reaksi anorganik dalam medium air 1

REAKSI ANORGANIK DALAM MEDIUM AIR:

REAKSI REDOX DAN ELEKTROKIMIA

Oxidation and reduction in terms of oxygen transfer

Definitions

Oxidation is gain of oxygen.

Reduction is loss of oxygen.

For example, in the extraction of iron from its ore:

Because both reduction and oxidation are going on side-by-side, this is known as a redox reaction.

Redox reaction

Oxidation and reduction in terms of electron transfer

This is easily the most important use of the terms oxidation and reduction at A' level.

Definitions

Oxidation is loss of electrons.

Reduction is gain of electrons.

Copper(II) oxide and magnesium oxide are both ionic. The metals obviously aren't. If you rewrite this as an ionic equation, it turns out that the oxide ions are spectator ions and you are left with:

Oxidation and reduction always occur simultaneously

Losing electrons is oxidation, and the substance that loses the electrons is called the reducing agent.Gaining electrons is reduction, and the substance that gains the electrons is called the oxidizing agent.

Mg is oxidized: loses e-, becomes a Mg2+ ion

S is reduced: gains e- = S2- ion

Mg is the reducing

agent

S is the oxidizing agent

Mg(s) + S(s) → MgS(s)

Not All Reactions are Redox ReactionsNot All Reactions are Redox Reactions

- Reactions in which there has been no change in oxidation number are NOT redox reactions.

Examples:

)()()()( 3

2511111

3

251

aqONNasClAgaqClNaaqONAg

)()()()(22

2

1

4

26

2

1

4

26

2

1121

lOHaqOSNaaqOSHaqHONa

O- I- O O

-

The I-O bonds are polar covalent, polarized toward the more electronegative oxygen. If these bonds were ionic, then the shared electron density would reside entirely with the oxygens

This leads to an oxidation number of -2 for each oxygen (eight valence electrons versus six in the free atom) and +5 for iodine (two valence electrons versus seven).

As is the case for formal charge, the algebraic sum of all oxidation numbers equals the charge on the species

Oxidation involves an increase in oxidation stateReduction involves a decrease in oxidation state

Oxidation Numbers that occurs in compounds

Rules for Assigning Oxidation NumbersRules for Assigning Oxidation Numbers

1. The oxidation number of any uncombined element is zero.

2. The oxidation number of a monatomic ion equals its charge.

11

2

00

22

ClNaClNa

3. The oxidation number of oxygen in compounds is -2, except in peroxides, such as H2O2 where it is -1.

4. The oxidation number of hydrogen in compounds is +1, except in metal hydrides, like NaH, where it is -1.

2

2

1

OH

5. The sum of the oxidation numbers of the atoms in the compound must equal 0.

2

2

1

OH2(+1) + (-2) = 0 H O

2

122

)(

HOCa(+2) + 2(-2) + 2(+1) = 0 Ca O H

6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its ionic charge.

3

2?

ONX + 3(-2) = -1N O

24

2?

OS

thus X = +5 thus X = +6

X + 4(-2) = -2S O

Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents

Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

With electon configuration 1s2 2s2 2p6 3s2,magnesium has two valence electrons.

Consequently,the first two energies of Mg are small relative to IE3 and beyond.

Thus,it is reasonable to think of a Mg2+ ion, but no cations of higher charge exist

Electron transfer in Aqueous Solution

• Recall the basic eqiation relating enthalpy, entropy and free energy:

• For aqueous solution, it is convenient to use the electrochemical potential in volts as a substitute for free energy

• Standard potential can be related to the equilibrium constant by the equation:

G = H-TS

G = - nFE

log K = nE/0,05916

Pada reaksi redoks dikenal potensial standar reduksi yaitu harga potensial sel standard dari reaksi setengah sel yang diukur dengan pembanding potensial standard reduksi dari hydrogen. Keadaan standar diukur pada temperatur 250C, tekanan 1 atm dan konsentrasi 1M. Reaksi reduksi H+ menjadi H2 dalam keadaan standard memiliki harga E0=0.

2H+ + 2e- H2 E0 = 0,000 V

Harga potensial standard reduksi lainnya adalah harga relatif dengan pembanding elektroda hidrogen standard.

