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 three. 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 reaction. Overall reaction : cathode Cu2+ + 2e- ----> Cu +0.34 V anode H2O ---> 1/2O2 + 2H+(10-7M) + 2e- -0.82 V ------------------------------------------ 2+ 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). 1 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: 1 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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