Transcript Chapter 17 Oxidation and Reduction

Chapter 17 Oxidation and Reduction
Notes One Unit Six
•Redox
•Oxidation Numbers
•Identifying what is
oxidized and what is
reduced
Oxidation Reduction Chemisty: Redox Chemistry
Oxidation and Reduction reactions always
take place simultaneously.
Loss of electrons – oxidation (Increase in Oxidation
Number)
Ex:Na ------> Na+1 + e-1
Gain of electrons - reduction ( Decrease in Oxidation
Number)
Cl2 + 2 e-1 ------> 2 Cl-1
Redox:
Reduction occurs when an atom gains one or more
electrons.
Ex: O + 2e-1  O2Oxidation Foccurs when an atom or ion loses one
or more electrons.
Ex: Fe
 Fe+3 + 3e-1
LEO goes GER
Copper metal reacts with silver nitrate to form silver
metal and copper nitrate:
Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.
Redox reactions involve electron transfer:
Lose e - =Oxidation
Cu (s) + 2 Ag+1 (aq)
Cu
+2
Gain e - =Reduction
(aq) + 2 Ag(s)
Oxidation occurs when a molecule
does any of the following:
Loses electrons
Loses hydrogen
Gains oxygen
If a molecule undergoes oxidation, it has
been oxidized and it is the reducing agent
(aka reductant).
Reduction occurs when a molecule does any
of the following:
Gains electrons
Gains hydrogen
Loses oxygen
If a molecule undergoes
reduction, it has been reduced
and it is the oxidizing agent (aka
oxidant).
zinc is being oxidized while the
copper is being reduced. Why?
Redox
Burning:
C6H12O6 + 6O2 6CO2 + 6H2O+ Heat
Rusting Iron:
4Fe + 3O22Fe2O3 + Heat
Oxidation - Loss of e-1.
Na(s)Na+1 +1e-1
Reduction - Gain of e-1.
Cl2+ 2e-1 2Cl-1
Number line (Oxidation…Left or right?)
Oxidation Numbers
•
•
•
•
Rules for Assigning Oxidation States
The oxidation state of an atom in an uncombined element is 0.
The oxidation state of a monatomic ion is the same as its charge.
Oxygen is assigned an oxidation state of –2 in most of its covalent
compounds. Important exception: peroxides (compounds containing the O2
2- group), in which each oxygen is assigned an oxidation state of –1)
•
In its covalent compounds with nonmetals, hydrogen is assigned an
oxidation state of +1
•
For a compound, sum total of Oxidation Numbers is zero.
•
For an ionic species (like a polyatomic ion), the sum of the oxidation states
must equal the overall charge on that ion.
Key Elements
•
•
•
•
•
•
•
•
(99%) H+1
H-1
(99%)O-2
O-1
(Always) Li+1, Na+1, K+1, Rb+1, Cs+1, Fr+1
(Always) Be+2, Mg+2, Ca+2, Ba+2, Sr+2, Ra+2
(Always) Al+3
(with only a metal) F-1, Cl-1, Br-1, I-1
(NO3-1) ion is always +5
(SO4-2) ion is always +6
Finding Oxidation Numbers
The sum of the oxidation numbers must
zero for a compound.
be equal to _____
Find Ox#’s for H2O?
+1 -2
H2O
2 (H)+ 1(O) = Zero
2 (+1)+ 1(-2) = Zero
Find Ox#’s for H3PO4?
+1 +5 -2
H3PO4 3 (H)+ 1 (P)+ 4(O) = Zero
3(+1)+1(+5)+ 4(-2) = Zero
Finding Oxidation #’s for Compounds
+1 -2
H2O
+1+5 -2
H3PO4
+1+5-2
HNO3
+1+6 -2
H2SO4
+1 +6 -2
Hg2SO4
+1 +6 -2
Na2Cr2O7
+1+4-2
H2CO3
-3 +1 +4-2
(NH4)2CO3
+2 +5 -2
Ca3(AsO4)2
+3 +6-2
Fe2(SO4)3
+2 +7 -2
Ba(ClO4)2
+3 +4 -2
Al2(CO3)3
Identifying OX, RD, SI
Species
• Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
• Oxidation = loss of electrons. The species becomes
more positive in charge. For example, Ca0  Ca+2,
so Ca0 is the species that is oxidized.
• Reduction = gain of electrons. The species becomes
more negative in charge. For example, H+1  H0, so
the H+1 is the species that is reduced.
• Spectator Ion = no change in charge. The species
does not gain or lose any electrons. For example, Cl1  Cl-1, so the Cl-1 is the spectator ion.
Oxidizing Agent and Reducing Agent:
Oxidizing agent gets reduced itself and reducing
agent gets oxidized itself, so a strong oxidizing agent
should have a great tendency to accept e and a strong
reducing agent should be willing to lose e easily. What
are strong oxidizing agents- metals or non metals?
Why?
Spectator Ions are ions that do NOT change their
oxidation number from the reactant side of a RXN to
