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Electrolysis: An Introduction VCE Chemistry Unit 4: Chemistry at Work Area of Study 2 – Using Energy. Electrolysis. During electrolysis , electrical energy is converted into chemical energy.

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slide1

Electrolysis: An Introduction

VCE Chemistry

Unit 4: Chemistry at Work

Area of Study 2 – Using Energy

slide2

Electrolysis

During electrolysis, electrical energy is converted into chemical energy.

This is the reverse of the process that occurs in a galvanic cell where spontaneous reactions are a source of electrical energy.

Electrolysis has a variety of industrial applications.

These include electroplatingandelectrorefiningas well as the recharging secondary cells.

Non-spontaneous reactions are forced to occur in electrolytic cells by passing an electric current from an external power source through an electrolyte.

The potential energy of the system is increased as electrical energy is forced into the cell.

slide3

Galvanic & Electrolytic Cells - A Comparison

In galvanic cells the reactions are separated so electrons can be used as they are transferred.

Electrolytic cells have two electrodes and an electrolyte.

Electrolytic cells differ from galvanic cells in that the anode is the positive electrode while the cathode is the negative electrode.

The polarities of the anode and cathode change because in galvanic cells the polarity is imposed by the reactions themselves.

At the anode, electrons are produced and at the cathode electrons are consumed.

In electrolytic cells the polarity is imposed by the power source.

This pulls electrons from the anode and forces electrons onto the cathode.

slide4

Electrolytic Cells

Regardless of which cell, electron flow is always the same – from anode to cathode!

Oxidation always occurs at the anode and reduction always occurs at the cathode!

slide6

Predicting ElectrolyticReactions

When electrons are forced into the cell, the oxidants present at the surface of the electrode compete with one another to accept them.

The strongest of the oxidants present preferentially accepts the electrons and is discharged, i.e. the oxidant highest on the electrochemical series accepts the electrons.

Similarly, at the anode where the electrons are being withdrawn, the reductants present at the anode surface compete to donate electrons.

The strongest of the reductants is preferentially discharged.

slide7

Predicting Electrolytic Reactions

Note that if two competitors have very similar E° values, they will probably both be discharged simultaneously.

In aqueous solutions, the water may be involved in either the oxidation or the reduction reaction or both.

slide8

Predicting Electrolytic Reactions - Questions

Predict the products at each electrode during electrolysis for the following 1 M solutions (unreactive electrodes used) . The nitrate ion is not involved in any of the reactions.

a. Copper(II) bromide

b. Sodium iodide

c. Lead nitrate

d. Zinc chloride

e. Aluminium nitrate

slide9

Predicting Electrolytic Reactions - Solutions

a. 1 M copper(II) bromide

Species present at the electrodes are Cu2+ (aq), Br- (aq) and H2O (l)

Possible reductions:

Cu2+ (aq) + 2e- Cu (s) +0.34 V

2H2O (l) + 2e- H2 (g) + 2OH- (aq) -0.83 V

Here the strongest oxidant (most positive Eo value) is Cu2+.

Thus, at the cathode Cu forms.

Possible oxidations:

Br2 (g) + 2e- 2Br- (aq) +1.09 V

O2 (g) + 4H+ (aq) + 4e- 2H2O (aq) +1.23 V

Here the strongest reductant (least positive Eo value) is 2Br-.

Thus at the anode Br2 forms.

Overall equation: Cu2+ (a) + 2Br- (aq)  Cu (s) + Br2 (g)

slide10

Predicting Electrolytic Reactions - Solutions

b. 1 M sodium iodide

Species present at the electrodes are Na+ (aq), I- (aq) and H2O (l)

Possible reductions:

Na+ (aq) + e- Na (s) -2.71 V

2H2O (l) + 2e- H2 (g) + 2OH- (aq) -0.83 V

The strongest oxidant (most positive Eo value) is H2O.

Thus at the cathode H2 and 2OH- form.

Possible oxidations:

I2 (g) + 2e- 2I- (aq) +0.54 V

O2 (g) + 4H+ (aq) + 4e-  2H2O (aq) +1.23 V

The strongest reductant (least positive Eo value) is 2I-.

Thus at the anode I2 forms.

Overall equation: 2H2O (l) + 2I- (aq)  H2 (g) + 2OH- (aq) + I2 (g)

slide11

Predicting Electrolytic Reactions - Solutions

c. 1 M lead(II) nitrate

Species present at the electrodes are Pb2+ (aq) and H2O (l)

Possible reductions:

Pb2+ (aq) + 2e- Pb (s) +0.13 V

2H2O (l) + 2e- H2 (g) + 2OH- (aq) -0.83 V

The strongest oxidant (most positive Eo value) is Pb2+.

