Chemical Education Today
Letters Redox Challenges In your December 1995 issue there were three Redox Balancing Challenges put forth by Roland Stout (J. Chem. Educ. 1995, 72, 1125). He stated that he had offered an “A” for the entire course if a student could balance equation 3 (the hardest), and that no one had ever achieved it. After reading this I decided to offer it to my Chemistry II A.P. class, but only for a grade of 90 on the semester exam. Surely no one would balance this monster equation. A week later Nathan Mettlach came in with this proof and the balanced equation. While talking with him I found out it had taken him only 2 hours. I had never even thought of taking his approach, but once I read through the proof it all made sense. This is a remarkable student who deserves recognition for his understanding of math and science, and how the two are related. I believe this should be printed in your Journal so other less-enlightened individuals can see really how easy this equation is to balance and to provide them with a new method for balancing equations. A Large Equation Balanced by Nathan Mettlach [Cr(N2 H4CO) 6]4[Cr(CN)6]3 + KMnO 4 + H2SO 4 → a b c K 2Cr 2O 7 + MnSO4 + CO2 + KNO3 + K 2SO4 + H2O d e f g h i To balance this equation, balancing for mass seems to be the easiest way. The number of each type of element should be the same on both sides. Since coefficients
for chromium for nitrogen hydrogen carbon oxygen (excluding O present in SO4 ion) potassium manganese sulfate ions (since it is not broken apart)
7a = 2d 66a = g 96a + 2c = 2i 42a = f 24a+4b = 7d+2f+3g+i b = 2d + g + 2h b=e c = e+h
Since all the equations need to be satisfied, combining them while trying to eliminate as many variables as possible is the next step. c = e+h b = e, b = 2d + g + 2h c = (2d + g + 2h) + h 96a + 2c = 2i 96a + 2(2d + g + 3h) = 2i 24a + 4b = 7d + 2f + 3g + [48a + (2d + g + 3h)] 4b = 24a + 9d + 2f + 4g + 3h d = 3.5a 4b = 24a + 31.5a + 84a + 264a + 3h h = 0.5b – d – 0.5g h = 0.5b – 3.5a – 33a 4b = 403.5a + 1.5b – 10.5a – 99a 2.5b = 294a (at this point, since d = 3.5a, the coefficient in front of “b” is even) 10b = 1176a (since “a” must be even, and the only factors of ten are 5 and 2, these numbers are factors of the original equation’s coefficients)
Using a = 10 and b = 1176, we can find the rest of the coefficients by substituting into the set of equations that needed to be satisfied. c = 1399, d = 35, e = 1176, f = 420, g = 660, h = 223, i = 1879
The balanced equation is as follows: 10[Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176KMnO4 + 1399H2SO4 → 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + 1879H2O
Corrections The following errors appear in my Letter (J. Chem. Educ. 1996, 73, A226) : 1. The conservation relationship for H should read 96a + 2c = 2i, rather than 96a + 2 = 2i. 2. The conservation relationship for O should read 24a + 4b + 4c = 7d + 4e + 2f + 3g + 4h + i, rather than 24a + 4b + 4c = 7d + 4e + 2f + 3e + 4h + i. 3. Equation 3 should read d = (7/84)f, rather than d = (7/84). David M. Hart Department of Chemistry University of Central Oklahoma Edmond, OK 73034 ❖ ❖ ❖ One of my students recently found an error in Table 2 of our article, “The pH of Sweat of Horses” (J. Chem. Educ. 1997, 74, 1135). The treatment J-Flex is incorrectly listed as a vitamin. J-Flex is instead chondroitin sulfate. John Tierney Department of Chemistry Penn State University–Delaware County Campus Media, PA 19063
1256
are sought, let the letters “a” through “i” stand for the coefficients of the different compounds. All the following equations need to be satisfied to balance the equation:
Rodger S. Nelson Harlingen High School 1201 E. Marshall Harlingen, TX 78550
✍ The problem of balancing redox equations has been under considerable discussion in the Journal lately (1–5). However, as was shown by Hart (3), chemical knowledge is not necessary for properly balancing chemical equations. Therefore, in the process of teaching redox reactions, the importance of proper balancing should not be overemphasized. Instead, it is worth concentrating on the properties of the reagents. Students should be able to predict the products for given substrates and next to balance the equation. Some examples, which represent a middle level of difficulty, are listed below (students know only the left-hand side of an equation). a. KMnO4 + KI + H 2SO4 → [MnSO4 + K 2SO4 + I2 + H2O] b. KMnO4 + Na2SO3 + H2O → [MnO2 + Na2SO 4 + KOH] c. KMnO4 + NaNO 2 + KOH → [K2MnO4 + NaNO3 + H2O] d. CuS + HNO3 (concd) → [CuSO 4 + NO2 + H2O]
