Transcript Oxidation/Reduction - Dallas High School

Oxidation/Reduction
Chapter 20
Two Types of Chemical Rxns
1. Exchange of Ions – no change in
charge/oxidation numbers
– Acid/Base Rxns
NaOH + HCl
Two Types of Chemical Rxns
–
Precipitation Rxns
Pb(NO3)2(aq) + KI(aq)
–
Dissolving Rxns
CaCl2(s) 
Two Types of Chemical Rxns
2. Exchange of Electrons – changes in oxidation
numbers/charges
Fe(s) + CuSO4(aq)  FeSO4(aq) + Cu(s)
Remove spectator ions
Fe + Cu2+  Fe2+ + Cu
Protons
Electrons
Review of Oxidation Numbers
Oxidation numbers – the charge on an ion or an
assigned charge on an atom.
Al Cl2 P4
Mg2+ Cl-
Review of Oxidation Numbers
Calculate the oxidation numbers for:
HClO
S8
Mn2O3
KMnO4
HSO4-
Cr3+
Fe2(SO4)3
SO32NO3-
Oxidation
1. Classical Definition –addition of oxygen
2. Modern Definition – an increase in oxidation
number
Fe + O2  Fe2O3
CO + O2  CO2
CH3CH2OH  CH3CHO  CH3COOH
Oxidation
Fe + O2 
(limited oxygen)
Fe + O2 
(excess oxygen)
Oxidation
C + O2 
(limited oxygen)
C + O2 
(excess oxygen)
Reduction
1. Classical –addition of hydrogen
2. Modern –decrease (reduction) in oxidation
number
N2 + 3H2  2NH3
(Haber process)
R-C=C-R
+ H2 
| |
H H
(unsaturated fat)
(saturated fat)
Oxidizing/Reducing Agents
Oxidation and Reduction always occur together.
Oxidizing Agents
– Get reduced
– Gain electrons
Reducing Agents
– Get oxidized
– Lose electrons
Oxidizing/Reducing Agents
0
Cu
Got oxidized, reducing agent
+
O2
0

+2
CuO
-2
Got reduced, oxidizing agent
Identify the Oxidizing/Reducing Agents in the
following (Calculate the ox. numbers also).
Cu + S8  Cu2S
H2 + O2  H2O
Cu + AgNO3  Cu(NO3)2 + Ag
H2O + Al + MnO4-  Al(OH)4- + MnO2
Identify the Oxidizing/Reducing Agents in the
following (Calculate the ox. numbers also).
Al + O2  Al2O3
Li + N2  Li3N
Cu(NO3)2 + Fe  Fe(NO3)3 + Cu
KMnO4 + FeSO4 Fe2(SO4)3 + Mn + K+
Balancing Redox Reactions
Half-Reaction Method
• Break eqn into oxidation half and reduction
half
• Easy Examples:
– Al + Fe2+  Fe + Al3+
– Cu + Zn2+  Cu2+ + Zn
– Mg + Na+  Mg2+ + Na
Balancing Redox Reactions
What’s really happening:
Cu2+ + Zn  Cu + Zn2+
Balancing Redox Reactions
Steps for more complicated examples
1. Balance all atoms except H and O
2. Balance charge with electrons
3. Balance O with water
4. Balance H with H+
===================================
5. (Add OH- to make water in basic solutions)
Balancing Redox Reactions
Example 1:
MnO4- + C2O42-  Mn2+ + CO2
1. Separate into half reactions
MnO4-  Mn2+
C2O42-  CO2
Balancing Redox Reactions
MnO4-  Mn2+
+7
+2
5e- + MnO4-  Mn2+
5e- + MnO4-  Mn2+ + 4H2O
8H+ + 5e- + MnO4-  Mn2+ + 4H2O
Balancing Redox Reactions
C2O42-  CO2
C2O42-  2CO2
+3
+4 (1 e- per carbon)
C2O42-  2CO2 + 2e-
Balancing Redox Reactions
C2O42-  2CO2 + 2e8H+ + 5e- + MnO4-  Mn2+ + 4H2O
(X 5)
(X 2)
5C2O42-  10CO2 + 10e16H+ + 10e- + 2MnO4-  2Mn2+ + 8H2O
16H++5C2O42-+2MnO4-  2Mn2+ + 10CO2 + 8H2O
Balancing Redox Reactions (Acidic
Solutions)
Cr2O72- + Cl-  Cr3+ + Cl2
14 H+ + Cr2O72- + 6Cl-  2Cr3+ + 3Cl2 + 7H2O
Cu + NO3-  Cu2+ + NO2
Cu + 4H+ + 2NO3-  Cu2+ + 2NO2 + 2H2O
Mn2+ + BiO3-  Bi3+ + MnO414H+ + 2Mn2+ + 5BiO3-  5Bi3+ + 2MnO4- + 7H2O
Balancing Redox Reactions (Basic
Solutions)
Add OH- AT THE VERY END ONLY!!!!!