Thus, at the cathode Pb forms.

Possible oxidations:

O2 (g) + 4H+ (aq) + 4e- 2H2O (aq) +1.23 V

The strongest reductant (least positive Eo value) is 2H2O.

Thus at the anode O2 and 4H+ form.

Overall equation: 2Pb2+ (aq) + 2H2O (l)  2Pb (s) + O2 (g) + 4H+ (aq)

slide12

Predicting Electrolytic Reactions - Solutions

d. 1 M zinc chloride

Species present at the electrodes are Zn2+ (aq), Cl- (aq) and H2O (l)

Possible reductions:

Zn2+ (aq) + 2e- Zn (s) -0.76 V

2H2O (l) + 2e- H2 (g) + 2OH- (aq) -0.83 V

The strongest oxidant (most positive Eo value) is Zn2+.

Thus, at the cathode Zn forms.

Possible oxidations:

Cl2 (g) + 2e- 2Cl- (aq) +1.36 V

O2 (g) + 4H+ (aq) + 4e- H2O (l) +1.23 V

The strongest reductant (least positive Eo value) is 2H2O.

Thus at the anode O2 and 4H+ form.

Overall equation: 2Zn2+ (aq) + 2H2O (aq)  2Zn (s) + O2 (g) + 4H+ (aq)

slide13

Predicting Electrolytic Reactions - Solutions

e. 1 M aluminium nitrate

Species present at the electrodes are Al3+ (aq) and H2O (l)

Possible reductions:

Al3+ (aq) + 3e- Al (s) -1.71 V

2H2O (l) + 2e-  H2 (g) + 2OH- (aq) -0.83 V

The strongest oxidant (most positive Eo value) is 2H2O.

Thus at the cathode H2 and 2OH- form.

Possible oxidations:

O2 (g) + 4H+ (aq) + 4e- 2H2O (aq) +1.23 V

The strongest reductant (least positive Eo value) is 2H2O.

Thus at the anode O2 and 4H+ form.

Overall equation: 6H2O (l)  2H2 (g) + 4OH- (aq) + O2 (g) + 4H+ (aq)

slide14

Applied Electrolysis - Electroplating

In this application, the cathode is coated with a thin layer of metal from a solution containing ions of the metal.

This is done to improve the appearance or to prevent corrosion by the application of a protective layer.

The object to be plated is attached to the negative terminal of a power supply and becomes the cathode.

It is then placed in an electrolyte solution containing ions of the metal that forms the plating.

The anode is made from the metal that is to form the plating.

To ensure the coating on the surface is strongly bonded and smooth, the electrolyte is a complex mixture of ions and the conditions (voltage, current, time) carefully controlled.

slide16

Applied Electrolysis - Electrorefining

Metals can be purified by electrolysis.

The impure metal is used as the anode and oxidised.

Metals that are stronger reductants will also be discharged at the electrode surface.

Any constituents of the anode that are weaker reductants are not oxidised but fall as uncharged metal atoms to the bottom of the cell.

This leaves an electrolyte solution containing ions of the metal to be refined and a small amount of other ions.

The mixture of metal ions in solution compete for electrons at the cathode surface.

As the dissolved impure ions are weaker oxidants, the metal to be refined is preferentially reduced at the cathode, and is in a much purer form.

slide18

Applied Electrolysis

Sodium from Molten Sodium Chloride – The Downs Cell

slide19

Applied Electrolysis

Sodium from Molten Sodium Chloride – The Downs Cell

slide20

Applied Electrolysis

Aluminium from Molten Alumina The Hall-Héroult Process

slide21

Applied Electrolysis

Aluminium from Molten Alumina The Hall-Héroult Process

slide22

Applied Electrolysis

Sodium Hydroxide and Chlorine from Brine The Membrane Cell

slide23

Applied Electrolysis

Sodium Hydroxide and Chlorine from Brine The Membrane Cell

slide24

Faraday’s Laws of Electrolysis

If you wished to electroplate an object, you might ask yourself the following questions:

How can I determine how much metal is being plated?

How long should I leave the object being plated in the electroplating cell?

What size electric current should be used?

slide25

Faraday’s Laws of Electrolysis

Michael Faraday was a 19th century English chemist.

His studies yielded two laws relating amount of electric current and the chemicals produced by the current or used to produce it.