Journal of Chemical Education • Vol. 74 No. 11 November 1997
Chemical Education Today
e. Cu + HNO3 (dil) → [Cu(NO3)2 + NO + H2O] f. Na2S2O3 + I2 → [Na2S4O6 + NaI] g. NaBr + NaBrO3 + H2SO 4 → [Na2SO 4 + Br2 + H2O] h. K2Cr2O7 + FeSO4 + H2SO4 → [Cr2(SO4)3 + Fe2(SO4)3 + K2SO 4 + H2O] i. SnCl2 + K2Cr2O7 + HCl → [SnCl4 + CrCl3 + KCl + H2O] j. KMnO4 + HCl → [MnCl2 + Cl2 + KCl + H2O] k. KMnO4 + H2O2 + H2SO4 → [MnSO4 + K2SO4 + O2 + H2O] l. H2O2 + KI → [I2 + KOH]
an example when an oxidant and a reductant give the same product. The oxidative property of dichromate is shown in reaction h. Reactions i and j indicate that in contrast to dichromate, permanganate can oxidize Cl{ to Cl2 and therefore, in the latter case, HCl cannot be used as a reaction medium. Oxidative and reductive properties of hydrogen peroxide are visualized by reactions k and l, respectively. Reaction m shows the oxidative properties of chlorates and reaction n shows the oxidative properties of nitrate, which are not always recognized by students. Finally, reaction o is an example of the dismutation reaction. Literature Cited
m. KI + KClO3 + H2SO4 → [I2 + KCl + K2SO4 + H2O] n. KNO2 + KI + H2SO4 → [NO + I2 + K2SO4 + H2O] o. Cl2 + KOH (hot) → [KClO3 + KCl + H2O] The solution of such formulated problems gives the chemical sense of the reactions and enables students to learn more about redox properties of chemical compounds. Thus, the first three reactions show the oxidative properties of permanganate in acidic, neutral, and basic media, respectively. Reactions d and e indicate the properties of nitric acid according to its concentration. Reaction f is frequently used in analytical chemistry, whereas reaction g is
1. 2. 3. 4. 5.
Stout, R. J. Chem. Educ. 1995, 72, 1125. Ludwig, O. D. J. Chem. Educ. 1996, 73, 507. Hart, D. M. J. Chem. Educ. 1996, 73, A226. Zaugg, N. S. J. Chem. Educ. 1996, 73, A226. Stout, R. J. Chem. Educ. 1996, 73, A227.
Andrzej Sobkowiak Faculty of Chemistry Rzeszów University of Technology 35-959 Rzeszów, Poland continued on page 1270
Vol. 74 No. 11 November 1997 • Journal of Chemical Education
1257
Chemical Education Today
Letters continued from page 1257 On Balancing “Redox Challenges” by Unconventional Oxidation Numbers Oliver G. Ludwig in his recent article (1) balances Stout’s “redox challenges” (2) by unconventional oxidation numbers. I agree that it is pedagogically useful to show to the students that the oxidation numbers are conventional tools for interpretation and balancing redox reactions. However, in my opinion, balancing chemical equations by nonconventional oxidation numbers is permitted in case of need only. It may be a powerful method for balancing complicated chemical equations such as double disproportionations or double redox reactions. For example, the equation P2 I4 + P4 + H2 O → PH4I + H 3PO4 published in this Journal (3), is a really serious “redox challenge” for balancing by conventional oxidation numbers (4– 6). Introducing unusual oxidation numbers we can convert this double disproportionation into a simple redox reaction as follows: Taking the oxidation number of P in P 2I4 , and PH4I equal to its conventional value in H3PO4 (P: +5), we get only P is oxidized in the reaction. If we select iodine as the other redox element among those kinds of atoms that occur in only one substance on each side of the equation (I, O) and assign H and O their usual oxidation numbers of +1, and –2, respectively, we get unconventional oxidation number of –2, 5 for I in P2 I4, and –9 in PH4I. Thus, the P is oxidized from 0 to +5, and I is reduced from –2, 5 to –9, giving stoichiometric coefficients of 6, 5/4, 5/4 and 5 for P4, P 2I4, and PH4I, respectively. After balancing in P, then O atoms, and multiplying through by 8, we obtain the balanced equation: 10P2 I4 + 13P4 +128H 2O = 40PH4 I + 32H3 PO4 Literature Cited 1. 2. 3. 4. 5.
