elect notes 2 by HC120913171550


									Uses for electrochemistry
       1. Breathalyzer test
              - Demonstration lab - see handout.

       2. Electrochemical cells (p. 862-866)

                If we could somehow separate the two half reactions and connect the two half
cells with a wire, we can give a path that the electrons can flow through. This flowing of
electrons through a wire we can put to our own use and force the electrons to pass through a
device that can use them. In other words, we can create an electrical current.
                But how do we separate the two half reactions ? We must set up a system such as
that on the next page.

Demonstration with diagram :

                       Side A                Side B
Electrode              Zn                    Cu
Solutions              Zn(NO3)2          Cu(NO3)2

Note : Nitrates are used for solutions because of their
        solubility and their lack of reactivity.

        The two electrodes are connected with a wire. In order to see if there is anything
happening, a galvanometer is introduced into the system. (A galvanometer is a like a very
sensitive voltmeter.)

        When the two electrodes are connected, nothing happens. Why ? If we take a look at
what is supposed to happen, it might explain this non-reaction.

       The reaction that is to occur in Side B is
                      Cu2+(aq) + 2e- ----> Cu (s)
       The reaction that is to occur in Side A is
                      Zn (s) ----> Zn2+(aq) + 2e-
If the reaction were to begin, an atom of the Zn metal would lose two electrons and become an
ion. The electrons will pass through the wire to the Cu electrode. The Zn2+ ion would fall into
beaker A. At this point in time, there would be more positive ions in beaker A than negative ions.
The solution would be electrically charged (called a polarized solution.) Nature does not allow
polarization to occur so this will not happen. At the same time, in the copper beaker, a copper ion
is accepting two electrons to become a copper atom. This reduces the total positive charge in
beaker B giving us a polarized solution as well. If we could allow the extra ions to flow between
the two beakers, polarization will be prevented.

        A glass tube filled with an ionic solution (KNO3) is placed between the two beakers. The
positive ions tend to migrate towards the copper electrode to balance off the electrically charged
solution. At the same time, the negative ions migrate towards the zinc electrode for the same
reason. The pathway for this migration, the glass tube, is called a SALT BRIDGE.
        Let's see what else is happening :

       1. The copper electrode is the electrode where
         reduction is occurring. This is also known as the
         CATHODE. Because more atoms of copper are being
         deposited on the electrode, it will increase in mass
         over a period of time.

       2. The zinc electrode is the electrode where
         oxidation is occurring. This is also known as the
         ANODE. Because more atoms of zinc are being
         removed from the electrode, it will decrease in mass
         over a period of time.

       3. The positive ions (Cu2+, K+, Zn2+) all make their
         way towards the copper electrode (cathode) through
         the salt bridge. Because the positive ions move
         towards the cathode, the positive ions are also
         referred to as CATIONS.

       4. The negative ions (NO3-) all make their way towards
         the zinc electrode (anode) through the salt bridge.
         Because the negative ions move towards the anode,
         the negative ions are also referred to as ANIONS.

       5. The electrons flow through the wire from the zinc
         (anode) to the copper (cathode).
Assignment :
      1) Complete the balanced chemical equation for the breathalyzer test.

       2) Draw and design an electrochemical cell using the following half-cells:
                              Au/Au3+ Ag/Ag+
         The cell should indicate a proper choice for solutions, which is the anode, which the
cathode, which way the electrons and ions flow.
Summary for electrochemical cells (so far)

       1. Oxidation occurs at the anode.
       2. Reduction occurs at the cathode.
       3. The anode loses mass while the cathode gains mass.
       4. Ions migrate towards the following electrodes
         through the salt bridge :
               positive ions (cations) towards cathode.
               negative ions (anions) towards anode.
       5. Electrons travel from anode to cathode through wire.