Standard reduction potentials

Standard reduction potentials

Define a standard half-cell of potential = 0V against which all other half-cell reduction potentials are measured. Each component in these standard cells having unit concentration. By convention: Standard (or Normal) Hydrogen Electrode is used

Pt(s) | H2(g) | H+(a) || Ag+(a) | Ag(s) |_______________|

NHEH+(aq) + e- H2(g)E0=0V

- +e-

AgSalt bridge

AAg+=1

V=+0.799V

AH+=1

AH2=1

K+ + e- K E0 = -2,970 V.

Reduksi K+ menjadi K memiliki harga potensial standard reduksi negatif, lebih rendah dibandingkan E0

H+/H2. Arah reaksi seperti tertulis secara thermodinamika tidak berjalan, sehingga reduksi K+ menjadi K bukanlah reaksi yang spontan.

Reaksi sebaliknya dengan harga potensial standard oksidasi 2,970 V lebih dapat berjalan jika ditinjau dari segi thermodinamika. Kespontanan reaksi yang secara kuantitatif diukur dengan G didukung dari harga E0 positif. Hubungan harga G dengan harga E0 adalah

G = -nF E0

F adalah konstanta Faraday (96,487 kJ/mol V) sedangkan n adalah

jumlah elektron yang ditransfer saat proses redoks.

Basic Cell for Measurement of Electrode Potentials

SHE Half Cell

The cell potential 

E = E+ - E- (written as reductions)

Standard Reduction PotentialsStandard Reduction Potentials

• Consider Zn(s) Zn2+(aq) + 2e-. We measure Ecell relative to the SHE (cathode):

Ecell = E+(cathode) - E-(anode)

0.76 V = 0 V - Ered(anode).

• Therefore, Ered(anode) = -0.76 V.

• Standard reduction potentials must be written as reduction reactions:

Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.

• Since Ered = -0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous.

Electrochemical series  

half-rxns oxidant reductant E0 (V)

   stronger oxidant F2(g) + 2e- <=> 2F- 2.890 Ce4+ + e- <=> Ce3+ 1.720 Ag+ + e- <=> Ag(s) 0.799 Fe3+ + e- <=> Fe2+ 0.771 O2 + 2H+ + 2e- <=> H2O2 0.695 Cu2+ + 2e- <=> Cu(s) 0.339 2H+ + 2e- <=> H2(g) 0.000 Cd2+ + 2e- <=> Cd(s) -0.402 Zn2+ + 2e- <=> Zn(s) -0.762 K+ + e- <=> K(s) -2.936 Li+ +e- <=> Li(s) -3.040 stronger reducer

G = -nFE

Therefore if Ecell is positive, the reaction is spontaneous

J/V.mol 96,500

C/mol 500,961

F

Hubungan entalpi, entropi dan energi bebas gibbs:

G = H-TS

Hubungan harga G dengan harga E0 :

G = -nF E0

F adalah konstanta Faraday (96,487 kJ/mol V) sedangkan n adalah jumlah elektron yang ditransfer saat proses

redoks.

Hubungan potensial standard reduksi dengan harga Q

adalah mengikuti persamaan Nernst

ln Q = n E0/RT

atau

log Q = n E0/0,05916

Pada keadaan non standard harga E (tidak ada tanda 0 pada E untuk keadaan non standard) adalah sebesar:

E = E0 - RT ln Q / n = E0 - 0,05916 log Q/n

Diagram potensial reduksi menunjukkan harga potensial standard reduksi pada beberapa harga tingkat oksidasi

E10= +0,682 E20 = +1,776

O2 H2O2 H2O

E30 = +1,229

Pada diagram potensial tersebut O memiliki 3 tingkat oksidasi yaitu 0, -1, dan –2. Dengan melihat hubungan antara G dengan E0 maka harga G akan sebanding dengan harga E0 sehingga jika G bersifat aditif maka E0 juga bersifat aditif. Pada contoh diagram potensial diatas maka hubungan besaran E1

0, E20 dan E3

0 adalah

n3E30 = n1E1

0 + n2E20 (‘volt equivalent’)

Volt equivalent

FeO42- Fe3+ Fe2+ Fe

+2,20 +0,771, n2 -0.440, n3

n4, E04?

E04 = -0,036 V

Potensial sel (overall cell potentials) merupaka driving force reaksi redoks.

Harga potensial sel positif menunjukkan reaksi berjalan sesuai dengan arah reaksi tertulis.

Driving force dari reaksi dicerminkan dari harga konstanta equilibrium, K, dan perubahan energi bebas Gibbs, G.

Dari hubungan log K dengan Esell maka harga K yang tinggi didapatkan dari harga Esell yang tinggi dan harga G yang negatif (sejumlah energi dibebaskan) didapatkan dari harga Esell positif.

