Electrochemistry: Oxidation-Reduction Reactions

Zn(s) + Cu+2 (aq) Zn2+(aq) + Cu(s)

loss of 2e-

gaining to 2e-

Zinc is oxidized - it goes up in charge and is the reducing agent

Copper is reducedreduced because it goes down in charge and is the oxidizing agent

Electrochemistry

Messing with electrons

What is the charge on each atom

KMnO4 Cr(OH)3 Fe(OH)2+1

-2

-8

+1

+1 +7

+7 -2 +1

-3+3

+3 -1

-2 = 1+3

+3

Electrochemistry: Oxidation-Reduction Reactions

By writing the oxidation number of each element under the reaction equation, we can easily see the oxidation state

changes that occur

Zn(s) + 2H+(aq) Zn2+(aq) + H2(aq)0 +1 +2 0

In any oxidation-reduction reaction (redox), both oxidation and reduction must occur.

Electrochemistry: Balancing Oxidation-Reduction Reactions

Oxidation Number Method

Al(l) + MnO2 Al2O3 + Mn 0 +4 -2 +3 -2 0

-3e’s

+4e’s

X 4 = -12

X 3 = +12

4 3 2 3

Electrochemistry: Balancing Oxidation-Reduction Reactions

Sample problem:

I2O5(s) + CO(g) I2(s) + CO2(g)

-2e’s

+5e’s

+2 +4

+5 0X 2 = 10

X 5 = -10

5 5

:

MnO4-(aq) + C2O4

2-(aq) Mn2+(aq) + CO2(g) (acid)

Half-Reaction Method

: Half-Reaction Method (in acid)

MnO4- Mn2+

C2O42- CO2

MnO4-(aq) + C2O4

2-(aq) Mn2+(aq) + CO2(g)

Step 1: Divide into two half-reactions,

Step 2: Balance main element

MnO4- Mn2+

C2O42- 2CO2

: : Step 3: Balance the O atoms by adding H2O

MnO4- Mn2+

C2O42- 2CO2

Step 5: Balance charge with electrons

8H++ MnO4- Mn2+ + 4H2O

C2O42- 2CO2

+ 4H2O

8H+ +

+7 +2

5e- + -2 0

+ 2e-

Step 4: Balance the H atoms by adding H+

MnO4- Mn2+ + 4H2O

C2O42- 2CO2

: Step 6: Make the electrons balance

5e- + 8H++ MnO4- Mn2+ + 4H2O

C2O42- 2CO2 + 2e-

10e- + 16H++ 2MnO4- 2Mn2+ + 8H2O

5C2O42- 10CO2 +10e-

Step 7: Cancel and add

16H+ + 2MnO4- + 5C2O4

2- 2Mn2+ + 10CO2 + 8H2O

10 16 2 2 8

5 10 10

+

Balancing in base

CN- + MnO4- CNO- + MnO2

CN- CNO-

MnO4- MnO2

H2O +

+ 2H2O

+ 2H+

4H+ +

-1 +1

+3 0

+ 2e-

3e-+

3 3 3 6 6

6 8 2 2 4

2H+ + 3CN- + 2MnO4- 3CNO- + 2MnO2 + H2O

2OH-+ 2OH-

2H2O1

Balancing in base

Cr(OH)3 + ClO CrO4-2 + Cl2

Have at it

4Cr(OH)3 + 6ClO + 8OH- 4CrO4-2 + 3Cl2 + 10H2O

Note that the activity series is simply the

oxidation half-reactions of the

metals ordered from the highest to lowest

Using the hydrogen electrode as a referenceUsing the hydrogen electrode as a reference2H2H++ (1 M) + 2e (1 M) + 2e-- H2 (1 atm, 25 °C) E°red = 0 V

Cell EMF or Voltage

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s) E°cell = 1.10 V

E ° ox + E ° red = E°cell

From the table:Zn+2 + 2e- Zn red= -.76VCu+2 + 2e- Cu red= .34V

Flip

Zn Zn+2 + 2e- ox = .76V

Voltaic or Galvanic Cells

-oxidation

+reduction

Sample Problem. Draw a voltaic cell using the following equation and label

all parts

CrCr22OO772-2-

(aq)(aq) + I + I--(aq)(aq) Cr Cr3+3+

(aq)(aq) + I + I2(s)2(s)

Spontaneity and Extent of Redox ReactionsA positive emf indicates a spontaneous process, and a negative emf indicates a nonspontaneous one.Sample Problem: Using the standard electrode potentials, determine whether the following reaction are spontaneous A. Cu(s) + 2H+

(aq) Cu Cu2+2+(aq)(aq) + H + H2(g)2(g)

B. Cl2(g) + 2I-(aq) 2Cl 2Cl--

(aq)(aq) + I + I2(s)2(s)

Cu(s) Cu Cu2+2+(aq)(aq) + 2e + 2e- - EEoxox

= -.34V= -.34V2H+

(aq) + 2e- H H2(g) 2(g) EEredred = 0.0V= 0.0V

EEcellcell = -.34V= -.34V

Cl2(g) + 2e- 2Cl 2Cl--(aq)(aq) EEredred

= 1.36V= 1.36V 2I-

(aq) 2e 2e- - + I+ I2(s)2(s) EEoxox = -.54V= -.54V

EEcellcell = .82V= .82V

EMF and Free Energy ChangeAny redox reaction involves free energy change

(G) which also may be used as a measure of spontaneity or work (max or min.)

n = # of moles of e-s transferred = 96500 C the charge of one mole of e-s

or 96,500 J/V-mol e-

Note that because n and are both positive values, a positive value in E leads to a negative value of G.

