21 Electrochemistry Electricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies.

21 Electrochemistry 1

Electricity

Ancient people noticed electricity 1746 B. Franklin demonstrated lightening as electric effect and performed the kite experiment in 1751. Two people tried to repeat his kite flying experiment were killed by thunder.

1767 L. Galvani inserted two different metals in frog fluid and constructed a electric cell 1800 A. Volta substituted frog fluid; made batteries, consisted of several cells.

1802 G. Romagnosi noticed magnetism related to electricity Michael Faraday 1791-1867 discovered many theories of electricity and magnetism 21 Electrochemistry 2

Galvani

Luigi Galvani

(1737-1798): erroneously concluded that the frog's nervous system generated an electrical charge, his work stimulated much research into the electrochemistry.

The depiction of his laboratory  21 Electrochemistry 3

Electrochemistry

A. Volta (1745-1827) experimented with different materials, and made voltaic piles (batteries) William Nicholson (1753-1815) observed bubbles forming on the surfaces of metals submerged in water when they are connected to a voltaic pile Humphry Davy (1778-1829) observe electrolysis of water and metal salts. Following that, … Michael Faraday (1791-1860) studied electrolysis, and discovered the relationship

between charges and chemical stoichiometry

21 Electrochemistry 4

Electrons

q

e

= –1.60217733e-19 C

F = 96485 C

m

e = 0.00054856 amu = 9.1093897e-31 kg

spin = ½ (two state)

magnetic moment

= 9.284770e–24 J/tesla Voltaic piles (batteries) made the following study possible W. Crookes (1832-1919) observed cathode rays in low-pressure tubes.

1897: J.J. Thomson determined the charge to mass ratio (e – cathode rays (electrons).

/ m

e ) of 1916 R. Millikan (1868-1953) measured the amount of charge of e – .

21 Electrochemistry 5

Redox reactions and electrons

Energy drives chemical reactions. Redox reactions involve the transfer of electrons.

L oss of e lectron (increase oxidation state) is o xidation, ( Leo ).

G ain of e lectron (decrease oxidation state) is r eduction, ( ger ).

net : Zn  Cu Zn 2+ Zn + Cu 2+ 2+  + 2 e + 2 e – –  Cu Cu + Zn 2+ leo ger redox Chemical energy in redox reactions may be convert to electric energy by applying electrochemistry.

In the meantime, we should learn to balance the redox reaction equations .

21 Electrochemistry 6

Galvanic Cell

A galvanic cell consists of two different metals inserted into a solution of an electrolyte (salt, acid or base), simulated Representation :

Zn | Zn 2+ || Cu 2+ | Cu

21 Electrochemistry 7

–3 NH 3 –2 N 2 H 4 –1 NH 2 OH 0 N 2 +1 N 2 O +2 NO +3 NO 2 – +4 NO 2 +5 NO 3 –

Assign oxidation states

0 for any element 1 for H in compounds, but –1 for LiH, NaH, etc – 2 for O in compounds, but –1 for H 2 O 2 , Na 2 O 2 +1 for alkali metals, +2 for alkaline earth metals The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion. 21 Electrochemistry –1 Cl – 0 Cl 2 +1 ClO – +3 ClO 2 – +4 ClO 2 +5 ClO 3 – +7 ClO 4 – 8

Half reaction equations

Oxidation and reduction can be written as half-reaction equations such as

Steps to balance half reaction

Zn  Zn 2+ + 2 e – leo Assign oxidation number Cu 2+ + 2 e –  Cu net : Zn + Cu 2+  Cu + Zn 2+ ger redox Figure out what is oxidized or reduced.

Demonstrate how to balance these Add electrons according to oxidation number change Fe 2+  Fe 3+ + __ e – C 2 O 4 2  2 CO 2 MnO 4 – + __ e –  + __ Mn e – 2+ Cr 2 O 7 2– + __ e – + __ H +  2 Cr 3+ + __ H 2 O 21 Electrochemistry Balance charge with H + OH – (base) (acid) or Balance atoms with H 2 O 9

More half-reaction equations

2 I –  I 2 + __ e – ClO 2 + __ OH –  ClO 3 + __ e – + H 2 O (in basic solution) 2 S 2 O 3 2–  S 4 O HS(= S )O 3 –  6 2 S + __ + __ e – e + HSO 4 – __ H 3 O + + __ e –  H 2 O 2 + __ e –  H 2 (g) + 2 H 2 O __ H 2 O ClO 2 + __ e –  NO 3 – + __ e –  ClO NH 4 + 2 – 21 Electrochemistry 10

Electrochemical Series

An electrochemical series is a list of metals in order of decreasing strength as reductant, or increasing strength as oxidant.