E = E0 - RT ln K / n = E0 - 0,05916 log K/n

G = -n F E0

Driving force reasi redoks

Note, however, that if we are combining two half reactions to obtain a third half reaction, the values are not additive, since we are not eliminating electrons. Free energies are always additive, so we combine them.

It is obvious that the iron reaction will have to happen twice for every chlorine molecule that reacts. Allow for that, and then add the two half-equations together.

The reaction between chlorine and iron(II) ions

The reaction between hydrogen peroxide and manganate(VII) ions

The oxidation of ethanol by acidified potassium dichromate(VI)

2 x [MnO42- + 4H+ +2e- MnO2 + 2H2O] E0=+2,26 V

2H2O O2 +4H+ + 4e- E0= -1,23 V

2MnO42- + 4H+ MnO2 + 2H2O + O2 E0 =1,03 V

Electrone gained in the reduction must equal the electrons lost in the oxidationPotential is independent of stoichiometry

Potential for overall process is positive reaction is spontaneous under standard condition

2 x [OCl- + 4H2O + 2e- ] Cl- + 2OH-] E0=+0,89 V

OCl- + 4OH- ClO3- +2H2O + 4e- E0= -0,5V

3OCl- Cl- + ClO3- E0 =0,39 V

ClO4- ClO3

- ClO2- OCl- Cl2 Cl-

Standard reduction potential diagram for Cl in basic solution:

0,36 0,33 0,66 0,40 1,36

0,50 0,89

The influence of non aqueous ligand on standard reduction potential of Fe3+/Fe2+ couple

• The presence of coordinating ligand can play an important role in stabilizing or destabilizing the oxidation state of transition metals.

Ligand E0, V Ligand E0, V

H2O

OH-

C2O42-

+0,77

-0,56

+0,02

CN-

Bipy

Phen

+1.10

+0.36

+1.12

Fe(H2O)6 2+ Fe(H2O)6 3+ E0 = -0,77

Fe(phen) 2+ Fe(phen) 3+ E0 = -1,12

Reduction of CoIII CoII

4 Co3+ + 2 H2O 4 Co2+ + 4 H+ + O2 Eo = +0,579 V

[Co(NH3)6]3+ + 1 e- [Co(NH3)6]2+ Eo = +0,11 V

[Co(CN)6]3+ + 1 e- [Co(CN)6]2+ Eo = -0,83 V

Application of a combined oxidation-nonaqueous ligand

2Au + 6H+ 2Au3+ + 3H2

NO3- 4H+ + 3e- NO + 2 H2O

Au + NO3- + 4 Cl- + 4 H+ AuCl4- +NO + 2 H2O

E0 =-1,52 V

E0 = 0,96 V

Au + 4 Cl- + 4H+ AuCl4- + 4H+ + 3e- E0 = -0,99 V

oksidasi

reduksi

E0 = -0,03

Non standard concentrations

Non standard temperature

Non standard pH

Non standard concentration

NiO2 Ni2+ Ni

+1,678 -0,250

E = -0,250 -

0,05916 log (1/[Ni2+])

2

= -0,250 + 0,02958 log [Ni2+]

Non standard Temperature

E = Eo - (1,985 x 10-4) T. Log Q

n

E = -0,250 + (9,925 x 10-5) T. Log [Ni2+]

The potential changes very slowly with changing temperature

Non standard pH: pH effect

NiO2 Ni2+ Ni

+1,678 -0,250

NiO2 + 4 H+ + 2e- Ni2+ + 2 H2O Eo = +1,678 V

E = Eo - log 0,05916

n

[Ni2+]

[H+]4

E = +1,678 - 0,1183 . pH

Jika konsentasi Ni2+ = 1M

Ni + 2 H+ Ni 2+ + H2 E0 = 0,250 V

Buktikan bahwa pada pH 5, E = -0,046

NiO2 Ni(OH)2 Ni

+0,490 -0,720

Solubility effect

Ni(OH)2 Ni2+ + 2 OH- Ksp = 1,5 x 10-16

Ni2+ + 2e- Ni E01 = -0,250 V

Ni(OH)2 + 2e- Ni + 2 OH- E02 = ?

0,05916.log Ksp + (2 x -0,250) = 2 x E02

E02 = -0,72 V

"Electron Activity" (pe)

• We can imagine that electrons have a measurable activity and use -log (e) = pe, instead of EH.

