AP Chemistry - Electrochemistry
Introduction
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. The rusting of iron, the photosynthesis that takes place in the leaves of green plants, and the conversion of foods to energy in the body are all examples of chemical changes that involve the transfer of electrons from one chemical species to another. 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.
The applications of electrochemistry are widespread. Batteries, which produce electrical energy by means of chemical reactions are in almost anything portable and electronic. In the laboratory, electrical measurements enable us to monitor chemical reactions of all sorts, even those in systems as tiny as a living cell. In industry, many important chemical - including liquid bleach (sodium hypochlorite) and lye (sodium hydroxide) are manufactured by electrochemical means. If it weren't for electrochemistry the important structural metals of aluminum and magnesium would only be laboratory curiosities and most people would see them only in museums.
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.

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.
Go to the Electrochemistry Oxidation Numbers Worksheet

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.
Go to the Electrochemistry Electron Transfer Balancing Worksheet

Balancing Redox Equations by the Ion-Electron Method or Half Cell Method
This method divides a reaction into two parts that are balanced separately. It provides a simple method for obtaining a net ionic equation for a redox reaction.
In this method the oxidation and reduction processes are divided into separate equations called half-reactions that are balanced separately. Then the half reactions are combined to give the overall net ionic reaction.
In many redox reaction in aqueous solution the H+ and OH- ions play an important role, as do the H2O molecules themselves. Because H+ and OH- can be consumed or produced by a redox reaction, and Because the products can change during a reaction if the solution changes from acidic to basic (or vice versa), redox reactions are generally carried out in solutions containing a substantial excess of either acid or base. Therefore, before you apply the ion-electron method, you have to know whether the reaction occurs in an acidic or basic solution.
In Acidic Solutions
Step 1 Divide the skeleton equation into half-reactions.
Step 2 Balance atoms other than H and O.
Step 3 Balance oxygen atoms by adding H2O to the side that needs O.
Step 4 Balance hydrogen by adding H+ to the side that needs H.
Step 5 Balance the charge by adding electrons.
Step 6 Make the electrons gained equal to the electrons lost and then add the two half-reactions.
Step 7 Cancel anything that is the same on both sides.
In Basic Solutions
Step 8 Add the same number of OH- as there are H+ to both sides of the equation.
Step 9 Combine OH- and H+ to form H2O
Step 10 Cancel any H2O that you can.
Go to the Electrochemistry Balancing by Half-Cell Worksheet

Galvanic, Voltaic and Daniell Cell Reactions
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, but only in an Electrolytic cell in and Galvanic cell the positions are reversed.

Oxidation occurs at the cathode and reduction 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)

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.
 

Go to the Standard Cell Potential Exercise

Cell Potentials and Thermodynamics
Standard cell potentials can be used to calculate free energy changes and equilibrium constants.

The fact that cell potentials allow us to predict the spontaneity of redox reactions is no coincidence.  There is a relationship between the cell potential and the free energy change for a reaction.  DG for a reaction is a measure of the maximum useful work that can be obtained from a chemical reaction.
 

-ΔG = maximum work


In an electrical system, work is supplied by the electric current that is pushed along  by the potential of the cell. It can be calculated from the equation
 

maximum work = nFEcell


where n is the number of moles of electrons transferred, F is the Faraday constant (96,500 coulombs per mole of electrons), and Ecell is the potential of the cell in volts.

maximum work = n     F Ecell    (unit analysis)
                           = ( mol e- ) X (coulomb) X (       J       )  =  joule
                                                      mol e-        coulomb
 

 Therefore ΔG = -nFEcell


If we are dealing with the standard cell potential, we can calculate the standard free energy change.

ΔGo = -nFEocell


ex.1   Calculate ΔGo for the following reaction. given that its standard cell potential is 0.320 V at 25oC.
 

NiO2  +  2 Cl-  + 4 H+   ---->  Cl2  +  Ni2+  +  2 H2O


Since two Cl- are oxidized to Cl2, two electrons are transferred.  Interpreting the coefficient as moles, 2 moles of electrons are transferred, so n=2.

