Unit 5 Electrochemistry
Are You Ready? Page 370

Electrochemistry involves oxidation reduction reactions that can be brought about by electricity or used to produce electricity.

Oxidation and reduction, which will be considered here as the loss and gain of electrons, occur in many chemical systems. When such reactions can be made to cause electrons to flow through a wire or when a flow of electrons makes a redox reaction happen, the processes are referred to as electrochemical changes. The study of these changes is called electrochemistry.

Oxidation-Reduction Reactions
Oxidation-reduction reactions or redox reactions involve the transfer of electron density from one atom to another. Two example are the reaction of sodium with chlorine and the reaction of hydrogen with oxygen.
2 Nao + Cl2 ------> 2 NaCl

2 H2 + O2 --------> 2 H2O
 

At first glance these reaction appear to be very different from each other. In NaCl ions have been formed, and that has certainly involved the transfer of electrons. The electron is transferred completely from a sodium to a chlorine atom as the Na+ and Cl- ions are created.   But how about the reaction that produces water? Here we have the formation of a molecule held together by covalent bonds - bonds in which electrons are shared. How does this reaction involve a transfer of electrons? To answer this question, we have to look closely at the bonds in both the reactants and products.

Hydrogen and oxygen molecules are non-polar. This is because both atoms in an H2 or O2 molecule are the same, so the electronegativity difference between them is zero. In a non- polar molecule, the electron pair in the bond is shared equally, and neither atom carries a partial charge. Stated another way, each atom in an H2 or O2 molecule is electrically neutral. Now let's look at the product, water. The electronegativities of hydrogen and oxygen are quite different, oxygen being more electronegative than hydrogen. this means that the O-H bonds are quite polar, with the hydrogen carrying a substantial positive partial charge and the oxygen carrying a substantial negative partial charge.

Now we can look at what happens to the electrons around an atom during the reaction. A hydrogen atom begins with a zero positive charge in H2 and finishes with a partial positive charge in H2O. Similarly, oxygen begins with a zero partial charge in O2 and finishes with a partial negative charge in H2O. Thus, during the reaction there is a shift of electron density from a hydrogen atom to an oxygen atom, and it is in this sense that the reaction of H2 and O2 is similar to the reaction of Na with Cl2.

Many chemical reactions involve (or at least appear to involve) a shift of electron density by one atom to another. Collectively, such reactions are called oxidation-reduction reactions, or simply redox reaction. The term oxidation refers to the loss of electrons by one reactant, and reduction refers to the gain of electrons by another. For example, the reaction between sodium and chlorine involves a loss of electrons by sodium (oxidation of sodium) and a gain of electrons of chlorine (reduction of chlorine).

Nao ------> Na+ + e- (oxidation)

Cl2 + 2 e- -------> 2 Cl- (reduction)

we say that the sodium is oxidized and the chlorine is reduced.
 

Oxidation and reduction always occur together. No substance is ever oxidized unless something else is reduced. Otherwise, electrons would appear as a product of the reaction, and this is never observed. During a redox reaction, then, some substance must accept the electrons that another substance loses. this electron-accepting substance is called the oxidizing agent because it helps something else to be oxidized. The substance that supplies the electrons is called the reducing agent because it helps something else be reduced. Sodium is a reducing agent, for example, when it supplies electrons to chlorine. In the process, sodium is oxidized. Chlorine is an oxidizing agent when it accepts electrons from the sodium, and when that happens, chlorine is reduced to chloride ion. One way to remember is:

The substance that is oxidized is the reducing agent.
The substance that is reduced is the oxidizing agent.

Redox reactions are very common. Whenever you use a battery, a redox reaction occurs. The metabolism of foods, which supplies our bodies with energy, also occurs by a series of redox reactions that use oxygen to convert carbohydrates and fats to carbon dioxide and water. Ordinary household bleach works by oxidizing substances that stain fabrics, making them easier to remove from the fabric or rendering them colourless.

Liquid bleach contains hypochlorite OCl-, as the oxidizing agent.