Comparing electrochemical cells

         As we can see on our standard reduction potential chart, each half cell has been given a
comparative half cell potential in volts. The symbol for this potential is Eo. This refers to the half
cell potential for solutions of 1.0 M concentrations and at 25oC. If gases are involved, the
pressure is at 101.3 kPa (standard pressure).
         An arbitrary decision was made to use the reduction potential for the hydrogen half cell as
a basis for comparison. It was decided to make this reaction 0.00 volts and compare everything
else to it. Thus we have the voltages listed on the reduction potential chart.
         Because oxidation is really the reverse of reduction, the voltage for the oxidation reaction
is the opposite of that for the reduction reaction. For example :

       Au3+ + 3e- ---> Au +1.50 V reduction
       Au ---> Au3+ + 3e- -1.50 V oxidation

        It should be stressed that none of these voltages can be made separately. They have to
occur in concert with another reaction.
        For homework, we designed a silver/gold electrochemical cell. Let's take a look at what
voltage we can expect from it. :
        Au3+ + 3e- ---> Au               +1.50 V
        Ag ----> Ag+ + e-                -0.80 V
        Au3+ + 3Ag ----> Au + 3Ag+ +0.70 V

Note #1: We multiplied the silver equation by three the balance the electrical charge, but this
does not change the potential for the reaction to occur, therefore the voltage is not multiplied by
Note #2: The voltage of the cell is positive. This makes sense because we are talking a
spontaneous reaction. If the voltage worked out to be negative, the reaction would not proceed
by itself the way it is written. What has probably happened is that someone has mixed up the
oxidation and reduction reactions.
Let's take another look at the demonstration we did the other day (the zinc/copper cell)
        Before we put a voltmeter into the circuit, what will be the expected voltage of the cell ?

               Cu2+ + 2e- ---> Cu             +0.34 V
               Zn ---> Zn2+ + 2e-             +0.76 V
               Cu2+ + Zn ---> Cu + Zn2+       +1.10 V

As we can see, the expected voltage will be 1.10 V. Now when we set up the cell and determine
the actual voltage, we find that it is less. Why ? Four reasons : 1. not 1 M solutions; 2. not 25oC ;
3. resistance ; 4. The voltages listed are the MAXIMUM theoretical voltages possible. Because
of all of the above reasons, we should not expect the voltage to be as high as 1.10 V.

      Let's change the concentrations of the ions in solution somewhat and see what happens.
We will do this by adding a solution of Na2S to each side, precipitating the zinc and copper ions.

       Addition of S2- ions to Zn side
              ZnS precipitates out.
              Voltage INCREASES. Why ? Even though the system is
              not at equilibrium, we can utilize Le Chatelier's
              principle to explain.
              Cu2+ + Zn ---> Cu + Zn2+
              A decrease in [Zn2+] concentration causes a shift
              to the right in the overall equation. Because of
              this, the voltage will increase.
       Addition of S2- ions to Cu side
              CuS precipitates out.
              Voltage DECREASES. Why ? Even though the system is
              not at equilibrium, we can utilize Le Chatelier's
              principle to explain.
              Cu2+ + Zn ---> Cu + Zn2+
              A decrease in [Cu2+] concentration causes a shift
              to the left in the overall equation. Because of
              this, the voltage will decrease.
Quantitative aspects of electrochemical cells

        In the above copper/zinc cell, we have seen that as the cell operates, there will be mass
lost by the zinc and gained by the copper. The actual amounts are not necessarily the same. For
instance, if we gain 1.00 gram of copper, we will not lose 1.00 gram of zinc. We must look at the
balanced chemical equation to see exactly what the MOLAR ratio between the two is.
                Cu2+ + Zn ---> Cu + Zn2+
According to the equation one mole of Cu was produced when one mole of Zn was used. To see
exactly what mass of Zn left the electrode,
                               1.00 g
                mol Cu =      ---------- = 0.0157 mol Cu
                              63.5 g/mol
                mol Zn used = mol Cu produced = 0.0157 mol Zn
                mass Zn = (0.0157 mol)(65.4 g/mol)
                         = 1.03 g Zn lost

Example :
                 In an earlier example, we used a gold/silver cell.
                 What will the mass change be at each electrode
                 when 0.0510 mol of electrons passes from one
                 electrode to the other ?
Solution :
                 Au3+ + 3e- ---> Au +1.50 V
                 Ag ----> Ag+ + e- -0.80 V
                 Au3+ + 3Ag ----> Au + 3Ag+ 0.70 V

