What is the importance of redox reactions? Their importance lies in the fact that we can use the transfer of electrons between species to do useful

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1 What is the importance of redox reactions? Their importance lies in the fact that we can use the transfer of electrons between species to do useful work. This is accomplished by constructing a voltaic cell. A voltaic cell relies in separating the reactants into two chambers and therefore forcing the electrons to travel through electrodes to perform the oxidation and the reduction. While the electrons are traveling from one electrode to another, we can make them do work.

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3 The Daniell Cell is an electrochemical cell named after John Frederic Daniell, the British chemist who invented it in 1836 flash

4 A voltaic cell spontaneously converts chemical energy to electrical energy. In voltaic cells the two half-reactions of a redox reaction, in an aqueous neutral solution, are separated but connected between each electrode with a conductive wire and between the two solutions with a salt bridge. Table J, Activity series, is used to find out which metal or nonmetal is more reactive than another or which one will lose e- and which one will gain e-. The reaction is spontaneous if the metal oxidized is above the one been reduced. Example Al(s) is oxidized and loses e- and Zn 2+ (aq) is reduced and gains the lost e- because Al is more reactive than Zn.

5 In an electrochemical cell, oxidation occurs at the anode and reduction at the cathode. Oxidation occurs at the anode therefore the anode is negative since e- are lost (ANode OXidation) Reduction occurs at the cathode therefore the cathode is positive since e- are gained (REDuction CAThode) Example The Al(s) electrode is the anode and is (-) The Zn(s) electrode is the cathode and is (+) The Al(s) electrode loses mass since the metal changes into aqueous ions and e- in the reaction Al(s) --> Al 3+ (aq) + 3e- The Zn(s) electrode or the cathode gains mass since the e- combine with the aqueous ions in the solution to produce more metal in the reaction Zn 2+ (aq) + 2e- --> Zn(s) When the oxidized electrode is completely used up the cell voltage is zero and the cell has reached equilibrium since the concentrations are not longer changing. The cell is dead. The salt bridge keeps both solutions neutral by allowing ions in the solutions to move through it. In the cell where oxidation occurs the concentration of the positive aqueous ion increases making the solution more and more positive. In the cell where reduction occurs the concentration of the positive aqueous ion decreases making the solution more and more negative. Example In the Al(s) cell, [Al 3+ (aq)] increases making the neutral solution increasingly positive. As the solution becomes increasingly positive the e- produced by the Al (s) electrode will be attracted to the positive solution and not to the Zn(s) electrode. In the Zn(s) cell, [Zn 2+ (aq)] decreases making the neutral solution increasingly negative. As the solution becomes increasingly negative the e- coming from the Al(s) electrode will be repelled by the negative solution back to the Al(s) electrode. The change of charge in cells is a problem because e- will no longer flow if they find another pathway and over time the cell will no longer function.

6 TEST YOUR UNDERSTANDING 1/07 Base your answers to questions 61 through 63 on the diagram below. The diagram shows a voltaic cell with copper and aluminum electrodes immediately after the external circuit is completed. 61 Balance the redox equation using the smallest whole-number coefficients. 62 As this voltaic cell operates the mass of the Al(s) electrode decreases. Explain, in terms of particles, why this decrease in mass occurs. 63 Explain the function of the salt bridge. 6/07 24 Which energy conversion occurs during the operation of a voltaic cell? (1) Chemical energy is spontaneously converted to electrical energy. (2) Chemical energy is converted to electrical energy only when an external power source is provided. (3) Electrical energy is spontaneously converted to chemical energy. (4) Electrical energy is converted to chemical energy only when an external power source is provided. 8/07 42 Given the balanced equation representing the reaction occurring in a voltaic cell: Zn(s) + Pb 2+ (aq) ---> Zn 2+ (aq) + Pb(s) In the completed external circuit, the electrons flow from (1) Pb(s) to Zn(s) (2) Pb 2+ (aq) to Zn 2+ (aq) (3) Zn(s) to Pb(s) (4) Zn 2+ (aq) to Pb 2+ (aq)