Homework Practice Page 376
Extension Exercise 5.1

Summary Page 377

Review of Oxidation Numbers
An oxidation number is the charge an atom in a compound would have if the electron pairs in the bonds belonged entirely to the more electronegative atoms.
In the example below, we assign the numbers of +1 to H and to the chlorine we assign -1 in a molecule of HCl. We know that in covalently bonded molecules such as HCl, that the atoms never carry more than partial positive or negative charges. Nevertheless, oxidation numbers are assigned as if each compound were ionic. It is important to remember, therefore, that the oxidation numbers assigned to atoms in compounds do not have to correspond to the actual charges of the atoms -- sometimes they do, but often they do not.
A redox reaction is a chemical reaction in which changes in oxidation numbers occur.
Any element, when not combined with atoms of a different element, has an oxidation number of zero.
e.g.. Na = 0, Cl2 = 0
Any simple monatomic ion (one-atom ion) has an oxidation number equal to its charge. 
Na+ = +1, S-2 = -2
The sum of the oxidation numbers of all of the atoms in a formula must equal the charge written for the formula.
e.g.. SO4-2 = S has an oxidation number of +6 in this ion. Each oxygen has a charge of -2. Therefore +6-2-2-2-2 = -2 is the overall charge on the ion. The oxidation number of oxygen is always -2. If you look at the oxidation number of S on the periodic table it can be in many different states. Its the +6 of the S that is the most important in this ion.
In compounds, the oxidation number of any Group IA metal is always +1, the oxidation number of any Group IIA elements is always +2, and the oxidation number os aluminum is always +3. Check your periodic table to verify this.
In binary compounds with metals, the oxidation number of a nonmetal is equal to the charge of its simple monatomic anion. e.g.. FeBr2 The bromine is -1. therefore the Fe must be +2 and is therefore the iron(II) cation or the ferrous cation.
In compounds F is always -1. O is almost always -2 and H is almost always +1.

Homework Section 5.1 Page 377 Questions 1 to 9
Go to the Electrochemistry Oxidation Numbers Worksheet
Homework Section 5.2  Read Pages 379 through 381
Extension Exercise 5.2
Homework Practice Page 381
5.1 - 5.2 Self Quiz

Using Oxidation Numbers to Balance Equations
Some equations can be balanced by sight. Others require more work on the part of the student. Some equations cannot be balanced by sight and ones involving many reactants and products can be almost impossible to solve. Below is a method of balancing using redox numbers that works for all reaction no matter how complex.
Step 1 Write the correct formulas for all the reactants and products
Step 2 Assign oxidation numbers to the atoms in the equation.
Step 3 Identify which atoms change their oxidation numbers.
Step 4 Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficients.
Step 5 Compute the total change in oxidation number for the oxidation and reduction that occur.
Step 6 Make the total increase in oxidation number equal the total decrease by multiplication using appropriate factors.
Step 7 Balance the remainder by inspection.
Homework Practice Page 383
Go to the Electrochemistry Electron Transfer Balancing Worksheet
Homework Section 5.2 Questions 1 through 6
Activity 5.3 Developing an Activity Series of Metals
Homework Section 5.4 Page 386
Homework Practice Page 388
Homework Section 5.4 Questions 1 through 10
Investigation 5.5 Testing The Activity Series
5.3 - 5.6 Self Quiz

Galvanic Cells
A galvanic cell is also referred to as a voltaic cell or Daniell cell.  The common household battery  is an example of a galvanic cell.  The flow of electrons from one chemical reaction to another happens through an outside circuit resulting in current.  Current is measured in amperes (A) and is a measure of the number of electrons that flow past a certain point in the circuit at any given moment.
This is a simple redox reaction in which both cells are combined into one.  There is a flow of electrons but not through an outside circuit.
One half of the cell is separated from the other by a porous barrier, sometimes called a semi-permeable barrier or by a physical salt bridge like in the diagram below.  The metal strips are referred to as electrodes and the dissolved salt solutions are electrolytes.
To help remember which electrode does what use the chemist's mneumonic term "REDCAT"
i.e..  Reduction(RED) occurs at the cathode(CAT), and oxidation occurs at the anode.
 This is the chemical way of looking at it.

