3: Assign oxidation numbers for the other atoms in
the compound in order to make the sum of the
oxidation numbers equal to zero
Example: Determine the oxidation number of
carbon in carbon dioxide
CO2
-2
2(-2) = -4
C + (-4) = 0
C = +4 (oxidation number for carbon)
Step 1: Write the oxidation number of each known atom below the atom in
the formula
Step 2: Multiply each oxidation number by the number of atoms of that
element in the compound
Step 3: Assign oxidation numbers for the other atoms in the compound in
order to make the sum of the oxidation numbers equal to zero
Example: Determine the oxidation number of
sulfur in sulfuric acid
Example: Determine the oxidation number of
chromium in Cr2O72-
H2SO4
+1
Cr2O72-
-2
2(+1) = +2
-2
4(-2) = -8
7(-2) = -14
+2 + S + (-8) = 0
2Cr + (-14) = -2 (the charge on the ion)
S = +6 (oxidation number for sulfur)
Cr = +6 (oxidation number for chromium)
Step 1: Write the oxidation number of each known atom below the atom in
the formula
Step 1: Write the oxidation number of each known atom below the atom in
the formula
Step 2: Multiply each oxidation number by the number of atoms of that
element in the compound
Step 2: Multiply each oxidation number by the number of atoms of that
element in the compound
Step 3: Assign oxidation numbers for the other atoms in the compound in
order to make the sum of the oxidation numbers equal to zero
Step 3: Assign oxidation numbers for the other atoms in the compound in
order to make the sum of the oxidation numbers equal to zero
Example: Determine the oxidation number of
potassium and nitrogen in KNO3
KNO3
K+
Oxidation - reduction (redox) reactions are chemical
processes that involve a transfer of electrons between
substances
NO3–
Recognize that this is an ionic compound between K+ and NO3The oxidation number of potassium in
For nitrogen:
K+
Oxidation - Reduction Reactions
is +1 (the charge on the ion)
-- this can be a complete transfer to form ionic bonds or
a partial transfer to form covalent bonds
In redox reactions, the oxidation numbers of the
elements involved in the reaction change
NO3–
• one substance is oxidized (loses electrons)
-2
-- oxidation number increases
3(-2) = -6
• one substance is reduced (gains electrons)
N + (-6) = -1 (the charge on the ion)
-- oxidation number decreases
N = +5 (oxidation number for nitrogen)
Reaction can be rewritten to
emphasize electron transfer
Reaction between zinc and sulfuric acid
Zn(s) + H2SO4(aq)
ZnSO4(aq) + H2(g)
Zn + 2 H+ + SO42-
Zn2+ + SO42- + H2
0
+2
0
+2
+1
0
+1
0
•zinc loses electrons
•the oxidation number of Zn increases
•zinc is oxidized
•zinc loses electrons
•the oxidation number of Zn increases
•zinc is oxidized
•hydrogen gains electrons
•the oxidation number of H decreases
•hydrogen is reduced
•hydrogen gains electrons
•the oxidation number of H decreases
•hydrogen is reduced
Electrons are transferred from zinc to hydrogen
Electrons are transferred from zinc to hydrogen
Reaction between zinc and sulfuric acid
Oxidizing and reducing agents
Oxidizing agent: The reactant that causes another
substance to be oxidized
– i.e., the reactant that causes an increase in
the oxidation state of another substance
Zn(s) + H2SO4(aq)
ZnSO4(aq) + H2(g)
0
+2
+1
The oxidizing agent is reduced in a redox reaction
Reducing agent: The reactant that causes another
substance to be reduced
– i.e., the reactant that causes a decrease in
the oxidation state of another substance
The reducing agent is oxidized in a redox reaction
0
•
•
•
•
zinc loses electrons
the oxidation number of Zn increases
zinc is oxidized
zinc is the reducing agent
(it causes hydrogen to be reduced)
•
•
•
•
hydrogen gains electrons
the oxidation number of H decreases
hydrogen is reduced
hydrogen is the oxidizing agent
(it causes zinc to be oxidized)
Example: Is the following a redox reaction?
Example: Is the following a redox reaction?
Neutralization reaction between hydrochloric acid and potassium hydroxide
Thermite reaction
HCl (aq) + KOH (aq)
+1
-1
Element
+1
H2O (l) + KCl (aq)
-2 +1
+1
-2
+1
-1
Oxidation number Oxidation number
before reaction
after reaction
H
+1
+1
O
–2
–2
K
+1
+1
Cl
–1
–1
Balancing redox equations
Balancing any chemical equation is based on the law of conservation
of mass
• the
number of atoms of each element must be the same on both
sides of the equation
2 Al (s) + Fe2O3 (s)
0
Element
+3
Al2O3 (l) + 2 Fe (l)
-2
+3
-2
Oxidation number Oxidation number
before reaction
after reaction
Al
0
+3
Fe
+3
0
O
–2
–2
0
Which element
is oxidized?