An example of an activity series of metals based on the Standard Potentials given would be: K > Ba > Ca > Na > Mg > Al > Mn > Zn > Fe > Ni > Sn > Pb > Cu > Ag In this series the most active metal is potassium (K) and the least active metal is silver (Ag) Reactions and cells are illustrated in 21-1 of Text (PHH).

21 Electrochemistry 11

Constructing half cells

A half cell consists of an oxidizing and its oxidized species Zn | Zn 2+ Cu | Cu 2+ Pt | H 2 Pt |Fe 2+ | H + , Fe 3+ Cl – | Cl 2 | Pt (Pt as conductor) Explain the cell convention and reactions of cells.

21 Electrochemistry Student cell set from School Master Science $30 12

Galvanic cells

Using corns for a galvanic cells is illustrated by an Internet site: (schoolnet.ca/general/electric-club/e/page9.html) This picture illustrates a way to make a pact of battery using coins of different metals. Apply the principles you have learned regarding electrochemical series and electrolytes to make such a pile will be an interesting exercise.

21 Electrochemistry 13

Cell convention

Oxidation takes place always at the anode Zn (s)  Zn 2+ (aq) + __ e – Zn(s) | Zn 2+ (aq) Fe 2+ (aq)  Fe 3+ (aq) + e – Pt | Fe 2+ , Fe 3+ H 2 (g)  2 H + (aq) + __ e – Pt | H 2 (g) | H + (aq) Reduction takes place at the cathode Cu 2+ (aq) + __ e –  Cu (s) Cu 2+ | Cu(s) Cl 2 (g) + __ e –  Cl – (aq) Cl 2 (g) | Cl – (aq) | Pt Fe 3+ (aq) + __ e –  Fe 2+ (aq) Fe 3+ (aq), Fe 2+ (aq)| Pt 2 H + (aq) + __ e –  H 2 (g) H + (aq) | H 2 (g)| Pt Concentration (1.1 M) and pressure (0.9 atm) are also included in cell notations. 21 Electrochemistry 14

Electric energy and work

Electric energy or electric work = charge * potential difference

W = q * V

compare W = m g h (1 J = 1 Coulomb Volt, C V) The Faraday constant F is the charge for one mole of electrons, F = 96485 C; F / N A = 96485 C / 6.022e23

= 1.602177e-19 C, charge per e – The maximum chemical energy of the cell that can be converted to electric work is the Gibbs free energy change, 

G

See slides in 16 Equilibria 

G

= –

n F



E

(

n F

= q is the charge) electromotive force (emf, or V) number of electrons in the reaction equation n F is the charge q 21 Electrochemistry 15

Standard cell emf’s and electrode potentials

The standard cell emf standard conditions is the emf of a voltaic cell operating under (1 M, 1 atm, 25 o C etc). E.g.

Zn (s) | Zn 2+ (aq, 1 M) || Cu 2+ (aq, 1 M) | Cu (s) 

E

o = 1.10 V Mg (s) | Mg 2+ (aq, 1 M) || Cu 2+ (aq, 1 M) | Cu (s) 

E

o = 2.90 V The absolute potential of the electrode cannot be determined. Only relative potential can be measured. The standard reduction potential measured against a SHE, for which E o = 0.0000 V is Zn (s) | Zn 2+ (aq, 1 M) Zn = Zn 2+ + 2 e – || H + (aq, 1 M) | H 2 (1 atm) (g) (oxidation, displace H + possible) 

E

o = 0.76 V

E

o = 0.76 V Zn 2+ + 2 e –  Zn (reduction)

E

o = – 0.76 V 21 Electrochemistry 16

Gibb’s Free Energy in a Cell

How much energy is available for the cell Zn | Zn 2+ || Ag+ | Ag operating at standard condition when one mole of Zn is consumed ?

