CHEMISTRY 534 CHAPTER 7 OXIDATION-REDUCTION Electrochemistry is that branch of chemistry which deals with electric currents associated with chemical reactions. It is divided into two main branches: • Electrochemical cells: The conversion of chemical energy into electrical energy. Electric current (energy) is obtained from an electrochemical system. • Electrolysis: The conversion of electrical energy into chemical energy. Electric current (energy) is put into an electrolysis system. Both branches are based on a chemical reaction known as oxidation-reduction or redox for short. Redox reactions consist of two separate yet dependent and simultaneous reactions. One reaction is oxidation, the other is reduction. The substance which causes oxidation is called a "reducing agent" while the substance which causes reduction is called an "oxidizing agent". Like acid-base reactions, oxidation-reduction reactions also form equilibrium systems. Indeed, there are a number of similarities between oxidation-reduction (redox) reactions and acid-base (neutralization) reactions. Just as an acid neutralizes a base, for example, an "oxidizing agent" neutralizes the "reducing agent". While in an acid-base reaction there is the transfer of protons (H+ ions), in an oxidation-reduction reaction there is the transfer of electrons (e- charges). Moreover, much like the acidity constant is an indication of the relative strength of an acid, a voltage is an indication of the relative strength of an oxidizing or reducing agent. Originally, oxidation was considered as a reaction in which a substance was "oxidized" thereby producing an "oxide" The rusting of iron, for example, is an oxidation reaction in which oxygen in the air (moisture) reacts with iron to produce rust (iron oxide). In fact, although slow, rusting is a burning process (an exothermic reaction) just like the burning of wood, natural gas, gasoline, etc. An oxidation reaction, therefore, was defined as a reaction in which a substance "gained" oxygen. Today, we know that during an oxidation reaction, an atom loses electrons while in a reduction reaction an atom gains electrons. As a result, the definition of oxidation has been generalized to describe any process in which there is a transfer of electrons between atoms regardless of the presence of oxygen. In this chapter, we study oxidation-reduction reactions. The study of such reactions provides us with the opportunity to use and relate the chemical concepts covered in previous chapters. Page 1 ▼ TOPIC-1: AN AFFAIR OF ELECTRONS In order to discover the characteristics of oxidation-reduction reactions, let's perform an experiment comparing the reaction of different metals. ➲ EXPERIMENT-33: Objective: The capacity to react To study the reactivity of lead, copper, and zinc. This experiment is divided into two parts • PART-A: Visible facts • PART-B: Hidden facts PART -A: Visible Facts Step-1: Place 3 mL of 0.50 M copper(II) nitrate, Cu(NO3)2, into three test tubes. ☞Hint: Identify each test tube as to its contents. Step-2: Place 3 mL of 0.50 M lead (II) nitrate, Pb(NO3)2, into three test tubes. Step-3: Place 3 mL of 0.50 M zinc nitrate, Zn(NO3)2, into three test tubes. Step-4: Into three test tubes each containing Cu(NO3)2, Pb(NO3)2, and Zn(NO3)2 respectively, add a piece of copper metal. Step-5: Into three test tubes each containing Cu(NO3)2, Pb(NO3)2, and Zn(N03)2 respectively, add a piece of lead metal. Step-6: Into the remaining three test tubes containing Cu(NO3)2, Pb(NO3)2, and Zn(NO3)2 respectively, add a piece of zinc metal. 1) Observe each test tube for about 2 minutes then record whether or not (yes or no) a reaction has occurred in each test tube. METAL Copper (Cu) Lead (Pb) Zinc (Zn) Page 2 Copper(II) nitrate Cu(NO3)2 SOLUTIONS Lead(II) nitrate Pb(NO3)2 Zinc nitrate Zn(NO3)2 2) Of the three metals (copper, lead and zinc), which metal reacted: a) With the greatest number of solutions? ______________ b) With the least number of solutions? ______________ 3) Classify the three metals in decreasing order of their reactiveness (tendency to react). ① ________________________________ ② ________________________________ ③ ________________________________ 4) Define a spontaneous reaction: ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ 5) Give an example of a spontaneous reaction: ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ PART-B: Hidden facts 6) For those solutions in which there was a reaction (there should be three), identify the reactants and the products: (Note: nitrate ions (NO3-) are spectator ions. They remain in the solution and are unreactive ) METAL Copper (Cu) Lead (Pb) Zinc (Zn) Copper(II) nitrate Cu(NO3)2 SOLUTIONS Lead(II) nitrate Pb(NO3)2 Zinc nitrate Zn(NO3)2 Reactants: Reactants: Reactants: Products: Products: Products: Reactants: Reactants: Reactants: Products: Products: Products: Reactants: Reactants: Reactants: Products: Products: Products: Page 3 7) Consider the transformation of zinc to zinc ions in the presence of copper(II) nitrate. At the start of the reaction, zinc is a solid. At the end of the reaction, the solid zinc has been transformed into zinc ions. At the same time, copper(II), Cu2+, ions have been transformed into solid copper. a) Write the equation which represents solid zinc being transformed into zinc ions: __________________________________________________________________________ b) Write the equation which represents copper(II) ions being transformed into solid copper. __________________________________________________________________________ c) The copper(II) ions have gained two electrons. Where did they come from? __________________________________________________________________________ d) Write the overall equation for this reaction: __________________________________________________________________________ 8) Now consider the transformation of zinc to zinc ions in the presence of lead(II) nitrate. In this case, as the solid zinc is transformed into ions, the lead ions are transformed into solid lead. a) Write the equation which represents the transformation of zinc: __________________________________________________________________________ b) Write the equation which represents the transformation of lead (Pb2+) ions: __________________________________________________________________________ c) Write the overall equation