21 Electrochemistry
This presentation is the property of its rightful owner.
Sponsored Links
1 / 52

21 Electrochemistry PowerPoint PPT Presentation


  • 48 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

21 Electrochemistry

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Electricity

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


Electricity

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


Galvani

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


Electrochemistry

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


Electrons

Electrons

qe = –1.60217733e-19 C F = 96485 C me = 0.00054856 amu = 9.1093897e-31 kgspin = 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– / me) of cathode rays (electrons).

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

21 Electrochemistry


Redox reactions and electrons

Redox reactions and electrons

Energy drives chemical reactions.

Redox reactions involve the transfer of electrons.

Loss of electron (increase oxidation state) is oxidation, (Leo).Gain of electron (decrease oxidation state) is reduction, (ger).Zn  Zn2+ + 2 e–leo Cu2+ + 2 e – Cu gernet: Zn + Cu2+ Cu + Zn2+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


Galvanic cell

Galvanic Cell

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

Representation:Zn | Zn2+ || Cu2+ | Cu

21 Electrochemistry


Assign oxidation states

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 H2O2, Na2O2

+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.

–3NH3

–2 N2H4

–1NH2OH

0N2

+1 N2O

+2 NO

+3 NO2–

+4 NO2

+5 NO3–

–1Cl–

0Cl2

+1 ClO–

+3 ClO2–

+4 ClO2

+5 ClO3–

+7 ClO4–

21 Electrochemistry


Half reaction equations

Half reaction equations

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

Zn  Zn2+ + 2 e–leo

Cu2+ + 2 e –  Cuger

net: Zn + Cu2+ Cu + Zn2+redox

Demonstrate how to balance these

Fe2+ Fe3+ + __ e–

C2O42- 2 CO2 + __e –

MnO4 – + __e– Mn2+

Cr2O72– + __e– + __ H+ 2 Cr3+ + __ H2O

Steps to balance half reaction

Assign oxidation number

Figure out what is oxidized or reduced.

Add electrons according to oxidation number change

Balance charge with H+ (acid) or OH – (base)

Balance atoms with H2O

21 Electrochemistry


More half reaction equations

More half-reaction equations

2 I– I2 + __e–

ClO2 + __ OH– ClO3- + __e– + H2O (in basic solution)

2 S2O32–  S4O62- + __e–

HS(=S)O3–S + __e- + HSO4–

__ H3O+ + __e – H2(g) + __ H2O

H2O2 + __e – 2 H2O

ClO2 + __e – ClO2–

NO3– + __e – NH4+

21 Electrochemistry


Electrochemical series

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


Constructing half cells

Constructing half cells

A half cell consists of an oxidizing and its oxidized species

Zn | Zn2+Cu | Cu2+Pt | H2 | H+(Pt as conductor)Pt |Fe2+ , Fe3+Cl– | Cl2 | Pt

Student cell set from School Master Science $30

Explain the cell convention and reactions of cells.

21 Electrochemistry


Galvanic cells

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


Cell convention

Cell convention

Oxidation takes place always at the anodeZn (s) Zn2+ (aq) + __ e– Zn(s) | Zn2+(aq)

Fe2+ (aq) Fe3+ (aq) + e– Pt | Fe2+ , Fe3+

H2 (g) 2 H+ (aq) + __ e– Pt | H2(g) | H+(aq)

Reduction takes place at the cathodeCu2+ (aq) + __ e– Cu (s) Cu2+ | Cu(s)

Cl2 (g) + __ e– Cl– (aq) Cl2(g) | Cl –(aq) | Pt

Fe3+ (aq) + __ e – Fe2+ (aq) Fe3+(aq), Fe2+(aq)| Pt

2 H+ (aq) + __ e– H2 (g) H+(aq) | H2(g)| Pt

Concentration (1.1 M) and pressure (0.9 atm) are also included in cell notations.

