NOTES ON THE ELECTRON METHOD OF TEACHING OXIDATION AND REDUCTION L. J. WALDBAUER AND W. E. Tmm, LEHIOH UNIVERSITY, BETHLEHEM, PA. The use of the electron concept in teaching the nature of oxidation and reduction, and in the balancing of equations is the only rational scheme. The old-fashioned step equation method involves absurdities and unwarranted assumptions, while the positive and negative valence method, though algebraically effiaent, does not present a clear view of the nature of oxidation and reduction. Professor Brinkleyl has presented the electronic conception in a manner that must please those desirous of giving a fundamentally correct idea of the subject to their classes. The writers have used a method similar to Professor Brinkley's in their elementary classes, and offer the following ideas as a supplement to his admirable outline. Before discussing the method of balancing equations, we have found i t of advantage to give type compounds for every possible state of oxidation of an atom, e. g., nitrog-en. Gained 3c NHa Lost 1. N2O (HNO) B. Carbon Gained 4r
Gained 2r NHlNHn Last 2r NO
Gained 3r Lost l r Net gain 20
Gained 2r 16 Lost Net gain 1. N&OH Lost 3e N10s HNOn Gained 2r Lost 2 c Net -
CH'
Neither gained nor lost C
Neither gained nor
lost r
Gained l r Lost 3t Net loss 2r OH H-C& Lost 26
Lost 4t
C. Sulfur Gained 2 c HzS
Neither gained nor last S
Lost 2t
Lost 4r
SO Gained 2r Lost 4s Net loss 2.
Son Gained Is Lost 5c Net lass 4s
"%o
H-o/
Lost 6r SO1
H
From this it may be seen that atomic carbon represents the same state of oxidation as is found in the carbon of formaldehyde, while CO represents 1 THISJOURNAL, 2, 576--87 (1925).
VOL. 3, No. 12 ELECTRON METHOD OP TEACHING OXIDATION m REDUCTION1431
the same state of oxidation as the carbon of formic acid. These facts are ordinarily not apparent. Balancing Equations We do not believe in the necessity of burdening students with a complicated system of balancing equations, such as that proposed by Griggs and Warner.% For the special cases, such as N%Oz and Na& oxidations, we use the following simple explanation: The NatOp molecule (or the NazSz molecule) as a whole can take up two electrons, which is a fact requiring no assumptions.
+
NaCrOz NaOs = NaaCrO, Cr bas lost 3r Cr has lost 6r N a 0 2 can take up 26. The least common multiple of 2 and 3 is 6, and therefore 4NaOH 2NaCr0~ 3Nan02 2HzO = 2NaL!rO,
+
+
+
Water may be used on either side of the equations to balance them. The reaction between KMnOa and HCl also presents some difficulties.
+
+
+
+
KMnOl HCI HCI = CIS KC1 MnClz. Mn gains 5t in going from KMn04 to MnCI2. C1 loses it in going from HCI to Ch. The L.C.M. is 5, but since chlorine is molecular. 10 must be used. Therefore, 10 HC1 molecules are oxidized and 2KMn04molecules must be reduced. 2KMnOn lOHCl 6HC1 = 5CL 2MnCL 2KC1 8H20. Obviously 6 HCI molecules are required to furnish anions for the K and Mn.
+
+
+
+
+
Balancing Equations Involving Compounds in Which One Kind of Atom Exists in Different States of Oxidation HNOI
+ Fe304= NO + Fe(NO&.
In Fea04,the Fe atoms have lost 8 electrons; the 3 Fe atoms can lose a maximum of 96. Therefore, one FeaOa molecule can lose one electron. The N in HNOa has lost 56, while that in NO has lost 2e; therefore, there has been a net gain of 3 in going from HN0a to NO, and we can write: HNOI
+ 3Fea01 + 27HNOa = NO + 9Fe(NO& + 28H10.
The 27 molecules of HN08 are required to furnish anions for the Fe. NanS108
+ Iz
=
NaI
+ NasS40e.
Each iodine atom can gain an electron, and in sodium iodide this has taken place. Two molecules of sodium thiosulfate can lose two electrons. Hence we can write: 2NazSsOs
+
12
=
2 NaI
+ NazS40s.
The old-fashioned structural formulas are frequently of great assistance in showing the electronic changes. 3,425--31 (1926). THISJOURNAL,