Physics people will tell you the cathode is positive(+ve) and the anode is negative (-ve) and electrons flow from the anode to the cathode through an external circuit.

Do you see the inherent conflict between these two statements.
If reduction occurs at the cathode, then the cathode must have the surplus of electrons, but its' labeled as being positive (+ve).  If the cathode was the surplus then the anode must be deficit in electrons yet it is labeled as negative (-ve).  Why is this???

It comes from the old cause and effect concept.  We are used to looking at the flow of electrons (electricity) that we almost never stop to see what caused the flow to happen.  This is where the science of chemistry comes in.
The reasons behind electron flow don't start with the flow of electrons.  It starts with 2 metals, one being stronger in it's desire for electrons than the other.   (i.e.. bending trees don't cause the wind to blow either)

Lets see if we can clarify this:
1.    The metal with the stronger desire for electrons; i.e. the higher electronegativity, is the one that will be reduced. The metal ions in the  electrolyte steal electrons from the metal strip.  This causes the metal ions to become reduced to the metal atom.  The strip of metal, having lost electrons becomes more positive.

2.  The deficit of electrons at the cathode means that there is now a surplus of electrons at the anode. i.e.. The anode is now negative when compared to the cathode.  (This is the physics point of view)

3.   Electrons flow from the -ve anode to the +ve cathode to replace those electrons lost to the reduction reaction.

4.  As the electron quantity at the anode drops there is an attraction for electrons in the electrode.  As electrons get removed from the electrode, metal atoms in the electrode give up their electrons, becoming positive ions, and these positive ions dissolve off into the electrolyte solution.

5.   If the salt bridge is not there, the cell that is performing the reduction would become very negative, because the negative anion must remain while all the positive cations are being reduced.  The cell that is performing the oxidation will become very positive, because of the formation of positive ions.  Eventually this buildup would stop the reactions since the positive cell would build up such a large positive charge that it would start to become more attractive to the electron flow that the original cathode metal electrode. At the same time the buildup of the large negative charge in the cathodic cell would start to repel or oppose the flow of electrons.   A salt bridge between the two cells allows a balancing of the electrolytes so that this buildup does not take place.  The negative anions from the reduction cell react with the positive cations produced in the oxidation cell neutralizing their charges.

To obtain the overall reaction that takes place in the electrochemical cell, the cell reaction, simply add the individual electrode reactions together. Before doing this make sure that the number of electrons gained is equal to the number of electrons lost. This is a requirement that every redox reaction must obey. Multiply the half-reactions by a common multiple in order to achieve this equal number.

Na+(aq) + e- --> Na(s) (cathode)
Cl-(l) --> Cl2(g) + 2e- (anode)
2 Na+(l) + 2 Cl-(l) + 2e- --> 2 Na(l) + Cl2(g) + 2e- (cell reaction)
Then finish the reaction by canceling out like terms on either side of the ----->. The overall cell reaction is therefore:   2 Na+(l) + 2 Cl-(l) --> 2 Na(l) + Cl2 (g)

Homework Practice Page 397

Standard Cell Reduction Potentials
The voltage across the electrodes of a galvanic cell can be attributed to the differnece in the tendencies of the two half-cells to undergo reduction.
Voltage or electromotive force (emf) can easily be thought of as the force with which an electric current is pushed through a wire.  It is measured in terms of an electrical unit called the volt (V).   Voltage is a measure of the amount of energy that can be delivered by a coulomb of electrical chrage as it passes through a circuit.   A current flowing under an emf of 1 volt can deliver 1 joule of energy per coulomb.
1 V = 1 J/C
The maximum emf of a galvanic cell is called the cell potential, Ecell.  It is measured with a device that draws negligible current during the experiment.  Ecell depends on the composition of the electrodes and the concentrations of the ions in each of the half-cells.
  For reference purposes, the standard cell potential, symbolized Eocell, is the potential of the cell when all of the ion concentrations are 1.00 M, the temperautre is 25oC, and any gases that are involved in the cell reaction are at a pressure of 1 atm.