G° = -nE °

Use the standard electrode potentials to calculate the standard free energy

change, G°, for the following reaction

2Br-(aq) + F2(g) Br2(l) + 2F-

(aq)

2Br-(aq) Br2(l) + 2e E°ox = -1.06

F2(g) + 2e- 2F-(aq) E°red = 2.87

2Br-(aq) + F2(g) Br2(l) + 2F-

(aq) E°cell = 1.81 V

G = -nEG = -(2 mol e- )(96,500 J/volt-mol e-)( 1.81 V)G° = -3.49 x 105 J = -349 kJ

EMF and Equilibrium Constant“Remember in Chapter 19, we related G° to

the equilibrium constant, K?”

G° = -RT lnKG ° = -nE °

-nE° = -RT lnK

E ° = 0.0591 logK n

Simplify

Calculate the K for the following reactionO2 (g) + 4H+

(aq) + 4Fe2+ (aq) 4Fe3+

(aq) + 2H2O (l)

O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E°red = 1.23 V

4Fe2+ (aq) 4Fe3+

(aq) + 4e- E°ox = -0.77 V

O2 (g) + 4H+ (aq) + 4Fe2+

(aq) 4Fe3+ (aq) + 2H2O (l)

E°cell = 0.46 V

log K = n E °

0.0591V 4(0.46V)

0.0591V= = 31.1

K = 1.36 x 1031

Calculate the equilibrium constant for the reaction

IO3-

(aq) + 5Cu (s) + 12H+ (aq) I2 (s) + 5Cu2+

(aq) + 6H2O (l)

2IO3-

(aq) + 12H+ (aq) + 10e- I2 (s) + 6H2O (l) red= 1.20V

5Cu (s) 5Cu2+ (aq) + 10e- ox= -.34V

cell= .86V = .0591 log k n .86V = .0591 log k

10Log K = 145.5

K = 10145.5

G = G° + RT lnQG° = -RT lnKStandard Conditions

Non- Standard Conditions

G = G° + 2.30 RT log Q

-nE = -nE ° + 2.30 RT log Q

E = E ° - 2.30 RT log Qn

Nernst Equation

E = E ° - 0.0591 log Qn

When T=298K

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) E °= 1.10 V

E = 1.10 ° - 0.0591 log

2

[Zn2+]

[Cu2+]

If [Zn2+]=0.05M, and [Cu2+] = 0.50 M

E = 1.10 ° - 0.0591 log

2

[0.05]

[.50]= 1.13V

2e- + Cu2+ (aq) Cu (s)

Zn (s) Zn2+ (aq) + 2e-

Therefore

Calculate the emf generated by the the following reaction

CrCr22OO772-2- + 14 H + 14 H ++ + 6I+ 6I-- 2Cr 2Cr3+3+ + 3I + 3I22 + 7H + 7H22OO

[Cr[Cr22OO772-2-] = 2.0 M] = 2.0 M

[H [H ++ ] = .05 M ] = .05 M

[I[I--] = .25 M] = .25 M

[Cr[Cr3+3+] = 1.0 x 10] = 1.0 x 10-5-5MM

red= 1.33V

ox= -.54V

cell= .79V

= .79V - .0591 log (1 x 10-5)2

6 (2.0)(.05)14(.25)6

= .68V

Concentration CellsWhat happens when both cells are identical except for

concentration. Nature will try to equalize the two cells.

E= Eº - .0591 log Qn

E = 0 - .591 log .01 = .591V.1

anode: Pb (s) + SO42-

(aq) PbSO4 (s) + 2e + 2e-- E °= 0.356 V

cathode:PbO2(s)+ SO42-

(aq)+ H+(aq)+ 2e- PbSO4(s)+ 2H2O(l)

E °= 1.685 V Pb(s)+ PbO2(s)+ 4H+ +2SO4(aq) 2PbSO4(s)+

2H2O(l) E °= 2.041 V

Note that one advantage of the lead storage battery is that it can be recharged because the PbSO4 produced during discharge

adheres to the electrodes

Lead Storage BatteryLead Storage Battery

Dry Cell alkalineDry Cell alkalineDry Cell alkalineDry Cell alkalineanode: Zn (s) Zn 2+ 2+