for this reaction: __________________________________________________________________________ 9) The last reaction is the transformation of lead in the presence of copper(II) nitrate. Write the three equations which describe this reaction: ➀ ______________________________________________________________________ ➁ ______________________________________________________________________ ➂ ______________________________________________________________________ ÄEnd of experiment-33 Page 4 An oxidation-reduction reaction consists of two "half-reactions" each forming a "half-cell". One half is the oxidation reaction, the other half is the reduction reaction. The two "halves" work together in that one releases (loses) electrons while the other takes in (gains) the lost electrons. Don't be confused! Electrons are negative charges. Note that if you lose electrons, you increase your total numeric value. Conversely, if you gain electrons, you decrease your total numeric value. Study these examples: INITIAL REACTION VALUE TRANSFER Reduction 10 + 3 e= 10 + (-3) = 10 - (-2) Oxidation 10 -2e Reduction 7 +4 e = 7 + (-4) = 12 - (-5) Oxidation 12 -5 e= 6 - (-4) Oxidation 6 -4 e Reduction 0 +1e = 0 + (-l) = 5 + ( -1) Reduction 5 +1e = = = = = = = 10 - 3 10 + 2 7-4 12 + 5 6+4 0-1 5-1 = = = = = = = FINAL VALUE 7 12 3 17 10 -1 4 ÄRemember: Oxidation is losing electrons thereby becoming more positive. Reduction is gaining electrons thereby becoming more negative. When we place a copper electrode in a solution of silver nitrate (AgNO3), the reaction proceeds slowly. Soon, we notice the formation of silver- colored "pins" as the solution turns blue. The blue color is an indication that copper(II) ions are present in the solution and the solid deposit is silver. The equations which represent these transformations are: Loss of 2 electrons by a copper atom: Cu(s) à Cu2+(aq) + 2e- Gain of 1 electron by a silver ion: { Ag+(aq) + le- à Ag(s) }x2 Overall equation: Cu(s) + 2 Ag+(aq) à Cu2+(aq) + 2 Ag(s) Note that in this redox reaction each copper atom loses two electrons while each silver ion gains one electron. Nonetheless, the electron exchange (gain and loss) occurs in such a way that none of the electrons appear in the overall equation. As a result, a reaction which liberates electrons is always accompanied by a reaction which absorbs the liberated electrons. Page 5 10) As you know, an oxidation reaction is one in which electrons are released. Write the halfreaction which represents the oxidation in the above transformations: ____________________________________________________________________________ 11) A reduction reaction is one in which electrons are absorbed. Write the half-reaction which represents the reduction in the above transformations: ___________________________________________________________________________ By definition, the cathode of an electrochemical cell is the electrode at which reduction occurs, while the anode is the electrode where oxidation occurs. AN OX RED CAT Again, an oxidation-reduction reaction is an exchange of electrons between a donor and a receiver. A cathodic reaction is the reduction half of a redox reaction. Cathodic reactions take place at the cathode and consist of receiving electrons. An anodic reaction, on the other hand, is the oxidation half of a redox reaction. Anodic reactions take place at the anode and consist of the release of electrons. Furthermore, in all redox reactions, there is no net loss or gain of electrons. What is lost is what is gained. For example, consider the following redox reaction: Cu(s) + 2 Ag+(aq) à Cu2+(aq) + 2 Ag(s) Copper atoms here undergo oxidation (since they lose electrons) while silver ions undergo reduction (since they gain the lost electrons). Because copper causes the Ag+ ions to gain electrons, copper is called the reducing agent or reductant. Conversely, because the Ag+ ions cause the copper to lose electrons, the Ag+ ions constitute the oxidizing agent or oxidant 12) For each oxidation-reduction reaction given below, complete the following: a) Page 6 Zn(s) + Cu2+(aq) à Cu(s) + Zn2+(aq) The oxidation reaction: __________________________________________ The reduction reaction: __________________________________________ The oxidizing agent: __________________________________________ The reducing agent: __________________________________________ b) Zn(s) + Pb2+(aq) à Pb(s) + Zn2+(aq) The oxidation reaction: __________________________________________ The reduction reaction: __________________________________________ The oxidizing agent: __________________________________________ The reducing agent: __________________________________________ c) Pb(s) + Cu2+(aq) à Cu(s) + Pb2+(aq) The oxidation reaction: __________________________________________ The reduction reaction: __________________________________________ The oxidizing agent: __________________________________________ The reducing agent: __________________________________________ 13) When a piece of aluminum (Al) is placed in a lead(II) nitrate solution, Pb(NO3)2, a deposit of lead (Pb) appears on the aluminum. Use this information to complete the following: 2 Al(s) + 3 Pb2+(aq) à 3 Pb(s) + 2 Al3+(aq) The oxidation half-reaction: __________________________________________ The reduction half-reaction: __________________________________________ Redox reaction: __________________________________________ The oxidizing agent: __________________________________________ The reducing agent: __________________________________________ Page 7 ▼ TOPIC-2: HIGH POTENTIAL PAIRS In the last experiment, we discovered that the capacity to react varies from metal to metal. Zinc, for example, acted as a good donor of electrons as it gave electrons both to lead ions and to copper(II) ions. Lead, however, supplied electrons to the Cu2+ ions but not to the Zn2+ ions. It seems that the ability to donate electrons varies for different metals. In other words, the capacity to donate electrons is not the same for all metals. Is the capacity to donate electrons a characteristic property of metals? Let's find out by an experiment investigating a few different metals. wExperiment -34 Classifying Metals Objective: To investigate the capacity of different metals to supply electrons thereby classifying them according to their ability to donate electrons. ➣Reminder: Solid metals