• For a half cell reaction.

oxidized + mH+ + ne = reduced

K (reduced)

(oxidized)(H )m en

"Electron Activity" (pe) (cont.)

at 25°C

• Let pe = -log(e-)

• then

logK nlog(e-) log

(reduced)

(oxidized) mlog(H

+)

"Electron Activity" (pe) (cont.)

• log K is sometimes called pe0

• ***Remember*** Electron activity is a convenient fiction, not a physical reality.

logK npe log(reduced)

(oxidized) mpH

We also included protons in the Nernst equation

EH E0 RT

2.3nFlog

(reduced)

(oxidized) mpH

Comparison of the Nernst and pe equations

log(reduced)

(oxidized) mpH logK npe

RT

2.3nFlog

(reduced)

(oxidized) mpH

E0 EH

Equate the two equations

2.3nF

RTE0 E H logK - npe

At 25 C

n

0.059E0 EH logK - npe

pe can easily be calculated from EH

pe EH

0.059

pe0 = logK nEH

0

0.059

Combining half cell reactions to describe a complete reaction.

• Question: What is the equilibrium constant for the oxidation of Fe2+ by O2?

log K

Fe2+ = e- + Fe3+ -13.04

H+ + 1/4O2 + e- = 1/2H2O 20.78

---------------------------------------------------------------

H+ + 1/4O2 + Fe2+ = 1/2H2O + Fe3+ 7.74

– This equation says that O2 can readily oxidize Fe2+ at acid pH values

Combining half cell reactions (cont.)

K = 107.7 (Fe3 )

(H )(Fe2 )(PO2)

1

4

log K = pH -1/4log Po2 + log(Fe3+)/(Fe2+)]

Combining half cell reactions (cont.)

• Find the pH at which (Fe3+) = (Fe2+) at Po2 = 0.21 atm

Answer

– 7.7 = pH + .17– pH = 7.57

Combining half cell reactions with solubility reactions

• If (Fe3+) is controlled by the solubility of Fe(OH)3 in a soil then under reducing (flooded soil) conditions Fe2+ can be calculated as a function of pH and EH.

• Fe(OH)3 + 3H+ = Fe3+ + 3H2O log K = 2.70 (Lindsay)

• Predict Fe2+ as a function of pe in soil

Combining half cell reactions with solubility reactions(cont.)

log KFe3+ + e- = Fe2+ 13.04

Fe(OH)3 + 3H+ = Fe3+ + 3H2O 2.70Soil

----------------------------------------------------------------------

Fe(OH)3 + e- + 3H+ = Fe2+ + 3H2O 15.74

EH = 0.93 V

Combining half cell reactions with solubility reactions(cont.)

logK pe + log(Fe2 )

(H )3 pe + log(Fe2) + 3pH

log(Fe2 ) logK - pe - 3pH

Combining half cell reactions with solubility reactions(cont.)

• If measured pH = 7.0 and

• (Fe2+) = 1.0 x 10-5 M, what is pe?

Answer

pe = 15.74 + 5 - 3(7)

pe = 15.74 - 16.0

pe = -0.26 and EH = -0.014

Combining half cell reactions with solubility reactions(cont.)

• Plot pe (or EH) vs. pH

• Example: Fe(OH)3 + 3H+ + e- = Fe2+ + 3H2O

pe = log K - 3pH - log(Fe2+)– at log (Fe2+) = -5

pe = 20.74 - 3pH

pe - pH plot for Fe

Precipitation of Siderite

» log K

Fe2+ + CO32- = FeCO3 10.8

H2O + CO2 = H++ HCO3- - 7.81

HCO3- = H+ + CO3

2- -10.33

-------------------------------------------------------

Fe2+ + H2O + CO2 = 2H+ + FeCO3 - 7.34

FeCO3 (cont.)

• From earlier slide

Fe(OH)3 + e- + 3H+ = Fe2+ + 3H2O +15.74

soil

Fe2+ + H2O + CO2 = 2H+ + FeCO3 - 7.34

--------------------------------------------------------------------

Fe(OH)3 + e- + CO2 + H+ = FeCO3 + 2H2O 8.40

K 1

(H)(PCO2)(e )

Example, FeCO3 (cont.)

• log K = pH + pe - log Pco2

• Thus at equilibrium pe is a function of pH and Pco2, only.