NO2- + Al  NH3 + Al(OH)45H2O + OH- + NO2- + 2Al  NH3 + 2Al(OH)4Cr(OH)3 + ClO-  CrO42- + Cl2
2Cr(OH)3 + 6ClO-  2CrO42- + 3Cl2 + 2OH- + 2H2O
Balance in Both Acidic and Basic Solutions
F- + MnO4-  MnO2 + F2
HNO2 + H2O2  O2 + NO
H is +1
F- + MnO4-  MnO2 + F2
8H+ + 6F- + 2MnO4-  2MnO2 + 3F2 + 4H2O
4H2O + 6F- + 2MnO4-  2MnO2 + 3F2 + 8OHHNO2 + H2O2  O2 + NO
2HNO2 + H2O2  O2 + 2NO + 2H2O
Voltaic (Galvanic) Cells
• Voltaic(Galvanic) Cells – redox reactions that
produce a voltage
– Spontaneous reactions (DG<0)
– Voltage of the cell (Eocell) is positive
– Batteries
• Electrolytic cells – redox reactions that must
have a current run through them.
– DG>0 and Eocell is negative.
– Often used to plate metals
Voltaic (Galvanic) Cells
History
• Galvani (died 1798)– uses static electricity to
move the muscles of dead frogs
• Volta (1800) – Created the first battery
Voltaic (Galvanic) Cells
Voltaic cell
1. Anode – Oxidation site
2. Cathode – Reduction site (RC cola)
3. Salt bridge – completes the circuit
Voltaic (Galvanic) Cells
• Cell Notation
Zn | Zn2+(aq) ||Cu2+(aq) | Cu
• Anode Zn
 Zn2+ + 2e• Cathode Cu2+ + 2e-  Cu
• Cell
Cu2+ + Zn  Cu + Zn2+
Voltaic (Galvanic) Cells
Hydrogen Electrode
1. Standard Electrode
2. Voltage(potential) = 0 Volts
2H+(aq) + 2e-  H2(g)
H2(g)  2H+(aq) + 2e-
0 volts
0 volts
3. Often used in electrodes (like pH)
Standard Reduction Potentials
• Rules
– Flipping an equation changes the sign of E
– Multiplying an equation does not change the
magnitude of E
Calculating Cell Potential
A cell is composed of copper metal and Cu2+(aq)
on one side, and zinc metal and Zn2+(aq) on
the other. Calculate the cell potential.
Zn2+ + 2e-  Zn
-0.76 V
Cu2+ + 2e-  Cu
+0.34 V
flip the zinc equation
Zn
 Zn2+ + 2e+0.76 V
Cu2+ + 2e-  Cu
+0.34 V
Zn + Cu2+  Zn2+ + Cu
+1.10 V
What is the cell emf of a cell made using Cu and
Cu2+ in one side and Al and Al3+ in the other?
Write the complete cell reaction.
ANS: 2.00 V
Calculate the standard emf for the following
reaction. Hint: break into half-reactions.
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
A voltaic cell is based on the following half
reactions.
In+(aq)  In3+(aq) + 2eBr2(l) + 2e-  2Br-(aq)
+1.06 V
If the overall cell voltage is 1.46 V, what is the
reduction potential for In3+?
Calculate the standard emf for the following
reaction.
Cr2O72- + 14H+ + 6I-  2Cr3+ + 3I2 + 7H2O
Two half reactions in a voltaic cell are:
Zn2+(aq) + 2e-  Zn(s)
Li+(aq) + e-  Li(s)
a) Calculate the cell emf.
b) Which is the anode? Which is the cathode?
c) Which electrode is consumed?
d) Which electrode is positive?
e) Sketch the cell, indicating electron flow.
