# Guide for Reading. Vocabulary. reduction potential reduction potential. standard cell potential standard hydrogen electrode.

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2 Section 21.2 (continued) Electrical Potential Use Visuals Figure 21.8 Check student understanding of electrochemical, or voltaic, cells by ask students to identify the anode and cathode in the lemon battery. (Zinc is the anode and copper is the cathode.) What process goes on at the anode? (oxidation) What process goes on the cathode? (reduction) In which process are electrons lost? (oxidation) What role does the lemon play in the battery? (It is the salt bridge.) Standard Cell Potential Use Visuals Figure 21.9 Have students study the diagram. Ask, What is the significance of the standard hydrogen electrode? How is it different from other electrodes? (It has been assigned a standard reduction potential of 0.00 V at 25 C.) What does the standard hydrogen half-cell consist of? (A platinum electrode with hydrogen gas at atmospheric pressure bubbling over its surface; the electrode is immersed in 1.00M acid solution at 25 C.) Students may be confused by the arbitrary assignment of 0.00 V to the hydrogen half-cell. Use the Celsius temperature scale as an analogy. When the Celsius temperature scale was invented, 0 C was arbitrarily assigned to the temperature at which water freezes. All measured temperatures are relative to that assigned value. Temperatures higher than the freezing point of water are positive temperatures. Temperatures lower than 0 C are negative. Salt bridge to other half-cell 1.00M acid solution 672 Chapter 21 Platinum wire Glass sleeve H 2 (g) (101 kpa) Platinum foil coated with platinum black Figure 21.9 The standard hydrogen electrode is arbitrarily assigned a standard reduction potential of 0.00 V at 25 C. Simulation 27 Simulate the operation of voltaic cells. Standard Cell Potential The standard cell potential ( ) is the measured cell potential when the ion concentrations in the half-cells are 1M, any gases are at a pressure of 101 kpa, and the temperature is 25 C. The symbols E 0 red and E 0 oxid represent the standard reduction potentials for the reduction and oxidation halfcells, respectively. The relationship between these values follows the general relationship for cell potential given on the previous page. E 0 cell E0 - E 0 Because half-cell potentials cannot be measured directly, scientists have chosen an arbitrary electrode to serve as a reference. The standard hydrogen electrode is used with other electrodes so the reduction potentials of the other cells can be measured. The standard reduction potential of the hydrogen electrode has been assigned a value of 0.00 V. The standard hydrogen electrode, which is illustrated in Figure 21.9, consists of a platinum electrode immersed in a solution with a hydrogen-ion concentration of 1M. The solution is at 25 C. The electrode itself is a small square of platinum foil coated with finely divided platinum, known as platinum black. Hydrogen gas at a pressure of 101 kpa is bubbled around the platinum electrode. The half-cell reaction that occurs at the platinum black surface is as follows. 