the product side of a RXN. They are just “hanging out”.
Notes Two Unit Six
•
•
•
•
Activity Series of Metals
Balancing by Redox
Electrolysis
Lab A-Electrolysis
The activity series of metals is an empirical tool used to predict products in
displacement reactions and reactivity of metals with water and acids in
replacement reactions and ore extraction. It can be used to predict the
products in similar reactions involving a different metal.
The activity series is a chart of metals listed in order of declining relative
reactivity. The top metals are more reactive than the metals on the bottom. For
example, both magnesium and zinc can react with hydrogen ions to displace
H2 from a solution by the reactions:
Mg(s) + 2 H+(aq) → H2(g) + Mg2+(aq)
Zn(s) + 2 H+(aq) → H2(g) + Zn2+(aq)
Both metals react with the hydrogen ions, but magnesium metal can also
displace zinc ions in solution by the reaction:
Mg(s) + Zn2+ → Zn(s) + Mg2+
Reduction: Cu+2(aq) + 2 e-  Cu(s)
The Cation becomes a solid metal (the + charge GAINS
ELECTRONS to become a zero charge.
Oxidation: Cu(s)  Cu+2(aq) + 2 eThe metal becomes a cation (the zero charge metal LOSES
ELECTRONS
To become a + charge.
Redox: Oxidation Reduction Reaction
3Cu+2 (aq) + 2Fe (s)  3 Cu (s) + 2Fe+3 (aq)
6 e-
The paired reduction and oxidation
Electrons transfer from the metal to the cation if the metal
Is above (ie higher) on the Activity Series Chart in your
packet.
Cu+2 (aq) + Mg (s)  Cu (s) + Mg+2 (aq)
Zn+2 (aq) + Ag (s)  No RXN
Balancing By Redox Example One
#1. Find the oxidation #’s.
#2. ID the element (i) oxidized and (ii) reduced.
#3. Find # of electrons lost or gained.
#4. Cross-multiply.
#5. Balance using whole # ratios.
#6. Find whole number coefficients.
+1 -2
0
+1 +6 -2
+1 +5 -2
5e-1 lost X8
+1 -2
12 H2O + 2 P4+ 5 H2SO4 8 H3PO4+ 5 H2S
8e-1 Gained X5
Multiply by 1.
12H2O + 2P4+ 5H2SO4 8H3PO4+ 5H2S
Balancing By Redox Example Two
#1. Find the oxidation #’s.
#2. ID the element (i) oxidized and (ii) reduced.
#3. Find # of electrons lost or gained.
#4. Cross-multiply.
#5. Balance using whole # ratios.
#6. Find whole number coefficients.
+1 +5 -2
0
0
5e-1 Gained X3
+1 -2
+1 +3 -2
3 K3PO4 +5/2Cl23/4P4+ 2 K2O+ 5 KClO2
3e-1 Lost X5
Multiply by 4.
12K3PO4 +10Cl2 3P4+ 8K2O+ 20KClO2
Applications of
Oxidation-Reduction
Reactions
Batteries
Alkaline Batteries
Hydrogen Fuel Cells
Corrosion and…
…Corrosion Prevention
How to use the ½ Reaction Resource:
Notice they are all written as Reductions (gaining of
electrons)
1. Find the highest one on the left hand side and write it in
the forward direction.
2. Find the lowest one on the right and write it backwards as
an oxidation ,along with changing the sign of the voltage.
3. Make sure to balance electrons lost and gained (you DO
NOT MULTIPLY the voltages !!!)
4. So you will have two ½ cell reactions and you can cancel
electrons and write the WHOLE CELL RXN.
Let’s Practice????? X
Stronger 1/2O2(g)+2H+(pH=7)+2e-H2O
+1.23
-2BrBr
(l)+2e
+1.06
2
Oxidizing
NO3-+4H++3e-NO(g)+2H2O
+0.96
Agent1/2O (g)+2H+(pH=7)+2e-H O
+0.82
2
2
Ag++e- Ag(l)
+0.80
Hg2++2e- Hg(l)
+0.78
NO3-+2H++e-NO2(g)+H2O
+0.78
3++e-Fe2+
Fe
+0.77
Gains e I2(s)+2e 2I
+0.53
Loses e +0.52
Cu++e-Cu(s)
Cu2++2e-Cu(s)
+0.34
SO42-+4H++2e-SO2(g)
+0.17
Sn4++2e-Sn(s)
+0.15
2H++2e-H2(g)
0.00
Pb2++2e-Pb(s)
-0.13
-0.14
Sn2++2e-Sn(s)
Ni2++2e-Ni(s)
-0.25
2+
-0.28
Co +2e Co(s)
-0.41
2H+(pH=7)+2e-H2(g)
-0.44
Fe2++2e-Fe(s)
Weaker
Cr3++3e-Cr(s)
-0.74
Oxidizing
2+
-0.76
Zn +2e 
Zn(s)2H2O+2e 2OH +H2(g)
-0.84
Agent
Weaker
Reducing
Agent
Stronger
Reducing
Agent
Electrolysis
Electrolysis -electric current
produced chemical reaction
Cathode -reduction
Anode -oxidation
Which is the…
Cathode=?
Anode=?
Electron flow?
Mass Gain=?
Cathode(spoon)
Mass Loss=?
Copper
e-1
 anode
Cathode
Electrolysis Lab- Demo KI (aq)
What is available to react?
+1
K
-1
I
Cathode Reaction highest reaction on left.
Anode Reaction lowest reaction on right.

