electrolysis.ppt

Houghton Mifflin Company and G. Hall. All rights reserved.

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Lecture 25 Electrolysis

Define electrolysis?Some examples.What are the values of G and

Ecell?Electrolysis of water.Some industrial applications.Corrosion


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Two Types of Cells

Cell 1: does work by releasing free energy from a spontaneous reaction to produce electricity such as a battery.

Cell 2: does work by absorbing free energy from a source of electricity to drive a non-spontaneous reaction such as electroplating.


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A voltaic (Galvanic) cell can power an electrolytic cell


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Electrolysis

The splitting (lysing) of a substance or decomposing by forcing a current through a cell to produce a chemical change for which the cell potential is negative.


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Electrolysis

Previously our lectures on electrochemistry were involved with voltaic cells i.e. cells with Ecell > 0 and G < 0 that were spontaneous reactions.

Today we discuss electrochemical cells where Ecell < 0 and G > 0 that are non-spontaneous reactions and require electricity for the reactions to take place. We can take a voltaic cell and reverse the electrodes to make an electrochemical cell.


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VoltaicElectrolytic


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Fig. 21.17


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Fig. 21.18: Car battery, both voltaic and electrochemical cell.


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Increase oxidizing power

Increase reducing power


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A standard electrolytic cell. A power source forces the opposite reaction


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Electrolysis


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(a) A silver-plated teapot. (b) Schematic of the electroplating of a

spoon.


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Schematic of the electroplating of a

spoon.

AgNO3(aq)


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The electrolysis of water produces hydrogen gas at the cathode (on the right) and oxygen gas at the anode

(on the left).


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Fig. 21.19 Electrolysis of

water


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Electrolysis of water

At the anode (oxidation):2H2O(l) + 2e- = H2(g) + 2OH-(aq) E=-0.42V

At the cathode (reduction):2H2O(l) = O2(g) + 4H+(aq) + 4e- E= 0.82VOverall reaction after multiplying anode

reaction by 2,2H2O(l) = 2H2(g) + O2(g) Eo

cell = -0.42 -0.82 = -1.24 V


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Electrolysis: Consider the electrolysis of a solution that is 1.00 M in each of CuSO4(aq) and NaCl(aq) Oxidation possibilities follow.2Cl–(aq) = Cl2(g) + 2e– E° = –1.358 V2SO4

2–(aq) = S2O82–(aq) + 2e– E° = –2.01 V

2H2O = 4H+(aq) + O2(g) + 4e– E° = –1.229 V

Reduction possibilities follow:Na+(aq) + e– = Na(s) E° = –2.713 VCu2+(aq) + 2e– = Cu(s) E° = +0.337 V2H2O + 2e– = H2(g) + 2OH–(aq) E° = +0.828 V


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Electrolysis

We would choose the production of O2(g) and Cu(s). But the voltage for producing O2(g) from solution is

considerably higher than the standard potential, because of the high activation energy needed to form O2(g).

The voltage for this half cell seems to be closer to –1.5 V in reality.

The result then is the production of Cl2(g) and Cu(s).anode, oxidation: 2Cl–(aq) = Cl2(g) + 2e– E° = –1.358 V

cathode, reduction: Cu2+(aq) + 2e– : Cu(s) E° = +0.337 V

overall: CuCl2(aq) : Cu(s) + Cl2(g) E = –1.021 V We must apply a voltage of more than +1.021 V to

cause this reaction to occur.


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E = -2.37 V


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E = 1.07 V

Ecell = -2.37-1.07 = -3.44V


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Prob. 21.9


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Stoichiometry of electrolysis: Relation between amounts of charge and

productFaraday’s law of electrolysis

relates to the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell (half reaction).

Each balanced half-cell shows the relationship between moles of electrons and the product.


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Application of Faraday’s law

1. First balance the half-reactions to find number of moles of electrons needed per mole of product.

2. Use Faraday constant (F = 9.65E4 C/mol e-) to find corresponding charge.

3. Use the molar mass of substance to find the charge needed for a given mass of product.1 ampere = 1 coulomb/second or 1 A = 1 C/sA x s = C


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Stoichiometry of Electrolysis

How much chemical change occurs with the flow of a given current for a specified time?

current and time quantity of charge

moles of electrons moles of analyte

grams of analyte


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Fig. 21.20


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Doing work with electricity.


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Industrial Applications of Electrolysis

http://academic.pgcc.edu/~ssinex/E_cells.ppt.. All rights reserved.

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What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)?

Na+ Cl-

Let’s examine the electrolytic cell for molten NaCl.

All rights reserved. http://academic.pgcc.edu/~ssinex/E_cells.ppt.