Solution

Zn = Zn 2+ + 2 e 2 Ag + + 2 e = 2 Ag Zn + 2 Ag + = Zn 2+ + 2 Ag

E

o 0.762 V 0.799 V (from table) 

E

o = 1.561 V 

G

o = –

n F



E

= – 2 * 96485 C = 301226 J (1J = 1 C V) = 301.2 kJ * 1.561 V How much silver is consumed?

How much energy is available if 6.5 g of Zn is consumed?

17

Table of standard reduction potential

Reaction E o (V)

Standard cell potentials Li + Na + + e + e – – = Li (s) – 3.04

= Na (s) – 2.71

Mg 2+ Zn 2+ + 2 e – + 2 e – = Mg (s) – 2.38

= Zn (s) – 0.76

(reference) (reference)

2 H H 2

Cu 2+

+ + 2 e (g) = 2 H +

+ 2 e –

– = H 2 (g) 0.000

+ 2 e

= Cu (s)

– 0.000

0.34

Br 2 Cl 2 F 2 I 2 Cu + (l) + 2 e + e (s) + 2e (g) + 2 e – (g) + 2 e – – – – = Cu (s) = 2 I = 2 Br = 2 Cl – = 2 F – – – (aq) (aq) (aq) (aq) 0.52

0.54

1.07

1.36

2.87

21 Electrochemistry

Cell

Li | Li + || Cu 2+ | Cu Mg | Mg 2+ || I 2 | I – | Pt ____ Zn | Zn 2+ || Br 2 | Br – | Pt ____ Cu | Cu 2+ || Zn 2+ | Zn Which is not spontaneous?

See 21-2 

E

o

____ ____ 18

Table of standard reduction potential

Reaction E o (V)

F 2 Cl 2 Br 2 (g) + 2 e (g) + 2 e – (l) + 2 e – – = 2 F = 2 Cl – = 2 Br – – I 2 (s) + 2e – = 2 I – Cu + + e – (aq) (aq) (aq) (aq) = Cu (s) (reference) (reference) 2.87

1.36

1.07

0.54

Cu 2+ + 2 e – = Cu (s) 0.52

0.34

H 2 (g) = 2 H 2 H + + 2 e – +

Zn 2+ + 2 e –

+ 2 e – 0.000

= H 2 (g) 0.000

= Zn (s) – 0.76

Mg 2+ + 2 e – Na + + e – Li + + e – = Mg (s) – 2.38

= Na (s) – 2.71

= Li (s) – 3.04

The listing order in the table may be different in different text books. However, the principles and methods of application remain the same.

This is the order given on the Exam Data Sheet, that is different from the text.

21 Electrochemistry 19

Strength of oxidation

The ability of a chemical to oxidize is its ability to take electrons from other species, Oxidizing agent + n e  Reduced species Strength of oxidation potential.

of an oxidizing agent is measured by its reduction Similarly, strength of reduction of a reducing agent is measured by its oxidation potential.

Oxidized species  Reducing agent + n e Be able to order the species according to oxidizing strength. _____ 21 Electrochemistry 20

Reaction direction and emf

What is the emf for the reaction, Zn 2+ (aq) + 2Fe 2+ (aq) = Zn (s) + 2Fe 3+ (aq)?

Solution:

Know what data to look for Zn 2+ (aq) + 2 Fe 2+ 2 (aq)  Fe 2+ Zn 2+ Fe 3+  2 + 2 e  + e Fe  3+ Zn (s) + 2Fe 3+ Fe + 2 Zn E 2+ e E o (aq)

E

o o = – 0.76 V = + 0.77 V = – 

E

o = 0.77 V – 1.53 V + non-spontaneous Pt | Zn 2+ | Zn || Fe 2+ | Fe 3+ | Pt impractical The reverse reaction is spontaneous , Zn (s) + 2Fe 3+ (aq)  Zn 2+ (aq) + 2Fe 2+ Zn | Zn 2+ || Fe 3+ (aq) | Fe 2+ | Pt 

E

o = + 1.53

V 21 Electrochemistry 21

Free energy and emf

What is the free-energy change Zn | Zn 2+ (aq) 1 M || Ag + for the cell, (aq) 1 M | Ag?