21 Electrochemistry


Electric energy and work

Electric energy and work

Electric energy or electric work = charge * potential difference

W = q * V (1 J = 1 Coulomb Volt, C V)compare W = m g h

The Faraday constant F is the charge for one mole of electrons,F = 96485 C; F / NA = 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G = – n F E (n F= q is the charge)electromotive force (emf, or V)number of electrons in the reaction equationn F is the charge q

See slides in 16 Equilibria

21 Electrochemistry


Standard cell emf s and electrode potentials

Standard cell emf’s and electrode potentials

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

Zn (s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 1.10 V

Mg (s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 2.90 V

The absolute potential of the electrode cannot be determined. Only relative potential can be measured. The standard reduction potential is measured against a SHE, for which Eo = 0.0000 V

Zn (s) | Zn2+(aq, 1 M) || H+(aq, 1 M) | H2(1 atm)(g) Eo = 0.76 V

Zn = Zn2+ + 2 e–(oxidation, displace H+ possible)Eo = 0.76 V

Zn2+ + 2 e– Zn (reduction)Eo = – 0.76 V

21 Electrochemistry


Gibb s free energy in a cell

Gibb’s Free Energy in a Cell

How much energy is available for the cell Zn | Zn2+ || Ag+ | Agoperating at standard condition when one mole of Zn is consumed ?

SolutionEo Zn = Zn2+ + 2 e0.762 V 2 Ag+ + 2e = 2 Ag 0.799 V (from table) Zn + 2 Ag+ = Zn2+ + 2 Ag Eo = 1.561 V

Go = – nFE = – 2 * 96485 C * 1.561 V = 301226 J (1J = 1 C V) = 301.2 kJ

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

21 Electrochemistry


Table of standard reduction potential

Table of standard reduction potential

Reaction E o (V)

Li+ + e– = Li (s) – 3.04Na+ + e– = Na (s) – 2.71Mg2+ + 2 e– = Mg (s) – 2.38Zn2+ + 2 e– = Zn (s) – 0.76(reference)2 H+ + 2 e– = H2 (g) 0.000(reference)H2 (g) =2 H+ + 2 e– 0.000Cu2+ + 2 e– = Cu (s) 0.34Cu+ + e– = Cu (s) 0.52I2 (s) + 2e– = 2 I– (aq) 0.54Br2 (l) + 2 e– = 2 Br– (aq) 1.07Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36F2 (g) + 2 e– = 2 F– (aq) 2.87

Standard cell potentials

Cell E o

Li | Li+ || Cu2+ | Cu____

Mg | Mg2+ || I2 | I– | Pt ____

Zn | Zn2+ || Br2 | Br– | Pt ____

Cu | Cu2+ || Zn2+ | Zn____

Which is not spontaneous?

See 21-2

21 Electrochemistry


Table of standard reduction potential1

Table of standard reduction potential

Reaction E o (V)

F2 (g) + 2 e– = 2 F– (aq) 2.87Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36Br2 (l) + 2 e– = 2 Br– (aq) 1.07I2 (s) + 2e– = 2 I– (aq) 0.54Cu+ + e– = Cu (s) 0.52Cu2+ + 2 e– = Cu (s) 0.34(reference)H2 (g) =2 H+ + 2 e– 0.000(reference)2 H+ + 2 e– = H2 (g) 0.000Zn2+ + 2 e– = Zn (s) – 0.76Mg2+ + 2 e– = Mg (s) – 2.38Na+ + e– = Na (s) – 2.71Li+ + e– = 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


Strength of oxidation

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 of an oxidizing agent is measured by its reduction potential.

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


Reaction direction and emf

Reaction direction and emf

What is the emf for the reaction, Zn2+ (aq) + 2Fe2+ (aq) = Zn (s) + 2Fe3+ (aq)?

Solution: Know what data to look forZn2+ + 2 e  ZnEo = – 0.76 VFe3+ + e  Fe2+Eo = + 0.77 V +2 Fe2+2 Fe3+ + 2 eEo = – 0.77 VZn2+ (aq) + 2 Fe2+ (aq)  Zn (s) + 2Fe3+ (aq) Eo = – 1.53 Vnon-spontaneous Pt | Zn2+ | Zn || Fe2+ | Fe3+ | Pt impractical

The reverse reaction is spontaneous, Zn (s) + 2Fe3+ (aq)  Zn2+ (aq) + 2Fe2+ (aq) Eo = + 1.53 VZn | Zn2+ || Fe3+ | Fe2+ | Pt

21 Electrochemistry


Free energy and emf

Free energy and emf

What is the free-energy change for the cell, Zn | Zn2+ (aq) 1 M || Ag+ (aq) 1 M | Ag?