       Silver electrode will lose mass.
       mol Ag = mol e- = 0.0510 mol Ag lost
       mass Ag lost = (0.0510 mol)(107.9 g/mol)
                        = 5.50 g Ag lost
       Gold electrode will gain mass
       mol Au = 1/3 mol e- = 0.0170 mol Au made
       mass Au made = (0.0170 mol)(197.0 g/mol)
                        = 3.35 g Au made

- Assignment :

       - Read Section 21-5 - 21-7 and 21-9
               (pages 633-642 and 645-646)
       - Electrochemical Cell Assignment
Uses for electrochemical cells (Types of cells)
        1. Dry Cells
            Dry cells are really not dry but contain a moist paste . The circuit could not be
completed and the cell would not work if it were really dry. Some sort of electrolyte is needed to
allow for ion movement. They are called dry cells because there is no exorbitant amount of water
in the cell like there was in the cells we built.

There are two types of dry cells we should discuss :

         a) Leclanche Cell, Zinc-Carbon Battery:

                 A Leclanche cell is a dry cell that is made with a zinc outer case. The zinc acts as
an anode and is oxidized. Inside the cell is a central cathode made of a combination of carbon
and manganese. A past made from water, MnO2 and ammonium chloride (NH4Cl) fills the cell. It
is the MnO2 that is reduced to various manganese compounds at the cathode. The NH4Cl acts as
the electrolyte.
                 A Leclanche cell does not last too long because the manganese compounds
produced tend to increase the electrical resistance (get in the way of) hence the quick demise of
the cell. Moreover, the cell has a short shelf life, and cannot be recharged.

Cathode:        2MnO2(s) + 2NH4+(aq) + 2e- ------ > 2MnO(OH)(s) + 2NH3(aq)
                ( Mn4+ + e- ------- > Mn3+)

Anode:          Zn(s) + 4NH3(aq) ------ > Zn(NH3)3+4(aq) + 2 e-
                (Zn ----- > Zn2+ + 2 e-)

         b) Alkaline Dry Cell:

                An alkaline battery is essentially the same design as a Leclanche cell but it does
have a few significant differences. The anode has a powdered zinc anode allowing for more
surface area to be exposed for use. The cathode is made of carbon and MnO2 . (The carbon is
there to increase conductivity of the cathode.) The main difference is that the electrolyte is a
basic solution, KOH. Because it is alkaline (basic) in nature, this is called an alkaline cell.

               The advantages of an alkaline cell over a Leclanche cell are that the alkaline cell
does not increase the resistance to electron flow ass the cell operates and it offers substantially
more current than a Leclanche cell.

Cathode:        2 MnO2(s) + H2O(l) + 2 e- ------- > Mn2O3(s) + 2OH-(aq)

Anode:          Zn(s) + 2OH-(aq ) ------- > ZnO(s) + H2O(l) + 2 e-

                The main reaction in an alkaline cell is:         Zn + 2MnO2 ----> ZnO + Mn2O3
       2. Lead-Acid Storage Battery

                 A lead-acid storage battery is commonly used in cars as a supply of electricity. It
is called a battery as opposed to a cell because it is composed of (generally) six cells all
connected in series.
                 A lead-acid storage battery consists of anodes of lead and cathodes of PbO2. The
electrolyte is sulfuric acid and not only does it help keep the charge balance of the cell, but it also
takes part in the reactions. This type of battery is definitely not dry. It must have a solution of
H2SO4 in it in order to operate. A look at the equations will show you why :

Cathode:                PbO2 + SO42- + 4H+ + 2e- ---> PbSO4 + 2H2O
Anode:                  Pb + SO42- ----> PbSO4 + 2e-
                        PbO2 + Pb + 4H+ + 2SO42- ---> 2PbSO4 + 2H2O

        The voltage of this cell is +2.04 V. The reason why a car battery is 12V is because there
are 6 of these cells in each battery, each producing about 2 V.

