Download Oxidation

The Columbia Disintegrates on Reentry (2003)

Reading for NEXT Monday


HOMEWORK – DUE Monday 2/13/17





BW 5b (Bookwork): CH 5 #'s 43, 46, 48, 50, 53, 54, 56, 58, 59, 62, 70, 73, 96, 116, 118
WS 7 (Worksheet): (from course website)
Lab Today/Tomorrow – EXP 5


BW 5a (Bookwork): CH 5 #'s 1, 14, 16, 20-24 all, 27, 32-37 all, 95, 103
WS 6 (Worksheet): (from course website)
HOMEWORK – DUE Wednesday 2/22/17


BW 4b (Bookwork): CH 4 #'s 42, 44, 46, 56, 62, 66, 78-81 all, 83-89 odd, 90, 93, 95, 104-112 even, 156
WS 5 (Worksheet): (from course website)
HOMEWORK – DUE Wednesday 2/15/17


Chapter 5 sections 4-6
Prelab!!!
Lab Monday/Tuesday

EXP 5 continued
Oxidation and Reduction

Oxidation is the process that occurs when






the oxidation number increases (gets more positive)
an element loses electrons
a compound adds bonds to oxygen
a compound loses bonds to hydrogen
a half–reaction has electrons as products
Reduction is the process that occurs when





the oxidation number of an element decreases (gets more negative)
an element gains electrons
a compound loses bonds to oxygen
a compound gains bonds to hydrogen
a half–reaction has electrons as reactants
Types of Reactions
Type
Look for
Generic
Redox?
Decomposition
one reactant
AB+C+…
depends
CaCO3(s)


CaO(s) + CO2(g)
2 KClO3(s)


2 KCl(s) + 3 O2(g)
Type
Look for
Generic
Redox?
Combination
multiple reactants/one product
reactants typically elements
A+B+…C
YES
3 Mg(s) + N2(g)  Mg3N2(s)
2 Na(s) + Se(s)  Na2Se(s)
Types of Reactions
Type
Look for
Combustion
Reactant with only C,
H, and sometimes O
2 CH4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(g)
Generic
Redox?
A + O2(g)  CO2(g) + H2O(g)
YES
C2H6O(l) + 3 O2(g)  2 CO2(g) + 3 H2O(g)
Type
Look for
Generic
Redox?
Single replacement
an element reacting with
a compound (or ion)*
A + BX  B + AX
YES
3 Mg(s) + 2 FeCl3(aq)  3 MgCl2(aq) + 2 Fe(s)
Al(NO3)3 (s) + Co(s)  N.R.
*check activity series
Types of Reactions
Type
Look for
Generic
Redox?
Double replacement
typically 2 aqueous
solutions reacting
AX + BY  AY + BX
NO
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Type
Look for
Neutralization
acid and base
HF(aq) + LiOH(aq)  HOH(l) + LiF(aq)
3 K2S(aq) + 2 CrCl3(aq)  6 KCl(aq) + Cr2S3(s)
Generic
HA(aq) + BOH(?)  HOH(l) + BA(?)
Redox?
NO
Ni(OH)2(s) + H2SO4(aq)  NiSO4(aq) + 2 HOH(l)
Acids and Bases

An acid is any substance that releases hydrogen ions, H+, into water.

Acids are in water and therefor need and (aq)

A base is a substance that releases hydroxide ions, OH –, into water.

An acid and a base react with each other in a neutralization reaction.

