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JOURNAL OF CHEMICAL EDUCATION
MAY.1930
THE STRUCTURE OF MATTER: BALANCING OXIDATION-REDUCTION EQUATIONS My dear Mr. Reinmuth: I have just been reading your series of articles on "The Structure of Matter"' and am trying the electron story with my pupils. But I am puzzled how to apply it in oxidation reductions involving radicals. Would you say that sulfur has a valence of four in SOa--? Or that chlorine has a valence of three in C103-? (Calculating each shared elec' THISJOURNAL. 5, 1152-63, 1312-20, 1473-9, 163946 (1928); 6, 117-22. 341-8. 527-35 (1929).
VOL. 7, No. 5
CORRESPONDENCE
1181
tron pair as a valence bond.) Wouldn't most non-metals have a valence of four? I have just finished your article in the March, 1929, JOURNAL.^ Thought I saw light, but apparently not enough, for am stuck on the old favorite: 8HN03+ 3Cu(N03)% 4He0 2N0. 3Cu How do you decide the nnmber of electrons gained or lost in these 4 0 ? Will you explain cases? Why not write: MnOi- -+ Mn++ in detail some of these equations and how electrons are of use in writing G. A. C. them ?
+
+
+
+
--
Dear Sir: Your letter of the 21st is a t hand and I shall attempt to offer you what suggestions are possible within the rather limited compass of a letter. The matter of valence is always puzzling, particularly in view of the fact that there is no universally accepted definition of valence. I have discussed this problem a t some length in the latter half of the third article of the series on "The Stricture of Matter." See pp. 1476-9, November, 1928. (Incidentally, the first word in the second line on p. 1479 should be tetravalent rather than tewalent.) Personally I find it best to avoid attempting to ascribe valences to the individual atoms in an ion. The ion as a whole has a valence, which is easily determined by counting the nnmber of electrons which the atoms would have in the uncombined state and cvmparing i t with the total number of electrons in the ion3 Thus the SOa ion has two more electrons than an atom of sulfur and four atoms of oxygen would have; consequently its valence is -2.
On the other hand, the ammonium ion has one less electron than an atom of nitrogen and four atoms of hydrogen; its valence is 4-1.
When the ion goes through a reaction unchanged no complications 2
%IS JOURNAL,
6, 52735 (Mar., 1929).
This point of view leads to two definitions of valence. In non-ionic compounds the valence of an atom is the nnmber of electron pairs shared with other atoms. The valence of an ion is its electrical charge. As applied to individual atoms in ions, the term is meaningless. 8
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JOURNAL OF CHEMICAL EDUCATION
MAY.1930
'
arise. When the ion takes part in an oxidation reduction it is still unnecessary to ascribe valences to individual atoms. The procedure I have found most satisfactory is that described in the article to which you refer.% That is, to write equations which include the initial reactants and the final products and add the number of electrons necessary to balance the equation electrically. The equations on p. 534, March, 1929, illustrate the procedure for K M ~ O and P Na2SO3. The second pair might be written as follows:
A count of the electrons on each side of the respective equations above makes it obvious why the balancing must take place as indicated in the article. 40. The There are two objections to writing MnOn- +Mn++ first is that KMnOa does not yield "nascent" (or atomic) oxygen. The second objection is that the equation as written is not balanced elec-
+
To balance, it would be necessary to add 3 electrons to the right-hand side of the equation, which would seem to indicate that the permanganate ion has a reducing power of 3, rather than an oxidizing power of 5 , as it actually has. I believe that the best way to determine the number of electrons gained or lost is to diagram the equations as I have done here and to count the number of electrons necessary to balance. It cannot he too strongly emphasized, however, that one must first know the actual reactants and the reaction products before one can balance equations by any system. The electronic system is no exception in this respect. It is well to keep in mind the fundamental principles that oxidation represents a loss of electrons, reduction a gain of electrons; an oxidizing agent must he reduced and hence must operate, either by itself or in cooperation with some other agent, as an acceptor of electrons; a reducing agent must give up electrons. Paradoxical as it may seem, the permanganate ion acts in conjunction with the hydrogen ion as an electron
VOL. 7. NO. 5
CORRESPONDENCE
+
1183
+
8H+ 5 e +Mn++ i4Hz0,takes "sink." The equation, MnOpaccount of that fact. It is, then, a general rule that in any properly written and properly balanced partial electronic equation, the electrons added for the purpose of balancing must appear as positive quantities on the same side of the equation as the oxidizing agent. When placed on the same side of the equation as the reducing agent, they must appear as negative quantities. As further examples, let us consider the respective reactions of dilute and concentrated nitric acid with metallic copper. The nitric ion is capable of undergoing a number of different reductions depending upon the nature of the other reactants present, the concentrations of the reactants, the temperature, etc. . The reduction of the nitric ion in concentrated nitric acid by metals takes the following course: NOz-
+ 2H+ + 1. +NO2 + H?O
(1)
This may, of course, be written:
The oxidation of copper under these conditions takes the following course: Cu" - 2.+cut+
*
(2)
which may be written: [:Cul - 2r
+ [Cult'
Equations (1) and ( 2 ) above are therefore our electronically balanced, partial equations for the reduction of NOa- ion and the oxidation of Cu0, respectively. Since equation ( 2 ) involves twice as many electrons as equation (1) we must multiply equation (1) by two before adding the partial equations together.
The sum, equation (3), is a completely balanced ionic equation for the actual reactants. If we desire to convert this into a molecular equation i t is necessary to add to each side of the equation the two Nos- ions which go through the reaction unchanged. 2N01-
Cu"
--t
2NOa-
+ 4N01- + 4H+ +Cu++ + 2NO.- + 2N02 + 2H20
(4)
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JOURNAL OF CHEMICAL EDUCATION
MAY.1930
Equation (4) is easily assembled into the familiar molecular equation (5). Cu
+ 4HNOs -+
Cu(N0s)n
+ 2N02 + 2H10
(5)
When copper reacts with dilute nitric acid the oxidation of copper takes the same course as before, but the reduction of nitric ion proceeds as follows: NOa-
[ ::o:N:o: :'d:::]-+
+ 4H+ + 3r +NO + 2 8 0 ,[HI+
+ 3.-
(1.4)
:ir:6: .. + z [ H : ~ : H ]
Multiplying the partial equations by appropriate factors and adding, we arrive a t equation (3A) as follows:
To convert to the molecular equation we must add six NOa- ions to each side of the equation.
You will find a number of other specific examples in the two articles by E. R. Jette and V. K. LaMer, THIS JOURNAL, 4, 1021-30 (Aug., 1927); 115&67 (Sept., 1927). OTTO REINMUTH,Associate Editor
