Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25...

16
25-1 Homework solution At the anode, oxidation: 2H 2 O(R) 6 4H + (aq) + O 2 (g) + 4e - , EE -1.23 V At the cathode, reduction: Ag + (aq) + e - 6 Ag(s), EE +0.80 V pH of 2.00 means [H + ] = 1.0 x 10 -2 M number of moles of H + (aq) (= n) in 0.500 L of solution = 5.0 x 10 -3 moles 6 which means that 5.0 x 10 -3 moles of electrons will be needed. 5.0 x 10 -3 = (Current x time)/F t = 5.0 x 10 -3 mol x 96485 C mol -1 /0.240 C s -1 t = 2.0 x 10 3 s

Transcript of Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25...

Page 1: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-1

Homework solution

At the anode, oxidation:

2H2O(R) 6 4H+(aq) + O2(g) + 4e-, EE -1.23 V

At the cathode, reduction:

Ag+(aq) + e- 6Ag(s), EE +0.80 V

pH of 2.00 means [H+] = 1.0 x 10-2 M

number of moles of H+(aq) (= n) in 0.500 L of solution =

5.0 x 10-3 moles 6 which means that

5.0 x 10-3 moles of electrons will be needed.

5.0 x 10-3 = (Current x time)/F

t = 5.0 x 10-3 mol x 96485 C mol-1/0.240 C s-1

t = 2.0 x 103 s

Page 2: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-2

WEEK 9

Lectures 25-27

Be well into Chapter 20 problems

Lecture 25

Monday: Industrial Applications. Corrosion

Lecture 26 Review : Nernst + Faraday

Lecture 27 Kinetics (Chapter 14)

(New (and last) topic

Page 3: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-3

Lecture 25

Industrial Application of Electrolysis

Isolation of Elements

Aluminum, Sodium, Fluorine, Chlorine

Electro-refining of Copper

First: Corrosion

Page 4: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-4

Corrosion is a major problem: rusting is caused by an

electrochemical reaction involving oxygen and water

A water droplet on an iron surface is a miniature electrochemical cell.

The oxidation of iron occurs in an interior region of the droplet, whereas

the reduction of oxygen preferentially occurs near the air-droplet

interface. Ionic charges carriers are required to complete the circuit and

allow the redox reactions to proceed.

Page 5: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

Corrosion is particularly a problem with iron and steel.

Simply:

Fe(s) 6 Fe2+(aq) + 2e, EE +0.44 V

O2(g) + 4H+(aq) + 4e 6 2H2O(R) EE +1.23 V

Once in solution the Fe2+ is oxidized to Fe3+ which is converted to hydrated

Fe2O3. This does NOT adhere to the metal surface and corrosion and pitting

occurs.

the half cell with the more

-ve E will be the anode

25-5

So if E is > than 1.23 no rusting

Idea for rust-protection

Page 6: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-6

Cathodic Protection:

Protect the metal surface from oxidation.

Paint, or coat with another metal, e.g. Zn

(galvanized steel)

Zn 6 Zn2+ + 2e EE = +0.76 V

O2(g) + 4H+(aq) + 4e 6 2H2O(R) EE = +1.23 V

Fe2+(aq) + 2e Fe(s) 6 , EE -0.44 V

Zinc forms the effective anode (and dissolves) and

the iron forms the cathode and does not react.

Page 7: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-7

Sacrificial Anodes

Instead of coating

(not always feasible with large objects) the

steel is

connected to a large block of zinc or

magnesium

this corrodes faster * Large Displacement Engine For Better Low- & Mid-

Range Power

* Low Oil Warning System

* Standard Fresh Water Flushing Port With Attachment

Included

* Includes 3.3 Gallon Fuel Tank & Fuel Hose

* Has Remote Control Capability

* Fold Down Handle For Easy Transportation

* Transistorized Ignition For Reliable, Easy Starting

* Built-In Start In Gear Protection

* Throttle Friction Lock

* Reverse Lock Mechanism

* Shallow Water Drive

* Large Sacrificial Anode For Corrosion

Protection* Longer Standard Shaft Lengths For More Reach On

Sail Boats

Zn2+ + 2e Zn 6 EE = -0.76 V

Fe2+(aq) + 2e Fe(s) 6 , EE -0.44 V

the half cell with the more

-ve E will be the anode

Also used on pipe lines

Page 8: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

Aluminum metal

Aluminum is the third most abundant element in the earth's crust.

It occurs naturally as Bauxite (hydrated aluminum oxide) Al2O3.x(H2O). This crude

ore is first purified by dissolving in base, then re-precipitating with CO2.

Al2O3(s) + 2(OH-)(aq) + 3H2O 6 [Al(OH)4]-(aq)

[Al(OH)4]-(aq) + CO2(g) 6 Al(OH)3(s) + HCO3

-(aq)

The Al(OH)3(s) is then heated to give Al2O3 which is electrolyzed at about

4.2 volts in a solution of molten fluoride salts at around 1000EC. Carbon

rods serve as the anodes and the steel holding tank as the cathode.