Cell potnetials are rarely larger than a few volts.  For example, the standard potential for the galvanic cell cunstructed from silver and copper electrodes produces only about 0.46 V, and one cell in an automotive battery produces only about 2 V.  Batteries that generate higher voltages contains a number of cells wired in series so that their emfs are additive.
 

Reduction Potentials
It is useful to imagine that the measured overall cell potential arises from a competition between the two half-cells for electrions.  If we think of each half-cell reaction as having a certain natural tendeancy to procedd as a reduction, the magnitude of which is expressed by its reduction potential or standard reduction potential when the temperature is 25oC, concentrations are 1 M, and the pressure is 1 atm.    When two half-cells are connected, the one with the larger reduction potential - the one with the greater tendency to undergo reduction - acquires electrons from the half-cell with the lower reduction potential, which is therefore forced to undergo oxidation.  The measured cell potential actually represents the magnitutue of the differnce between the reduction potential of one half-cell and the reduction potential of the other.  In general,

Eocell = (standard reduction                          (standard reduction
                 potential of the                   -              potential of the
                substance reduced)                          substance oxidized)

Let's look at the silver-copper cell.  From the cell reaction

2 Ag+(aq) + Cu(s)  ---->  Cu2+(aq)  +  2 Ag(s)


we can see that the silver ion is reduced and copper is oxidized.  If we compare the twow possible reduction half-reactions,
 

     Ag+(aq) + e- ----> Ag(s)
Cu2+(aq) + 2 e- ----> Cu(s)


the one for Ag+ must have a greater tendency to roceed than the one for Cu2+, because it is the silver ion that is actually reduced.  This means that the standard reduction potential of Ag+, EoAg+, must be larger than the standard reduction potential of Cu2+, EoCu2+.  If we knew the values of  EoAg+  and EoCu2+  , we could calculate Eocell thusly: 

                                                   Eocell   =  EoAg+ -  EoCu2+

Assigning  Standard Reduction Potentials
Potentials are assigned  when a cell is compared against a hydrogen half cell.  Some cells like the fluorine half cell are reduced more easily than hydrogen therefore the reduction potential of fluorine compared to hydrogen as +2.86 V.  At the other end of teh scale, lithium does not like to rduce compared to hydrogen, hydrogen is more easily reduced so Li is oxidixed and has a potential voltage of -3.05 V. 

Using a list of standard reduction potentials, you can quickly and easily calcualte the Eocell  for any pair of possible half-cells under normal conditions.

Using Standard Reduction Potentials
The key to this using standardreduction potentials is to remember that:
 

The more positive reduction potnetial occurs as a reduction


The half-cell reactions andtheri standard reduction potnetials from a Standard Reduction Table can be used in a number of ways to give information about galvanic cells.  Equally useful is the fact that they can provide information about the outcome of redox reactions, whether they occur in a galvanic cell or not.  The example below shows hoe reduction potenitals can be sued to predict the standard cell potential and the overall reaciton in a galvanic cell.

e.x.  A galvanic cell was constructed using eletrodes made of lead and lead(IV) oxide (PbO2) with sulfuric acid as the electrolyte.  The half-cell reactions and their reduction potentials in this system are:

PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- <===>  PbSO4(s)  + 2 H2O     EoPbO2 = 1.69 V
                                    PbSO4(s)  + 2e- <====>  Pb(s)  +  SO42-(aq)  EoPbSO4 = -0.36 V

What is the cell reaction and what is the standard potential of the cell?
 