• Set the PC02 then calc. pH• At pH = 7.0 and Pco2 = 0.10

– pe = log K - pH + log Pco2

– pe = 8.40 - 7.0 - 1.0– pe = 0.4– EH = (0.059)( 0.4) = - 0.024 V

pe - pH plot for Fe

Reduction Fe(III) phosphate (strengite)

Fe3+ + e- = Fe2+ 13.04

FePO4•2H2O + 2H+ = Fe3+ + H2PO4- - 6.85

---------------------------------------------------------------

FePO4•2H2O + e- + 2H+ = Fe2+ + H2PO43- 6.19

pe 6.2 log(Fe2 )(H2PO2

)

(H )2

Redox reaction of strengite (cont.)

pe = 6.2 - log(Fe2+) - log(H2PO4) - 2pH

• Let pH = 7.0, (Fe2+) = 10-5 M, and

(H2PO43-) = 10-5 M

– pe = 6.2 + 5 + 5 -14 = 2.2

– EH = 0.129v

Redox reaction of strengite (cont.)

• Comparison with previous computations shows strengite is reduced at a similar pe as Fe(OH)3

• Reduction of iron oxides releases adsorbed P.

• Drainage and restoration of oxic conditions can tie up P.– Following draining of rice soils, phosphate can be

tied up causing P deficiency for rotation crops.

Half-Cell Reactions of Common Electron Acceptors in Soils

Oxidized + mH+ + n electrons = reduced + H2O

Nernst equation

At 25 0C

EHE

o RT

nFln

(reduced)

(oxidized)(H)m

EH Eo 0.059

nlog

(reduced)

(oxidized)(H + )m

In Volts

EH E o 0.059

nlog

(reduced)

(oxidized)

0.059

nmpH

In millivolts

EH E o 59

nlog

(reduced)

(oxidized)

59m

npH

Applications of Galvanic Cells

Voltaic or Galvanic CellsVoltaic or Galvanic Cells

• The energy released in a spontaneous redox reaction can be used to perform electrical work.

• Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit.

• Voltaic cells use spontaneous reactions.

• If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+.

• Zn is spontaneously oxidized to Zn2+ by Cu2+.• The Cu2+ is spontaneously reduced to Cu0 by Zn.

• Galvanic cells consist of– Anode: Zn(s) Zn2+(aq) + 2e2

– Cathode: Cu2+(aq) + 2e- Cu(s) – Salt bridge (used to complete the electrical circuit): cations move

from anode to cathode, anions move from cathode to anode.

• The two solid metals

are the electrodes

(cathode and anode).

• As oxidation occurs,

Zn is converted to

Zn2+ and 2e-. The

electrons flow towards

the anode where they are

used in the reduction reaction.

A shorthand convention exists for describing batteries. For the Cu/Zn battery, it would be described as follows:

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

The ANODE... The CATHODE...-supplies electrons to the external circuit (wire) -accepts electrons from the external circuit (wire) -is the negative pole of the battery -is the positive pole of the battery-is the site of OXIDATION -is the site of REDUCTION-is written on the left hand side if the convention -is written on the right hand side if the is followed convention is followed-is the half-cell with the lowest electrode potential -is the half-cell with the highest electrode

potential

Applications of Galvanic Cells

Potentiometry and Ion Selective Electrodesthe measure of the cell potential to yield chemical information (conc., activity, charge ....)

A difference in the activity of an ion on either side of a selective membrane results in a thermodynamic potential difference being created across that membrane

C a 2 + C a 2 + 0 . 0 1 M C a 2 +

0 . 0 2 M C l -

0 . 1 M C a 2 +

0 . 2 M C l -

( 0 . 0 1 + ) M C a 2 + ( 0 . 1 - ) M C a 2 +

0 . 0 2 M C l - 0 . 2 M C l -

+

+

+

+

-

-

-

-

Calcium selective molecular recognition ligand

G RT lnA1A2

nFE

E RT

nFlnA1A2

0. 05916

nlog

A1A2

(à 25C)

The glass pH electrode

int ref

int+H

analyte

ext ref

Ag(s)|AgCl(s)|(aq)-Clint),(aq,Hext)(aq,H||(aq)-Cl|AgCl(s)|Ag(s)

Ag

Soln. aq. satdin KCl + AgCl

AgCl(s) + KCl(s)

AgCl porousglass

+ -

0.1M HCl inAgCl sat.

Corrosion

Fe2+ +2e Fe E0=-0.44V

2H+ + 2e H2 E0=0V

2H2O + O2 =4e4OH- E0=1.23V

Iron is oxidized in water or humid conditions to give rust.