Given the following half-reactions:
Pb2+ + 2e- Pb
Ni2+ + 2e-  Ni
a.
b.
c.
d.
e.
f.
Calculate the cell potential (Eo).
Label the cathode and anode.
Identify the oxidizing and reducing agents.
Which electrode is consumed?
Which electrode is plated?
Sketch the cell, indicating the direction of
electron flow.
Strengths of Oxidizing and
Reducing Agents
• larger the reduction potential, stronger the
oxidizing agent
– Wants to be reduced, can oxidize something else.
• lower the reduction potential, stronger the
reducing agent
– Would rather be oxidized
Stronger
Oxidizing
Agents
F2 + 2eCl2 + 2e-
 2F 2Cl-
+2.87 V
+1.36 V
.
.
.
.
.
Li+ + e-
 Li
-3.05V
Example 1
Which of the following is the strongest oxidzing
agent? Which is the strongest reducing
agent?
NO3Cr2O72Ag+
Which of the following is the strongest reducing
agent? Which is the strongest oxidizing
agent?
I2(s)
Fe(s)
Mn(s)
Can copper metal (Cu(s)) act as an oxidizing
agent?
Spontaneity
Voltaic Cells
•Positive emf
•Spontaneous
•Can produce
electric current
•Batteries
Electrolytic Cells
•Negative emf
•Not spontaneous
•Must “pump”
electricity in
•Electrolysis
Example 1
Are the following cells spontaneous as written?
a) Cu + 2H+  Cu2+ + H2
b) Cl2 + 2I-  2Cl- + I2
c) I2 + 5Cu2+ + 6H2O  2IO3- + 5Cu +
12H+
d) Hg2+ + 2I-  Hg + I2
EMF and DGo
DG = -nFE
n = number of electrons transferred
E = Cell emf
F = 96,500 J/V-mol (Faraday’s Constant)
Positive Voltage gives a negative DG (spont)
Calculate the cell potential and free energy
change for the following reaction:
4Ag + O2 + 4H+  4Ag+ + 2H2O
ANS: +0.43 V, -170 kJ/mol
Calculate DG and the EMF for the following
reaction. Also, calculate the K.
3Ni2+ + 2Cr(OH)3 + 10OH-  3Ni + 2CrO42- + 8H2O
ANS: +87 kJ/mol, -0.15 V, 6 X 10-16
EMF and K
DGo = -RTlnK
-nFEo = -RTlnK
lnK = nFEo
RT
log K = nEo
0.0592
(DGo = -nFEo)
(assume 298 K)
Example 1
Calculate DG, cell voltage and the equilibrium
constant for the following cell:
O2 + 4H+ + 4Fe2+  4Fe3+ + 2H2O
ANS: -177 kJ/mol, 0.459 V, 1 X 1031
Example 2
If the equilibrium constant for a particular
reaction is 1.2 X 10-10, calculate the cell
potential. Assume n = 2.
Concentration Cells: Nernst Equation
DG = DGo + RT lnQ
-nFE = -nFEo + RT lnQ
E = Eo -
RT lnQ
(assume 298 K)
nF
E = Eo 0.0592 log Q
n
Can adjust the voltage of any cell by changing
concentrations
Using the Nernst Eqn
Suppose in the following cell, the concentration
of Cu2+ is 5.0 M and the concentration of Zn2+
is 0.050 M. Calculate the cell voltage.
Zn(s) + Cu2+(aq)Zn2+(aq) + Cu(s)
+1.10 V
E = Eo -
0.0592 V log Q
n
E = 1.10V - 0.0592 V log [Cu][Zn2+]
2
[Cu2+][Zn]
E = 1.10V - 0.0592 V log [Zn2+]
2
[Cu2+]
E = 1.10V - 0.0592 V log [0.050]
2
[5.0]
E = +1.16 V
Example 1
Calculate the emf at 298 K generated by the
following cell (Eo= 0.79 V) where: [Cr2O72-]=
2.0 M, [H+ ]=1.0 M, [I-]=1.0 M and [Cr3+ ]= 1.0
X 10-5M.