        As the battery winds down, the H2SO4 is constantly being used up and water is being
produced. Because battery acid is much more dense than water, a simple way of testing the
battery's charge is to measure the density of the electrolyte. The denser the electrolyte, the more
charged the battery is. Also, as the battery runs down, there is more and more water being
produced. The freezing point of this solution becomes closer and closer to 0oC. In the winter, this
could mean problems with the battery freezing in cold weather so it is important to keep the
battery fully charged in cold weather.

       3. Fuel Cells

        In recent years, scientists have developed fuel cells that convert chemical energy stored in
fuels directly into useable electrical energy. The best example is the hydrogen-oxygen fuel cell.

      When hydrogen is used as a fuel, it is combined with oxygen (burned) in the following
manner :          2 H2 (g) + O2 (g) ----> 2 H2O (l)

       In this situation, most of the energy is converted into heat energy and lost. This can be an
advantage to a space craft for supplying propulsion but not great for making electricity.

        In a fuel cell, two porous inert carbon electrodes impregnated with a catalyst are placed in
an electrolytic solution of KOH. Gaseous hydrogen and oxygen are allowed to flow through
these. The following reactions occur :

anode :        2 H2 + 4 OH- ----> 4 H2O + 4e-
cathode :      O2 + 2 H2O + 4e- ----> 4 OH-
                  2 H2 (g) + O2 (g) ----> 2 H2O (l)
        Notice we get the same reaction as what we got when we burned the hydrogen. However,
when we do the reaction electrochemically, we can get upwards of 75% of the stored chemical
energy converted into useable electrical energy. In fact, on recent space missions, the product of
this electrochemical reaction has been used as drinking water.

       There are however disadvantages to this type of cell. The most noticeable of the
disadvantages is the high cost. NASA can get away with it (besides they use the reactants for
another purpose as well) but we can't afford this type of cell at the present time.

        There are several other types of cells on the market today and there will doubtlessly be
many more to come in the future. As with the fuel cell, there are advantages and disadvantages to
all types of cells. One must be careful to choose the correct cell type for the function desired.

Other uses of electrochemistry : Electrolysis
        We have been talking over the last little while about spontaneous redox reactions. These
reactions turn stored chemical energy into useable electrical energy.
        If we supplied an outside source of electrical energy to a electrochemical cell, in the right
manner, we can force the electrons to move in the opposite direction. This means that the
reaction can be made to go the other way. This process is known as ELECTROLYSIS.
Electrolysis takes place in slightly different types of cells, known as electrolytic cells. There are
three major types of electrolytic cells :

       Type 1 cell :

                 In a type 1 electrolytic cell (see diagram), an ionic compound is heated up to its
molten state. Inert electrodes are placed in the molten salt and a power supply is added to connect
the two. It is important to connect the negative terminal of the battery to the cathode, for the
electrons supplied by the battery are required for reduction.
       The reactions for the reaction on the following page are as follows :
              K+ + e- ----> K                 -2.92 V
              2Br- ----> Br2 + 2e-             -1.06 V
              2K+ + 2Br- ---> 2K + Br2        -3.98 V

        As we can see, the voltage of this cell is negative. This is an indication to us that the
reaction would not take place spontaneously but indeed must have an external source of
electrons. In fact, we must have a power source supplying MORE than 3.98 volts to counter the
natural tendency for the electrons to flow. If just 3.98 Volts were added to the system, there
would be no reaction at all. There would be no net flow of electrons.

       Uses for type one cells:
              One of the main uses for this type of cell is to produce elements. For example, on
page ______ there is a diagram of a commercial type one cell, where molten NaCl is
decomposed into Na and Cl2 to be used for a number of purposes.

       Type 2 cell:

               Because heating ionic solid to high temperatures requires a great deal of energy,
another type of cell has been developed to reduce cost. However, the result has become a more
complicated cell.

       In this type of cell, inert electrodes are placed in an AQUEOUS solution of the ionic
compound to be electrolyzed and a power supply is connected. At first glance, this doesn't appear
to make the cell any more difficult to understand, however, there is now an additional component
to consider when we are gaining or losing electrons - WATER.

       Because water is present in all aqueous solutions and because water can undergo
oxidation and reduction, we must consider this as a possible reactant in our discussion. It is
suggested that we list all the possible oxidation reactions and determine which will happen. We
must also do the same for reduction reactions. In this manner, we will have considered all the
possible reactions and chemically determined which ones will happen.

