1. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 23.1 Enthalpy Changes in a Converter Which produces more heat in a converter, oxidation of 1 mol of C or oxidation of 1 mol of Si? (a) What would you expect as the product when any dissolved manganese in the molten mixture is oxidized? (b)Would the manganese oxidation be exothermic or endothermic? Answers: (a) MnO 2, (b) exothermic Practice Exercise Solution Analyze: We must compare the enthalpy change for 1 mol of C oxidized to form CO 2 with the enthalpy change for 1 mol of Si oxidized to form SiO 2. Plan: The reactions occur under conditions far different from the standard-state conditions for the substances. Nevertheless, we can estimate the enthalpy changes by using the thermodynamic values of Appendix C. Solve: The oxidation reactions and enthalpy changes under standard conditions are The actual numbers will differ substantially from these because the temperature is far from 298 K, because both C and Si are dissolved in the molten iron, and because SiO 2 will be incorporated in the slag. Nevertheless, the differences in the enthalpies of reaction are so great that it seems certain that the enthalpy change will be larger for Si.

2. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 23.2 Impurities in Electrorefining Nickel is one of the chief impurities in the crude copper that is subjected to electrorefining. What happens to this nickel in the course of the electrolytic process? Zinc is another common impurity in copper. Using standard reduction potentials, determine whether zinc will accumulate in the anode sludge or in the electrolytic solution during the electrorefining of copper. Answer: It is found in the electrolytic solution because the standard reduction potential of Zn 2+ is more negative than that of Cu 2+. Practice Exercise Solution Analyze: We are asked to predict whether nickel can be oxidized at the anode and reduced at the cathode during the electrorefining of copper. Plan: We need to compare the standard reduction potentials of Ni 2+ and Cu 2+. The more negative the reduction potential, the less readily the ion is reduced but the more readily the metal itself is oxidized. (Section 20.4) Solve: The standard reduction potential for Ni 2+ is more negative than that for Cu 2+ : As a result, nickel is more readily oxidized than copper, assuming standard conditions. Although we do not have standard conditions in the electrolytic cell, we nevertheless expect that nickel is preferentially oxidized at the anode. Because the reduction of Ni 2+ occurs less readily than the reduction of Cu 2+, the Ni 2+ accumulates in the electrolyte solution, while the Cu 2+ is reduced at the cathode. After a time it is necessary to recycle the electrolyte solution to remove the accumulated metal ion impurities, such as Ni 2+.

3. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 23.3 Writing Half-Reactions for a Oxidation-Reduction Reaction Write the two balanced half-reactions for the reaction in Equation 23.26. Calculate the standard potential for the reaction of Equation 23.22. Answer: 1.64 V Practice Exercise Solution Analyze: We are asked to write two oxidation-reduction half-reactions that together make up a given oxidation- reduction reaction. Plan: We need to separate the two half-reactions that make up the given balanced reaction. We can best start by noting that Fe appears in the +2 and +3 states and that oxygen appears in the 0 and –2 oxidation states. Solve: We note that iron is oxidized from the +2 to +3 oxidation state. The half-reaction for this process is Oxygen in acidic medium is reduced: To achieve a balanced equation, we need four Fe 2+ for each O 2. Therefore, all we need to do to obtain Equation 23.26 is to multiply the oxidation half-reaction through by 4. Comment: As noted in the text, the potential for the oxidation of Fe 2+ (aq) is negative, but the potential for the reduction of O 2 (g) in acidic aqueous solution is sufficiently positive to overcome that, leading to an overall positive potential for the reaction of Equation 23.26.

4. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Integrative Exercise Putting Concepts Together The most important commercial ore of chromium is chromite (FeCr 2 O 4 ). (a) What is the most reasonable assignment of oxidation states to Fe and Cr in this ore? (b) Below 74 K the FeCr 2 O 4 magnetically orders so that the unpaired electrons on chromium point in the opposite direction as those on iron. What type of magnetic state describes FeCr 2 O 4 below 74 K? Would you expect the magnetism to increase or decrease as FeCr 2 O 4 is cooled through the 74 K transition? (c) Chromite can be reduced in an electric arc furnace (which provides the required heat) using coke (carbon). Write a balanced chemical equation for this reduction, which forms ferrochrome (FeCr 2 ). (d) Two of the major forms of chromium in the +6 oxidation state are CrO 4 2– and Cr 2 O 7 2–. Draw Lewis structures for these species. (Hint: You may find it helpful to consider the Lewis structures of nonmetallic anions of the same formula.) (e) Chromium metal is used in alloys (for example, stainless steel) and in electroplating, but chromium is not widely used by itself, in part because it is not ductile at ordinary temperatures. From what we have learned in this chapter about metallic bonding and properties, suggest why chromium is less ductile than most metals. Solution (a) Because each oxygen has an oxidation number of –2, the four oxygens represent a total of –8. If the metals have whole-number oxidation numbers, our choices are Fe = +4 and Cr = +2, or Fe = +2 and Cr = +3. The latter choice seems the more reasonable because a +4 oxidation number for iron is unusual. (Although one alternative would be that Fe is +3 and the two Cr have different oxidation states of +2 and +3, the properties of chromite indicate that the two Cr have the same oxidation number.) (b) Fe 2+ and Cr 3+ have electron configurations of [Ar]3d 6 and [Ar]3d 3, respectively. Therefore, Fe 2+ will have four unpaired electrons, and Cr 3+ will have three unpaired electrons. Because they line up in opposite directions, FeCr 2 O 4 will either be antiferromagnetic or ferrimagnetic. Because the ions possess different numbers of unpaired electrons as well as the fact that there are twice as many Cr 3+ ions as there are Fe 2+ ions, their spins will not cancel each other out, and FeCr 2 O 4 will be ferrimagnetic below 74 K. The magnetic behavior of a ferrimagnet is similar to that of a ferromagnet, so we would expect to see a large increase in the magnetism below 74 K.

5. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Integrative Exercise Putting Concepts Together Solution (Continued) (c) The balanced equation is (d) We expect that in CrO 4 2–, the Cr will be surrounded tetrahedrally by four oxygens. The electron configuration of the Cr atom is [Ar]3d 5 4s 1, giving it six electrons that can be used in bonding, much like the S atom in SO 4 2–. These six electrons must be shared with four O atoms, each of which has six valence-shell electrons. In addition, the ion has a 2– charge. Thus, we have a total of 6 + 4(6) + 2 = 32 valence electrons to place in the Lewis structure. Putting one electron pair in each Cr—O bond and adding unshared electron pairs to the oxygens, we require precisely 32 electrons to achieve an octet around each atom: In Cr 2 O 7 2– the structure is analogous to that of the diphosphate ion ( P 2 O 7 4– ), which we discussed in Section 22.8. We can think of the Cr 2 O 7 2– ion as formed by a condensation reaction as shown in Equation 23.24.

6. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Integrative Exercise Putting Concepts Together Solution (Continued) (e) Recall that chromium, with six electrons available for bonding, has relatively strong metallic bonding among the metals of the transition series, as evidenced by its high melting point (Figure 23.14). This means that distortions of the metallic lattice of the sort that occur when metals are drawn into wires will require more energy than for other metals with weaker metallic bonding.