2H (aq, 1M) 2e - m H 2 (g, 101 kpa) E 0 H 0.00 V The double arrows in the equation indicate that the reaction is reversible. The symbol E 0 represents the standard reduction potential of H H. The standard reduction potential of H is the tendency of H ions to acquire electrons and be reduced to H 2 (g). Whether this half-cell reaction occurs as a reduction or as an oxidation is determined by the reduction potential of the half-cell to which the standard hydrogen electrode is connected. Checkpoint What components and conditions are required for a standard hydrogen electrode? Standard Reduction Potentials A voltaic cell can be made by connecting a standard hydrogen half-cell to a standard zinc half-cell, as shown in Figure To determine the overall reaction for this cell, first identify the half-cell in which reduction takes place. In all electrochemical cells, reduction takes place at the cathode and oxidation takes place at the anode. A voltmeter gives a reading of 0.76 V when the zinc electrode is connected to the negative terminal and the hydrogen electrode is connected to the positive terminal. The zinc is oxidized, which means that it is the anode. Hydrogen ions are reduced, which means that the hydrogen electrode is the cathode. You can now write the half-reactions and the overall cell reaction. Oxidation: Zn1 s 2 Zn 2 1aq 2 2e - 1at anode2 Reduction: 2H 1aq 2 2e - H 2 1 g 21at cathode2 Cell reaction: Zn1 s 2 2H 1aq 2 Zn 2 1aq 2 H 2 1 g 2 Special Needs Give students small cards with e printed on them in large letters. Divide students into pairs. Let one student role play being an anode and the other a cathode. The anode should hand one or more cards to the cathode 672 Chapter 21

3 e 0.76V e Standard Reduction Potentials Anode ( ) Zn Salt bridge Cotton plugs Cathode ( ) 1.00M Zn M H Zinc half-cell Hydrogen half-cell H 2 (g) (101 kpa) You can determine the standard reduction potential of a halfcell by using a standard hydrogen electrode and the equation for standard cell potential. In the zinc hydrogen cell, zinc was oxidized and hydrogen ions were reduced. Let E 0 red E0 H and in the standard cell E0 oxid E0 Zn 2 potential equation. E 0 - E 0 E 0 cell E0 - H E0 Zn 2 The cell potential ( ) was measured at 0.76 V. The reduction potential of the hydrogen half-cell is a defined standard: E 0 H always equals 0.00 V. Substituting these values into the preceding equation will give the standard reduction potential for the zinc half-cell V 0.00 V - E 0 Zn 2 E 0 Zn V 2 The standard reduction potential for the zinc half-cell is 0.76 V. The value is negative because the tendency of zinc ions to be reduced to zinc metal in this cell is less than the tendency of hydrogen ions to be reduced to hydrogen gas (H 2 ). Consequently, the zinc ions are not reduced. Instead, the opposite occurs: Zinc metal is oxidized to zinc ions. Many different half-cells can be paired with the hydrogen half-cell in a similar manner. Using this method, the standard reduction potential for each half-cell can be obtained. For a standard copper half-cell, for example, the measured standard cell potential is 0.34 V. Copper is the cathode, and Cu 2 ions are reduced to Cu metal when the cell operates. The hydrogen half-cell is the anode, and H 2 gas is oxidized to H ions. You can calculate the standard reduction potential for copper as follows. E 0 cell E0 - E 0 E 0 cell E0 - E 0 Cu 2 H 0.34 V E 0 2 Cu V 2 This value is positive because the tendency for copper ions to be reduced in the cell is greater than the tendency of hydrogen ions to be reduced. Figure This voltaic cell consists of zinc and hydrogen half-cells. Interpreting Diagrams Where does reduction occur, and what is reduced in this cell? TEACHER Demo The Corrosion of Iron Purpose Students observe the galvanic corrosion of iron compared to copper. Materials 2 pennies, metal wires from twist ties, table salt, paper towel Procedure Thoroughly clean the pennies and the wires. Moisten a paper towel with tap water and sprinkle it with salt. Wrap one wire around a penny and place it on the paper towel. Place