Ludwig, O. G. J. Chem. Educ. 1996, 73, 507. Stout, R. J. Chem. Educ. 1995, 72, 1125. Carrano, S. A. J. Chem. Educ. 1978, 55, 382. Kolb, D. J. Chem. Educ. 1979, 56, 181. Cardinali, M. E., Giomini, C., Marrsu, G. J. Chem. Educ. 1995, 72, 716. 6. Cardinali, M. E.; Giomini, C.; Marrosu, G. Educ. Chem. 1996, 51.
Zoltán Tóth Lajos Kossuth University H-4010 Debrecen, P. O. Box 66 Hungary Unconventional oxidation numbers If oxidation numbers (ON’s) merely serve as bookkeeping numbers for balancing equations, then Ludwig’s method (J. Chem. Educ. 1996, 73, 507) would be acceptable. However ON’s have a chemical reality in that they can show, albeit in an exaggerated fashion, the direction of charge distribution and its change in a reaction. The claim, for example, that reversing the electron distribution in the cyanide ion from II {III |C ( ≡ N|
1270
to
{IV III |C ≡ ) N|
is chemically reasonable is not supported by any evidence (see J. Chem. Educ. 1988, 65, 45 for conventions used). In fact semiempirical or ab initio calculations show that nitrogen is the negative partner, as does the experimentally derived electron distribution obtained from the X-ray structure of LiCN. (N varies from 0.046e more negative than C on PM3 to 0.295e on 6.13G**). Students are likely to be more confused than enlightened when cations can have variable negative and sometimes nonintegral ON’s. They need a systematic procedure to generate consistent integral numbers linked to concepts such as electronegativities and Lewis structures taught in their courses. Also, Ludwig’s method is hardly a great time saver over conventional balancing. For example, in dealing with the third equation he has to assign conventional ON’s to all but one of the molecules. The conventional numbers required in the remaining molecule are shown.
[( III
IV
Cr O
C
NH2 -III
NH2
)] [
II
II
Cr ( C
6
4
-III
]
N )6
3
The increase in overall oxidation number has to be balanced by the 5-electron reduction Mn(VII) → Mn(II). Thus 558/5 permanganates are required to oxidize each molecule of the chromium complex. Mean ON's in reactant
ON's in products
No. of atoms
Increase in ON 24
Cr
18/7
6
7
C
132/42
4
42
36
N
{3
5
66
528
Interestingly in the second equation Ludwig considers the CN group on the thiocyanate and cyanide ions to be unchanged. In terms of conventional ON’s this is only true if the former is formulated as 0 II {III — — S| = C (= N — — since the electronegativities of C and S are so close; otherwise the ON’s of C and S both vary and only N remains constant. A.A. Woolf Faculty of Applied Sciences University of the West of England Frenchay Campus Bristol BS16 1QY, UK
The author replies: A. A. Woolf takes exception to my note on balancing redox equations (J. Chem. Educ. 1996, 73, 507), remarking that oxidation numbers do have a “chemical reality” in the CN– ion, for example. As Woolf mentions, Gaussian orbital and perturbation calculations do show that the N end of the ion is the more negative, but hardly by a full electronic charge. My comment was that further in-
Journal of Chemical Education • Vol. 74 No. 11 November 1997
Chemical Education Today
formation than would be available to a freshman learning to balance equations would be needed to lead to a preference in oxidation numbers. The beginning student has seen N in the {3 and +3 oxidation states, and carbon in the +2 and {4, but ordinarily would not have met the concepts of either Lewis structures or electronegativity so early in the course. I consider it important that oxidation number not be confused with charge; only the latter has physical reality. (Surely the Mn atom in MnO4{ does not have a charge of +7; Coulomb’s law would certainly not permit such charge separation.) Students would get even more confused later in the course when they see formal charges if they were to consider them to be real as