which are dipped in solutions are called electrodes. ÄNote: To be sure that the metals react readily, remove the corrosion (oxidation) on their surface by sand papering the electrodes. Step-1: Pour 100 mL of a 0.50 mol/L sodium chloride (NaCl) solution into a 250 mL beaker. Step-2: Place two different metal electrodes into the solution and connect the electrodes to a voltmeter. Follow the diagram below and make sure that the metals do not touch each other. ÄImportant: In using an analog voltmeter, reverse the polarity of the probes if the needle moves against the zero mark. If you are using a digital voltmeter, reverse the probes if you obtain a negative voltage. Step-3: Page 8 Record the voltage readings for each pair of electrodes in the table below. (Ä Note: The voltage is independent of the surface area of the electrodes) 14) Record the voltages for each metal pair here: Electrodes (-) â/ (+) à Zinc (Zn) Cu Mg Ni Pb Sn Zn Copper (Cu) Magnesium (Mg) Nickel (Ni) Lead (Pb) Tin (Sn) 15) How can we classify the metals in order of their ability to donate electrons? __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ 16) Classify the metals in decreasing order to their ability to donate electrons: • ________________________ ‚ ________________________ ƒ ________________________ „ ________________________ … ________________________ † ________________________ 17) Identify the metal "pair" which produced the highest potential difference (voltage): _________________________________________________________________ Page 9 18) With reference to the list of STANDARD OXIDATION POTENTIALS, write the oxidation half-reaction for each metal analyzed. Arrange them in decreasing order of their ability to release electrons. • ____________________________________________________________________ ‚ ____________________________________________________________________ ƒ ____________________________________________________________________ „ ____________________________________________________________________ … ____________________________________________________________________ † ____________________________________________________________________ ➢End of experiment-34 19) Here are more oxidation half-equations: Al Cr Ni Pb à à à à A13+ + 3 eCr3+ + 3 eNi2+ + 2 ePb2+ + 2 e- With reference to the above list, identify the pair which produces the greatest difference in potential (voltage). Why? Al/ Cr Al/ Ni Al/Pb Cr/Ni Cr/Pb Ni /Pb ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Experimentally, we can obtain the difference in potential (voltage) between two metals thus comparing one metal to another. However, this comparison gives us only the difference between two metals. This is similar to comparing the difference in heights of a group of people. For example, we may say, the difference between Jack and Jill is 8 cm, the difference between Jack and Tom is 3 cm, the difference between Jane and David is 2 cm, etc. Instead, it would be more useful to simply list the height of each person (from some "zero" reference point such as the floor). Page 10 Likewise with metals. It would be more useful to determine and list the difference in potential of all metals from some zero reference point. To do this, we must select an arbitrary potential of 0.00 volts to be used as a reference. To establish a potential of 0.00 V as the reference point, chemists have selected the dissociation of gaseous hydrogen at a pressure of 101.3 kPa and at a temperature of 25 0C. The potential of this reaction is fixed to be 0.00 V. All potentials relative to this zero voltage point are designated as E0. Thus, we have: H2(g) à 2 H+(aq) + 2 eE0 = 0.00 V Using hydrogen as the standard reference, the potential of a great many metals can be measured. Indeed, chemists have already done this and call the list the "Standard Oxidation Potentials". w SOME STANDARD OXIDATION POTENTIALS w HALF-REACTION POTENTIAL Mg Al Zn Cr Co Ni Pb H2 Sn Cu Ag à à à à à à à à à à à Mg2+ + 2 eA13+ + 3 eZn2+ + 2 eCr3+ + 3 eCo2+ + 2 eNi2+ + 2 ePb2+ + 2 e2 H+(aq) + 2 eSn2+ + 2 eCu2+ + 2 eAg+ + 1 e- E0 E0 E0 E0 E0 E0 E0 E0 Eo E0 E0 = +2.37 V = +1.66 V = +0.76 V = +0.74 V = +0.28 V = +0.26 V = +0.13 V = 0.00 V = -0.13 V = -0.34 V = -0.80 V The oxidation potential of a metal is the tendency of its half-reaction to give up electrons (at standard conditions) as compared to the tendency of the hydrogen half-reaction. A positive sign is given to metals whose half-reaction releases electrons more readily than the hydrogen halfreaction, and a negative sign is given to metals whose half-reaction releases electrons less readily than the hydrogen half-reaction. 20) Explain the arbitrary nature of the oxidation potential table: ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ Page 11 21) Find the difference in potential for each of the following metal pairs: a) Chromium and nickel _________________________________ b) Chromium and lead _________________________________ c) Chromium and hydrogen _________________________________ d) Chromium and copper _________________________________ e) Chromium and silver _________________________________ 22) Derive a formula for calculating the difference in potential between two metals: ________________________________________________________________________ ________________________________________________________________________ 23) Identify the metal pair which has the highest difference in potential. Explain: ________________________________________________________________________ ________________________________________________________________________ Given any two metal combinations, the oxidation potentials can be used to predict which metal will undergo oxidation and which will undergo reduction. It's simple, the metal with the higher oxidation potential undergoes oxidation while the metal with the lower oxidation potential undergoes reduction. Also, if the difference in potential between a metal and the metal ion in solution is positive, then the reaction will take place spontaneously (proceeds by itself). On the other hand, if the difference in potential between a metal and the metal ion in solution is negative, the reaction will not occur spontaneously. Sample Problem-1 Zinc is placed in a solution of nickel sulfate. a) Is the reaction spontaneous? b) Which metal will undergo oxidation? To find out, we