In a competition for electrons, the half-reaction having the higher (more positivie) reduction potential undergoes reduction.  The half-reaction with the lower reduction potential is therfore forced to undergo oxidation.  This means that here is the first half-reaction given will occur as reduction and the second will be forced to reverse its direction.  In the cell, the reactions are:

PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- ------->  PbSO4(s)  + 2 H2
                              Pb(s)  +  SO42-(aq) -------> PbSO4(s)  + 2e-
 

Adding the two half-reactions and canceling electrons gives the cell reaction,

        PbO2(s) + 4H+(aq) + Pb(s) + 2 SO42-(aq)  -------> 2 PbSO4(s)  + 2 H2

(This is the reaction that takes place in lead storage batteries used to start cars, motorcycles and boats.)  The cell potential can be obtained by using :

Eocell = (standard reduction                          (standard reduction
                 potential of the                   -              potential of the
                substance reduced)                          substance oxidized)

Since the first half-reaction occurs as a reduction and the second as an oxidation,

Eocell =   EoPbO2  -  EoPbSO4
          =   1.69 V - (-0.36 V)
          =    2.05 V
 

ex.1. 2  Predicting the Cell Reaction and Cell Potential
What would be the cell reaction and the standard cell potential of a galvanic cell employing the following half-reactions?

Al3+(aq) + 3e- <==>  Al(s)         EoAl3+ = -1.66 V
Cu2+(aq) + 2e- <==>  Cu(s)       EoCu2+ = 0.34  V

Which half-cell would be the anode?

The reaction with the more positive reduction potential occurs as a reduction; the other occurs as an oxidation.  In this particular half-cell pair, Cu2+ is reduced and Al is oxidized.
To obtain the cell reaction,  leave the reduction reaction alone and flip the oxidation reaction.
 

Cu2+(aq) + 2e- <==>  Cu(s)
Al(s)  <==>   Al3+(aq) + 3e-


The electrons must cancel so multiply so that electrons cancel out. (i.e. cross multiply the reactions by the electron coefficients), then add the result
 

3  [Cu2+(aq) + 2e- <==>  Cu(s)]  (reduction)
2 [Al(s)  <==>   Al3+(aq) + 3e-]  (oxidation)
3 Cu2+(aq)  + 2 Al(s) <==> 3 Cu(s) + 2 Al3+(aq)


the anode in the cell is aluminum because that is where oxidation takes place.  to obtain the cell potential, we substitute into the cell potential equation.

Eocell =   EoCu2+ -  EoAl3+
          =   0.34 V - (-1.66 V)
          =    2.00 V

Notice that we multiply the half-reaction by factors to make the electrons cancel, but we do not multiply the reduction potentials by these factors.  (Reduction potentials are intensive quantities; they have the units volts, which are joules per coulomb.  The same number of joules are available for each coulomb of charge regardless of the total number of electrons shown in the equation.  Therefore, reduction potentials are never multiples by factors before they are subtracted to give the cell potential.)

Predicting a Spontaneous Reaction
Reduction potentials can also be used to predict the spontaneous reaction between the substances given in two half-reactions, even when these substances are not in a galvanic cell.  The procedure is very simple because we know that the half-reaction having the more positive reduction potential always undergoes reduction while the other is forced to undergo oxidation.

ex.1  What spontaneous reaction will occur if Cl2 and Br2 are added to a solution containing Cl- and Br-?

There are two possible reduction reactions.

Cl2 + 2e- ----> 2 Cl-
Br2 + 2e- ----> 2 Br-


Refer to a standard reduction potential table.
Cl2 has the more positive potential at 1.36 V.  Br2 has a somewhat less positive potential at 1.07 V.  This means the Cl2 will be reduced and the Br2 will be oxidized.  When the reaction occurs the half-reactions will be:
 

Cl2 + 2e- --->  2 Cl-
2 Br- --->  Br2 + 2e-


The net reaction will be Cl2 + 2 Br- ---->  Br2 + 2 Cl-

Experimentally, we find that chlorine does indeed oxidize bromide ion to bromine, and this fact is used in the synthesis of bromine.

ex.2   Predict the reaction that will occur when Ni and Fe are added to a solution that contains both Ni2+ and Fe2+.

Ni2+ has a more positive (less negative) reduction potential than Fe2+.  Therefore, Ni2+ will be reduced and Fe will be reduced.