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+-battery

Na (l)

electrode half-cell

electrode half-cell

Molten NaCl

Na+

Cl-

Cl-

Na+

Na+

Na+ + e- Na 2Cl- Cl2 + 2e-

Cl2 (g) escapes

Observe the reactions at the electrodes

NaCl (l)

(-)

Cl-

(+)

http://academic.pgcc.edu/~ssinex/E_cells.ppt. All rights reserved.

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+-battery

e-

e-

NaCl (l)

(-) (+)

cathode anode

Molten NaCl

Na+

Cl-

Cl-

Cl-

Na+

Na+

Na+ + e- Na 2Cl- Cl2 + 2e-

cationsmigrate toward

(-) electrode

anionsmigrate toward

(+) electrode

At the microscopic level

http://academic.pgcc.edu/~ssinex/E_cells.ppt. All rights reserved.

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Molten NaCl Electrolytic Cell

cathode half-cell (-)REDUCTION Na+ + e- Na

anode half-cell (+)OXIDATION 2Cl- Cl2 + 2e-

overall cell reaction2Na+ + 2Cl- 2Na + Cl2

X 2

Non-spontaneous reaction!


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The Downs Cell for the Electrolysis of Molten Sodium Chloride


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If the products are mixed, the result is household bleach.2 NaOH(aq) + Cl2(g) = NaCl(aq) + NaOCl(aq) + H2O


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The Mercury Cell for Production of Chlorine and Sodium Hydroxide


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A schematic diagram of an electrolytic cell for producing aluminum by the

Hall-Heroult process.


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Fig. 22.19 A schematic diagram of an electrolytic cell for producing aluminum by the Hall-Heroult

process.

http://academic.pgcc.edu/~ssinex/E_cells.ppt.

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The Hall Process for Aluminum

Electrolysis of molten Al2O3 mixed with cryolite – lowers melting point

Cell operates at high temperature – 1000oC

Aluminum was a precious metal in 1886.

A block of aluminum is at the tip of the Washington Monument!

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carbon-lined steel vesselacts as cathode

CO2 bubbles

Al (l)Al2O3 (l)

Drawoff Al (l)

-

+

Cathode: Al+3 + 3e- Al (l)

Anode: 2 O-2 + C (s) CO2 (g) + 4e-

frompowersource

Al+3

O-2O-2

Al+3

O-2

graphite anodes

e-

e- From: http://academic.pgcc.edu/~ssinex/E_cells.ppt.

http://academic.pgcc.edu/~ssinex/E_cells.ppt

http://academic.pgcc.edu/~ssinex/E_cells.ppt.

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The Hall Process

Cathode: Al+3 + 3e- Al (l)

Anode: 2 O-2 + C (s) CO2 (g) + 4e-

4 Al+3 + 6 O-2 + 3 C (s) 4 Al (l) + 3 CO2 (g)

x 4

x 3

The graphite anode is consumed in the process.


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Fig. 22.21: Production of solid Mg


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Corrosion

Electrochemistry plays a major role in corrosion and protection against it.

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Calculating the cell potential, Eocell, at

standard conditions

Fe+2 + 2e- Fe Eo = -0.44 v

O2 (g) + 2H2O + 4e- 4 OH- Eo = +0.40 v

This is corrosion or the oxidation of a metal.

Consider a drop of oxygenated water on an iron objectFe

H2O with O2

Fe Fe+2 + 2e- -Eo = +0.44 v2x

2Fe + O2 (g) + 2H2O 2Fe(OH)2 (s) Eocell= +0.84 v

reverse

From: http://academic.pgcc.edu/~ssinex/E_cells.ppt.

http://academic.pgcc.edu/%7Essinex/E_cells.ppt

http://academic.pgcc.edu/%7Essinex/E_cells.ppt


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Cathodic Protection Against Corrosion

Underground steel pipes offer the strength to transport fluids at high pressures, but they are vulnerable to corrosion driven by electrochemical processes. A measure of protection can be offered by driving a magnesium rod into the ground near the pipe and providing an electrical connection to the pipe. Since the magnesium has a standard potential of -2.38 volts compared to -.41 volts for iron, it can act as a anode of a voltaic cell with the steel pipe acting as the cathode. With damp soil serving as the electrolyte, a small current can flow in the wire connected to the pipe. The magnesium rod will be eventually consumed by the reactionMg(s) -> + Mg2+(aq) + 2e-

while the steel pipe as the cathode will be protected by the reactionO2(g) + 2H2O(l) + 4e- -> 4OH-(aq).

From: http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/corrosion.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/corrosion.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/electrochem.html#c2


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Lecture summary

Electrolysis is often the reverse of voltaic cell in that Ecell < 0, and G >0 and reaction is non-spontaneous.

Electrolysis of water and to produce O2 and H2.

Faraday’s law allows us to determine how much current is needed to produce a certain amount of an element.

Industrial applications are numerous for producing a variety of solid elements (Al, Mg, Na, etc).

electrolysis.ppt

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