look up the oxidation potential of both metals. In this case, we have +0.76 V for zinc metal and +0.26 V for nickel ions. Since the difference in potentials is positive [(+0.76)- (+0.26) = +0.50], reaction is spontaneous. Moreover, since the oxidation potential of zinc is higher that of nickel, zinc undergoes oxidation while nickel ions undergo reduction. Page 12 Sample Problem-2 Silver is placed in a solution of copper (II)sulfate. a) Is the reaction spontaneous? b) Which metal will undergo oxidation? We look up the oxidation potential of both metals. For silver, the potential is -0.80 V and for copper it is -0.34 V. In this case, the difference in potentials is negative [(-0.80)- (-0.34) = -0.46 V], we know that the reaction will not occur spontaneously. As a result, we conclude that oxidation will not occur to any appreciable extent. Conclusion, neither metal undergoes oxidation. Page 13 ▼ TOPIC-3: GIVING AND RECEIVING A QUESTION OF POTENTIAL We know that an oxidation-reduction reaction involves the movement of electrons from one substance to another. Recall that an electric current is the movement of electrons in a wire. How may we utilize an oxidation-reduction reaction in order to produce an electrical current in a wire? Electrons are negatively charged particles which form an integral part of matter. The movement of electrons is responsible for the operation of all electrical devices, from the light bulb to the computer. It would be useful to generate electrons on demand. While there are a number of ways to generate electrons, two particularly convenient methods are by using a generator and by using a battery. Whereas a generator produces electrons by physical interaction, a battery produces electrons by chemical reaction. Note that the generation and control of electrons is so important that a branch of science is dedicated exclusively towards that end. Electronics is the science of generating and controlling electrons. In experiment-33, we saw that an oxidation-reduction reaction is a chemical method of generating and moving electrons What is the mechanism of this electronic movement? Zn 2+ Cu Direct exchange of electrons Since an electrical current is the flow of electrons in a conductor, to obtain an electric current, we can use a redox reaction. The oxidation process causes the electrons to be generated while the reduction process attracts the electrons. By connecting an external wire between the two processes, we have an electric current (electron flow). Page 14 Review of electric current Recall that current is the rate of flow of charge and is designated by the letter "I" (Intensity). The unit for current is the amp•re. One amp•re equals one coulomb per second or 1.04 x 10-5 mol of electrons flowing per second. In a wire (solid metal), current consists of the flow of electrons and is called metallic conduction. In a solution , current consists of ions (both anions and cations ) and is called ionic conduction. Anions are negative ions (attracted to the anode) and cations are positive ions (attracted to the cathode). But what causes charges to flow? The answer is the potential difference or voltage. Do not confuse electrical potential (voltage) with current. By way of analogy, you can think of potential as height, and current as water flow. Note that it is the height, or more precisely, the difference in height between two points, which causes water to flow. Likewise with current, it is the potential difference which causes charge to flow (electrons in a wire and ions in a solution). Where there is no potential difference (no difference in height), no current can flow. An electrical "load" is any device which requires electricity to operate such as a lamp, a motor, a heater, etc. All loads require the flow of electrons for their operation. However, in order for electrons (current) to flow, a voltage must be applied to the load. In effect, the voltage creates a difference in potential (similar to a difference in height) so that electrons can flow. The two points where a voltage is applied to a load are called the terminals. Recall from chapter-6 that all systems have a natural tendency to lower their energy level as much as possible. When we raise a book in the air, for example, we place the book at a higher energy level. When we let go of the book, it spontaneously falls down towards a lower energy level. Likewise with electricity, current always flows from a higher to a lower potential (energy level). When a potential difference exists between two points, one point is at a higher potential than the other point. Just as potential energy is a measure of the energy between two heights, voltage is a measure of the difference in electrical potential energy between two points. Thus, when a voltage is applied to a wire, one end of the wire is at a higher potential than the other end. Electrons at the higher end have more potential energy than the electrons at the lower end (just as a book has more potential energy at a higher level than at a lower level). As a result, electrons flow from the higher to the lower level. Work is defined as the energy required to perform a useful task. When work is being done, energy is being used. As electrons flow from a higher to a lower potential, we can use (convert) part of their energy to do work such as producing heat or light. In doing so, we are in effect lowering the potential energy of the electrons (just as a falling book can be made to do something useful while falling and in the process lower its potential energy). A voltmeter is used to measure the difference in potential. The voltmeter is placed across the two points whose potential (voltage) is to be measured; that is, in parallel with the terminals of a load. An ammeter is used to measure the intensity of the current (rate of flow of electrons in a wire). Since an ammeter measures the flow of charges (electrons), it must be connected along the path of the electron flow; that is, in series with the terminals of a load. Page 15 In order to generate a current. a potential difference is required. The potential difference can be generated with an oxidation-reduction reaction. However, to prevent the electrons from flowing directly from the reducing agent to the oxidizing agent, we must keep the two reactions separated. We do this by placing the oxidation