     ΔGo    = ( 2 mol e- ) X ( 96,500 coulomb) X (    0.320  J 
                                                mol e-                     coulomb
                            = -61,800 J
                            = -61.8 kJ
 

Determining Equilibrium Constants
One of the most useful applications of electrochemistry is the determination of equilibrium constants.  ΔGo is related to the equilibrium constant (Kp for gaseous reactions and Kc for reactions in solution.)
ΔGo = -RT ln Kc

ΔGo is also related to Eocell

ΔGo = - nFEocell

Therefore we should be able to combine the equations and obtain:

              Eocell =RT ln Kc
                           nF

This equation is usually expressed as a base 10 logarithm:

Eocell0.0592 log Kc
                 n

The constant 0.0592 has the units joules/coulombs, which is volts.  As before n is the number of moles of electrons that are transferred in the cell reaction as written.

Calculate Kc for the reaction below.  It has a standard cell potential of 0.320 V at 25oC. 
 

NiO2  +  2 Cl-  + 4 H+   ---->  Cl2  +  Ni2+  +  2 H2O


The value of n = 2

Using the equation and substituting we get:
 

0.320 V = 0.0592 log Kc
                      2                  

2(0.320) = log Kc
(0.0592)             

10.8 = log Kc

Taking the antilog   Kc = 6 X 1010

Go to the Cell Potentials and Thermodynamics Worksheet

Nernst Equation
Altering the concentrations of the ions in the half-cells affects the cell potential in a way that depends upon the logarithm of the mass action expression for the cell reaction.
When all of the ion concentrations in a cell are 1 M, the cell potential is equal to the standard potential.  When the concentrations change, however, so does the potential.  For example, in an operating cell or battery, the potential gradually drops as the reactants are consumed.  the approaches equilibrium, and when it gets there the potential has dropped to zero - the battery is dead.

The effect of concentration on the cell potential can be calculated from the Nernst equation.
 

Ecell = Eocell2.303 RT log Q
        nF 


where Ecell is the actual cell potential, Eocell is the standard cell potential, R is the ideal gas law constant in energy units, T is the absolute temperature, F is the faraday andn is the number of moles of electrons transferred in the cell reaction and Q is the reaction quotient for the reaction.  This equation is often written in terms of base 10 logarithms so:
 

Ecell = Eocell0.0592 log Q
                 n 


ex.1  What is the potential of a copper-silver cell is the Ag+ concentration is  0.001 M and the Cu2+ concentration is 0.0001 M?  The standard potential of the cell is 0.46 V and the cell reaction is
 

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


First, rewrite the correct form of the Nernst equation for this particular cell.  Since two electrons are lost when one Cu atom is oxidized to give Cu2+, n = 2.  The correct mass action expression for this reaction, as usual, has the product concentration in the numerator.
 

Q = [Cu2+]
        [ Ag+]2


Notice that the concentrations have been omitted.  The revised Nernst equation is:
 

Ecell = Eocell0.0592 V log [Cu2+]
                             n               [ Ag+]2

Ecell = 0.46 V - 0.0592 V log [0.0001]
                              2              [0.001]2

Ecell = 0.46 V - 0.0296 V log [0.01]

  = 0.46 V - 0.0296 V (2.00)
 = 0.46 V - 0.0592 V 
= 0.40 V 
 
The cell potential is 0.40 V.

ex.2.   A cell employs the following half-reactions
Ni2+(aq) + 2e- <===>  Ni(s)    EoNi2+ = -0.25 V
Cr3+(aq) + 3e- <===>  Cr(s)    EoCr3+ = -0.74 V

Calculate the potential if [Ni2+] = 0.0001 M and the [Cr3+] = 0.002 M


First we need the cell reaction.  Nickel is reduced because it has the higher reduction potential.  This means, of course, that chromium is oxidized.  Making the electron gained by the nickel equal to the electrons lost by the chromium we get.
 