H2O
1/2O2(g)+2H+(pH=7)+2e-H2O
Br2(l)+2e-2BrNO3-+4H++3e-NO(g)+2H2O
-H O
K+1 I-1 H2O 1/2O2(g)+2H++(pH=7)+2e
2
+e Ag(l)
Cath: highest on left. HgAg
2++2e-Hg(l)

NO3-+2H++e-NO2(g)+H2O
Fe3++e-Fe2+
-2II
(s)+2e
We see pink.
2
We sawCu
bubbles
++e-Cu(s)
An: lowest on right.
2++2e-Cu(s)
Cu

2I-I2(s)+2e- SO42-+4H++2e-SO2(g)
Sn4++2e-Sn(s)
2H++2e-H2(g)
We see brown:I2(s)
Pb2++2e-Pb(s)
Sn2++2e-Sn(s)
Ni2++2e-Ni(s)
Co2++2e-Co(s)
2H+(pH=7)+2e-H2(g)
Fe2++2e-Fe(s)
Cr3++3e-Cr(s)
Zn2++2e-Zn(s)
-2OH-+H (g)
2H2O+2e
2
K++e-K(s)
KI(aq)
+1.23
+1.06
+0.96
+0.82
+0.80
+0.78
+0.78
+0.77
+0.53
+0.52
+0.34
+0.17
+0.15
0.00
-0.13
-0.14
-0.25
-0.28
-0.41
-0.44
-0.74
-0.76
-0.84
-2.92
1/2O2(g)+2H+(pH=7)+2e-H2O
CuSO4(aq)
Br2(l)+2e-2BrNO3-+4H++3e-NO(g)+2H2O
1/2O2(g)+2H+(pH=7)+2e-H2O
++e-Ag(l)
+2
-2
Ag
Cu SO4 H2O
Hg2++2e-Hg(l)
Cath: highest(left) NO -+2H++e-NO (g)+H O
3
2
2
3+
2+

Fe +e Fe
I2(s)+2e-2I-Cu(s)
Cu++e
We saw copper on the pencil
tip!
Cu2++2e-Cu(s)
An: lowest(right) SO42-+4H++2e-SO2(g)

