The Chemical Equation Part II: Oxidation-Reduction Reactions Doris Kolb Illinois Central College
middle 1600's with the work of Robert Boyle. He found that when air was removed most materials would not hurn. On the other hand, gunpowder would hurn even when placed under water, which led him to believe that the saltpeter ( K N 0 3 in gunpowder could serve as a substitute for air. Whereas wood and other organic materials hurned away to ashes, Boyle noted that metals gained weight when they burned in air (a process referred since calx). ~ - to as calcination. ~ ~ the ash ~ was called ~ ~ Hoyle's assistant. ~ o h e i Houke, t also observed that air was necessnrv for hurnine. It was his belief that some "sulfureous" s&tanEe had to he present in combustible material and that the air acted as a solvent for this sulfureous material, the combination of the two producing heat. His idea was not completely new. Two thousand years earlier Plato had sug.gested that everything burnable must contain some flammable suhstance; and the alchemists had long held some vague ideas relating fire to the presence of sulfur, because of its combustildity. John Mayow shared Hwke's helief in the sulfweouu nature of combustible materials. He also published a book in 1674 discussing the importance of air in comhustion. Mayow believed that there was something present in air (and also in saltpeter) which was necessaty for comhustion and which he referred to as spiritus nitro-aerius. I t is interesting that the discovery of oxygen did not come until a full century later. One reason for the long delay must have been the wide acceptance of theorv.. which was introduced about this .-the ~ - ohloeiston ~ same time and which explained combustion in a quite different wav. dohann'~oachirnBecher considered the basic elemenw of the uni\.erse to be three "earths", one fluid (terra mercurialis). one vitreous (terra lapidea), and one fatty (terra pinguis). According to Becher the fatty earth, terra pinguis, was a part of all living things, and he considered it to he the combustible substance that was lost by flammahle materials when they hurned. Becher's idea was expanded by Georg Ernst Stahl, who replaced the terrapinguis term with the word "phlogiston" (from the Greek phlogistos, meaning flammable). The hurning process, according to Stahl's theory, involved the loss of this mysterious suhstance phlogiston from a flammable material.
If chlorination means "reaction with chlorine", then why lesn't oxidation mean "reaction with oxygen"? Well it does, course. Over the years, however, oxidation has come to have much broader meaning, one that need not involve oxygen all. When calcium undergoes oxidation, the product is calcium :ide.
~~
2CatOz-ZCaO hen the CaO is written in the form of ions, 2CaCOz-2CaZ++ZO2becomes more obvious that the calcium atoms are losing sctrons and taking on a positive charge, while the oxygen oms are picking up electrons and becoming negatively targed. The reaction is a simple electron transfer. Calcium giving up electrons, and oxygen is accepting them. When calcium reacts with chlorine,
-
Ca t Clz CaZ+t 2 C1le reaction is properly referred to as chlorination. As far as le calcium atom is concerned, however, this reaction is no fferent than the reaction with oxygen. The calcium atom is sine " two electrons. If calcium is being oxidized in the first se, then why not describe it as being oxidized in the second tse as well? The onlv diflerence ii that the oxidizing agent the second reaction is chlorine rather than oxygen. Oxidation is an extremely common and also a highly imxtant kind of chemical reaction. Probably the most familiar r m of oxidation is combustion (burning), a very rapid oxitio on process that occurs with much evolution of heat and ght. I t is a chemical reaction much older than recorded hisfry. re Nature of Combustion Perhaps the greatest discovery made by prehistoric man was lat of fire and how to use it. The ancients placed great value I fire: manv even worshi~ved .. it. Always they were fascinated y it, hut they could not explain it. The scientific study of fire seems to have begun during the ~
"The Chemieal Equation" is I: Simple Reactions" is part of a series of substantive reviews of chemical principles taught first in high school chemistry courses. Dr. Kolh received a BS degree from the University of Louisville and both MS and PhD degrees from The Ohio State University. She has been employed as a chemist at the Standard Oil Company and as a television lecturer in a series "Sootlieht on Research." She hna .. a&ve$on staffs -~ ~ ~ ~ --. ~ the ~-~~ ~ of ~Corning Community College and Bradley University. Since 1967, she has been Professor of Chemistry at Illinois Central College. ~~~
Doris Kolb Illinois Central College East Peoria, Illinois 61635
326 / Journal of Chemical Education
A
~
~
~
~
~~~
~
. ~
~
-
~
-
Wood Ash (Calx)+ Phlogiston f When wwd hurned, it lost weight because of its loss of phlogiston; hut when phosphorus or metals hurned, their gain in weight was explained by the fact that their phlogiston had negative weight. (Its loss i:$erefore resulted in a weight increase.) Whenever something hurned, i t was said to he "dephlogisticated". When metal ores were smelted, on the other hand, they were "phlogisticated".