When an acid and a base react, water and a salt are produced.
Acids and Bases
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H2SO4(aq) + 2 KOH(aq)  K2SO4(aq) + 2 H2O(l)
2 H3PO4(aq) + 3 Ba(OH)2(aq)  Ba3(PO4)2(s) + 6 H2O(l)
HNO3(aq) + NH3(aq)  NH4NO3(aq)
2 HBr(aq) + K2CO3(aq)  2 KBr(aq) + H2CO3(aq)
+ H2O(l) + CO2(g)
HI(aq) + NaHCO3(aq)  NaI(aq) + H2CO3(aq)
+ H2O(l) + CO2(g)
Oxidation Numbers/States
A Bit of Review
Monatomic ions

Some charges can be predicted based on group number
2+
Be
A Bit of Review
Variable Oxidation State metals

Charge can NOT be predicted based on periodic table
IA
1
VIIIA
18
(ALMOST) ANYTHING IN HERE CAN BE A VOS METAL
IIA
2
IIIA
13
IIIB
3
IVB
4
Ti
VB
5
VIB
6
VIIB
7
VIII
8
VIII
9
VIII
10
IB
11
IIB
12
Cr
Mn
Fe
Co
Ni
Cu
Zn
IVA
14
VA
15
+2
+1
+2
Ag
Cd
Sn
Sb
Au
Hg
Pb
Bi
VIA
16
VIIA
17
A Bit of Review
Variable Oxidation State metals

Charge can NOT be predicted based on periodic table

Can assume multiple ion charges

non VOS metal –
potassium ion = K+
VOS metal –
lead ion = Pb2+ or Pb4+
The charge of VOS metals MUST be indicated in the
name with the use of roman numerals
Sn2+ = tin (II) ion
Sn4+ = tin (IV) ion
Co2(CO3)3
cobalt (III) carbonate
Total cation charges + total anion charges = 0
CO32–
Co
Co
CO32–
CO32–
2(Co)
+
3(–2)
x = 3 = oxidation state of Co
=0
Pb(CrO4)2
lead (IV) chromate
Total cation charges + total anion charges = 0
CrO42–
Pb
1(Pb)
CrO42–
+
2(–2)
x = 4 = oxidation state of Pb
=0
MnO4–
Total cation charges + total anion charges = 0
Total atom “A” charges + total atom “B” charges = charge
O2–
O2–
O2–
O2–
Mn
1(Mn)
+
4(–2)
x = 7 = oxidation state of Mn
= –1
Cr2O7–2
Total atom “A” charges + total atom “B” charges = charge
O2– 2–
O
O2– 2–
O
O2– 2–
O
O2–
Cr
Cr
2(Cr)
+
7(–2)
x = 6 = oxidation state of Cr
= –2
Rules for Assigning Oxidation Numbers
1) The sum of the oxidation numbers will always equal the particle’s charge
Co2(CO3)3
2(Co) + 3(–2) = 0
CO32–
1(C) + 3(–2) = –2
MnO4–
1(Mn) + –8 = –1
P2O7–4
2(P) + –14 = –4
Co = +3
C = +4
Mn = +7
P = +5
Rules for Assigning Oxidation Numbers
1) The sum of the oxidation numbers will always equal the particle’s charge
2) The oxidation number for a neutral atom is always zero
2 K(s) + 2 H2O(l)  2 KOH(aq) + H2(g)
Oxidation states are zero
Rules for Assigning Oxidation Numbers
1) The sum of the oxidation numbers will always equal the particle’s charge
2) The oxidation number for a neutral atom is always zero
3) Oxidation numbers for non–VOS metals depend on their group
Li = +1
Ba = +2
Al = +3
Rules for Assigning Oxidation Numbers
1) The sum of the oxidation numbers will always equal the particle’s charge
2) The oxidation number for a neutral atom is always zero
3) Oxidation numbers for non–VOS metals depend on their group
4) Oxidation numbers for VOS metals are found based on anion
5) Oxidation numbers for nonmetals are typically found based on their group
P = –3
Se = –2
Cl = –1
Rules for Assigning Oxidation Numbers
5) Oxidation numbers for nonmetals are typically found based on their group
typically?!?!
Fluorine = –1 always!
OF2 (F = –1, O = +2)
Hydrogen = +1 unless paired with a metal NaH (Na = +1, H = –1)
Oxygen = –2 unless peroxide
BaO2 (Ba = +2, O = –1)
Other nonmetals: the element closest to fluorine on the PT gets to keep its “usual” O.S.
SCl4
ClO3–1
P2S5
O.S. chlorine = –1
O.S. oxygen = –2
O.S. sulfur = –2
O.S. sulfur = +4
O.S. chlorine = +5
O.S. phosphorus = +5
Practice!!
Assign oxidation numbers to all atoms in each of the following:
a) H2CO
b) S2O32–
c) NH4+
d) NO3–
e) Br2
f) Ca3(PO4)2
g) 2 K(s) + 2 H2O(l)  2 KOH(aq) + H2(g)
Oxidation and Reduction