4++2e-Sn(s)
Sn
+
H2O1/2O2(g)+2H (pH=7)+2e
2H++2e-H2(g)
Pb2++2e-Pb(s)
We saw bubbles!
Sn2++2e-Sn(s)
Ni2++2e-Ni(s)
Co2++2e-Co(s)
2H+(pH=7)+2e-H2(g)
Fe2++2e-Fe(s)
Cr3++3e-Cr(s)
Zn2++2e-Zn(s)
-2OH-+H (g)
2H2O+2e
2
Na++e-Na(s)
+1.23
+1.06
+0.96
+0.82
+0.80
+0.78
+0.78
+0.77
+0.53
+0.52
+0.34
+0.17
+0.15
0.00
-0.13
-0.14
-0.25
-0.28
-0.41
-0.44
-0.74
-0.76
-0.84
-2.71
1/2O2(g)+2H+(pH=7)+2e-H2O
Na2SO4(aq)
Br2(l)+2e-2BrNO3-+4H++3e-NO(g)+2H2O
1/2O2(g)+2H+(pH=7)+2e-H2O
++e-Ag(l)
+1
-2
Ag
Na SO4 H2O
Hg2++2e-Hg(l)
Cath: highest(left) NO -+2H++e-NO (g)+H O
3
2
2

3+
2+
Fe +e Fe
I2(s)+2e-2I++e-Cu(s)
Cu
We saw pink. We saw bubbles
Cu2++2e-Cu(s)
An: lowest(right) SO42-+4H++2e-SO2(g)

4++2e-Sn(s)
Sn
+
H2O1/2O2(g)+2H (pH=7)+2e
2H++2e-H2(g)
Pb2++2e-Pb(s)
We saw bubbles.
Sn2++2e-Sn(s)
Ni2++2e-Ni(s)
Co2++2e-Co(s)
2H+(pH=7)+2e-H2(g)
Fe2++2e-Fe(s)
Cr3++3e-Cr(s)
Zn2++2e-Zn(s)
-2OH-+H (g)
2H2O+2e
2
Na++e-Na(s)
+1.23
+1.06
+0.96
+0.82
+0.80
+0.78
+0.78
+0.77
+0.53
+0.52
+0.34
+0.17
+0.15
0.00
-0.13
-0.14
-0.25
-0.28
-0.41
-0.44
-0.74
-0.76
-0.84
-2.71
Electrolysis lab A
Notes Three Unit Six
• Electrolysis Lab Results
• Concept Check Assignment
• Quiz-Balancing/Electrolysis
Cl2(g)2e-2Cl-1
1/2O2(g)+2H+(pH=7)+2e-H2O
-2BrBr
(l)+2e
+2
-2
2
Cu
SO4 H2O NO -+4H++3e-NO(g)+2H O
3
2
+
H2O
Na2SO4(aq) 1/2O2(g)+2H+ (pH=7)+2e
Ag +e-Ag(l)
Hg2++2e-Hg(l)
Na+1 SO4-2 H2O NO3-+2H++e--NO
2(g)+H2O
I2(s)+2e 2I
HCl(aq)
Cu++e-Cu(s)
Cu2++2e-Cu(s)
2-+4H++2e-SO (g)
SO
4
2
H+1 Cl-1 H2O
4+
Sn +2e Sn(s)
2H++2e-H2(g)
Mg(s) Cu(s)
Pb2++2e-Pb(s)
Sn2++2e-Sn(s)
Cath: highest(left)
Ni2++2e-Ni(s)

Co2++2e-Co(s)
2H+(pH=7)+2e-H2(g)
An: lowest(right)
Fe2++2e-Fe(s)
3++3e-Cr(s)