These laws enable us to:

Work out the amount of energy required to discharge a metal ion and place it out on an object or

To calculate the amount of metal produced in an electrolytic cell.

slide26

Faraday’s Laws of Electrolysis – First Law

The mass of metal produced at the cathode is directly proportional to the quantity of electricity passed through the cell

m  Q

Electric charge, Q, is measured using the unit coulomb.

The electric charge passing through a cell may be calculated from measurements of the current, I, through the cell and the time, t, for which the current flows.

Charge (coulombs) = current (amps) x time (seconds)

Q = I t

slide27

Faraday’s Laws of Electrolysis – Second Law

In order to produce one mole of metal, one two or three moles of electrons must be consumed.

Faraday found that there was a certain charge associated with one mole of electrons.

This amount of charge is now called the Faraday and is equivalent to 96 487 (96 500) Coulomb.

  • Faraday = 96 500 Coulomb
slide28

Faraday’s Laws of Electrolysis – Second Law

In order to produce one mole of metal, one two or three moles of electrons must be consumed.

Ag+ (aq), Cu2+ (aq) and Cr3+ (aq) require 1, 2 and 3 moles of electrons for discharge.

The quantity of electricity required will be:

Amount of charge = no. of moles of metal ions x charge on an ion x 1 Faraday

Q = n z F and Q = I t

I t = n z F

slide29

Faraday’s Laws of Electrolysis – Questions

  • Calculate the number of mole of copper produced in an electrolytic cell if a current of 5.0 A at a voltage of 6.0 V flows through a solution of copper ions for 10 minutes.
  • Calculate the time taken to deposit 1.00 g of copper onto an object that is placed in a solution of copper nitrate, Cu(NO3)2, and has a current of 2.50 A flowing through it.
  • In an operating Hall-Héroult cell, a current of 150 000 A is used at 5.0 V. Calculate the mass of aluminium that would be produced if this cell operates continuously for 1 day.
  • The electrolysis of a solution of chromium ions using a current of 2.2 A for 25 minutes produced 0.60 g of chromium. Calculate the charge on the chromium ion.
  • Calculate the masses of metal produced when 600 Faraday of charge is used to reduce the ions of aluminium, silver and zinc.
slide30

Faraday’s Laws of Electrolysis – Solutions

1. n (Cu) = I t / z F

n (Cu) = 5.0 x 10 x 60 / 2 x 96 500

n (Cu) = 1.6 x 10-2 mol

That is, 1.6 x 10-2 mole of copper would be produced

2. t = n (Cu) z F / I

t = (1.00 / 63.5) x 2 x 96 500 / 2.5

t = 1216 seconds

That is, it takes 20 minutes 15 seconds to deposit 1.00 g of copper

3. n (Al) = I t / z F

n (Al) = 150 000 x 24 x 60 x 60 / 3 x 96 500

n (Al) = 44 767 mol

m (Al) = 44 767 x 27 = 1 208 705 g

That is, 1.2 tonne of aluminium is produced per day

slide31

Faraday’s Laws of Electrolysis – Solutions

4. z (Cr) = I t / n F

z (Cr) = 2.2 x 25 x 60 / (0.60 / 52) x 96 500

z (Cr) = 2.96

As the charge on an ion is a small integer, it must be 3+.

The ion is Cr3+.

5. Q = m z F / M

As Q = 600 F, then m = 600 M / z

m (Al) = 600 x 27 / 3 = 5400 g

m (Ag) = 600 x 107.9 / 1 = 64 740 g

m (Zn) = 600 x 65.4 / 2 = 19 620 g

The same 600 F of charge would produce different masses of these metals; 5.4 kg of Al, nearly 20 kg of Zn and almost 65 kg of Ag.

slide32

Faraday’s Laws of Electrolysis – Application

Q. Australia leads the world in the per capita recycling of aluminium. Explain why the production of aluminium uses so much energy and therefore why it is so much better to recycle aluminium.

A. Aluminium is a reactive metal - it’s low on the electrochemical series. The only method of reducing Al ions is electrolytically, hence the energy costs will be quite high. Because Al is so low on the electrochemical series, most other reactions will occur more easily - any water present would be reduced in preference to Al.

From Faraday’s Second Law, no. of moles of Al is inversely related to the charge of Al. Thus more charge is required to produce Al than almost any other metal. Mass of Al produced per mole of charge depends directly on the molar mass of Al. Al has the lowest molar mass of the commonly used industrial metals.