Oxidation is the process that occurs when






the oxidation number increases (gets more positive)
an element loses electrons
a compound adds bonds to oxygen
a compound loses bonds to hydrogen
a half–reaction has electrons as products
Reduction is the process that occurs when





the oxidation number of an element decreases (gets more negative)
an element gains electrons
a compound loses bonds to oxygen
a compound gains bonds to hydrogen
a half–reaction has electrons as reactants
Redox Reactions
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
Ox #
0
+1
+2
0
Lost Electrons OXIDIZED
reducing agent
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
Gain Electrons REDUCED
oxidizing agent
Redox Reactions
2 I–(aq) + Br2(aq)  I2(aq) + 2 Br–(aq)
Ox #
–1
0
0
–1
Lost Electrons OXIDIZED
reducing agent
2 I–(aq) + Br2(aq)  I2(aq) + 2 Br–(aq)
Gain Electrons REDUCED
oxidizing agent
Redox Reactions
3 Cl2(g) + 2 Al(s)  2 AlCl3(s)
Ox #
0
+3 –1
0
Gain Electrons REDUCED
oxidizing agent
3 Cl2(g) + 2 Al(s)  2 AlCl3(s)
reducing agent
Lost Electrons OXIDIZED
Redox Reactions
Cr2O72–(aq) + IO–(aq)  IO3–(aq) + Cr3+(aq)
Ox # +6
+1
+5
+3
Lost Electrons OXIDIZED
reducing agent
Cr2O72–(aq) + IO–(aq)  IO3–(aq) + Cr3+(aq)
Gain Electrons REDUCED
oxidizing agent
Half–Reactions

We can split a RedOx reaction into two separate
half–reactions (or half–cell), each involving only
an oxidation OR a reduction.
Oxidation half–cells have electrons as products (OIL)
 Reduction half–cells have electrons as reactants (RIG)

3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+
0
–1
+1 –2
–1 +5 –2
+1
Reduction: Cl2 + 2 e− → 2 Cl−
Oxidation: I− → IO3− + 6 e−
Balancing Redox Reactions – Half–Cells
1. Write oxidation and reduction half–cells
a)
Assign oxidation numbers to determine what was oxidized and what was reduced.
Lost Electrons OXIDIZED
Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq)
+2
+7
+3
+2
Gain Electrons REDUCED
ox.:
5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− )
red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)
Balancing Redox Reactions – Half–Cells
1. Write oxidation and reduction half–cells
a)
Assign oxidation numbers to determine what was oxidized and what was reduced.
2. Balance half–cells by mass.
a)
First balance elements other than H and O.
Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq)
+2
ox.:
+7
+3
+2
5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− )
red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)
Balancing Redox Reactions – Half–Cells
1. Write oxidation and reduction half–cells
a)
Assign oxidation numbers to determine what was oxidized and what was reduced.
2. Balance half–cells by mass.
a) First balance elements other than H and O.
b) Add H2O where O is needed.
Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq)
+2
ox.:
+7
+3
+2
5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− )
red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)
Balancing Redox Reactions – Half–Cells
1. Write oxidation and reduction half–cells
a)
Assign oxidation numbers to determine what was oxidized and what was reduced.
2. Balance half–cells by mass.
a) First balance elements other than H and O.
b) Add H2O where O is needed.
c) Add H+ where H is needed.
Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq)
+2
ox.:
+7
+3
+2
5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− )
red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)