reaction in one container and the reduction reaction in another container. The set up consists of a large container which contains a zinc electrode in a solution of zinc nitrate, Zn(NO3)2. The zinc nitrate supplies zinc ions, Zn2+(aq). In another smaller container within the larger one, we have a copper electrode in a copper nitrate solution, Cu(NO3)2. Because the two solutions are in separate containers, the zinc ions, Zn2+(aq), are not in direct contact with copper ions, Cu2+(aq). As we saw in experiment-33, the reaction consists of a spontaneous exchange of electrons from the solid zinc metal to the copper(II) ions. There was no exchange of electrons between solid copper and the zinc ions. Due to the set up, the spontaneous tendency of the electrons to go from the zinc to the copper(II) ions takes place via the connecting wire. ÄNote: It is important to realize that the receiver of the electrons is not the copper metal (electrode) but the copper(II) ions in the solution. Metals have a tendency to give off electrons and in the process become more stable. In fact, when metals gain electrons they become less stable. To review, an electrochemical cell is an oxidation-reduction reaction. The oxidation process, called a half-reaction, generates electrons and forms 'half' of the electrochemical cell (called a half-cell). The reduction process, also called a half-reaction, receives the electrons generated by the oxidation process and forms the other half of the electrochemical cell. Page 16 w EXPERIMENT-35: Objective: Potential and work To study electrochemical cells. [micro approach see page 42 ] In this experiment, we'll construct two types of electrochemical cells. In one type, the oxidation and reduction reactions will be linked by a porous material (cup); in the other, a salt bridge will be used to link the two half reactions. Both set-ups form an electrochemical cell. Note that although these electrochemical cells do not resemble familiar batteries, they work on the same principle. Our laboratory cells are intended to make it easier to understand their operation. PART-A: An electrochemical cell using a porous cup Step-1: Fill a porous cup with 1.0 M copper(II) nitrate, Cu(N03)2. (To approximately 1 cm from the top) Step-2: Step-3: Place the porous cup into a 250 mL beaker. Fill the beaker with a 1.0 M zinc nitrate solution, Zn(N03)2. (To about the same level as the copper(II) nitrate solution) Step-4: Place a strip of copper metal (called the copper electrode) into the copper(II) nitrate solution. Ä Note: Sand paper the electrode so as to remove any oxidation. Page 17 Step-5: Place a strip of zinc metal into the zinc nitrate solution. (Again, sand paper the electrode so as to remove any oxidation) Step-6: Connect a voltmeter to the electrodes. (Ä Note: If alligator clips are used, make sure they make a tight grip) Step-7: Record the voltage of the circuit as well as your observations. 24) Record your observations of the Cu-Zn porous cup cell (including the cell voltage): ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Next, we'll make a magnesium-zinc electrochemical cell. Step-8: Replace the porous cup containing copper(II) nitrate with one containing magnesium nitrate. Step-9: Place a strip of magnesium ribbon into the magnesium nitrate solution. Step-10: Connect a voltmeter to the zinc and magnesium electrodes. Step- 11: Record the voltage of the circuit as well as your observations. 25) Record your observations of the Mg-Zn porous cup cell: __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ PART-B: An electrochemical cell using a salt bridge Page 18 Step-1: Fill a 100 mL beaker with about 80 mL of 1.0 M copper(II) nitrate solution. Step-2: Fill a 100 mL beaker with about 80 mL of 1.0 M zinc nitrate solution. Step-3: Place a salt bridge containing a solution of 1.0 M potassium nitrate between both beakers. (Be sure to fill the ends of the bridge with cotton) Step-4: Place a copper electrode in the copper(II) nitrate solution and a zinc electrode into the zinc nitrate solution. Step-5: Connect a voltmeter to the electrodes. Step-6: Record the voltage of the circuit as well as your observation. 26) Record your experimental observations regarding the salt bridge cell: ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ 27) Describe the similarities between the two electrochemical cells: ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ 2e- 2e- ! ; (Oxidation) (Reduction) Page 19 28) Describe the differences between the two electrochemical cells: ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ 29) Explain what happens if: a) We remove an electrode from solution: _______________________________________________________________________ _______________________________________________________________________ b) We remove the porous barrier: _______________________________________________________________________ _______________________________________________________________________ c) We remove the salt bridge: _______________________________________________________________________ _______________________________________________________________________ 30) Provide the following information concerning electrochemical cells used in this experiment: a) Oxidation reaction: ________________________________________________________________________ b) Reduction reaction: ________________________________________________________________________ c) Oxidation-reduction reaction: ________________________________________________________________________ d) Oxidizing agent: __________________________ e) Reducing agent: __________________________ f) The electrode which increased in mass: ________________________________________ g) The electrode which decreased in mass: _______________________________________ 31) Explain why the mass of the electrodes varies: ____________________________________________________________________ ____________________________________________________________________ Page 20 32) With reference to the STANDARD OXIDATION POTENTIALS table, determine the potential difference of this cell. __________________________________________________________________________ __________________________________________________________________________ 33) How does the calculated value compare to the experimental