Ni2+(aq) + 2e-  ---> Ni(s)    (reduction)
Fe(s) --->  Fe2+(aq) + 2e-     (oxidation)
Ni2+(aq) + Fe(s)  ---> Ni(s)  Fe2+(aq)    (net reaction)
 

Determining Whether a Reaction is Spontaneous
Since we can predict the spontaneous redox reaction that will take place among a mixture of reactants, it should be possible to predict whether or not a particular reaction will occur spontaneously as written.  This can be done by calculating the cell potential that corresponds to the reaction in question.

For any functioning galvanic cell, the measured cell potential has a positive value.

If we compute the cell potential for a particular reaction based on the way the equation is written and the potential comes out with a positive value, then the reaction is spontaneous.  If the calculated cell potential comes out negative, however, the reaction is nonspontaneous.  In fact, it really is spontaneous but in the opposite direction.

ex.1  Determine whether the following reactions are spontaneous as written.  If they are not, give the reaction that is spontaneous.
a)  Cu(s)  +  2 H+(aq) ---> Cu2+(aq)  +  H2(g)
b)   3 Cu(s)  +  2 NO3-(aq)  + 8 H+(aq) ---->  3 Cu2+(aq)  +  2 NO(g)  +  4 H2O

Solution
a)  the half-cell reactions involved in this reaction are:
        Cu(s)  ---> Cu2+(aq)   + 2 e-
        2 H+(aq)  + 2 e-  ---->  H2(g)

From the way the reaction is written, H+ is reduced and Cu is oxidized.
Using the values from the standard reduction potential table we get

Eocell = EoH+ - EoCu2+
          =  (0.00 V) - (0.34 V)
          =  -0.34 V

The calculated cell potential is negative and therefore the reaction above will not be spontaneous in the direction written. The spontaneous reaction will be:

                 Cu2+(aq)  +  H2(g)  ---> Cu(s)  +  2 H+(aq)

b)   The halfcell reactions are:
  Cu(s)  ---->   Cu2+(aq) + 2e-
   NO3-(aq)  + 4 H+(aq) + 3e-  ---->   NO(g)  +  2 H2O

The Cu is oxidized while the NO3- is reduced.
Eocell = EoNO3- - EoCu2+
          =  (0.96 V) - (0.34 V)
          =  0.62 V
Since the calculated cell potential is positive, this reaction is spontaneous in the forward direction.
Homework Section 5.7 Questions 1 through 9

Go to the Standard Cell Potential Exercise

Consumer Cells and Batteries
Homework Section 5.9 Pages 403 through 407
Homework Section 5.9 Page 408 Questions 1 through 10


Hydrogen Fuel Cells

Homework Read Page 411
Homework page 411 Questions 1 through 4


Corrosion
Homework Read Pages 412 through 415
Homework Page 416 Questions 1 through 10


Preventing Corrosion
Homework Read Pages 422 through 424
Homework Page 425 Questions 1 through 10

5.7 - 5.11 Self Quiz

Electrolytic Cells
Electricity can provide the necessary energy to cause otherwise nonspontaneous reactions to occur.

When electricity is padded through a molten ionic compound or through a solution containing ions - an electrolyte - a chemical reaction called electrolysis occurs. A typical electrolysis apparatus, referred to as an electrolysis cell or electrolytic cell, is shown below.
 

This particular cell contains molten sodium chloride. A substance undergoing electrolysis must be molten or in solution so that its ions can move freely and conduction can occur. Nonreactive, inert electrodes are dipped into the cell and then connected to a source of direct current (DC) electricity.

When electricity starts to flow, chemical changes begin to take place. At the positive electrode, the anode, oxidation occurs as electrons are pulled from negatively charged chloride ions. The DC source pumps these electrons through the external electrical circuit to the negative electrode, the cathode. At the cathode, reduction takes place as the electrons are picked up by positively charged sodium ions.

Anode is the name of the electrode at which oxidation occurs.

Cathode is the name of the electrode at which reduction occurs.
 

The chemical changes that take place at the electrodes can be expressed in the form of chemical equations.