an unwrapped penny and a coil of wire on the paper towel as a control. Observe what happens after at least 24 hours. Expected Outcomes Iron corroded; copper did not. Ask students to explain. (In galvanic corrosion, the corrosion rate (oxidation) of one metal accelerates while that of the other is decreased. Because iron is more active than copper, its corrosion was accelerated.) Section 21.2 Half-Cells and Cell Potentials 673 Less Proficient Readers Have pairs of students look at the diagram of the voltaic cell in Figure Have them trace the movement of the electrons from one electrode to the other. They should be able to explain that the electrons travel up from the zinc anode into the external wire, through the voltmeter, and down through the wire into the hydrogen/platinum cathode. Have them identify where oxidation is taking place and where reduction is taking place. Refer them to Table 21.2 to find the appropriate half-cell reactions and have them write the overall reaction for the cell. Answers to... Figure Reduction occurs at the cathode. In this cell hydrogen ions are reduced to hydrogen gas. Checkpoint The standard hydrogen electrode consists of a platinum electrode immersed in a 1M acid solution at 25 C. Electrochemistry 673

4 Section 21.2 (continued) Explain that the two half-cells of a voltaic cell are competing for electrons and that oxidation or reduction could occur in either cell. The half-cell with the more positive reduction potential will win the competition and undergo reduction; oxidation will occur in the other half-cell. The potential produced by the electrochemical cell is the difference in the reduction potentials of the two half-cell reactions. Explain that the quantitative value of any half-cell potential is obtained by measuring it against the standard hydrogen electrode. Point out that Table 21.2 contains tabulated values for some halfcells. These can be used to calculate cell potentials. Table 21.2 Reduction Potentials at 25 C with 1M Concentrations of Aqueous Species Electrode Half-reaction E 0 (V) Li /Li Li e Li 3.05 K /K K e K 2.93 Ba 2 /Ba Ba 2 2e Ba 2.90 Ca 2 /Ca Ca 2 2e Ca 2.87 Na /Na Na e Na 2.71 Mg 2 /Mg Mg 2 2e Mg 2.37 Al 3 /Al Al 3 3e Al 1.66 H 2 O/H 2 2H 2 O 2e H 2 2OH 0.83 Zn 2 /Zn Zn 2 2e Zn 0.76 Cr 3 /Cr Cr 3 3e Cr 0.74 Fe 2 /Fe Fe 2 2e Fe 0.44 H 2 O/H 2 (ph 7) 2H 2 O 2e H 2 2OH 0.42 Cd 2 /Cd Cd 2 2e Cd 0.40 PbSO 4 /Pb PbSO 4 2e 2 Pb SO Co 2 /Co Co 2 2e Co 0.28 Ni 2 /Ni Ni 2 2e Ni 0.25 Sn 2 /Sn Sn 2 2e Sn 0.14 Pb 2 /Pb Pb 2 2e Pb 0.13 Fe 3 /Fe Fe 3 3e Fe H /H 2 2H 2e H AgCl/Ag AgCl e Ag Cl 0.22 Hg 2 Cl 2 /Hg Hg 2 Cl 2 2e 2Hg 2Cl 0.27 Cu 2 /Cu Cu 2 2e Cu 0.34 O 2 /OH O 2 2H 2 O 4e 4OH 0.40 Cu /Cu Cu e Cu 0.52 I 2 /I I 2 2e 2I 0.54 Fe 3 /Fe 2 Fe 3 e Fe Hg 2 2 /Hg 2 Hg 2 2e 2Hg 0.79 Ag /Ag Ag e Ag 0.80 O 2 /H 2 O (ph 7) O 2 4H 4e 2H 2 O 0.82 Hg 2 /Hg Hg 2 2e Hg 0.85 Br 2 /Br Br 2 2e 2Br 1.07 O 2 /H 2 O O 2 4H 4e 2H 2 O 1.23 MnO 2 /Mn 2 MnO 2 4H 2e Mn 2 2H 2 O 1.28 Cr 2 O 2 7 /Cr 3 2 Cr 2 O 7 14H 6e 2Cr 3 7H 2 O 1.33 Cl 2 /Cl Cl 2 2e 2Cl 1.36 PbO 2 /Pb 2 PbO 2 4H 2e Pb 2 2H 2 O 1.46 MnO 4 /Mn 2 MnO 4 8H 5e Mn 2 4H 2 O 1.51 PbO 2 /PbSO 4 PbO 2 4H 2 SO 4 2e PbSO 4 2H 2 O 1.69 F 2 /F F 2 2e 2F Chapter 21 Special Needs Some students may have trouble remembering which reactions in Table 21.2 have the greatest tendency to occur as reductions. Have them compare the table with the activity series of metals in Table Point out that the most active metals are at the top of Table 21.1; ions of these same metals are at the top of Table Because active metals lose electrons easily, they are most easily oxidized. By the same reasoning, their ions are least likely to be reduced. 674 Chapter 21