7 45 A student collects the materials and equipment below to construct a voltaic cell. two 250-mL beakers wire and a switch one strip of magnesium one strip of copper 125 ml of 0.20 M Mg(NO 3 ) 2 (aq) 125 ml of 0.20 M Cu(NO 3 ) 2 (aq) Which additional item is required for the construction of the voltaic cell? (1) an anode (2) a battery (3) a cathode (4) a salt bridge 8/06 25 Which conversion of energy always occurs in a voltaic cell? (1) light energy to chemical energy (2) electrical energy to chemical energy (3) chemical energy to light energy (4) chemical energy to electrical energy 1/05 45 Which half-reaction can occur at the anode in a voltaic cell? (1) Ni e Ni (2) Sn + 2e Sn 2+ (3) Zn Zn e (4) Fe 3+ Fe 2+ + e Base your answers to questions 80 and 81 on the information below. The outer structure of the Statue of Liberty is made of copper metal. The framework is made of iron. Over time, a thin green layer (patina) forms on the copper surface. 80. When copper oxidized to form this patina layer, the copper atoms became copper (II) ions (Cu 2+ ). Write a balanced half-reaction for this oxidation of copper. 81. Where the iron framework came in contact with the copper surface, a reaction occurred in which iron was oxidized. Using information from Reference Table J, explain why the iron was oxidized. 6/05 26 Where does oxidation occur in an electrochemical cell? (1) at the cathode in both an electrolytic cell and a voltaic cell (2) at the cathode in an electrolytic cell and at the anode in a voltaic cell (3) at the anode in both an electrolytic cell and a voltaic cell (4) at the anode in an electrolytic cell and at the cathode in a voltaic cell

8 Base your answers to questions 71 through 73 on the diagram of a voltaic cell and the balanced ionic equation below. 71. What is the total number of moles of electrons needed to completely reduce 6.0 moles of Ni 2+ (aq) ions? 72. Identify one metal from Reference Table J that is more easily oxidized than Mg(s). 73. Explain the function of the salt bridge in the voltaic cell. 1/04 Base your answers to questions 59 through 61 on the diagram of the voltaic cell below. 59. When the switch is closed, in which half-cell does oxidation occur? 60. When the switch is closed, state the direction that electrons will flow through the wire. 61. Based on the given equation, write the balanced half-reaction that occurs in half-cell 1. 6/04 27 A voltaic cell spontaneously converts (1) electrical energy to chemical energy (2) chemical energy to electrical energy (3) electrical energy to nuclear energy (4) nuclear energy to electrical energy 8/04 47 Which metal reacts spontaneously with a solution containing zinc ions? (1) magnesium (2) nickel (3) copper (4) silver

9 Base your answers to questions 76 through 78 on the diagram below, which represents a voltaic cell at 298 K and 1 atm. 76. In which half-cell will oxidation occur when switch S is closed? 77. Write the balanced half-reaction equation that will occur in half-cell 1 when switch S is closed. 78. Describe the direction of electron flow between the electrodes when switch S is closed. 1/03 1. According to Reference Table J, which of these metals will react most readily with 1.0 M HCl to produce H 2 (g)? (1) Ca (2) K (3) Mg (4) Zn 2. Base your answers on the diagram of a voltaic cell and on your knowledge of chemistry. 1. On the diagram provided, indicate with one or more arrows the direction of electron flow through the wire. 2. Write an equation for the half-reaction that occurs at the zinc electrode. 3. Explain the function of the salt bridge.

10 6/ Which statement is true for any electrochemical cell? (1) Oxidation occurs at the anode, only. (2) Reduction occurs at the anode, only. (3) Oxidation occurs at both the anode and the cathode. (4) Reduction occurs at both the anode and the cathode. 46. A diagram of a chemical cell and an equation are shown below. Pb(s) + Cu 2+ (aq) --> Pb 2+ (aq) + Cu(s) When the switch is closed, electrons will flow from (1) the Pb(s) to the Cu(s) (2) the Cu(s) to the Pb(s) (3) the Pb 2+ (aq) to the Pb(s) (4) the Cu 2+ (aq) to the Cu(s) 8/03 49 What is the purpose of the salt bridge in a voltaic cell? (1) It blocks the flow of electrons. (2) It blocks the flow of positive and negative ions. (3) It is a path for the flow of electrons. (4) It is a path for the flow of positive and negative ions.