Which element
is reduced?
Balancing redox equations
For many simple chemical reactions, balancing the overall equation
“automatically” balances the gain/loss of electrons
• i.e., the equation can be balanced without explicitly considering the
transfer of electrons
Example: Thermite reaction
When balancing redox equations, there is an additional requirement
•
the total gains and losses of electrons must balance each other
The increase in oxidation number for the oxidized substance must be
equal to the decrease in oxidation number for the reduced substance
Al is oxidized (oxidation number: 0 to +3)
Each Al atom loses 3 electrons
Total change in oxidation number of Al: +3 x 2 atoms of Al = +6
2 Al (s) + Fe2O3 (s)
Al2O3 (l) + 2 Fe (l)
Fe is reduced (oxidation number: +3 to 0)
Each Fe atom gains 3 electrons
Total change in oxidation number of Fe: -3 x 2 atoms of Fe = -6
Balancing redox equations
Method for balancing complex redox equations
For complex redox equations, it can be difficult to balance the equation
without explicitly accounting for the numbers of electrons lost and gained
Example:
Example:
+6
2 Fe+2(aq)
+
2 Fe+3(aq)
Sn+4(aq)
+
Sn+2(aq)
+2
+3
Cr2O7-2(aq) + Fe+2(aq)
+3
Cr+3(aq) + Fe+3(aq)
-2
The numbers of atoms of each element are the same on each side
Step 1: Identify atoms undergoing oxidation and reduction
But the total ionic charge is not balanced
• +6 on the left
• +5 on the right
Fe undergoes oxidation (
oxidation number: +1)
Electron gains/losses are also not balanced
• 1 electron lost from Fe (oxidation)
• 2 electrons gained by Sn (reduction)
Cr undergoes reduction (
oxidation number: -3)
Method for balancing complex redox equations
Method for balancing complex redox equations
Example:
Example:
+1 x 1 Fe atom = +1
+6
+2
+3
Cr2O7-2(aq) + Fe+2(aq)
+3
2 Cr+3(aq) + Fe+3(aq)
Cr2O7-2(aq) + Fe+2(aq)
2 Cr+3(aq) + Fe+3(aq)
-3 x 2 Cr atoms = -6
Step 2: Adjust coefficients to balance atoms undergoing oxidation/reduction
Step 3: Write arrows connecting the atoms being oxidized and reduced
• indicate the change in oxidation indicated above the arrow
Be sure to multiply the change in oxidation number for each atom by
the number of atoms undergoing the change
Method for balancing complex redox equations
Example:
Method for balancing complex redox equations
Example:
+1 x 1
6 Fe atom
atoms==+1
+6
Cr2O7-2(aq) + 6 Fe+2(aq)
2 Cr+3(aq) + 6 Fe+3(aq)
-3 x 2 Cr atoms = -6
Step 4: Check to see if the oxidation balances reduction
• i.e.,
total change in oxidation number for oxidized substance
equals total change in oxidation number for reduced substance
If not, adjust the reaction coefficients again to balance oxidation and
reduction
Total charge on left side: -2 + +2(6) = +10
-2(aq)
+2(aq)
Cr2O7-2Cr
(aq)
Fe+2(aq)
+ 6 Fe
+14
H+(aq)
2O+
7 6
+3(aq)
2 Cr
Cr+3
(aq) + +6 6Fe
Fe+3+3(aq)
(aq)+ 7 H2O(l)
Total charge on right side: +3(2) + +3(6) = +24
Step 5: Balance elements and charge by adding available chemical
species to the reactants or products sides of the equation
a. Acid solutions: Use H2O and H+ to balance the equation
It is best to add H+ ions first to obtain charge balance, and then add H2O
to the equation to obtain material balance
Add 14 H+ to left side to obtain charge balance
Add 7 H2O to right side to obtain material balance
Method for balancing complex redox equations
Example:
Method for balancing complex redox equations
Note: For redox reactions taking place in basic solutions, replace Step 5a
with Step 5b.
Cr2O7-2(aq) + 6 Fe+2(aq) + 14 H+(aq)
2 Cr+3(aq) + 6 Fe+3(aq) + 7 H2O(l)
Step 6: Check your final equation to verify charge and material balance
Step 5b. Basic solutions: Use H2O and OH– to balance the equation
It is best to add OH– ions first to obtain charge balance, and then add H2O
to the equation to obtain material balance
Material Balance
Element
Left
Right
Cr
2
2
O
7
7
Fe
6
6
H
14
14
Charge Balance
Net Charge on Left = (-2) + 6(+2) + 14(+1) = +24
Net Charge on Right = 2(+3) + 6(+3) =
+24
For redox reactions taking place in neutral solutions, use either Step 5a
or Step 5b.