Download + 2 I

Aqueous Reactions and
Solution Stoichiometry
(continuation)
1. Oxidation-Reduction Reactions
2. The Activity Series
3. Balancing REDOX Equations
Terminology for Redox Reactions
• OXIDATION—loss of electron(s) by a
species; increase in oxidation number.
• REDUCTION—gain of electron(s);
decrease in oxidation number.
• OXIDIZING AGENT—electron acceptor;
species are reduced.
• REDUCING AGENT—electron donor;
species are oxidized.
Oxidation-Reduction Reactions
- reaction where electrons are exchanged.
2 Na(s) + 2 H2O(l) Æ 2 NaOH(aq) + H2(g)
Æ
+ 2
oxidation – loss of electrons
2Na0(s)
Na+(aq)
e-
These processes
occur
simultaneously –
you cannot have an
oxidation without a
2 H+(g) + 2 e- Æ H20(g)
reduction and vicereduction – gain of electrons versa!!!
GeneralRules for Assigning an Oxidation Number (O.N.)
O.N
1) For an atom in its elemental neutral form (Na, O2 Cl2,
etc.): O.N. = 0
2) For a monoatomic ion: O.N. = ion charge
3) The sum of O.N. values for the atoms in a compound
equals zero.
zero
4) The sum of the O.N. values for the atoms in a
polyatomic ion equals the ion’s charge.
charge
Rules for specific atoms or periodic table groups:
1) For Group IA(1): O.N. = +1 in all compounds
2) For Group IIA(2): O.N. = +2 in all compounds
3) For hydrogen:
O.N. = +1 in combination with nonmetals
O.N. = -1 in combination with metals and boron
4) For fluorine:
O.N. = -1 in all compounds
5) For oxygen:
O.N. = -1 in peroxides
O.N. = -2 in all other compounds (except with F )
6) For Group VIIA(17) O.N. = -1 in combination with metals, nonmetals
(except O), and other halogens lower in the group.
+1
-1
1
Period
Highest and Lowest Oxidation Numbers of
Chemically Reactive Main-Group Elements
H
1A 2A
+1 +2
2 Li
Be
3A 4A 5A 6A 7A
+3 +4-4 +5-3 +6-2 +7-1
B
C
N
O
F
Al
Si
P
S
Cl
Ga Ge As Se
Br
metalloids 5 Rb Sr
In
Sn Sb Te
I
6 Cs Ba
Tl
Pb Bi
3 Na Mg
non-metals 4
K Ca
metals
7 Fr Ra
Po At
Transition Metals
Possible Oxidation States
VIIIB
IIIB IVB VB VIB VIIB
IB IIB
Sc
Ti
V
Cr +2Mn Fe
Co Ni
Cu Zn
+3 +4,+3 +5,+4 +6,+3 +7,+6+3,+2 +3,+2 +2 +2,+1 +2
+4,+3
+2
+3+2 +2
Y
Zr
Nb Mo Tc Ru Rh Pd Ag
+3 +4,+3 +5,+4 +6,+5 +7,+5 +8,+5 +4,+ +4,+2 +1
+2
+4,+3 +4
+4,+3 3
Cd
+2
La Hf
Ta W
Re +2 Os Ir
Pt Au Hg
+3 +4,+3 +5,+4 +6,+5 +7,+5 +8,+6 +4,+3 +4,+2 +3,+1 +2,+1
+3
+4
+4
+4,+3 +1
Determining the Oxidation Number of an
Element in a Compound
Problem: Determine the oxidation number (O.N.)
O.N of each
element in sulfuric acid
H2SO4: The O.N. of H is +1, so the SO42- group must sum to
-2.
The O.N. of each O is -2 for a total of -8. So the Sulfur
atom is +6.
Recognizing Oxidizing and Reducing Agents
NiO(s) + CO(g) →
C[+2]
C[+4]
C is oxidized
-2
-2
+2
+2
NiO(s) + CO(g)
Ni(s) + CO2 (g)
Ni[+2]
Ni[0]
Ni is Reduced
0
Ni(s)
+4 -2
+ CO2 (g)
CO is the reducing agent and NiO is the oxidizing agent.