3 [Ni2+(aq) + 2e- <===>  Ni(s)]    (reduction)
2 [Cr(s) <===>  Cr3+(aq) + 3e-]    (oxidation)
   3 Ni2+(aq) + 2 Cr(s) <===> 3 Ni(s)  + 2 Cr3+(aq)  (cell reaction)


Notice that six electrons are transferred overall, so n=6.

Next we can calculate Eocell.

Eocell = EoNi2+  -  EoCr3+
= (-0.25 V) - (-0.74 V)
= 0.49 V 


Now we can write the Nernst equation.

Ecell = Eocell0.0592 V log [Cr3+]2
                            n             [ Ni2+]3

Ecell = 0.49 V - 0.0592 V log [0.002]2
                               6             [0.0001]3

Ecell = 0.49 V - 0.010 V log  (4.0 X 106)
=  0.49 V - 0.010 V (6.60) 
= 0.49 V - 0.066 V 
= 0.42 V 


Determination of Concentrations from Cell Potentials
One of the principle uses of the relationship between concentration and cell potential is in the measurement of concentrations.  experimental determination of cell potentials combines with modern developments in electronics has provided a means of monitoring and analyzing the concentrations of all sorts of substances in solution, even some that are not themselves ionic and that are not involved directly in electrochemical changes.  The basic concept is illustrated by the example below.

A chemist wanted to measure the concentration of Cu2+ in a large number of samples of water in which the copper ion concentration was expected to be quite small.  The apparatus that was used consisted of a silver electrode dipping of AgNO3.  This half-cell was connected by a salt bridge to a second half-cell containing a copper electrode that was able to be dipped into each water sample, one after another.  In the analysis of one of the samples, the cell potential was measured to be 0.62 V, with the copper serving as the anode.  What was the concentration of copper ion in this particular sample of water?

The first step is to write the proper chmeical equation, because we need it to compute Eocell and to construct the mass action expression (reaction quotient) that will be used in the Nernst equation.  Since copper is the anode, it is being oxidized.  This also means that Ag+ is being reduced.  Therefore, the correct equation for the cell reaction is:
 

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


and the Nernst Equation for the reaction is:
 

Ecell = Eocell0.0592 V log [Cu2+]
                            n               [ Ag+]2


The value of Eocell can be obtained from the standard reduction potential table.
 

Eocell = EoAg+ - EoCu2+
          = 0.80 V - 0.34 V
          = 0.46 V 


Now we can substitute values into the Nerst equation and solve for the concentration ratio in the mass action expression.
 

0.62 V = 0.46 V - 0.0592 log [Cu2+]
                                  2           [Ag+]2
-5.4 = log [Cu2+]
                 [Ag+]2
 
Taking the antilog gives us the value of the mass action expression.
[Cu2+]   =  4 X 10-6
 [Ag+]2

Since we know the concentration of the [Ag+] = 1.00 M then we can solve for Cu2+.
 

[Cu2+]   =  4 X 10-6
[1.00]2

Therefore  [Cu2+]  =  4 X 10-6

Notice that this is indeed a very small concentration, and it can be obtained very easily by simply measuring the potential generated by the electrochemical cell.  Determining the concentration in many samples is also very simple - just change the water sample and measure the potential again.

Go to the Cell Potentials and Cell Concentrations Worksheet

Electrolysis
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.

Electrolysis Reactions in Aqueous Solutions
When electrolysis is carried out in aqueous solution, the net reaction is more difficult to predict because the oxidation and reduction of water can also occur. This happens when electrolysis is carried out in a solution of potassium nitrate. the products of the electrolysis are hydrogen and oxygen, as shown below.

At the cathode, water is reduced.

2 H2O(l) + 2e- -----> H2(g) + 2 OH- (cathode)

At the anode, water is oxidized.

2 H2O(l) -----> O2(g) + 4 H+(aq) + 4e- (anode)
 

If acid-base indicators are placed in the solution, colour changes confirm that the solution becomes basic around the cathode and acidic around the anode as the electrolysis proceeds. In addition, the gases H2 and O2 can be collected.

