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+ + 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

Oxidation

Fe + O2 (limited oxygen)

Fe + O2 (excess oxygen)

Oxidation

C + O2 (limited oxygen)

C + O2 (excess oxygen)

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

What’s really happening:Cu2+ + Zn Cu + Zn2+

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- CO2

C2O42- 2CO2

+3 +4 (1 e- per carbon)

C2O42- 2CO2 + 2e-

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

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 + 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+

Voltaic (Galvanic) Cells

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+?

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- 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

F2 + 2e- 2F- +2.87 V

Stronger Cl2 + 2e- 2Cl- +1.36 V

Oxidizing .Agents .

.

.

.Li+ + e- Li -3.05V

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)

Can copper metal (Cu(s)) act as an oxidizing agent?

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

EMF and K

Go = -RTlnK (Go = -nFEo)-nFEo = -RTlnKlnK = nFEo (assume 298 K)

RTlog K = nEo

0.0592

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)

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) -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

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- 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

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) +0.799VFe3+(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.

Do SO3 and SO32- have the same molecular

shape? How about SO2?

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.