5 Table 21.2 on the preceding page lists some standard reduction potentials at 25 C. The half-reactions are arranged in increasing order of their tendency to occur in the forward direction that is, as a reduction. Thus, the half-reactions at the top of the table have the least tendency to occur as reductions. The half-reactions at the bottom of the table have the greatest tendency to occur as reductions. For example, lithium ions (at the top) have very little tendency to be reduced to lithium metal. Calculating Standard Cell Potentials To function, a cell must be constructed of two half-cells. The half-cell reaction having the more positive (or less negative) reduction potential occurs as a reduction in the cell. With this in mind, it is possible to write cell reactions and calculate cell potentials for cells without actually assembling them. You can simply use the known standard reduction potentials for the various half-cells (from Table 21.2) to predict the half-cells in which reduction and oxidation will occur. This information can then be used to find the resulting value. If the cell potential for a given redox reaction is positive, then the reaction is spontaneous as written. If the cell potential is negative, then the reaction is nonspontaneous. This latter reaction will be spontaneous in the reverse direction, however, and the cell potential will then have a numerically equal but positive value. CONCEPTUAL PROBLEM 21.1 Determining Reaction Spontaneity Many hearing aid batteries contain zinc and silver oxide. Show that the following redox reaction between Zn and Ag is spontaneous. Zn(s) 2Ag (aq) Zn 2 (aq) 2Ag(s) 1 2 Analyze Identify the relevant concepts. Identify the half-reactions, and calculate the standard cell potential ( E 0 cell E0 - E 0 ). If is positive, the reaction is spontaneous. Solve Apply concepts to this situation. The half-reactions from the equation are: Oxidation: Zn(s) Zn 2 (aq) 2e Reduction: Ag (aq) e Ag(s) Practice Problems Use Table 21.2 to look up the standard reduction potentials for the half-cells. Zn 2 (aq) 2e - Zn(s2 E 0 Zn V 2 Ag (aq) e - Ag(s2 E 0 Ag 0.80 V Calculate the standard cell potential. E0 red - E0 oxid E0 Ag - E0 Zn V - (-0.76 V) 1.56 V 0, so the reaction is spontaneous. Calculating Standard Cell Potentials CLASS Activity Determining Cell Voltages Purpose Students become familiar with combining reduction halfreactions. Procedure Divide the class into groups of three or four. Have them use Table 21.2 to combine half-reactions and create cells having potentials that fall in the following ranges: >2.5 V, 2.4 V to 1.4 V, 1.0 V to 0.5 V, and 0.4 V to 0.1 V. Expected Outcome Several cells could be created in each category. Typical cells might be Na/Na with Sn 2 / Sn, 2.57 V; Mg 2 /Mg with Pb 2 /Pb, 2.24 V; Ca 2 /Ca with Al 3 /Al, 1.21 V; PbSO 4 /Pb with Ni 2 /Ni, 0.11 V. CONCEPTUAL PROBLEM 21.1 Answers 9. nonspontaneous; = 0.02 V 10. yes; = 0.16 V Practice Problems Plus Chapter Assessment problem 41 is related to Conceptual problem Determine whether the following redox reaction will occur spontaneously. 3Zn 2 (aq) 2Cr(s) 3Zn(s) 2Cr 3 (aq) 10. Is this redox reaction spontaneous as written? Co 2 (aq) Fe(s) Fe 2 (aq) Co(s) Problem-Solving 12.9 Solve Problem 9 with the help of an interactive guided tutorial. Section 21.2 Half-Cells and Cell Potentials 675 Gifted L3 Have students investigate the practice of attaching large pieces of magnesium to the hulls of ocean going ships having steel (iron) hulls. Ask them to explain the reason for this practice and provide any supporting figures they can. (The more active metal, magnesium, is oxidized rather than the iron. The reduction potential for magnesium (Mg 2 /Mg) is 2.37 V and the reduction potential for iron (Fe 3 /Fe) is V). Electrochemistry 675