Solution: Reduction potential required,Zn  Zn2+ + 2eEo = 0.76 Ag+ + e  AgEo = 0.80 2 Ag+ + 2e  2AgEo = 0.80

Cell reaction 2 Ag+ + Zn  2Ag + Zn2+ Eo = 1.56 V

Go = –n F Eo = – 2 * 96485 * 1.56 = – 3.01e5 J or – 302 kJ

Negative indicate energy is released.

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

Condition for spontaneous reaction is – G or + E.

21 Electrochemistry


General cell emf

General cell emf

G o is the standard energy change. G is for non-standard conditions.

G = – n F E

Similarly, Eo is the standard emf whereas E is general emf.

G = Go + R T ln Q E = Eo – R T / n Fln Qreaction quotient

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

G = Go + R T lnK= 0 E = Eo – R T / n Fln K = 0equilibrium constant

Go = – R T lnKE o = R T / n Fln K

orGo = – ln(10) R T logKE o = 2.303 R T / n FlogK

Text uses Ecell instead of E

At 298 K 0.0592E o = ———— log K n

21 Electrochemistry


The nernst equation

The Nernst equation

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

R T[C]c [D]dE = E o – ——— 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.

See 21-4

21 Electrochemistry


Evaluating e

R T[C]c [D]dE = E o – ——— ln ————n F[A]a [B]b

Evaluating E

At 300 K, evaluate the cell emf for Zn | Zn2+ (0.100 M) || H+ (0.200 M) | H2 (1.111 atm) | Pt

Solution:Look up:Zn  Zn2+ (aq) + 2e, E o = 0.76 V2 H+ (aq) + 2 e  H2 (g), E o = 0.00 V (R = 8.314 J mol-1 K-1, F = 96485 C mol-1)E o = 0.76 VThe reaction is Zn(s) + 2 H+ (aq)  Zn2+ (aq) + H2(g)

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

See example 19.12

21 Electrochemistry


Concentration cell

Concentration cell

Problem: At 298 K, evaluate the emf of the cellCu | Cu2+ (0.10 M) | | Cu2+ (1.0 M) | Cu Cu(s)  Cu2+(0.1 M) + 2e; Cu2+ (1.0 M) + 2e  Cu(s)

Solution:The standard emf (Eo = 0.00)for Cu | Cu2+ || Cu2+ | Cu The reaction is actually Cu (s) + Cu2+ (1.0 M) = Cu2+ (0.1 M) + Cu (s)

R T [Cu2+] R T 0.10E = 0.00 – ——- ln ——— = – –––– ln ––––– 2 F [Cu2+] 2 F 1.00

8.3145 * 298 1.0 = +–––––––––– ln –––– = 0.0295 V 2*96485 0.1The voltage is purely due to concentration difference. Solutions in the two compartment try to become equal.

When 2 [ ]’s are equal, E = 0

See p. 841

21 Electrochemistry


Equilibrium constant k and e o cell

Equilibrium Constant K and Eocell

Calculate the solubility product of AgCl from data of standard cell

Solution: Look up desirable data

Ag+Cl– (s) + e  Ag0(s) + Cl–E° = 0.2223 V

Ag+ (aq) + e  Ag (s) E° = 0.799 V

Ag (s)  Ag+ (aq) + e E° = – 0.799 V

Get the desirable eq’n AgCl (s)  Ag+ (aq) + Cl– (aq)DE° = – 0.577 V

Ksp = [Ag+][Cl– ]

log Ksp = – 0.577 / 0.0592 = – 9.75Ksp = 10– 9.75 = 1.8e–10

Show that for this cellAg | Ag+,1 M || Cl–,1 M | AgCl | AgEo = – 0.577 Vbut for this cellAg | AgCl |Cl–, 1 M || Ag+,1 M | AgEo = +0.577 V

At 298 K 0.0592Eo = ———— log K n

21 Electrochemistry

See p. 837


Evaluate free energy change

Evaluate free-energy change

Evaluate G o for the reaction

Zn (s) + 2 Ag+ (aq) = Zn2+(aq) + 2 Ag (s)

Solution:

Required to look up: Eo VAg+ + e = Ag 0.80

Zn2+ + 2 e = Zn – 0.76

2 Ag+ + 2 e  2Ag 0.80+

Zn  Zn2+ + 2 e + 0.76Zn (s) + 2 Ag+ (aq)  Zn2+(aq) + 2 Ag (s)1.56 (= E o)