-
~
~
Ore (Calx) + Phlogiston Metal The phlogiston theory originated during the latter part of the ~ ~ 17th century, and by the middle of the 18th century it had achieved almost complete acceptance, in spite of the fact that it was altogether in error. The element oxygen was discovered independently by Karl Wilhelm Scheele and Joseph Priestley (1774). Scheele's work was actually done before Priestley's, hut i t was not published
~
until 1777, so the credit for the discovery went to Priestley. In soite of their im~ortantdiscovew. neither of these men ever re& at~andonedhis belief in the phi~gist*,ntheory. he name Priestlev ea\,e to the eas he had disrwcred was "de~hloris. ticated ai?. It was Antoine Lavoisier who first realized the significance of Priestley's new gas (to which Lavoisier gave the name "oxygen"). He became convinced that the burning of flammable materials was really just a combination of these suhstances with oxygen. He showed that when anything was burned, and all the products were recovered and weighed, the total weight of the products was always greater than the weight of the material burned, and the increase in weight was exactly equal to the weight of the oxygen used up during the hurnine nrocess. He sueeested the term "oxidation" (combination with oxygen) to replace dephlogistication and "reduction" (removal of oxygen) as a substitute for phlogistication. Not only did Lavoisier look upon combustion as a combination with oxveen. the calcination .- . hut he also recoenized " of metals and the respiration of animals to be oxidation reactions.
--
e.
Definitions of OxIdalion and Reduction The term oxidation, as first used by Lavoisier, was limited to reaction with oxveen.. and the term reduction meant s i m.~.l v the removal of oxygen. Later it was recognized that oxidation could he brought about hv certain comoounds of oxveen (such as nitric acid7 as well as-by oxygen itself, but its-iefinition remained linked to oxygen. The basic meanings of the terms oxidation and reduction did not reallv change a~oreciahlvuntil well into the 20th century. ~ f t & the disco%y of the electron (1897) and the introduction of Bohr's model of the atom (19131, chemical reactions were gradually reinterpreted in terms of electron activitv. In view of the fundamental similarities of many electron transfer reactions, the term oxidation began to take on more generalized meaning as a reaction type. Oxidation, in the broader sense of the term, came to mean a loss of electrons, and reduction a gain of electrons. Defining oxidation and reduction as electron loss or gain eliminates the need for any oxygen involvement at all. 0 -
t Clz Ca element oxidized (oxidizing agent) (reducingagent) element reduced)
-
Ca2+t 2 CI-
The oxidizing agent simply accepts electrons, and undergoes reduction in the process, while the reducing agent, which gives up electrons, must itself be oxidized. Thus oxidation and reduction are complementary processes. They must always occur simultaneously and in precisely the same amounts. Electrons must he given up by one substance in order that they may he accepted by another. So inseparable are the processes of reduction and oxidation that the combination word redox is often used to refer to reactions in which they are occurring. Defining oxidation as electron loss and reduction as electron gain implies that all redox reactions involve electron transfer, but this is not strictly true. When carbon is oxidized, its valence electrons are not transferred to oxygen. ctoz-CO2 In C o n the bonds are covalent and valence electrons are shared between the carbon and oxveen atoms. Since oxygen is more electronegative, however, theshared electrons shodd he closer to oxveen than to carbon, on the average. We might say that there"Gas been a partial loss of 4 electrons by the carbon atom and apartial gain of 2 electrons by each oxygen atnm
Those who object to the incongruity of defining reduction as again of something, albeit something negative, should find these definitions much more satisfactory:
Oxidation is increase in oxidation numher. Reduction is decrease in oxidation numher These definitions are easy to learn, and they are rarely confused. since reduction is loeicallv defined in terms of decreasing number. They are simple tb apply to all kinds of redox reactions. Thev.do..however. assume a knowledee of what is meant by "oxidation number". Oxldatlon Number (Oxidation State) The terms oxidation number and oxidation state mean essentially the same thing and are often used interchangeably. They refer to the degree of oxidation of a given element in a particular substance. The word valence has been used in the same way, but valence has a variety of meanings (combining capacity, ionic charge, covalence, coordination numher) and lacks the snecificitv of oxidation number. Oxidation numbers are positive or negative numbers assiened to individual atoms in substances for ourooses of ecctron "bookkeeping". Although the assignment is somewhat arbitrary, it provides a consistent method for determining what elements have been oxidized or reduced and deciding how many electrons per atom have been transferred (completely or partially). Oxidation numbers are especially useful in balancing redox equations. For simple ionic compounds, made up of monatomic ions, oxidation numbers are the same as ionic charges. t 2 -2 +2 -1(2) +I -1 Na C1 Ca 0 Ba Fz In the case of covalently bonded atoms, the oxidation numbers are imaginary charges the atoms would have if all shared electrons were given over completely to the more electronegative partner in each case. Let us take nitrous acid, for example, and mentally assign all shared electrons to the oxygen atoms.
The charges written above the atoms reflect the number of valence electrons that would be gained or lost. These fictitious rharges are oxidation numbers. The rules for determiningoxidation number may belisted as follows: 1) For any free (uncombined)element the oxidation numher = 0. 2) For any neutral compound, the sum of the oxidation numbers = 0.
3) For any monatomic ion, the oxidation numher = the ionic charge. 4) For any polyatomic ion, the sum of the oxidation numbers = the ionic charge. 5) Far hydrogen, the oxidation numher in compounds is +1 (except in metal hydrides, in which it is -1). 6) For oxygen, the oxidation number in compounds is -2 (exceptin peroxides, in which it is -1). (In OF2 oxygen has an oxidation numher of t 2 . since fluorine is the single element more electronegative than oxygen.) 71 In rhtir compounds Group IA elementsare always +I, Group IIA elements arc nlnn).s +2, and Group llIA elements are almost always +3. In order to determine the oxidation numher of chlorine in KC103, of sulfur in HzS04, or of carbon in C0$ ion, first consider the oxidation numbers of the other elements (+1 for potassium, -2 for oxygen, +1 for hydrogen). + l ? -2(3) (2)+1 ? -2(4) ? 4 3 ) K ~ 1 0 3 ' Hz SO4 C 0z2In the case of KClO3, +1 and -6 total -5, so C1 must be +5 if the sum is to he 0. For H 8 0 a +2 and -8 give -6, so S must be +6. In the carbonate ion -2 X 3 = -6, hut since the total ionic charge is -2, a charge of +4 must be assigned to carbon. 8
Volume 55, Number 5, May 1978 / 327
sometimes the oxidation state can turn out to be a fractional lumber, as in the caw of Na,Sr06, in which sulfur hasanoxdation numher of 2 112. In trying to identify elements being oxidized or reduced, limply look for any changes in oxidation number. When MnOz ,s converted to MnOa- ions, the oxidation numher of manpnese goes from +4 to +I. Its oxidation number is increasing, lo Mn is heing oxidized. When NaHSOa is changed to elenentary sulfur, the oxidation numher of sulfur goes from +4 to 0; and since i t is decreasing, S is heing reduced. Whenever 1 free element participates in a reaction, either as reactant or product, i t is almost sure to be undergoing oxidation or rehction. Writing Redox Equations The general rules for writing a chemical equation are the lame for all reactions regardless of type. First: make sure you hnow what the reactants and products are. Next: use chemical symbolism to list the reactants on the left of the arrow and produds on the right. Finally: after checking each formula for :orrectness. balance the eauation. Space does not permit a discussion here of how products can be oredicted in comolex redox reactions, since that requires :oisiderahle knowlLdge of the elements in their various oxi*ation states olus an understandinn- of electrode potentials. However, a few generalities might be mentioned: 1) Substances that are most easily reduced are strong oxidizing wents, and those that are most easily oxidized are strong reducing . ~gents. 