Cr2O72- + 14H+ + 6I-  2Cr3+ + 3I2 + 7H2O
ANS: 0.89 V
Example 2
Calculate the emf at 298 K generated by the
following cell (Eo= 2.20 V) where: [Al3+]= 0.004
M and [I- ]=0.010 M.
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
ANS: +2.36 V
Example 3
If the voltage of a Zn-H+ cell is 0.45 V at 298 K
when [Zn2+]=1.0 M and PH2=1.0 atm, what is
the concentration of H+? Note that atm can
be used just like molarity.
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
ANS: 5.2 X 10-6M
Example 4
What pH is required if we want a voltage of
0.542 V and [Zn2+]=0.10 M and PH2=1.0 atm?
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
ANS: 5.84 X 10-5M, pH = 4.22
Batteries
Lead Acid Battery
• 12 Volt DC
• Discharges when starting the
car, recharges as you drive
(generator). Running
reaction backward.
PbO2(s) + Pb(s) +2HSO4-(aq) + 2H+(aq) 2PbSO4(s) +2H2O(l)
Alkaline Batteries
• Basic
• Zinc can acts as the anode
2MnO2(s)+2H2O(l)+2e-2MnO(OH)(s) + 2OH-(aq)
Zn(s) + 2OH-(aq) Zn(OH)2(s) + 2eRechargable uses Ni-Cd
Corrosion
• Iron rusts in acidic solns (not above pH=9)
• Water needs to be present
• Salts accelerate the process
O2 + 4H+ + 4e-  2H2O
Fe  Fe2+ + 2e(The Fe2+ eventually goes to Fe3+, Fe2O3)
Preventing Corrosion
• Paint
• Sometimes oxide layer(Al2O3)
• Galvanizing (coating Fe with Zn)
Fe2+ + 2e-  Fe E = -0.44 V
Zn2+ + 2e-  Zn E = -0.76 V
Zinc is more easily oxidized
(Zn  Zn2+ + 2eE = +0.76 V)
• Cathodic
protection
(sacrificial anode)
• Magnesium used
in water pipes
• Magnesium rods
used in hot water
heaters
An iron gutter is nailed using aluminum nails.
Will the nail or the iron gutter corrode first?
Fe2+ + 2e-  Fe
Al3+ + 3e-  Al
E = -0.44 V
E = -1.66 V
Al will corrode first (Al  Al3+ + 3e- ,E = +1.66 V)
Which of the following metals could provide
cathodic protection to iron:
Al
Cu
Ni
Zn
Electrolysis and Electroplating
• Plating of silver on silverware
Electrolytic cells
• Must run electricity through them
• Running a voltaic cell backwards
• Used to produce sodium metal
Na+(aq) + e-  Na (s)
Cl2(g) + 2e-  2Cl-(aq)
-2.71 V
+1.36 V
• As a voltaic cell
2Na(s)
 2Na+(aq) + 2e- +2.71 V
Cl2(g) + 2e-  2Cl-(aq)
+1.36 V
2Na(s) + Cl2 (s)2NaCl(aq)
+4.07 V
• As an electrolytic cell
2Na+(aq) + 2e-  2Na (s)
-2.71 V
2Cl-(aq)  Cl2(g) + 2e-1.36 V
2NaCl(aq)  2Na(s) + Cl2 (s) -4.07 V
Quantitative Electrolysis
• Electric current = Amperes
• 1 ampere = 1Coloumb
1 second
• 1 F = 96,500 C/mol
I = Q
t
– One mole of electrons has a charge of 96,500 C
– One electron has a charge of 1.602 X 10-19 C
What mass of aluminum can be produced in
1.00 hour by a current of 10.0 A?
Al3+ + 3e-  Al
Moles of e- = (36,000C)(1 mol e-) = 0.373 mol e(96,500 C)
Al3+ + 3e-  Al
0.373 mol
Al3+ + 3e- 
0.373 mol
Al
0.124 mol Al  3.36 g Al
Example 2
What mass of magnesium can be produced in 4000
s by a current of 60.0 A?
Mg2+ + 2e-  Mg
ANS: 30.2 g Mg
What current is required to plate 6.10 grams of
gold in 30.0 min?