6 Section 21.2 (continued) As students work Sample Problems 21.1 and 21.2, they may think that when a half-cell reaction must be multiplied by a factor such as 2 or 3, the E 0 value must also be multiplied by that factor. Point out that even though two or three times as many electrons are present, the tendency for the electrons to flow is not two or three times greater. The tendency, which is measured by the E 0 value, remains the same. Point out that until now it has been assumed that reactions in standard half-cells always go in the spontaneous direction. For example, a Ag/Cu cell can be constructed in such a way that the only substance available for oxidation is silver metal and the only substance available for reduction is copper ion. In this case, the reaction is nonspontaneous and the cell potential is negative. Thus, the sign of the cell potential is an indicator of reaction spontaneity. Students will learn in the next section that nonspontaneous reactions can be driven by an outside source of energy. SAMPLE PROBLEM 21.1 Writing the Cell Reaction Determine the cell reaction for a voltaic cell composed of the following half-cells. Fe 3 1aq2 e - Fe 2 1aq2 E 0 3 Fe 0.77 V Ni 2 1aq2 2e - Ni1s2 E 0 2 Ni Analyze List the knowns and the unknowns. Knowns Unknowns E 0 Fe 0.77 V cell reaction? 3 E 0 Ni E 0? V 2 cell The half-cell with the more positive reduction potential is the one in which reduction occurs (the cathode). That means that the oxidation reaction occurs at the anode. Add the half-reactions, making certain that the number of electrons lost equals the number of electrons gained. Calculate Solve for the unknowns. The Fe 3 half-cell, which has the more positive reduction potential, is the cathode. So in this voltaic cell, Fe 3 is reduced and Ni is oxidized. The half-cell reactions, written in the direction in which they actually occur, are as follows. Oxidation: Ni(s) Ni 2 (aq) 2e (at anode) Reduction: Fe 3 (aq) e Fe 2 (aq) (at cathode) Before adding the half-reactions, be sure that the electrons cancel that is, that they are present in equal number on both sides of the equation. Because Ni loses two electrons as it is oxidized, while Fe 3 gains Practice only Problems one electron as it is reduced, you must multiply the Fe 3 half-cell equation by 2. Ni1s2 Ni 2 1aq 2 2e - 2 3Fe 3 1aq 2 e - Fe 2 1aq 24 Ni1s2 2Fe 3 1aq 2 Ni 2 1aq 2 2Fe 2 1aq 2 Evaluate Does the result make sense? The electrons cancelled when the half-reactions were added. The equation for the cell reaction is correctly written. Sample Problem 21.1 Practice Problems Answers 11. 2Al(s) 3Cu 2 (aq) 2Al 3 (aq) 3Cu(s) 12. Cu(s) 2Ag (aq) Cu 2 (aq) 2Ag(s) Practice Problems Plus A voltaic cell is constructed using the following half-reactions: Li (aq) e Li(s) E 0 Li = 3.05 V Mg 2 (aq) 2e Mg(s) E 0 Mg 2 = 2.37 V Determine the standard cell reaction and the standard cell potential. [2Li(s) Mg 2 (aq) Mg(s) 2Li (aq); 0.68 V] Problem-Solving Solve Problem 12 with the help of an interactive guided tutorial. 676 Chapter A voltaic cell is constructed using Cu 2 1aq2 2e - Cu1s V 2 Al 3 1aq2 3e - Al1s2 E 0 Al V 3 Determine the cell reaction. 12. A voltaic cell is constructed using Ag 1aq2 e - Ag1s2 E 0 Ag 0.80 V Cu 2 1aq2 2e - Cu1s V 2 Determine the cell reaction. Math Handbook For a math refresher and practice, direct students to significant figures, page R Chapter 21

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