Solution: Reduction potential required, Zn  Zn 2+ + 2e

E

o = 0.76

Ag + + e  Ag

E

o = 0.80

2 Ag + + 2e  2Ag

E

o = 0.80

Cell reaction 2 Ag + + Zn  2Ag + Zn 2+ 

E

o = 1.56 V 

G

o = – n F 

E

o = – 2 * 96485 * 1.56

= – 3.01e5 J or – 302 kJ Negative indicate energy is released.

Condition for spontaneous reaction is –  G or + 

E.

The free energy for the cell is –301 kJ per mole of Zn, what is the emf?

21 Electrochemistry 22

General cell emf



G o

is the standard energy change .  G is for non-standard conditions .  G = – n F 

E

Text uses E cell instead of  E Similarly, 

E

o is the standard emf whereas  E is general emf. 

G

= 

G

o + R T ln Q 

E =



E o – R T / n F

ln Q

When a system is at equilibrium (Q = K),  G = 0. Therefore,

reaction quotient



G

= 

G

o + R T ln K = 0 

E =



E o – R T / n F

ln K = 0

equilibrium constant

 

G

o = – R T ln K or 

G

o = – ln(10) R T log K 

E

o =

R T / n F

ln K



E

o = 2.303 R T

/ n F

log

K

21 Electrochemistry At 298 K

0.0592

E

o = ———— log K

n

23

The Nernst equation

For a general reaction, a A + b B = c C + d D  E = 

E

o

R T

[C] c [D] d – ——— ln ————

n F

[A] a [B] b This Nernst equation correlates cell emf with [ ] or reactivities of reactants and products as well as T Units for [ ]: mol L-1 for aqueous solution, atm (for gas), and constant for solid and liquid.

21 Electrochemistry See 21-4 24

Evaluating



E

 E = 

E

o

R T

[C] c [D] d – ——— ln ————

n F

[A] a [B] b

At 300 K, evaluate the cell emf for Zn | Zn 2+ (0.100 M) || H + (0.200 M) | H 2 (1.111 atm) | Pt

Solution: L oo k up: (R = 8.314 J mol -1 2 H Zn +  (aq) + Zn 2+ 2 (aq) + 2 e, e  H 2 (g), K -1 , F = 96485 C mol -1 ) The reaction is Zn(s) + 2 H + (aq)  Zn 2+ (aq) + H 2 (g)

E E

o o 

E

o = 0.76 V = 0.00 V = 0.76 V 8.314 J mol -1 K -1 * 300 K (0.100) (1.111)  E = 0.76 – —————————— ln —————— 2 * 96485 C mol -1 (0.200) 2 = 0.76 – 0.0129 * (1.02) = 0.76 – 0.013 = 0.75 V 21 Electrochemistry See example 19.12

25

Concentration cell

Problem

: At 298 K, evaluate the emf of the cell Cu | Cu 2 + (0.10 M) | | Cu 2+ (1.0 M) | Cu Cu(s)  Cu 2+ (0.1 M) + 2e; Cu 2+ (1.0 M) + 2e  Cu(s)

Solution

: The standard emf ( 

E

o = 0.00)for The reaction is actually Cu (s) + Cu 2+ Cu | Cu 2+ (1.0 M) = Cu 2 + || Cu 2+ | Cu (0.1 M) + Cu (s)  E = 0.00 –

R T

[ Cu 2 + ] ——- ln ——— = – 2 F [Cu 2+ ]

R T

0.10

–––– ln ––––– 2 F 1.00

= + 8.3145 * 298 –––––––––– ln –––– = 0.0295 V 2*96485 1.0

0.1

The voltage is purely due to concentration difference. Solutions in the two compartment try to become equal.