Types of Redox Reactions
Redox Combination Reactions (Intermolecular)when reducing and oxidizing agents are in different
compounds
0
0
+3 -1
P4(s) + 6 Cl2(g) → 4 PCl3(l)
0
0
+5 -1
P4(s) + 10 Cl2(g) → 4 PCl5(l)
0
0
+3 -2
Fe(s) + O2 (g) → Fe2O3
(s)
Rusting
Redox Decomposition Reactions (Intramolecular
reactions) -when both reducing and oxidizing agents
are in the same compound
+2 -2
0
0
2 HgO (s) → 2 Hg(l) + O2(g)
+1 -2
0
0
2 Hg2O(s) → 4 Hg(l) + O2(g)
+1 -1
+1 -2
0
2 H2O2 (l) → 2 H2O(l) + O2(g)
Oxygenation and Hydrogenation reactions
0
0
+1 -2
4 Li (s) + O2(g) → 2 Li2O(s)
+2 -2
0
+2 -2
+4 -2
2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)
0
0
+1 -1
2 Na(l) + H2(g) → 2 NaH(s)
+3 -2
0
0
+1 -2
Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g)
Disproportionation reactions
(self-reduction/self oxidation)
+1 -1
+1 -2
0
2 H2O2 (l) → 2 H2O(l) + O2(g)
+1
+2
0
2 Cu+(aq) → Cu2+(aq) + Cu(s)
(net ionic equation)
Displacement Reactions
+1 +5 -2
0
+2 +5 -2
2 AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2 Ag(s)
+1
0
+2
2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s)
0
+1 -1
0
-1
0
(overall equation)
0
(net ionic equation)
+1 -1
Cl2(g) + 2 KI(aq) → I2(s) + 2 KCl(aq)
0
0
(overall equation)
-1
Cl2(g) + 2 I-(aq) → I2(s) + 2 Cl-(aq)
(net ionic equation)
electrolyte
electrode
-
+
+
-
+
+
+
Redox potential depends on the
atomic structure and hydratation
tendency of the element
The Activity Series is based on the ability
of element to lose electrons in aqueous solutions
- Some metals are more easily oxidized than others.
Activity series - a list of metals arranged in decreasing
ease of oxidation
- The higher the metal on the activity series, the more
active that metal.
- Any metal can be oxidized by the ions of elements
below it
Hydrogen is in the activity series even though it is a
nonmetal
EN<1.4
1.4< EN <1.9
1.9< EN <2.54
most can be
dissolved by
oxidizing acids
(HNO3)
Tips on Balancing Equations
• Never add O2, O atoms, or O2- to
balance oxygen.
• Never add H2 or H atoms to balance
hydrogen.
• Be sure to write the correct charges on
all the ions.
• Check your work at the end to make
sure mass and charge are balanced.
Balancing REDOX Equations:
The Oxidation Number (O.N.)Method
Step 1) Assign oxidation numbers to all elements in the equation.
Step 2) From the changes in oxidation numbers, identify the oxidized and
reduced species.
Step 3) Compute the number of electrons lost in the oxidation and gained in
the reduction from the oxidation number changes. Draw tie-lines between
these atoms to show electron changes.
Step 4)
Multiply one or both of these numbers by appropriate factors
to make the electrons lost equal the electrons gained, and use
the factors as balancing coefficients.
Step 5) Complete the balancing by inspection, adding states of matter.
REDOX Balancing Using O. N.
+2
-1e-
+3
Fe+2(aq+ MnO4-(aq)+ H3O+(aq)
+7
+5 e-
Fe+3(aq)+ Mn+2(aq)+H2O(aq
+2
Multiply Fe+2 & Fe+3 by five to correct for the electrons
gained by the manganese.
5 Fe+2(aq) + MnO4-(aq) + H3O+(aq)
5 Fe+3(aq) + Mn+2(aq) + H2O(aq)
Make four water molecules from the 4 oxygen atoms from the MnO4- and
protons from the acid, and this will require 8 protons, or 8 hydronium
ions. This will give a total of 12 water molecules formed.