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

See slides 17 and 22

Write the cell for this rxn

21 Electrochemistry


Summary of thermodynamics

Summary of thermodynamics

Chemical energyHo, So

Stoichiometry

n

Electric energyGo = – n F Eo

Go = Ho – T So

Reaction quotient &equilibrium constant

G = G o + R T ln QG o = – R T ln K

Reaction quotient &equilibrium constant

E = E o – R T/n F ln QE o = R T/n F ln K

21 Electrochemistry


Insight from cold denaturation and a two state water structure

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 change at each step at high, and at low, temperatures can be conveniently explained by a two-state model of the water structure….

  • Biochem Mol Biol. 2002;37(2):55-69

21 Electrochemistry


Ph and emf

pH and emf

Consider the cell,

Zn | Zn2+ (1.00 M) || H+ (x M) | H2 (1.00 atm) | Pt

From table data, Zn2+ + 2e = Zn Eo = – 0.762 H+ + 2 e = H2Eo = 0.00Zn = Zn2+ + 2e Eo = 0.76Zn + 2H+ = Zn2+ + H2Eo = 0.76

R T [Zn2+] PH2E = Eo – —— ln ————2 F [H+]2

At 298 K (pH meters)

0.76 – EpH = —————0.0592

= Eo + 0.0592 log [H+]

= 0.76 – 0.0592 pH

21 Electrochemistry


Ph electrodes

pH electrodes

pH Range: 0-14Temp. Range: 0-100 CInternal Ref: ROSSJunction: CeramicDimensions: 120 mm x 12 mmSlope: 92 - 102%Temp. Accuracy: 0.5 CCatalog Number:8202BN (BNC Connector, 1 meter cable)

21 Electrochemistry


Ion selective electrode

Ion selective electrode

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

  • Concentration;

  • Temperature;

  • Electrode surface conditions;

  • Number of charges of ions (8);

  • Stirring (6);

  • Suspension (7);

  • Zwitterionic nature, net charge density;

  • Anything changing ionic adsorption;

  • Isoelectric nature of surface material;

  • The Nernst equation deals only with concentration and temperature

21 Electrochemistry


Battery technology

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.

By material: anode material, cathode material, electrode, etc.

Aluminum for battery manufacture

See 21-5

21 Electrochemistry


Lead storage battery for autos

Lead Storage Battery for Autos

Anode – Negative platePb + SO42-  PbSO4 +  2e-

Separator

Cathode – Positive plate

A 12-V battery consists of 6 such cells

H2SO4

PbO2 +  4H+  +  SO42- +  2e-   PbSO4 + 2 H2O

Net reaction: Pb + PbO2 + 2 H2SO4 = 2 PbSO4 + H2O

21 Electrochemistry


Electricity

21 Electrochemistry


A mercury battery

A mercury battery.

21 Electrochemistry


Corrosion unwanted voltaic cells

Corrosion: Unwanted Voltaic Cells

Fe(s)  Fe2+ (aq) + 2e–O2 + H2O (l) + 4e– 4 OH– (aq)

2 Fe(s) + O2 + H2O  2 Fe2+ (aq) + 4 OH– (aq)

What are effective corrosion prevention methods?

CoatingUse sacrifice electrode

21 Electrochemistry

See 21-6


Cathodic protection of an underground pipe

Cathodic protection of an underground pipe.

21 Electrochemistry


Ion displacement reactions corrosion

Ion displacement reactions (corrosion)

Reaction E o (V)

Li+ + e– = Li (s) – 3.04Na+ + e– = Na (s) – 2.71Mg2+ + 2 e– = Mg (s) – 2.38Zn2+ + 2 e– = Zn (s) – 0.76(reference)2 H+ + 2 e– = H2 (g) 0.000(reference)H2 (g) =2 H+ + 2 e– 0.000Cu2+ + 2 e– = Cu (s) 0.34Cu+ + e– = Cu (s) 0.52I2 (s) + 2e– = 2 I– (aq) 0.54Ag+ (aq) + e- = Ag (s) 0.80Br2 (l) + 2 e– = 2 Br– (aq) 1.07Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36F2 (g) + 2 e– = 2 F– (aq) 2.87

What metal will react with certain ions?