Au3+ + 3e-  Au
How long would it take to plate 50.0 g of
magnesium from magnesium chloride if the
current is 100.0 A?
Given the following:
Ag+(aq) + e-  Ag(s)
+0.799V
Fe3+(aq) + e-  Fe2+(aq)
+0.771 V
a. Write the reaction that occurs.
b. Calculate the standard cell potential.
c. Calculate DGrxn for the reaction from the cell
potential.
d. Calculate K for the reaction.
e. Predict the sign of DSrxn.
f. Sketch the cell, labeling anode, cathode, and
the direction of electron flow.
Do SO3 and SO32- have the same molecular
shape? How about SO2?
16. a) Not redox
b) I oxidized (-1 to +5) , Cl reduced (+1 to -1)
c) S oxidized (+4 to +6), N reduced (+5 to +2)
d) Br oxidized (-1 to 0), S reduced (+6 to +4)
20 a. Mo3+ + 3e-  Mo
b. H2O + H2SO3  SO42- + 2e- + 4H+
c. 4H+ + 3e- + NO3-  NO + 2H2O
d. 4H+ + 4e- + O2  2H2O
e. 4OH- + Mn2+  MnO2 + 2e- + 2H2O
f. 5OH- + Cr(OH)3  CrO42- + 3e- + 4H2O
g. 2H2O + 4e- + O2  4OH-
22. a. 3NO2- + Cr2O72- +8H+  3NO3- + 2Cr3+ +
4H2O
b. 2HNO3 + 2S +H2O  2H2SO3 + N2O
c. 2Cr2O72- + 3CH3OH + 16H+  4Cr3+ 3HCO2H +
11H2O
d. 2MnO4- + 10Cl- + 16H+  2Mn2+ + 5Cl2 + 8H2O
e. NO2- + 2Al + 2H2O  NH4+ + 2AlO2f. H2O2 + 2ClO2 + 2OH-  O2 + 2ClO2- + 2H2O
26. a) Al oxidzed, Ni2+ reduced
b) Al  Al3+ + 3e-Ni2+ + 2e-  Ni
c) Al anode, Ni cathode
d) Al negative, Ni positive
e) Electrons flow towards the Ni electrode
f) Cations migrate towards Ni electrode
34 a) Cd is anode, Pd is cathod
b) Ered = 0.63 V
36 a) 2.87 V b) 3.21 V
c) -1.211 V d) 0.636V
38 a) 1.35 V
b) 0.29 V
41 a) Mg
b) Ca
c) H2 d) H2C2O4
42 a) Cl2
b) Cd2+ c) BrO3- d) O3
44. a) Ce3+ (weak reductant)
b) Ca (strong reductant)
c) ClO3- (strong oxidant)
d) N2O5 ( oxidant)
46 a) H2O2 strongest oxidizing agent
b) Zn strongest reducing agent
50.a) 3.6 X 108
b) 1041
c) 10103
52. 0.292 V
54 a) 4 X 1015 b) 2 X1065 c) 7.3 X1049
62a) 2.35 V
b) 2.48 V c) 2.27 V
64. a) 0.771 V b) 1.266 V
88. a) 173 g
b) 378 min
90.E = 1.10 V Wmax = -212 kJ/mol Cu
W = -1.67 X 105 J
1a) 14H+ + Cr2O72- + 3Fe  2Cr3+ + 3Fe2+ + 7H2O
b) 2Br- + F2  2F- + Br2
c) 4OH- + 2Cr(OH)3 + ClO3-  2CrO42- + Cl- + 5H2O
2b) 0.463 V
c) -89.4 kJ/mol d) 4.4 X 1015
e) 0.442 V
3) F2 is str. oxidizing agent, Li, str. reducing agent
4) b) 78 minutes c) 1.19 g d) 0.695 g
In a measuring cup:
• 5 mL of oil
• 5 mL of ethanol
• 5 mL of 50% NaOH solution (approximately 30
drops).
Place in beaker
• Heat the mixture, stirring with popsicle stick.
• Remove from heat. After ~5 minutes, add 10 mL of
saturated salt solution.
• Collect some of the solid and test the pH of your
soap. Compare the pH to that of commercial bar
soap and liquid detergent solution. See if it
lathers.