Cr
-1
+2
Mg(s)2e +Mg
-2OH-+H (g)
+2.37
2H2O+2e
2
+2
Mg +e Mg(s)
+2
+2
Cu + Mg(s)Cu(s)+ Mg Na+1+2.71
+e-Na(s)
CuSO4(aq)
+1.36
+1.23
+1.06
+0.96
+0.82
+0.80
+0.78
+0.78
+0.53
+0.52
+0.34
+0.17
+0.15
0.00
-0.13
-0.14
-0.25
-0.28
-0.41
-0.44
-0.74
-0.84
-2.37
-2.71
Lab C Voltaic Cell
e-1
Mg(s)
H+1
Cl-1
SO4-2
Na+1
Anode
Mg(s)2e +Mg
+
-
rxn
-1
+2
Cathode rxn
Cu(s)
Cu+2
SO4-2
Cu2++2e-Cu(s)
Spectator Ions
SO4-2 H+1 Na+1 Cl-1
Current Flow
Notes Four Unit Six
• Faraday’s Law Lab B
• This as a chemical process that uses
electricity to produce industrial quantities
of specific chemicals.
Application of Faraday’s law
F = 96500 C/mol e-)
Axs=C
Faraday’s Law: Lab B
Fe(s)
Cathode?
Anode?
e-1
Fe(s)
Fe+3
3e-1
Fe+3
NO3-1
NO3-1
NO3-1
F = 96500 C/moleAmp x second = C
Lose Mass?
Gain Mass?
Faraday’s Law Calculation One
How many grams of Gold will be plated, using
a current of 3.0 amps for 1.5 hours?
1. Balanced Equation Au+3+3e-1Au(s)
2. Calculate Coulombs.
60
min
60
Sec
3.0 amp x 1.5 Hour x
x
=16000C
1 hour 1 min
3. Calculate moles e-1.
-1
1mole
e
16000C x
= 0.17 mol e-1
96500C
4. Calculate moles of substance.
1 m Au(s)
-1
0.17 mol e x
=0.056mol Au(s)
-1
3mol e
5. Calculate grams.
197.0gAu
0.056mol Au(s)x
=
11g
Au
1mol Au
Faraday’s Law Calculation Two
How many grams of Silver will be plated,
using a current of 2.0 amps for 45 minutes?
1. Balanced Equation Ag+1+1e-1Ag(s)
2. Calculate Coulombs.
60
Sec
2.0 amp x 45 Hour x
=5400C
1 min
3. Calculate moles e-1.
-1
1mole
e
5400C x
= 0.056 mol e-1
96500C
4. Calculate moles of substance.
1 m Ag(s)
-1
0.056 mol e x
=0.056mol Au(s)
-1
1mol e
5. Calculate grams.
107.9gAg
0.05596mol Au(s)x
1mol Ag = 6.0 g Ag
Salt Bridge
e-1 ?
1.08
Na+1
-2
SO
+1
4
Na
Cu(s)
Cu+2
-2
SO
4
Cu+2
-2
SO
+2
4
Cu
-2
SO
+2
4
Cu
SO4-2
Cu+2
SO4-2
Cu+2 SO4-2
Cathode rxn: +0.34
6 -1
+2
3Cu + 2e 3Cu x3
Anode rxn:
+0.74
6
2Cr 2 Cr+3+3e-1
x2
Overall rxn:
2Cr+3Cu+22Cr+3+3Cu
emf= 1.08volts
Cr(s)
SO4-2+3 3e-1 Salt Bridge
Cr
Cr+3
3e-1
Cr+3
SO4-2
Cr+3
SO4-2
Ion-Electron Method for Balancing
Ion-Electron Method for Balancing
#1. Separate Half-rxn.
#2. Bal Non-O Elem.
#3. + H2O.
#4. + H+1 .
#5. + e-1 to bal +/-.
UO2+2 + I2  U+4 + IO3-1 (Acid)
5X 1UO2+2+ 4 H+1 + 2 e-1 1U+4 + 2 H2O
1X 1 I2 + 6 H2O 2IO3-1 + 12 H+1 + 10 e-1
8
4
+2
5UO2 + 20 H+1 +10e-1 5U+4 + 10 H2O
1I2 + 6 H2O 2IO3-1 + 12 H+1 + 10 e-1
5UO2+2+ 8 H+1+ 1I2 2IO3-1 + 5 U+4+ 4 H2O
Ion-Electron Method for Balancing
#1. Separate Half-rxn.
#2. Bal Non-O Elem.
#3. + H2O.
#4. + H+1 .
#5. + e-1 to bal +/-.
IO3-1 + Ti+3 I2 + TiO2+1 (Acid)
1X 2 IO3-1 +12H+1+ 10 e-1 1I2 + 6 H2O
5X 1 Ti+3+ 2 H2O 1TiO2+1+ 4 H+1 + 2 e-1
2 IO3-1 + 12 H+1 +10e-1 1I2 + 6 H2O
8
4
5 Ti+3 +10H2O 5TiO2+1 + 20 H+1+ 10 e-1
2 IO3-1 + 4 H2O +5Ti+3 5 TiO2+1 + 1 I2 + 8 H+1