Zn + 2 Ag+ Zn2+ + 2 Ag

Zn + 2 Cu2+ Zn2+ + Cu

Zn2+ + 2 Ag  Zn + 2 Ag+

Zn2+ + Cu  Zn + 2 Cu2+

See 21-1

21 Electrochemistry


Electrolysis of molten salts

Electrolysis of molten salts

Eo=-2.71V;2Na+ + 2e– 2Na

Battery

Eo=-1.36 V; 2 Cl– Cl2 + 2e–

e

e

oxidation

reduction

ANODE

CATHODE

805o C

Molten salts consists of Na+ and Cl– ions

2 Cl– Cl2 + 2e–

2 Na+ + 2 e– 2 Na

Net2 NaCl  2 Na + Cl2

Charges required to produce 1 mole Cl2 and 2 moles Na = 2F

Energy required = 2 FE; (E > 4.07 V)

21 Electrochemistry

See 21-7


Electrometallurgy of sodium

Electrometallurgy of Sodium

21 Electrochemistry


Electrolysis of nacl solution

Electrolysis of NaCl solution

2 Cl– = Cl2 + 2e–Cl2 + H2O = HCl + ½ O2

2Na+ + 2e– = 2Na2Na + 2H+ = H2 + 2Na+

Battery

e

e

oxidation

reduction

ANODE

CATHODE

Salt solution consists of Na+ and Cl– ions

21 Electrochemistry


Refining copper by electrolsis

Refining Copper by Electrolsis

Copper can be purified by electrolysis. Raw copper is oxidized

Cu = Cu2+ + 2e

and purer copper deposited on to the cathode from a solution containing CuSO4

Cu2+ + 2e = 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 New65.5 1 mol 2 C _______

21 Electrochemistry


Production of aluminum

Aluminum (Al), the third most abundant elements on Earth crust as bauxite or alumina Al2O3, remain unknown to man until 1827, because it is very reactive. By then, Wohler obtained some Al metal by reducing Al2O3 with potassium vapore.

In 1886, two young men electrolyzed molten cryolite Na3AlF6 (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.

AlF63– + 3 e– Al + 6 F– . . . Cathode 2 Al2OF62– + C(s) + 12 F– + 4 AlF63– + CO2 + 4 e– . . . Anode

2 Al2O3 + 3 C  4 Al + 3 CO2. . . Overall cell reaction

Charge required for each mole Al = 3 F

Energy required = 3 FE

21 Electrochemistry


Electrometallurgy of aluminum

Electrometallurgy of Aluminum

21 Electrochemistry


Electrolysis of acid solution

Electrolysis of acid solution

Battery

H2O = ½ O2 + 2e– + 2 H+

2H+ + 2e– = H2

e

e

oxidation

reduction

ANODE

CATHODE

Solutions containing H+ and SO42– ions

Charges required to produce 1 mole H2 and ½ moles O2 = 2F

Energy required = 2 FE

21 Electrochemistry


Electrolysis of h 2 so 4 solution

Electrolysis of H2SO4 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 O2– in H2O.

In an H2SO4 solution,cathode reductions are2 H2O (l) + 2 e– = H2 (g) + 2 OH–(same as 2H+ + 2e– = H2)

Anode oxidation:2 H2O (l) = 4 e– + O2 (g) + 4 H+E o = – 1.23 V (observed)2 SO42– = [SO3O–OSO3]2– + 2 e–E o = – 2.01 V (not observed)

21 Electrochemistry


Electrolysis of h 2 so 4 solution1

Electrolysis of H2SO4 solution

E o = – 1.23 V2 H2O (l) = 4 e– + O2 (g) + 4 H+E o = – 2.01 V2 SO42– = [SO3O–OSO3]2– + 2 e–

Battery

2H+ + 2e = H2

e

reduction

e

oxidation

ANODE

CATHODE

Solution consists of H+ and SO42– ions

21 Electrochemistry


Electroplating of metals

Electroplating of metals

GalvanizingZn2+ + 2 e– Zn onto metal surface

Copper purification Cu2+ + 2 e– Cu onto pure Cu electrode

Silver platingAg+ + 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


Summary

Summary

The 20th century belongs to electrons. They continue affecting our lives the 21st 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 Chem1235 to understand what is happening around you and may your live be full of happiness.

21 Electrochemistry


Skills for electrochemistry review

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 Eo

21 Electrochemistry


  • Login