oxidation/reduction chapter 20. two types of chemical rxns 1.exchange of ions – no change in...
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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
2. Exchange of Electrons – changes in oxidation numbers/charges
Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)Remove spectator ions
Fe + Cu2+ Fe2+ + CuProtonsElectrons
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 Cr3+
S8 Fe2(SO4)3
Mn2O3 SO32-
KMnO4 NO3-
HSO4-
Oxidation
1. Classical Definition –addition of oxygen2. Modern Definition – an increase in oxidation
number
Fe + O2 Fe2O3
CO + O2 CO2
CH3CH2OH CH3CHO CH3COOH
Reduction
1. Classical –addition of hydrogen2. Modern –decrease (reduction) in oxidation
numberN2 + 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 +2Cu + O2 CuO
0 -2
Got oxidized, reducing agent
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
Steps for more complicated examples1. Balance all atoms except H and O2. Balance charge with electrons3. Balance O with water4. Balance H with H+
===================================5. (Add OH- to make water in basic solutions)
Balancing Redox Reactions
Example 1:
MnO4- + C2O4
2- Mn2+ + CO2
1. Separate into half reactionsMnO4
- 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- 2CO2 + 2e- (X 5)
8H+ + 5e- + MnO4- Mn2+ + 4H2O (X 2)
5C2O42- 10CO2 + 10e-
16H+ + 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+ + MnO4
-
14H+ + 2Mn2+ + 5BiO3- 5Bi3+ + 2MnO4
- + 7H2O
Balancing Redox Reactions (Basic Solutions)
Add OH- AT THE VERY END ONLY!!!!!
NO2- + Al NH3 + Al(OH)4
-
5H2O + OH- + NO2- + 2Al NH3 + 2Al(OH)4
-
Cr(OH)3 + ClO- CrO42- + Cl2
2Cr(OH)3 + 6ClO- 2CrO42- + 3Cl2 + 2OH- + 2H2O
F- + MnO4- MnO2 + F2
8H+ + 6F- + 2MnO4- 2MnO2 + 3F2 + 4H2O
4H2O + 6F- + 2MnO4- 2MnO2 + 3F2 + 8OH-
HNO2 + H2O2 O2 + NO
2HNO2 + H2O2 O2 + 2NO + 2H2O
Voltaic (Galvanic) Cells
• Voltaic(Galvanic) Cells – redox reactions that produce a voltage– Spontaneous reactions (G<0)– Voltage of the cell (Eo
cell) is positive
– Batteries
• Electrolytic cells – redox reactions that must have a current run through them.– G>0 and Eo
cell 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 cell1. Anode – Oxidation site2. 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+
Hydrogen Electrode
1. Standard Electrode2. Voltage(potential) = 0 Volts
2H+(aq) + 2e- H2(g) 0 volts
H2(g) 2H+(aq) + 2e- 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 VCu2+ + 2e- Cu +0.34 Vflip the zinc equationZn Zn2+ + 2e- +0.76 VCu2+ + 2e- Cu +0.34 VZn + 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) + 2e-
Br2(l) + 2e- 2Br-(aq) +1.06 V
If the overall cell voltage is 1.46 V, what is the reduction potential for In3+?
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- PbNi2+ + 2e- Ni
a. Calculate the cell potential (o).b.Label the cathode and anode.c. Identify the oxidizing and reducing agents.d.Which electrode is consumed? e.Which electrode is plated?f. 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
Example 1
Which of the following is the strongest oxidzing agent? Which is the strongest reducing agent?
NO3- Cr2O7
2- Ag+
Which of the following is the strongest reducing agent? Which is the strongest oxidizing agent?
I2(s) Fe(s) Mn(s)
Spontaneity
Voltaic Cells Electrolytic Cells
•Positive emf•Spontaneous•Can produce electric current•Batteries
•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 Go
G = -nFE
n = number of electrons transferredE = Cell emfF = 96,500 J/V-mol (Faraday’s Constant)
Positive Voltage gives a negative G (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 G 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
Example 1
Calculate G, 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
G = Go + 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 1Calculate 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 2Calculate 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 3If 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) + 2e-
Rechargable 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- 2H2OFe 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 VZn2+ + 2e- Zn E = -0.76 V
Zinc is more easily oxidized (Zn Zn2+ + 2e- E = +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 E = -0.44 V Al3+ + 3e- Al E = -1.66 V
Al will corrode first (Al Al3+ + 3e- ,E = +1.66 V)
Electrolytic cells
• Must run electricity through them• Running a voltaic cell backwards• Used to produce sodium metal
Na+(aq) + e- Na (s) -2.71 VCl2(g) + 2e- 2Cl-(aq) +1.36 V
• As a voltaic cell2Na(s) 2Na+(aq) + 2e- +2.71 VCl2(g) + 2e- 2Cl-(aq) +1.36 V2Na(s) + Cl2 (s)2NaCl(aq) +4.07 V
• As an electrolytic cell2Na+(aq) + 2e- 2Na (s) -2.71 V2Cl-(aq) Cl2(g) + 2e- -1.36 V 2NaCl(aq) 2Na(s) + Cl2 (s) -4.07 V
Quantitative Electrolysis
• Electric current = Amperes• 1 ampere = 1Coloumb I = Q
1 second t• 1 F = 96,500 C/mol
– One mole of electrons has a charge of 96,500 C– One electron has a charge of 1.602 X 10-19 C
Moles of e- = (36,000C)(1 mol e-) = 0.373 mol e- (96,500 C)
Al3+ + 3e- Al0.373 mol
Al3+ + 3e- Al0.373 mol 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
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) +0.799VFe3+(aq) + e- Fe2+(aq) +0.771 Va. Write the reaction that occurs.b.Calculate the standard cell potential.c. Calculate Grxn for the reaction from the cell
potential.d.Calculate for the reaction.e.Predict the sign of Srxn.
f. Sketch the cell, labeling anode, cathode, and the direction of electron flow.
16. a) Not redoxb) 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- Mob. H2O + H2SO3 SO4
2- + 2e- + 4H+
c. 4H+ + 3e- + NO3- NO + 2H2O
d. 4H+ + 4e- + O2 2H2Oe. 4OH- + Mn2+ MnO2 + 2e- + 2H2Of. 5OH- + Cr(OH)3 CrO4
2- + 3e- + 4H2Og. 2H2O + 4e- + O2 4OH-
22. a. 3NO2- + Cr2O7
2- +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
+ + 2AlO2-
f. H2O2 + 2ClO2 + 2OH- O2 + 2ClO2- + 2H2O
26. a) Al oxidzed, Ni2+ reducedb) Al Al3+ + 3e- Ni2+ + 2e- Nic) Al anode, Ni cathoded) Al negative, Ni positivee) Electrons flow towards the Ni electrodef) Cations migrate towards Ni electrode
34 a) Cd is anode, Pd is cathodb) Ered = 0.63 V
36 a) 2.87 V b) 3.21 V c) -1.211 V d) 0.636V38 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)
46a) H2O2 strongest oxidizing agent
b) Zn strongest reducing agent50.a) 3.6 X 108 b) 1041 c) 10103
52. 0.292 V54 a) 4 X 1015 b) 2 X1065 c) 7.3 X1049
62a) 2.35 V b) 2.48 V c) 2.27 V64. a) 0.771 V b) 1.266 V88. a) 173 g b) 378 min90.E = 1.10 VWmax = -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- 2CrO4
2- + Cl- + 5H2O
2b) 0.463 V c) -89.4 kJ/mol d) 4.4 X 1015 e) 0.442 V3)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.