value? __________________________________________________________________________ __________________________________________________________________________ Ä End of experiment-34 e- Pb Zn Pb2+ Zn2+ 34) An electrochemical cell is made by placing a lead electrode in a lead(II) nitrate solution and a zinc electrode in a zinc nitrate solution. Fill in the following information concerning this Zn-Pb cell: a) Oxidation reaction: ___________________________________________________ b) Reduction reaction: ___________________________________________________ c) Oxidation-reduction reaction: ___________________________________________ d) Oxidizing agent: ________________ e) Reducing agent: ________________ f) The electrode which increases in mass: ___________________________________ g) The electrode which decreases in mass: __________________________________ h) The cell voltage: ____________________________________________________ Page 21 35) An electrochemical cell is formed with a copper and a silver electrode. Each electrode is placed in its respective cation solution. a) Complete the sketch below by labeling the electrodes and the cations (positive ions) in each solution. Then fill-in the pertinent information below: b) Oxidation reaction: ______________________________________________________ c) Reduction reaction: ______________________________________________________ d) Oxidation-reduction reaction: ______________________________________________ e) Oxidizing agent: _____________________ f) Reducing agent:___________________ g) The electrode which increases in mass: ______________________________________ h) The electrode which decreases in mass: ______________________________________ i) The cell voltage: _______________ Page 22 36) A cell is made by using a solution of sodium nitrate as a bridge between two half cells. One half cell consists of a copper electrode in a copper(II) nitrate solution, while the other half cell is made with a magnesium electrode in a magnesium nitrate solution. Here is a diagram of the electrochemical cell: Mg Cu Cu(NO3)2(aq) Mg(NO3)2(aq) a ) Oxidation reaction: _________________________________________________________ b ) Reduction reaction: ________________________________________________________ c ) Oxidation-reduction reaction: ________________________________________________ d ) Oxidizing agent: _______________________ e ) Reducing agent: ___________________ f ) The electrode which increases in mass: _________________________________________ g ) The electrode which decreases in mass: ________________________________________ h ) The cell voltage: _______________ Page 23 Although a sketch or drawing of an electrochemical cell is a convenient way of representing an electrochemical cell, it is inconvenient to make a drawing each time a cell is to be specified. Rather than draw the actual cell, we can write the two half reactions instead. In using this method, a slash (or a vertical line "") is used to separate an electrode from its solution, and a double slash is used to separate one half cell from the other. Study the example below. Pb Ni Pb(NO3)2(aq) Ni(NO3)2(aq) Pb(s) / Pb(NO3)2(aq) // Ni(NO3)2(aq) / Ni(s) In using the notation illustrated above note that the electrodes are written at the two extremities. The slash represents the physical contact between the electrode, Pb(s), and its solution, Pb(NO3)2(aq). The double slash represents the link between the two half cells (in this case, a salt bridge). Finally, the last slash represents the physical contact between the electrode, Ni(s), and its solution, Ni(N03)2(aq). Each barrier (or physical contact) produces a small difference in potential (voltage). Thus, we have the following potential differences: Lead and lead(II) nitrate solution Lead(II) nitrate solution and salt bridge Salt bridge and nickel nitrate solution Nickel nitrate solution and nickel electrode In effect, the sum total of these tiny potential differences equals the cell voltage. Certain "electric fish", such as the eel, have naturally occurring chemical reactions which generate electric voltages. In effect, they possess a battery of electrochemical cells connected in series producing voltages up to 500 volts. They use this electrical energy as a weapon to shock and stun their enemies as well as a type of radar for orientation purposes. Page 24 37) An electrochemical cell is depicted by the following notation: Al(s) / Al(NO3)3(aq) // Co(NO3)2(aq) / Co(s) Sketch this cell and include the direction of electron flow. Fill in the following pertinent information concerning this cell: a ) Oxidation reaction: ________________________________________________________ b ) Reduction reaction: ________________________________________________________ c ) Oxidation-reduction reaction: ________________________________________________ d ) Oxidizing agent: _______________________ e ) Reducing agent: __________________ f ) The electrode which increases in mass: ________________________________________ g ) The electrode which decreases in mass: _______________________________________ h ) The cell voltage: _______________ Page 25 38) Sketch the following cell then fill-in the pertinent information: Cr(s) / Cr(NO3)2(aq) // Cu(NO3)2(aq) / Cu(s) a ) Oxidation reaction: _______________________________________________________ b ) Reduction reaction: _______________________________________________________ c ) Oxidation-reduction reaction: _______________________________________________ d ) Oxidizing agent: _______________________ e ) Reducing agent: _________________ f ) The electrode which increases in mass: _______________________________________ g ) The electrode which decreases in mass: ______________________________________ h ) The cell voltage: _______________ Page 26 ▼ Topic-4: A Different Point of View The list of electron donors provides much information. It tells us, for example, that aluminum is a better donor than zinc, chromium, cobalt, etc. HALF-REACTION Mg Al Zn Cr Co Ni Pb H2 Sn Cu Ag à à à à à à à à à à à POTENTIAL Mg2+ + 2 eA13+ + 3 eZn2+ + 2 eCr3+ + 3 eCo2+ + 2 eNi2+ + 2 ePb2+ + 2 e2 H+(aq) + 2 eSn2+ + 2 eCu2+ + 2 eAg+ + 1 e- E0 E0 E0 E0 E0 E0 E0 E0 Eo E0 E0 = +2.37 V = +1.66 V = +0.76 V = +0.74 V = +0.28 V = +0.26 V = +0.13 V = 0.00 V = -0.13 V = -0.34 V = -0.80 V The list also tells us that an electrochemical cell made with, say, aluminum and nickel, will