21 Electrochemistry When 2 [ ]’s are equal,  See p. 841 E = 0 26

Equilibrium Constant

K

and



E

o cell Calculate the solubility product of AgCl from data of standard cell At 298 K 

Solution

: Look up desirable data Ag + Cl – (s) + e  Ag + (aq) + e  Ag (s)  Ag 0 (s) + Cl – Ag + E° = 0.2223 V Ag (s) E° = 0.799 V (aq) + e E° = – 0.799 V Get the desirable eq’n AgCl (s)  Ag + (aq) + Cl – (aq)  E° = – 0.577 V

K

sp log K

sp K sp

= [Ag + ][Cl – ] = – 0.577

/ 0.0592

= 10 – 9.75

= – 9.75

= 1.8e–10

E

o 0.0592

= ———— log K

n

21 Electrochemistry Show that for this cell Ag | Ag + , 1 M || Cl – , 1 M | AgCl | Ag E o but for this cell Ag | AgCl |Cl = – 0.577 V –, 1 M || Ag + , 1 M E o = +0.577 V | Ag See p. 837 27

Evaluate free-energy change

Evaluate 

G

o for the reaction Zn (s) + 2 Ag + (aq) = Zn 2+ (aq) + 2 Ag (s)

Solution

: Required to look up : Ag + + e = Ag Zn (s) + 2 Ag + (aq)  Zn 2+ + 2 e = Zn 2 Ag + + 2 e  Zn  Zn 2+ 2Ag + 2 e Zn 2+ (aq) + 2 Ag (s) See slides 17 and 22

E

o – V 0.80

0.76

+ 0.80

0.76

+ 1.56 (= 

E

o ) 

G

o = – n F = – 2 * 96485 C * 1.56 V = – 301000 J = – 301 kJ 

E

o Write the cell for this rxn 21 Electrochemistry 28

Summary of thermodynamics

Chemical energy 

H

o , 

S

o Stoichiometry

n



G

o = 

H

o – T 

S

o Reaction quotient & equilibrium constant  G = 

G

o + R T ln Q 

G

o = – R T ln K 21 Electrochemistry Electric energy 

G

o = – n F 

E

o Reaction quotient & equilibrium constant  E = 

E

o – 

E

o =

R T / R T n F / n F

ln Q ln K 29

insight from cold denaturation and a two-state water structure

By Tsai CJ, Maizel JV Jr, Nussinov R. …The exposure of non-polar surface reduces the entropy and enthalpy of the system, at low and at high temperatures. At low temperatures the favorable reduction in enthalpy overcomes the unfavorable reduction in entropy , leading to cold denaturation. At high temperatures, folding/unfolding is a two-step process: in the first, the entropy gain leads to hydrophobic collapse, in the second, the reduction in enthalpy due to protein-protein interactions leads to the native state. The different entropy and enthalpy contributions to the Gibbs energy can be conveniently explained by a two-state model of the water structure….

change at each step at high, and at low, temperatures 21 Electrochemistry

pH and emf

Consider the cell,

Zn | Zn 2+ (1.00 M) || H + (x M) | H 2 (1.00 atm) | Pt

From table data , Zn 2+ + 2e = Zn 2 H + + 2 e = H 2 Zn = Zn 2+ + 2e Zn + 2H + = Zn 2+ + H 2

E

o = – 0.76

E

o = 0.00

E

o = 0.76



E

o = 0.76

 E = 

E

o

R T

[Zn 2+ ] P H2 – —— ln ———— 2

F

[H + ] 2 = 

E

o + 0.0592 log [H + ] At 298 K (pH meters) 0.76 – 

E

pH = ————— 0.0592

= 0.76 – 0.0592 pH 21 Electrochemistry 31

pH electrodes

pH Range: Temp. Range: Internal Ref: Junction: Dimensions: Slope:

0-14 0-100 C ROSS Ceramic 120 mm x 12 mm 92 - 102%

Temp. Accuracy: Catalog Number:

0.5 C 8202BN (BNC Connector, 1 meter cable) 21 Electrochemistry 32

Ion selective electrode

More research has gone into pH measurements. Nernst started it.

1. Concentration; 2. Temperature; 3. Electrode surface conditions; 4. Number of charges of ions (8); 5. Stirring (6); 6. Suspension (7); 7. Zwitterionic nature, net charge density; 8. Anything changing ionic adsorption; 9. Isoelectric nature of surface material; 10. The Nernst equation deals only with concentration and temperature 21 Electrochemistry 33

Battery technology

By use: automobile, flash light, radio, computer, camera, watch, emergency equipment, artificial heart machine, pace makers, hearing aids, calculators, … (portable energy) By type: alkaline, dry, wet, storage, rechargeable, etc By chemistry: alkaline, carbon zinc, dry, cell, lithium, lithium ion, lithium polymer, NiCd, etc.