The overall cell reaction can be obtained as before. Since the number of electrons lost has to equal the number of electrons gained, the cathode reaction must occur twice each time the anode reaction occurs once.

4 H2O(l) + 4e- -----> 2 H2(g) + 4 OH- (cathode)

2 H2O(l) -----> O2(g) + 4 H+(aq) + 4e- (anode)

After adding, cancel the redundant species.

6 H2O(l) -----> 2 H2(g) + O2(g) + 4 H+(aq) + 4 OH-(aq)

The overall net ionic reaction is:

2 H2O(l) -----> 2 H2(g) + O2(g)
 

If the electrolysis is attempted without the presence of KNO3 nothing happens. If the KNO3 were not present and the electrolysis were to occur, the solution around the anode would become filled with H+ ions, with no negative ions to counter the charge. Similarly, the solution surrounding the cathode would become filled with hydroxide ions with no nearby positive ions. When KNO3 is in the solution, K+ ions move toward the cathode where they mingle with the OH- ions as they are formed. The NO3- ions move toward the anode and mingle with the H+ ions as they are produced there. In this way, at any moment, each small region of the solution contains the same number of positive and negative charges.

Predicting the outcome of electrolysis reactions in aqueous solutions can be tricky because the reactions at the electrode surfaces are complex, especially when they involve the evolution of hydrogen and oxygen. For example, the oxidation of H2O to give O2 is thermodynamically easier (in terms of the sign and magnitude of Go) then the oxidation of Cl- to give Cl2. Therefore, it would be expected that the electrolysis of a solution of NaCl, where there are these competing reactions at the anode,

2 H2O -----> O2(g) + 4 H+(aq) + 4e-

2 Cl-(aq) -----> Cl2(g) + 2e-

oxygen should be evolved. However, because of the complexity of the electrode reactions, it is actually chlorine that is formed (provided that the chloride ion concentration is reasonably high). Thus, even though evolution of O2 is thermodynamically preferred, we don't see this happen because of other complicating factors.
 

Although it can be hard to anticipate beforehand what will happen in the electrolysis of aqueous solutions, it is possible to use the results of one electrolysis experiment to predict what will happen in others. For instance, when a solution of CuBr2 is electrolyzed, the cathode becomes coated with a reddish brown deposit of copper and the blue colour of copper ion in the surrounding solution fades. At the same time, the solution around the anode acquires a reddish brown colour. these observations tell us that copper ion is being reduced at the cathode and that bride ion is being oxidized to bromine at the anode. The electrode reactions are

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

2 Br-(aq) -----> Br2(aq) + 2e- (anode)

and the net cell reaction is

Cu2+(aq) + 2 Br-(aq) -----> Cu(s) + Br2(aq)

The behavior of a particular ion in solution toward oxidation or reduction by electrolysis is the same regardless of the source of the ion. Therefore, once you know what happens in the electrolysis of CuBr2, you can predict at least partially what will happen in the electrolysis of other salts that contain either of these ions. Solutions of CuCl2, CuSO4, Cu(NO3)2, and Cu(C2H3O2)2 all contain Cu2+ ion, and when they are electrolyzed we can expect that in each case Cu2+ will be reduced at the cathode. (Although we can't say at this point what will happen at the anode.) Similarly, solutions of KBr, NaBr, CaBr2 and FeBr2 all contain Br- so we expect that at the anode bromide ion will be oxidized to give Br2 when solutions of these salts are electrolyzed.
 

Stoichiometric Relationships in Electrolysis
Much of the early research in electrochemistry was performed by a British scientist named Michael Faraday (1791-1867). It was he who coined the terms anode, cathode, electrode, electrolyte, and electrolysis. In about 1833, Faraday discovered that the amount of chemical change that occurs during electrolysis is directly proportional to the amount of electricity that is passed through an electrolysis cell. For example, the reduction of copper ion at a cathode is given by the equation

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

To deposit one mole of metallic copper requires tow moles of electrons. To deposit two moles of copper requires four moles of electrons, and that takes twice as much electricity.