       Example :      electrolysis of aqueous CuSO4 solution

       Possible reduction reactions :
                       Cu2+ + 2e- ----> Cu                              +0.34 V
                       2H2O + 2e- ---> H2 + 2OH-(10-7M)                 -0.41 V

       We can see that the reduction of the copper ion will occur before the reduction of the
water molecule. (the copper ion is HIGHER on the reduction potential chart.) By the way, the
reduction reaction for water is highlighted on your chart for ease of finding it. It is the lower
highlighted reaction. The (10-7M) is included in the equation to show that we are talking about
using neutral water rather than having an acidic or basic solution to start with.

       Possible oxidation reactions :
                      2SO42- ----> S2O82- + 2e-                  -2.05 V
              H2O ---> 1/2O2 + 2H+(10-7M) + 2e-                  -0.82 V

For the same reasons talked about above, the oxidation reaction that will occur will be the water

       Overall reaction :

cathode             Cu2+ + 2e- ----> Cu                                 +0.34 V
anode               H2O ---> 1/2O2 + 2H+(10-7M) + 2e-                   -0.82 V
               Cu + H2O ----> Cu + 1/2O2 + 2H+(10-7M)                   -0.48 V

  Now that we have determined the actual reaction that is taking place, we can ask a huge
variety of questions based on this.

       What voltage is required ?                                more than 0.48 V
       What will be produced at the cathode ?                    copper
       What will be produced at the anode ?                      oxygen and H+
       Will the resulting solution be acidic or basic ?          acidic

Summary: The half reaction having the greatest tendency to reduce and the greatest
tendency to oxidize are preferred.

Important things to remember regarding WATER:

   Most Electrolitic reactions you will encounter will be in NEUTRAL solution. The few cases
    where of ACIDIC solutions you need to simply substitute H+ for any metal ions in solution.

   In neutral aqueous solutions, you have two water equations: (the 10-7M refers to the starting
    concentration of the ion).
       oxidation:          /2O2(g) + 2H+(aq)(10-7M) + 2e- --------- > H2O           Eo= +0.82V
       reduction:      2H2O(l) +2e- ------- > H2(g) + 2OH- (10-7M)                     Eo= -0.41V

   In acididc solutions there are also two water equations:
       oxidation:       /2O2(g) + H+ + 2e- -------- > H2O                              Eo= +1.23V
       reduction:      2H+ + 2e- ----------- > H2(g)                                   Eo=0.00V

   You will not need to consider cells operating in basic conditions.

Exception to the Rule!

         The overpotential effect
                In actuality, it is found that there can be a difference between the calculated
voltage required for an electrolytic cell and the actual voltage needed. The difference between
these is known as the overpotential. Each half cell has a different overpotential but most are
insignificant. However, any reaction that is producing hydrogen gas or oxygen gas does have a
significant effect. If we take a look at the reduction potential chart, we see that the reactions are
listed in numerical order with respect to voltage of the half cell. However, because of the
overpotential of the half cells involving the oxidation and reduction of water, the placement of
the water reactions should be in the positions indicated by the arrows.
         What this means is that even though water is above Cr3+ on the table, it really should be
below. The Cr3+ ion will undergo reduction before the water will. For the same reason, the Br-
ion will undergo oxidation before water.

       Example :

In the electrolysis of aqueous AlCl3, what will be produced at the anode ?

Solution :
       Because the question does not ask us about the overall reaction, let's only concern
       ourselves with the oxidation reaction :

Possible oxidation reactions :
               H2O ---> 1/2O2 + 2H+(10-7M) + 2e-              -0.82 V
                   -               -
               2Cl ----> Cl2 + 2e                             -1.36 V
       Even though the voltage for the water is higher than that for the chloride ion, because of
the overpotential effect, the chloride ion will undergo oxidation before the water. In answer to the
question, chlorine will be produced at the anode.

Just remember the following exception to the rule:

Electrolysis of aqueoous solutions containing Cl- or Br- will produce Cl2 or Br2 at the anode.

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