2) Free elements of high electronegativity (e.g. Fz, 02, Cln) are :xcellent oxidizing agents, while those of low electronegativity (e.g. Na, K, Ba) are excellent reducing agents. (The element nitrogen is 1 notable exception to this rule.) 3) Elements of multiple oxidation state are found mainly among the transition metals and the nonmetals. In most cases the maximum oxidation numher an element can exhibit is the same as its group number in the periodic table. Nitrogen, for instance, can exist in 9 different oxidation states, from +5 (its group numher) td -3 (the numher of electrons needed to complete its valence shell octet). 4) Elements in their highest oxidation state can usually act as oxidizing agents (e.g. Fe3+,NOa-, Mn04-), although they need not he good ones (Nat and P O P , for example, are difficultto reduce);elements in their lowest oxidation state- usually act asreducingagents (e.g., H-, N3-, Mg), although they are sometimes poor ones (02-and Pt, for example, are not easily oxidized); in any of their middle oxidation states elements can he either oxidized or reduced (e.g. Cu', NOI-, S0s2-).
Ordinarily students in introductory courses are provided with complete information as to what the reactants and products are in complex redox reactions, so their problem is primarily one of balancing the equation. Balancing Redox Equations Some redox equations are already balanced and need no added coefficients. s+o2-SO2 Zn + CuSOa Cu + ZnSOa Others are not self-balancing, hut they are easily balanced by inspection.
-
CHa+20z-COa+2HzO 2Al+3Br~-2AlBrs Many redox equations are more complicated, however, and we not readily balanced by the usual method of trial and error. For these equations there are some special techniques that can greatly simplify the balancing process. The mast important thing to remember in balancing a redox equation is that the amounts of oxidation and reduction must be equal. The number of electrons gained hy the oxidizing 328 1 Journal of Chemical Education
agent must be exactly the same as the number lost by the reducing agent. Once the amounts of oxidation and reduction are made equal, any further balancing required can usually be done easily by inspection. The two most widely used methods for balancing redox equations are the oxidation number and ion-electron methods. Let us compare these methods by using each of them to balance the same equation:
-
+
S + HNO3 SO2 NO + Hz0 Although this is a relatively simple redox reaction, i t is not easily balanced using the ordinary trial and error technique.
Oxidation Number Method The oxidation number method consists of determining what increases and decreases in oxidation number are occurring, and then makine them eaual. Given a choice of several different methods fir balancing redox equations, studenu almwt invariahlv orefer this one. 'l'hev find it fater. easier. and more consistentiy successful than oiher methods: Upon being given a redox equation to halance
-
S + HNOa SO2 + NO + Hz0 we must first look for any elements that are undergoing change in oxidation number. Sulfur appears on the left as the free element, so we know it is being either oxidized or reduced. Since it isaoinn from oxidationstate0 insulfur to t4 in SO?, its oxidation number is increasing, and therefore sulfur is being oxidized. We can indicate this by means of a bridge, as follows: +4 O n + 4 S + HN03 SO2 + NO + Hz0 We can rule out hydrogen and oxygen as being oxidized or reduced, since they occur only in compounds (in which hydrogen is +1 and oxygen is -2). The remaining element, nitrogen, is in a +5 oxidation state in HN08 and +2 in NO. Since its oxidation numher is decreasing by 3, nitrogen is being reduced. Let us indicate this change by the same technique we used before, but for convenience let us draw the nitrogen bridge below the equation.