react spontaneously. The aluminum acts as an electron donor and the nickel acts as an electron acceptor. Al ➝ Al3+ + 3e- E0 = + l.66 V Ni ➝ Ni2+ + 2e- E0 = + 0.26 V In a nickel-lead cell, however, it is the nickel which acts as the electron donor and the lead as an electron acceptor. Ni ➝ Ni2+ + 2e- E0 = + 0.26 V Pb ➝ Pb2+ + 2e- E0 = + 0.13 V Obviously, the behavior of nickel is determined by the metal with which it is paired. Note that while we may classify metals by their ability to donate electrons, it is also possible to classify them instead by the ability of the cations (positive ions) in solution to accept electrons. Page 27 In classifying a metal as a good donor of electrons, we are in effect classifying its cations as a weak acceptor of electrons. For example, since aluminum metal is a good donor of electrons, the aluminum ion, Al3+(aq), is a poor acceptor of electrons. Thus, it is not surprising to discover that the reducing potential of aluminum is negative: Capacity to donate electrons: Al ➝ Al3+ + 3e- E0 = + l.66 V Capacity to receive electrons: Al3+ + 3e- ➝ Al E0 = - l.66 V Indeed, it is possible to use either an oxidation potential list (that is, the ability to donate electrons), or conversely, a reduction potential list (that is, the ability to receive electrons). Both lists are valid. In the Oxidation Potential table, the equations indicate the donation of electrons. The best donor, in this scale, is magnesium. This means that silver, at the bottom of the scale, although the worst donor of electrons is the best receiver of electrons. On the other hand, in the Reduction Potential table, the equations indicate the acceptance of electrons. The best acceptor is silver which is at the top of the list. This means that magnesium, which here is at the bottom of the list, is the worst acceptor of electrons. Remember: Both tables give the same information but from a different point of view. Page 28 39) Using the Oxidation Potential table, identify the best donor of electrons for the following pairs of metals: a) Zn and Pb _____________________________________ b) Co and Cr _____________________________________ c) H2 and Ni _____________________________________ 40) Using the Oxidation Potential table, identify the best acceptor of electrons for the following pairs of metals: a) Zn2+ and Pb2+ _____________________________________ b) Co2+ and Cr3+ _____________________________________ c) H+ and Ni2+ _____________________________________ 41) Using the Reduction Potential table, identify the best donor of electrons for the following pairs of metals: a) Zn and Pb _____________________________________ b) Co and Cr _____________________________________ c) H2 and Ni _____________________________________ 42) Using the Reduction Potential table, identify the best acceptor of electrons for the following pairs of metals: a) Zn2+ and Pb2+ _____________________________________ b) Co2+ and Cr3+ _____________________________________ c) H+ and Ni2+ _____________________________________ 43) Compare your answers for number 40 with those of number 42, and compare your answers of number 39 with those of number 41. Explain: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Page 29 44) Here are two situations. In one beaker, we place a cobalt electrode in a solution of copper(II) nitrate. In another beaker, we place a copper electrode in a cobalt nitrate solution. Co Cu Cu2+ Cu(N03)2 Co2+ Co(N03)2 Identify the spontaneous reaction and explain your answer: ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 45) Consider the following situation: Al Mg2+ Mg (N03)2 After a certain time, is it possible to find aluminum ions (Al3+) in the solution? Explain your answer: __________________________________________________________________________ __________________________________________________________________________ 46) A copper electrode is placed in an aqueous solution of hydrochloric acid (HCl). Is a gas released? Explain: __________________________________________________________________________ __________________________________________________________________________ Page 30 47) A piece of silver jewelry falls into a solution of cobalt nitrate, Co(NO3)2. Is there reason to be upset? Explain: ________________________________________________________________________ ________________________________________________________________________ 48) To identify the presence of an acid, pieces of magnesium metal may be used. Explain why: _______________________________________________________________________ _______________________________________________________________________ 49) How may we predict if a reaction is spontaneous? ________________________________________________________________________ ________________________________________________________________________ Page 31 t TOPIC 5: Everything is Relative In constructing an electrochemical cell, we put together two half cells. When the cells are connected together, a competition begins between the electron donors. In the example below, magnesium is a better electron donor than lead. Thus, magnesium acts as the donor of electrons and lead as the receiver. Electrons travel from the magnesium electrode to the lead electrode. As a consequence, in the half-cell on the left, the Pb2+(aq) cations are the acceptors of the electrons. E0 = + 0.13 V E0 = + 2.37 V Pb Mg Pb(NO3)2 Mg(NO3)2 We can represent this system by the following equations: Oxidation: Mg(s) ➝ Mg2+(aq) + 2 e- Reduction: Pb2+(aq) + 2 e- ➝ Pb(s) E0 = +2.37V +E0= -0.13V Oxidation-reduction:Mg(s) + Pb2+(aq) ÝMg2+(aq) + Pb(s) E0 = +2.24 V In determining the potential values, we look them up in either the oxidation or the reduction table. However, it is important to note that the oxidation and reduction reactions are reversed as are the signs of their values. We can analyze a system from either point of view. That is, from the point of view of the donors of electrons or from the point of view of acceptors of electrons. Page 32 In the Reduction Potential table, lead cations, Pb2+(aq), are better acceptors of electrons than magnesium cations, Mg2+(aq). The lead cations, therefore, cause the electrons to move from the cell on the right towards the cell on the left. E0 = -2.37 V Eo = -0.13 V Pb Mg Pb(NO3)2 Mg(NO3)2 The lead cation, Pb2+(aq), acts as an electron acceptor