11 USING ELECTRICITY TO DECOMPOSE COMPOUNDS ELECTROLYSIS - Electro-the use of electricity Lysis - separate a thing into its parts Electrolysis involves melting the ionic solid that contains the element you want to extract and passing electric current through it. This current has to be DIRECT CURRENT (DC), the kind produced directly by batteries or by AC passed through a converter. The electrons must only go one way, which is the only way that the individual elements in the compound can be completely separated. The electrode from the DC power supply provides a source of electrons that can REDUCE the positive metal ion to form pure metal: Na e - (from the DC power supply) Na 0 (pure sodium metal). The + electrode from the DC power supply removes electrons from the negat ion, causing the negative nonmetal ions to be reduced to form pure nonmetal 2 Cl -1 Cl 0 (pure chlorine gas) + 2 e - (which are pulled into the side of the DC supply)

12 NOT SPONTANEOUS What is Happening Here??? Electrons are coming out of the negative end of the DC power supply. These travel down into the liquid NaCl, into the electrode. Since electrons are negative, they attract the positively charged Na +1 ions to it. The Na +1 is forced to eat the electrons, reducing them to metallic Na0, which forms around the electrode. Na e - Na 0 Electrons are being yanked towards the positive side of the DC power supply. The Cl -1 ions are forced to give up their electrons and be oxidized into Cl 2 0 (g) molecules, which bubble up and are trapped for later use. The electrons ripped from the Cl -1 are pulled back into the positive side of the DC power supply, completing the circuit. 2 Cl -1 Cl e - This same setup can be used to decompose any binary compound containing an alkali metal and halogen. The metal ion will always be reduced at the negative cathode and the nonmetal ion will always be oxidized at the positive anode. This has to be carried out in the equivalent of a blast furnace... NaCl melts at 1074 K! Since the reaction is not spontaneous, DC current must be added to make this electrolysis occur.

13 NOT SPONTANEOUS What is Happening Here??? Electrons are coming out of the negative end of the DC power supply. These travel down into the liquid water (which has electrolyte like sulfuric acid dissolved in it to conduct electricity, since water does not conduct on its own), into the electrode. Since electrons are negative, they attract the positively charged H +1 ions to it. The H +1 is forced to eat the electrons, reducing them to H 0 gas, which bubbles up displacement. 2 H e - H 2 0 Electrons are being yanked towards the positive side of the DC power supply. The O -2 ions are forced to give up their electrons and be oxidized into O 2 0 (g) into the test tube and is trapped by water molecules, which bubble up and are trapped by water displacement. The electrons ripped from the O 2-2 are pulled back into the positive side of the DC power supply, completing the circuit. 2 O -2 O e-

14 Electroplating One Metal Onto Another What is Happening Here? Silver is being oxidized at the + electrode as its electrons get stripped off and pulled into the + end of the DC power supply. The silver dissolves into solution as silver ions. Ag 0 Ag +1 +1e - Meanwhile, over at the electrode, electrons are being pumped out of the DC power supply and into the ring. Metal atoms don t gain electrons, so the electrons just hang out on the surface of the ring, giving the ring a negative charge. This attracts the positive silver ions in the solution, which migrate to the ring and touch it. As soon as a silver ion touches the ring, an electron jumps off the ring onto the silver ion, reducing it to solid silver metal, which freezes on to the ring. Ag e - Ag 0 As more and more silver ions reduce and plate on to the ring, the thickness of the plating increases. The final thickness depends on how much current is being used, and how long you let the process continue.

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16 The VOLTAIC CELL versus the ELECTROLYTIC CELL

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