Aluminum for battery manufacture By material: anode material, cathode material, electrode, etc.

21 Electrochemistry See 21-5 34

Lead Storage Battery for Autos

Anode – Negative plate Pb + SO 4 2  PbSO 4 + 2e Separator Cathode – Positive plate A 12-V battery consists of 6 such cells H 2 SO 4 PbO 2  + 4H + PbSO 4 + SO 4 2 + 2 H 2 O + 2e Net reaction: Pb + PbO 2 + 2 H 21 Electrochemistry 2 SO 4 = 2 PbSO 4 + H 2 O 35

21 Electrochemistry 36 10

A mercury battery.

21 Electrochemistry 37

Corrosion: Unwanted Voltaic Cells

Fe(s) O 2  + H 2 Fe 2+ (aq) + 2e – O (l) + 4e –  4 OH – (aq) 2 Fe(s) + O 2  + H 2 O 2 Fe 2+ (aq) + 4 OH – (aq) What are effective corrosion prevention methods?

Coating Use sacrifice electrode 21 Electrochemistry See 21-6 38

Cathodic protection of an underground pipe.

21 Electrochemistry 39

Ion displacement reactions (corrosion)

Reaction E o (V)

What metal will react with certain ions?

Zn + 2 Ag Zn + 2 Cu Zn Zn 2+ 2+ + 2+   Zn 2+ + 2 Ag  + Cu  See 21-1 Zn 2+ + 2 Ag + Cu Zn + 2 Ag Zn + 2 Cu 2+ + (reference) (reference)

2 H H 2

Cu 2+

+ + 2 e (g) = 2 H +

+ 2 e –

– = H 2 (g) 0.000

+ 2 e

= Cu (s)

– 0.000

0.34

21 Electrochemistry Br 2 Cl 2 F 2 I 2 Li + Na + + e + e – – = Li (s) – 3.04

= Na (s) – 2.71

Mg 2+ Zn 2+ Cu + + 2 e – + 2 e (s) + 2e Ag+ (aq) + e- = Ag (s) (l) + 2 e + e (g) + 2 e – (g) + 2 e – – – – – = Mg (s) – 2.38

= Zn (s) – 0.76

= Cu (s) = 2 I = 2 Br = 2 Cl – = 2 F – – – (aq) (aq) (aq) (aq) 0.52

0.54

0.80

1.07

1.36

2.87

40

Electrolysis of molten salts

2 Cl – 2 Na + Net  Cl 2 + 2 e – 2 NaCl  E o =-1.36 V; 2 Cl –  + 2e  – 2 Na 2 Na + Cl 2 oxidation Cl 2 + 2e – e A N O D E Battery e E o =-2.71V; 2Na + + 2e –  2Na reduction C A T H O D E 805 o C Molten salts consists of Na + and Cl – ions Charges required to produce 1 mole Cl 2 and 2 moles Na = 2F Energy required = 2 F  E; (  E > 4.07 V) 21 Electrochemistry See 21-7 41

Electrometallurgy of Sodium

21 Electrochemistry 42

Electrolysis of NaCl solution

Cl 2 2 Cl – = Cl 2 + 2e + H 2 O = HCl + ½ O 2 – oxidation A N O D E e Battery e 2Na + + 2e – 2Na + 2H + = 2Na = H 2 + 2Na + reduction C A T H O D E Salt solution consists of Na + and Cl – ions 21 Electrochemistry 43

Refining Copper by Electrolsis

Copper can be purified by electrolysis. Raw copper is oxidized Cu = Cu 2+ + 2 e and purer copper deposited on to the cathode from a solution containing CuSO 4 Cu 2+ + 2 e = Cu If a current of 2 amperes pass through the cell, how long will it take to deposit 5.00 g of copper on the cathode? (1 ampere = 1 C s –1 )

Solution

5.00 2 *96485 C 1 s ------ mol ----------- ------ = work out your answer 65.5 1 mol 2 C _______ 21 Electrochemistry New 44

Aluminum (Al), the third most abundant elements on Earth crust as bauxite or alumina Al Al metal by reducing Al 2 O 3 with potassium vapore. In 1886, two young men electrolyzed molten cryolite Na 3 AlF 6 2 O 3 , remain unknown to man until 1827, because it is very reactive. By then, Wohler obtained some (melting point 1000° C), but did not get aluminum.