25-8Bauxite

Page 9: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-9

The Al(OH)3(s) is then heated to give Al2O3 which is electrolyzed at about

4.2 volts in a solution of molten fluoride salts at around 1000EC. Carbon

rods serve as the anodes and the steel holding tank as the cathode.

NOTE: Al cannot be plated out from aqueous soln. Because water is more

easily reduced

Al 3+ + 3e Al E0 = -1.66

2H2O(R) + 2e- H2(g) + 2OH-(aq), EE -0.83 V

Reactions: Hall-Heroult process

Cathode: 4 (Al3+ + 3e 6Al(R) )

Anode: 3 (C(s) + 2O2- 6 CO2(g) + 4e)

Overall 4Al3+ + 3C(s) + 6O2- 6

4Al(R) + 3 CO2(g)

The Carbon anodes are consumed in the

process and must be replaced.

The process is very demanding of

electrical energy.

Aluminum smelters are situated near

cheap power sources.

Page 10: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-10

The 9-inch aluminum pyramid, which

completes the top of the Washington

monument as it narrows to a point, is 100-

ounces of solid aluminum, part of the

monument's lightning protection system. In

the 1880s, aluminum was a rare metal, selling

for $1.10 per ounce and used primarily for

jewelry. The pyramid was the largest piece of

aluminum of its day and was such a novelty

that it was displayed at Tiffany's jewelry store

in New York before it was placed at the top.

In the SRP table we see that:

Al3+ + 3e 6Al(s), EE = -1.66 V

which might lead you to think that it would corrode rapidly in air. It does

not do so because it forms an impervious layer of Al2O3 which protects the

metal form further attack.

Page 11: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-11

How much energy in GJ to manufacture 1.00 tonne of Aluminum at

a voltage of 4.2 volts?

Three moles of electrons are needed per mole of Aluminum.

1.00 Tonne = 1.0 x 103 kg = 1.00 x 106 g.

Moles Al = 1.00 x 106 g /(27.0 g mol-1)

= 3.70 x 104 moles

Moles of electrons = 3 x 3.70 x 104

= 1.11 x 105 mol e

Coulombs used = Q = F x (mol of e)

= 1.11 x 105 mol e) x 9.65 x 104 C/mol e)

= 1.07 x 1010 C

4.20 volts was used ( = 4.2 J/C)

Energy demand is then

Q x V = (1.07 x 1010 C) x (4.20 J/C)

= 4.50 x 1010 J = 45.0 GJ

( 1 GJ = 109 J )

m = MIt/Fn

m = MQ/Fn

Q = (1 x 106 x 96485 x 3)/27

= 1.07 x 1010 C

volts = J/C

Joules = 4.2 x 1.07 x 1010

Page 12: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-12

Sodium is made by electrolysis of molten sodium

chloride in a Downs Cell

The Downs Cell is used to manufacture sodium. This is achieved

by electrolysis of molten sodium hydroxide. A mixture of NaCl

and CaCl (added to reduce the melting point. The cell has a

central graphite anode and a steel cathode surrounding it. Chlorine

is liberated at the anode and bubbles up to be collected by a

central trap. Sodium (melts at 98°C) floats up the side of the

container from the circular steel anode and is collected by another

trap. A steel mesh diaphragm separates the anode and the cathode.

Page 13: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-13

Fluorine is made by electrolysis of a molten mixture of HF and KF

cathode: 2HF + 2e 6 H2 + 2F-

anode: 2HF 6 F2 + 2H+ + 2e

Anode (+) Cathode (-)

H+ H2F2 F-

Page 14: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-14

Chlorine is made by electrolysis of Brine.

"Chlor-Alkali" Membrane Cell

Cathode: 2H2O + 2e 6 H2(g) + 2(OH-)

Anode: 2Cl-(aq) 6 Cl2(g) + 2e

Overall: 2NaCl(aq) + 2H2O(l) H2(g) + Cl2(g) +2NaOH(aq)

Major industrial process

Page 15: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-15

Electrorefining of Copper.This uses impure Cu ingots as ANODES and plates out PURE copper

metal at a copper cathode

The impurities (precious metals such as Pt, Au, Ag etc.) fall out as an

"anode sludge" and are recovered.

Refinery with thousands of tanks

When copper is first obtained by reduction of its ores, it is

cast as impure slabs, called blister copper. In electrorefining,

the blister is used as the anode in an acidic solution of copper

(II) sulphate Initially, the cathodes consist of thin sheets of

pure copper.

During electrolysis, copper passes into solution from the anodes,

(leaving the impurities, normally containing silver, gold and platinum)

as an anode slime, which sinks to the bottom of the cell. The anode

reaction is

Page 16: Homework solution - Home - University of · PDF file · 2017-03-0625-3 Lecture 25 Industrial Application of Electrolysis Isolation of Elements Aluminum, Sodium, Fluorine, Chlorine

25-16

At the cathode, copper (II) ions are discharged and the pure copper sheet

becomes coated with an increasingly thick layer of very pure copper:

Impurities (silver nickel etc

are readily oxidized at anode

but don’t

reduce at cathode (redox pots

are more neg.

than copper -so fall to bottom