A unit normally used in electrochemistry to mean one mole of electrons is the Faraday (), named in honour of Michael Faraday.

1 = 1 mol e-

The half-cell reaction for an oxidation or reduction, therefore, related the amount of chemical substance consumed or produced to the number of Faradays that the electric current must supply.
 

To use the Faraday it must be related to an electrical measurement that can be used commonly in the laboratory. The SI unit of electric current is the ampere (A) and the SI unit of charge is the coulomb (C). A coulomb is the amount of charge that passes by a given point in a wire when an electric current of one ampere flows for one second. This means that coulombs are the product of amperes or current multiplied by seconds. Therefore

1 coulomb = 1 ampere X 1 second

1 C = 1 A.s

For example, if a current of 4 A flows for 10 sec, 40 C pass by a given point in the wire.

(4 A)X(10 s) = 40 A.s
                     = 40 C

Experimentally, it has been determined that 1 Faraday (1 mole of electrons) carried a charge of 96,485 C. Therefore

1 = 96,485 C

Now you have a way to relate laboratory measurements to the amount of chemical change that occurs during an electrolysis. Measuring current and time allows us to calculate the number of coulombs. From this we can get Faradays, which we can then use to calculate the amount of chemical change produced.

ex.1 Problem: How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00 A is run through a solution of CuSO4 for a period of 20.0 min?

Solution: First convert minutes into seconds.

20.0 min X 60 s = 1200 s
                  1 min

Then multiply the current by the time to obtain the number of coulombs (1 A.s = 1 C)

1200 s X 2.00 A = 2400 A.s = 2400 C

Since 1 = 96,485 C then

2400 C x      1      = 0.02487
               96,485 C

The equation for the reduction of copper ion to copper is

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

The equation relates the stoichiometric relationship of 1 Cu2+ reacting with 2 moles of e-.

So 1 Cu2+ ions = 2 e-
         x                0.02487 e-/mol (since 1 = 1 mol of e-)

x = 0.01244 mol of Cu2+ ions are going to get reduced.
 

Since 1 mole of Cu2+ produces 1 mole of Cuo then you can expect to get 0.01244 moles of Cu from this much current.

This would be g = mol x mm
                         = 0.01244 mol X 63.45 g/mol
                         = 0.791 grams of Cu atoms

The electrolysis will deposit 0.791 g of copper on the cathode.

ex.2 Electroplating is an important application of electrolysis. How much time will it take in minutes to deposit 0.50 g of metallic nickel on a metal object using a current of 3.00 A? The nickel is reduced from a solution of Ni(NO3)3.

Solution:
The deposit of 0.500 grams of Ni must be converted into moles.

moles = g/mm = 0.500 g / 58.69 g/mol = 0.00852 moles

The equation for the reduction is:

Ni2+ + 2e- -----> Nio

This gives the relationship that, for every mole of Ni2+ reduced or Nio plated, there is a need for 2 moles of e-.

Since we will plate out 0.00852 moles of Ni then we'll need twice the number of e- or 0.0170 moles of e- or 0.0170 .

The number of coulombs will be 0.0170 X 96485 C = 1640 C
                                                                      1

This 1640 C is really 1640 A.s which is the product of the current multiplied by time. The current is 3.00 A therefore the seconds must be = 1640 A.s/3.00 A

= 546.67 sec = 9.11 min

ex3. What current is needed to deposit 0.500 g of chromium metal from a solution of Cr3+ in a period of 1 hr?

Solution:

Cr3+ + 3e- -----> Cro

From the equation we will need three of e- for every mole of Cr3+ reduced or Cro plated out.

= 0.500 g X 1 mol Cr    3       = 0.0288
                     52.0 g       1 mol Cr

Then we convert to coulombs

0.0288 X 96485 C = 2778.77 C = 2778.77 A.s
                     1
The current will run for 1 hr (3600 s) so we need a current of 2778.77 A.s = 0.772 A
                                                                                                        3600 s

Go to the Electrolysis Worksheet
 

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.

Go to the Electrochemistry Review Worksheet
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