well. Woolf rejects my use of nonintegral oxidation numbers yet uses mean oxidation numbers, which are nonintegral. I fail to see a difference. I do mention using oxidation numbers as tools, irrespective of whether they make “chemical sense” or not. Similarly, almost everyone speaks of electrons in orbitals, although, strictly, the orbital concept fails for more than one electron system. Later in the course, when the sophistication of the student has increased, we can talk about models versus reality. By the way, I use the ion–electron/half-reaction method when I teach balancing redox equations. I consider it closer to reality. Oliver Ludwig Department of Chemistry Villanova University Villanova, PA 19085
obtained in this way is actually correct, and that the others can somehow be “negated”. The fact is that all of the equations that he presents are correctly balanced, and that none can be negated on this basis. He suggests that there is a constraint beyond the balancing of atoms—that is, the balancing of the transfer of electrons. However, if charge is balanced (it is in all these cases), and if each of the atom types in the equation is correctly balanced (they are), then the number of electrons in the equation is trivially balanced. Based on the mathematics of equation balancing, none of these equations is more correct than any of the others. There is frequently more than one correct answer to any given question, and often new insights into a problem come from the “unconventional” answer. To answer the question of what ratios of reactant and product masses are actually obtained requires experimental data that are not part of the algebraic equation-balancing theory. Wade A. Freeman Department of Chemistry University of Illinois at Chicago 845 W. Taylor St. Chicago, IL 60607 N. K. Goh and L. S. Chia Division of Chemistry National Institute of Education Nanyang Technological University 469 Bukit Timah Road Singapore 259756 David M. Hart Department of Chemistry University of Central Oklahoma Edmond, OK 73034-0177
How Do I Balance Thee?…Let Me Count the Ways! We read with interest “How Do I Balance Thee?…Let Me Count the Ways!” (J. Chem. Educ. 1996, 73, 1129). The author points out an interesting dilemma in balancing equations, namely, situations in which the number of variables (in this case, the number of coefficients of the species involved) is more than the number of independent constraints (in most cases, the number of elements to be balanced) plus one. This situation has been described previously in this Journal (1991, 68, 984; 1994, 71, 490; 1995, 72, 894). It results in an infinite number of algebraically balanced equations, each of which is a combination of two equations. One such pair of equations is 4 HCl + ClO 2 → 2 H2O + 2.5 Cl2 2 KClO3 + 1.5 Cl2 → 2 KCl + 3 ClO2 Ferguson correctly points out that the number of degrees of freedom in choosing a balanced equation for this system is two, and that one way to choose these degrees of freedom is to set the coefficients of KClO3 and ClO2. From these, all other coefficients are uniquely determined. [Note: There were two misprints in the table of coefficients in Ferguson’s paper. Line 3 should have coefficients 4, 16, 4, 8, 7, 2 and line 5 should have 5, 22, 5, 11, 10, 2.] However, in his final paragraph, he goes one step too far and incorrectly implies that only one of the equations
Eric A. Lucas 219 Altamont Place Somerville, NJ 08876 David J. Peery 4978 Wayland Ave. San Jose, CA 95118 R. Subramaniam Singapore Science Centre Science Centre Road Singapore 609081 Marten J. ten Hoor J. W. Frisolaan 40 9602 GJ Hoogezand The Netherlands Sidney Toby Department of Chemistry Rutgers University P. O. Box 939 Piscataway, NJ 08855-0939 Richard S. Treptow Chicago State University Chicago, IL 60628
Vol. 74 No. 11 November 1997 • Journal of Chemical Education
1271