thereby causing the magnesium metal to act as an electron donor. The equations are the same: Oxidation: Mg(s) ➝ Mg2+(aq) + 2 e- Reduction: Pb2+(aq) + 2 e- ➝ Pb(s) Oxidation-reduction: Mg(s) + Pb2+(aq) ➝ Mg2+(aq) + Pb(s) E0 = +2.37V +E0= -0.13V E0 = +2.24 V 50) Determine the potential of the following cell by writing the oxidation, reduction, and the overall (oxidation-reduction) equations. Cr(s) / Cr(NO3)3(aq) // Cu(NO3)2(aq) / Cu(s) Oxidation reaction: _______________________________________________________ Reduction reaction: _______________________________________________________ Overall reaction: _______________________________________________________ Page 33 t Activity Series of Metals with Half-Reactions 1. Fill in the blanks to complete the half-reactions on the table. Activity Series of Metals, with Half-Reactions for Oxidation Processes Elements Oxidation half-reactions Most active Lithium Li(s) Õ Li+(aq) + ______ and most Potassium K(s) Õ ______ + eeasily _____________ ____ Õ Ba2+(aq) + 2eoxidized Calcium Ca(s) Õ Ca2+(aq) + _____ Sodium Na(s) Õ Na+(aq) + _____ Magnesium Mg(s) Õ ______ + 2eAluminum Al(s) Õ Al3+(aq) + _____ Zinc Zn(s) Õ ______ + 2eIron Fe(s) Õ ______ + 2eNickel Ni(s) Õ Ni2+(aq) + _____ ____________ ____ Õ Sn2+(aq) + 2eLead Pb(s) Õ ______ + 2e* Hydrogen H2(g) Õ 2H+(aq) + _____ Least active Copper Cu(s) Õ Cu2+(aq) + _____ and least Mercury Hg(s) Õ ______ + 2eeasily Silver Ag(s) Õ Ag+(aq) + _____ oxidized Gold Au(s) Õ ______ + 3e* 2. Hydrogen is included for reference purposes. Use the table to determine if the following statements are true or false. ________ a. Tin is more easily oxidized than magnesium. ________ b. Aluminum is more easily oxidized than mercury. ________ c. Zinc becomes copper-plated when dipped in a copper sulfate solution. ________ d. Silver metal is oxidized by lead ions. ________ e. Iron metal is oxidized by nickel ions. ________ f. An acid will release H2 gas when copper is placed in the acid. Page 34 ▼ Voltaic Cell 1. Determine the location where each process occurs in this electrochemical cell. Write the letter for that location in the blank provided. Locations a. in the wire and light bulb c. in the salt bridge b. at the cathode d. at the anode Processes ______ Electrons pass through the external circuit to the copper strip. ______Positive and negative ions move through the aqueous solution to maintain electrical neutrality. ______Electrons are passed to copper ions, and reduction takes place. ______Electrons are produced by oxidation. 2. Write the half-reactions and overall cell reaction for this voltaic cell. Oxidation half-reaction: _______________________________________________________ Reduction half-reaction: _______________________________________________________ Overall cell reaction: _________________________________________________________ Page 35 t Electrochemical Cells Zn(s) / Zn2+(aq) // Cr2O72-(aq), H+(aq), Cr3+(aq) / C(s) Cell 1 2. reduction half cell: 3. oxidation half reaction: 4. net reaction: 5. spontaneous: yes/no Fe(s) / Fe2+(aq) // Cr2O72-(aq), H+(aq), Cr3+(aq) / C(s) Cell 2 2. reduction half cell: 3. oxidation half reaction: 4. net reaction: 5. spontaneous: yes/no Page 36 Cu(s) / Cu2+(aq) // Cr2O72-(aq), H+(aq), Cr3+(aq) / C(s) Cell 3 2. reduction half cell: 3. oxidation half reaction: 4. net reaction: 5. spontaneous: yes/no Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s) Cell 4 2. reduction half cell: 3. oxidation half reaction: 4. net reaction: 5. spontaneous: yes/no Page 37 Zn(s) / Zn2+(aq) // Fe2+(aq) / Fe(s) Cell 5 2. reduction half cell: 3. oxidation half reaction: 4. net reaction: 5. spontaneous: yes/no Fe(s) / Fe2+(aq) // Cu2+(aq) / Cu(s) Cell 6 2. reduction half cell: 3. oxidation half reaction: 4. net reaction: 5. spontaneous: yes/no Page 38 ▼ Battery Technology Examples (1) Dry Cell (LeclanchŽ) Battery characteristics: -not rechargeable -low cost -low energy density -low efficiency under high current drain (2) Alkaline Battery characteristics: -3x cost of LeclanchŽ cell -3x capacity of LeclanchŽ cell -longer shelf life -steadier voltage under high current Page 39 (3) Mercury Battery characteristics: -not rechargeable -constant voltage -long shelf life used in hearing aids, watches, calculators, pacemakers, etc. (4) Lead-acid Storage Battery Battery characteristics: -rechargeable by inputting energy -low cost vehicle usage -high rate performance -good charge retention -long lifespan (6 yrs.) (5) Nicad (Nickel-cadmium) Battery characteristics: -only water is consumed on discharge resulting in little change in electrolyte concentration -stable voltage -long shelf life used in calculators, flashlights, tools, etc. Page 40 (6) The Molicel Battery characteristics: -high-energy -rechargeable -3x energy density and 5x the power capability of the lead-acid cell in a car battery. invented in British Columbia (7) Aluminum-air fuel cell Battery characteristics: -reactants supplied externally (usually gaseous) redox reductions are used to convert chemical energy directly to electrical energy -requires electrolyte (ion conductor) -compact, lightweight, mobile, safe -high energy, high power density (8-10x energy density of lead-acid battery) -refuelable (aluminum anode oxidized by circulating alkaline electrolyte, air breathing cathode results in oxygen gas being reduced) - ideal supply of reserve and/or emergency power long shelf life (10 yrs.) Page 41 w Mirco Approach to Redox Experiment on page Objective: To study electrochemical cells. In this experiment, we'll construct electrochemical cells. In the oxidation and reduction reactions a salt bridge (consisting of a strip of filter paper dipped in a salt solution)will be used to link the two half reactions. This will form an electrochemical cell. Each of the following cells will be built in adjoining reaction wells with the salt bridge suspended in both wells. The two separate electrodes, each connected to the appropriate terminal of the voltmeter, will then be inserted into their respective well. Salt bridge Possible half cells to be used: 1.0 M Zn(NO3)2 (aq) with Zn electrode 1.0 M Cu(NO3)2 (aq) with Cu electrode 1.0 M Mg(NO3)2 (aq) with Mg electrode 1.0 M Al(NO3)3 (aq) with Al electrode 1.0 M Pb(NO3)2 (aq) with Pb electrode Anode Half-cell/ Cathode Half-cell Zn/Zn2+ Cu/Cu2+ Mg/Mg2+ Al/Al3+ Page 42 Zn/Zn2+ Cu/Cu2+ Mg/Mg2+ Al/Al3+