(aq)(aq) + 2e + 2e--

cathode:2NH4+

(aq)+2MnO2(s)+ 2e-

Mn2O3(s) + 2NH3(aq) + H2O (l)

Fuel CellsFuel CellsFuel CellsFuel Cellsanode: 2H2(g)+ 4OH-

(aq) 4H2O(l) + 4e- cathode: 4e- + O2(g) + H2O(l) 4OH 4OH--

(aq)

2H2(g) + O2(g) 2H 2H22OO(l)(l)

Electrolytic Cells: Electrolytic Cells: ElectrolysisElectrolytic Cells: Electrolytic Cells: Electrolysis

Electrolysis is driven by an outside energy Electrolysis is driven by an outside energy sourcesource

voltage source actsvoltage source acts like an electronlike an electron pumppump

voltage source actsvoltage source acts like an electronlike an electron pumppump(- red)

These electrodesare inertThese electrodesare inert

Notice

(+ ox.)

Eº = ( -)

Electrolysis of Aqueous SolutionsElectrolysis of Aqueous SolutionsElectrolysis of Aqueous SolutionsElectrolysis of Aqueous SolutionsSodium cannot be prepared by electrolysis of aqueous solutions of NaCl, because wateris more easily reduced than Na+:

2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 NaNa++ + e + e- - Na Na(s)(s) E° E°redred = -2.71 = -2.71 NaNa++ + e + e- - Na Na(s)(s) E° E°redred = -2.71 = -2.71 The possible Anode reactions are:

2Cl2Cl- - Cl Cl22 E° E°oxox = -1.36 = -1.362Cl2Cl- - Cl Cl22 E° E°oxox = -1.36 = -1.362H2H22O O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23 2H2H22O O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23

2Cl2Cl--(aq)(aq) Cl Cl2(g)2(g) E° E°oxox = -1.36 = -1.362Cl2Cl--(aq)(aq) Cl Cl2(g)2(g) E° E°oxox = -1.36 = -1.36

2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 E°E°cellcell >> -2.19 -2.19 E°E°cellcell >> -2.19 -2.19

Possible Cathode reactions

Therefore

(Note: Not in table on AP exam)

Electrolysis of With Active ElectrodesElectrolysis of With Active Electrodes

Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e-- E° E°oxox = 0.28 = 0.28Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e-- E° E°oxox = 0.28 = 0.28

2H2O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23 2H2O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23

Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e--Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e--

NiNi2+2+(aq)(aq) + 2e + 2e-- Ni Ni(s)(s)NiNi2+2+(aq)(aq) + 2e + 2e-- Ni Ni(s)(s)

anode:anode:

cathode:cathode:

Electroplating creates a silver liningElectroplating creates a silver lining

Possibilities

Quantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysis

Calculate the mass of aluminum produced in 1.00 hr. by the electrolysis of molten AlCl3 if the current is 10.0 A. (C = amperes x seconds)

= 3.36 g of Al

(10.0A)(1.00 hr)(3600 sec)

(1 hr) (1 A-s)

(1C)

(96,500C)(1)

(3)(1molAl)

(1 mol Al)(27.0 g Al)

Quantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysis

The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e- Mg. Calculate the mass of magnesium formed Mg. Calculate the mass of magnesium formed upon passage of 60.0 A for a period of 4000 supon passage of 60.0 A for a period of 4000 s

(60.0A)(4000s)(1C)(1) (mole Mg)(24.3g) (A•s)(96,500C) (2) (mole Mg)

30.2g Mg

Electrical WorkElectrical WorkElectrical WorkElectrical Work

since G= wmax

and G = - nEthen wmax= - nE

The unit employed by electric The unit employed by electric utilities is kilowatt-hour utilities is kilowatt-hour (kWh =3.6 x 10(kWh =3.6 x 1066J)J)

Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound.

Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound.

All metals except gold and platinum are thermodynamicallycapable of undergoing oxidation in air at room temperature

The rusting of iron is known to require oxygen; iron does not rust in water unless O2

is present.

E°red = 0.44 V

E°red = 1.23 V

Note that as the pH increases, the reduction of O2 becomes less favorable

The corrosion of Iron:Protecting the surface with tinThe corrosion of Iron:Protecting the surface with tin The corrosion of Iron:Protecting the surface with tinThe corrosion of Iron:Protecting the surface with tin

Fe(s) Fe Fe2+2+ + 2e + 2e- - E°ox = 0.44 V

Sn(s) Sn Sn2+2+ + 2e + 2e- - E°ox = 0.14 V

Tin has lower E°ox so less likely to oxidize until the

surface is broken then it accelerates as it makes a voltaic cell with iron.

The corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic Protection

Fe(s) Fe Fe2+2+(aq) + 2e(aq) + 2e- - E°ox = 0.44 V

Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e- - E°ox = 0.76V

GalvanizationGalvanization

sacrificialanode

Zinc is more positive andgoes first and hasan oxide coatthat seals.

The corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic Protection