Na+(aq) + e- ----- Na(l) (cathode)

2 Cl-(l) ----- Cl2(g) + 2e- (anode)

When sodium chloride undergoes electrolysis (when it is electrolyzed), no electrons actually pass through the molten NaCl from one electrode to another. The electrical conduction here is quite different from that in a metal, where electrons carry the charge. In a molten sat such as sodium chloride, or in a solution of an electrolyte, it is the ions that move through the liquid that carry the charge. The transport of electrical charge by the ions is called electrolytic conduction. In the case of molten NaCl, for example, negatively charged chloride ions gradually move toward the positive electrode, and positively charged sodium ions gradually move toward the negative electrode. Around each electrode, a layer of ions of opposite charge accumulates, and electrolysis is able to continue only because reactions of these ions deplete the layers and make room for more ions from the surrounding liquid. If the redox changes at the electrodes cease, the flow of electricity in the external circuit also stops.

Applications of Electrolysis
Electroplating
The application by electrolysis of a thin ornamental or protective coating of one metal over another. It is a common technique for improving the appearance and durability of metal objects. For instance, a thin, shiny coating of metallic chromium is applied over steel automobile bumpers to make them attractive and to prevent rusting of the steel. Silver and gold plating is applied to jewelry made from less expensive metals, and silver plating is common on eating utensils. These thin metallic layers, generally 0.03 to 0.05 mm thick, are applied by electrolysis.

The diagram to the right illustrates a typical apparatus for plating silver. Silver ions in the solution are reduced at the cathode where it is deposited as metallic silver on the object to e plated. At the anode, silver from the metal bar is oxidized, replenishing the supply of the silver ion in the solution. As time passes, silver is gradually transferred from the bar at the anode onto the object at the cathode.

The exact composition of the electroplating bath varies, and is often a trade secret, but it usually depends on the metal to be deposited, and can affect the appearance and durability of the finished surface. For example, silver deposited from a solution of silver nitrate does not stick to other metal surfaces very well. However, if it is deposited from a solution is silver cyanide containing Ag(CN)2- ions, the coating adheres well and is bright and shiny. Other metals that are electroplated from a cyanide bath are gold and cadmium. Nickel, which can also be applied as a protective coating, is plated from a nickel sulphate solution, and chromium is plated from a chromic acid, H2CrO4 solution.

Production of Aluminum
Until the latter part of the nineteenth century, aluminum was an uncommon metal - only the very rich could afford aluminum products. A student at Oberlin College, 21 year old Charles M. Hall, learned of this and began a series of experiments in an attempt to invent a cheap method of extracting the metal fro its compounds. The difficulty that he faced was that aluminum is a very reactive element. It is difficult to produce as a free element by usual chemical reactions. Efforts to produce aluminum by electrolysis were unproductive because its anhydrous salts were difficult to prepare and its oxide, Al2O3, has such a high melting point, > 2000oC, that no, practical method of melting it could be found. In 1886 Hall discovered that Al2O3 dissolves in a mineral called cryolite, Na3AlF6, to give a conducting mixture with a relatively low melting point from which aluminum could be produced electrolytically.

A diagram of the apparatus used to produce aluminum is shown below. Aluminum ore, called bauxite, contains Al2O3. The ore is purified and the Al2O3 is then added to the molten cryolite electrolyte in which it dissolves and dissociates. At the cathode, the aluminum ions are reduced to produce the free metal, which forms as a layer of molten aluminum below the less dense electrolyte. At the carbon anodes, oxide ion is oxidized to give free O2.

Al3+ + 3e- -----> Al(l) (cathode)

2 O2- -----> O2(g) + 4e- (anode)
 

The net cell reaction is

4 Al3+ + 6 O2- -----> 4 Al(l) + 3 O2(g)

The oxygen produced at the anode attacks the carbon electrodes, producing CO2, so the electrodes must be replaced frequently.
The production of aluminum consumes enormous amounts of electricity and is therefore very costly, not only in terms of dollars but also in terms of energy resources.


Unit 5 Summary
Unit 5 Review Page 444 Questions 1 through 32
Unit 5 Self Quiz