5 Fe+2(aq) + MnO4-(aq) + 8 H3O+(aq)
5 Fe+3(aq) + Mn+2(aq) + 12 H2O(aq)
5 FeCl2 + KMnO4 + 8HCl=5 FeCl3 + MnCl2 +KCl + 4 H2O
Balancing Redox Equations by half-reactions method
•Step 1 Write two unbalanced half-equations, one for the species that is oxidized
and its product and one for the species that is reduced and its product
•Step 2 Insert coefficients to make the numbers of atoms of all elements except
oxygen and hydrogen equal on the two sides of each half-equation.
•Step 3 Balance oxygen by adding H2O to the side deficient in O in each halfequation
•Step 4 Balance hydrogen.
–For half-reaction in acidic solution, add H+ on to the side deficient in hydrogen.
–For a half-reaction in basic solution, add H2O to the side that is deficient in
hydrogen and an equal amount of OH- to the other side
•Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each
half-reaction.
•Step 6 Multiply the two half-equations by numbers chosen to make the number of
electrons given off by the oxidation equal to the number taken up by the reduction.
Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or
H2O appears on both sides of the final equation, cancel out the duplication
•Step 7 Check for balance
Example: Balancing Redox Reactions
Dithionate ion reacting with chlorous
acid in aqueous acidic conditions
S2O62-(aq) +HClO2(aq) --> SO42-(aq) + Cl2(g)
Step 1 Write two unbalanced half-equations, one for the
species that is oxidized and its product and one for the
species that is reduced and its product
5+
6+
S2O62- --> SO42OXIDATION
3+
0
HClO2 --> Cl2
REDUCTION
Step 2 Insert coefficients to make the numbers
of atoms of all elements except oxygen and
hydrogen equal on the two sides of each
half-equation.
S2O62- -->
2 SO42-
2 HClO2 --> Cl2
Step 3 Balance oxygen by adding H2O to
the side deficient in O in each halfequation
2 H2O +
(Need 2 Os)
S2O62- --> 2 SO42(6Os)
2 HClO2 --> Cl2
(4Os)
(8Os)
+ 4 H2O
(Need 4 Os)
Step 4 Balance hydrogen. For half-reaction in acidic
solution, add H+ on to the side deficient in hydrogen.
For a half-reaction in basic solution, add H2O to the
side that is deficient in hydrogen and an equal amount
of OH- to the other side
2H2O + S2O62- → 2SO42- + 4H+
(4 Hs)
Need 4 Hs
6H+ +
Need 6 Hs
2HClO2 → Cl2 + 4H2O
(2 Hs)
(8 Hs)
Step 5 Balance charge by inserting e(electrons) as a reactant or product in each
half-reaction.
2H2O + S2O62- --> 2SO42- + 4H+ + 2 eOxidation reaction, electrons are lost on the reactant
side (gained on the product side)
6H+ + 2HClO2 + 6 e- --> Cl2 + 4H2O
Reduction reaction, electrons are gained on the reactant side
Step 6 Multiply the two half-equations by numbers
chosen to make the number of electrons given off
by the oxidation equal to the number taken up by
the reduction.
Oxidation Reaction has lost 2 electrons
Reduction reaction has gained 6 electrons
∴ Multiply the oxidation
reaction by 3.
This will give 6 electrons
on both sides of the
reaction
6H2O + 3S2O62- → 6SO42- + 12H+ + 6e-
Step 6 Then add the two half-equations and cancel out the
electrons. If H+ ion, OH- ion, or H2O appears on both
sides of the final equation, cancel out the duplication
6H2O + 3S2O62- --> 6SO42- + 12H+ + 6e+
6H+ + 2HClO2 + 6e- --> Cl2 + 4H2O
2H2O + 3S2O62- + 2HClO2 --> 6SO42- + 6H+ + Cl2
Step 7 Check for balance
2H2O + 3S2O62- + 2HClO2 --> 6SO42- + 6H+ + Cl2
Left side
Right side
H
6
H
6
O
24
O
24
S
6
S
6
Cl
2
Cl
2