Inhibit this by coating with another material (Zn for example that forms a protective oxide on the iron), or by providing a sacrificial anode (b).

Batteries-providing electricity from chemistry

The Lead Acid Storage Battery was developed in the late 1800's and has remained the most common and durable of the battery technologies (in vehicles).

Ecell ?

Lead-acid batteries

When the battery is used as a voltage supply, electrons flow from the Pb metal to the Pb(IV)oxide. The reactions aren't quite the reverse of the formation reactions, because now the sulfate ions in the solution begin to play a role. The two reactions are:

PbO2 + 4H+ + 2e + SO4-2 PbSO4 + 2H2O

Pb + SO4-2 PbSO4 + 2e

The overall reaction if we combine the hydrogen ions and the sulfate: PbO2 + Pb + 2H2SO4 2 PbSO4 + 2 H2O Lead sulfate is fairly insoluble so that as soon as Pb(II) ions are formed by either reaction, the ions immediately precipitate as lead sulfate. The beauty is that this lead sulfate stays attached to the grids so that it is there for recharging of the battery.

Pada recharge batery tipe ini

bagaimana prosedurnya?

Other batteries

Primary battery-non rechargeable

Longer shelf-life

Rechargeable

Offer higher efficiencies compared to burning fuels

Images of batteries

Leclanche Alkaline Fuel Cell

Electrolysis

Use Faraday’s Laws to evaluate the number of moles of a substance oxidised or reduced by passage of charge (current over a given period of time = I.t) through an electrode

Faraday: Q (charge) = nF

N=number of moles of electrons

F=constant of 96500 Coulomb/mole

Example (try it): What current is needed to deposit 0.500g of chromium from a solution containing Cr3+ over a one hour period (MW for Cr=52)? (Ans=0.77A)

Electrolysis

Use Faraday’s Laws to evaluate the number of moles of a substance oxidised or reduced by passage of charge (current over a given period of time = I.t) through an electrode

Faraday: Q (charge) = nF

n=number of moles of electrons

F=constant of 96500 Coulomb/mole

1 coulomb = 1 C = 1 Ampere. s

How many grams of copper are deposoted on the cathode of an electrolytic cell if an electric current of 2 A is run through a solution of CuSO4 for a period 20 min?

number of coulomb = 2 A x 2 min x 60 s/min = A C

mole e- = A x 1/(9,65 x 104 C) = B mol e-

Gram Cu = B x ½ mol Cu/mol e- x 63,5 g Cu/1 mol Cu = C gram Cu

Applications of Electrolytic Cells

Aluminium refining: The major ore of aluminium is bauxite, Al2O3.

Anhydrous Al2O3 melts at over 2000°C. This is too high to permit its use as a molten medium for electrolytic formation of free aluminium. The electrolytic process commercially used to produce aluminium is known as the Hall process, named after its inventor, Charles M. Hall. Al2O3 is dissolved in molten cryolite, Na3AlF6, which has a melting point of 1012oC and is an effective conductor of electric current. Graphite rods are employed as anodes and are consumed in the electrolysis process. The cell electrolytic reaction is:

2Al2O3 + 3C 4Al(l) + 3CO2(g)

Electrolysis of brine

Chlorine and sodium hydroxide are both manufactured by electrolysis of brine (aqueous sodium chloride) using inert electrodes. Chlorine is evolved at the anode, Cl- 1/2Cl2 + e

Hydrogen is evolved at the cathode: H+ + e 1/2H2

The removal of chloride ions and hydrogen ions leaves sodium ions and hydroxide ions in solution.

Chlorine is used to disinfect municipal water supplies and water in swimming pools. It is used to manufacture household bleaches and disinfectants. It is used to manufacture plastics (e.g. PVC), pesticides, anaesthetics, CFCs etc.

Sodium hydroxide is used in the manufacture of synthetic fibres, soaps and detergents.

ElectroplatingIn all aspects of our lives we are surrounded by products with electroplated surfaces. Whether we are looking at a silver-plated watch through gold-plated glasses, watching television, using the washing machine, getting into a car or boarding a plane: electroplating plays an important part in all of these situations. The objective is to prevent corrosion and wear,produce hardness and conductivity, and give products an attractive appearance.

Silver electroplating was the first large scale use of electrolysis for coating base metal objects with a higher value decorative finish.

The principle: thin metallic layers with specific properties are deposited on base materials including steel, brass, aluminium, plastic and die-cast parts.