Production of aluminum

Hall and Heroult tried to mix about 5% alumina in their molten cryolite, and obtained Al metal. This is the Hall process.

2 Al 2 OF 6 2– AlF 6 3– + 3 e + C(s) + 12 F – –   Al + 6 F – + 4 AlF 6 3– . . . Cathode + CO 2 + 4 e – . . . Anode 2 Al 2 O 3 + 3 C  4 Al + 3 CO 2 . . . Overall cell reaction Charge required for each mole Al = 3 F Energy required = 3 F 

E

21 Electrochemistry 45

Electrometallurgy of Aluminum

21 Electrochemistry 46

Electrolysis of acid solution

H 2 O = ½ O 2 + 2e – oxidation + 2 H + e A N O D E Battery e 2H + + 2e – = H 2 C A T H O D E reduction Solutions containing H + and SO 4 2– ions Charges required to produce 1 mole H 2 and ½ moles O 2 = 2F Energy required = 2 F 

E

21 Electrochemistry 47

Electrolysis of H

2

SO

4

solution

Pure water is not a good electric conductor. In the presence of electrolytes, water can be decomposed by electrolysis.

On the other hand, electrolysis of electrolyte solutions may reduce H + and oxidize O 2– in H 2 O.

In an H 2 H 2 2 SO 4 solution, cathode reductions are O (l) + 2 e – = H 2 (same as 2H + + 2e – = H 2 ) (g) + 2 OH – Anode oxidation: 2 H 2 O (l) = 4 e 2 SO 4 2– – + O 2 (g) + 4 H = [SO 3 O–OSO 3 ] 2– + + 2 e –

E E

o o = – 1.23 V (observed) = – 2.01 V (not observed) 21 Electrochemistry 48

Electrolysis of H

2

SO

4

solution

E

o

E

o

= – 1.23 V

= – 2.01 V

2 H 2 O (l) = 4 e

2 SO 4 2–

– + O 2 (g) + 4 H

= [SO 3 O–OSO 3 ] 2– + 2 e

+

– oxidation A N O D E e Battery e 2H + + 2e = H 2 reduction C A T H O D E Solution consists of H + and SO 4 2– ions 21 Electrochemistry 49

Galvanizing

Electroplating of metals

Zn 2+ + 2 e –  Zn onto metal surface Copper purification Cu 2+ + 2 e –  Cu onto pure Cu electrode Silver plating Ag + + e –  Ag onto metal surface Over a half century of extensive and innovative research has made us one of the nation's leading experts in plating on magnesium Since 1971, Cal-Aurum has provided electroplating services to the electronic component industry with the highest standards of quality, performance, and competitive pricing. Miller specializes in plating metals such as magnesium, aluminum, zinc, copper, powdered metals, steel and various other substrates.

21 Electrochemistry 50

Summary

The 20 th century belongs to electrons. They continue affecting our lives the 21 st century.

Chemistry studies the drama played by electrons, and electrochemistry is the finale.

Energy directs and produces the show, but you set the magic stage for a great performance. Leo and Ger tell electrons to get in and out of your stage, and you must skillfully provide paths to balance the flow.

Cells are the stages for the performance, you must construct, represent, figure out the potentials, and control the show.

Chemical reaction, equilibrium, (acid, base, heterogeneous, and complex formation) and electrochemistry guide us using simple rules.

Apply rules you have learned in Chem12 3 5 to understand what is happening around you and may your live be full of happiness .

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Skills for Electrochemistry (review)

Make up a Daniel cell using Pb and Ag as the electrodes. Draw a diagram for it.

Use short notation to represent the cell for the spontaneous reaction

Write half reaction equations for both cathode and anode and explain the reactions

Write balanced redox equations

Calculate emf for a nonstandard cell and its energy

Calculate equilibrium constant K from E

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