-
0 1 + 4 S+HNO -SOz+NO+HzO +51 \+2 -3 We have now established that each sulfur atom is going up 4 in oxidation number while each nitrogen atom is going down 3. We can make both of these numbers eaual to 12 (thus equalizing the changes in oxidation numbe;) if we multiply the sulfur oxidation by 9 and the nitrogen reduction by 4. +4 (3) = +12
3 + ~ H N o ~ - ~ ~ o ~ + ~ N o + H ~ o
-3 (4) = -12 The numbers in parentheses tell us the ratio of sulfur to nitrogen atoms that must he involved in this reaction. Notice that these numbers have been inserted into the eauation as roeffirients'l'he equation is not yet completely halnnced, but t h hard ~ onrt has been accomdished. Now that the sulfur and nitrogenare balanced, let us check the other elements, the hydrogen and oxygen. The 4 hydrogen atoms on the left side of the equation tell us that we need a 2 in front of HzO. Not only did we balance the hydrogen with that last coefficient, but a check of the oxygen atoms shows that they are also balanced (12 on each side). There are other modifications of this method, but the rules
are basically the same. They may be summarized as follows: 1) Assign oxidation numbers to identify the elements oxidized and
reduced. 2) Determine the number of units of increase and decrease in oxi-
dation numher. 3) Make the increase and decrease in oxidation number equal by using appropriate coefficients. 4) After the oxidation and reduction are balanced, complete the balancing of the equation hy inspection.
+
-
+
Oxidation: 6 H20 3 S 3 SOz 12 HC+ 12 eReduction: 1 2 e- + 16 H+ + 4 Nos4 NO + 8 Hz0 The half-equations are now ready to be added, but first let us cancel out all those terms that appear on both sides of the arrow. (In this case 12 e-, 6 H20, and 12 H+ can all he omitted from both sides.) 0 x i d a t i o n : W + 3 S 3 SOz+ +%W+ e e = 4 2 :Reduction:* +, W H + + 4 NOa- 4 NO + 8 H 2 0
-
-
-
3S+4Hf+4NOs--3SOz+4NO+2Hz0 Ion Electron Method
The ion-electron method is also called the half-reaction method, since it involves writing individual "half-equations'' for the oxidation and reduction Drocesses. Usina the same equation that we used before, lei us now balance it by the ion-electron or half-reaction method. This method requires that we start with the ionic form of the equation. The only electrolyte in the reaction is HN03, so i t must be written as ions: S+H++NOs--SOz+NO+HzO Sulfur is oxidized in this reaction and nitrogen is reduced, so let us begin by writing the two half-reactions separately.
-
Oxidation: S SO2 Reduction: Nos- -NO T o balance the oxidation half-reaction, note that the sulfur atoms are already balanced, but there are 2 oxygen atoms on the right and none on the left. Oxygen atoms can he supplied on the left by adding 2 molecules of H20. Oxidation: 2 Hz0 + S SO2 The oxygen atoms are balanced, but now there are 4 hydrogen atoms on the left that must he balanced on the right. We balance the hydrogen by adding 4 H+ions to the right side.
-
-
Oxidation: 2 Hz0 + S SO2 + 4 H+ The half-reaction is now balanced with respect to atoms, but the charges are not balanced. There is no charge on the left, but there is a +4 charge on the right. The charges are balanced by adding electrons to one side or the other. In this case we need to add 4 electrons to the right side.
+
-
The sum of the two half-equations is the complete balanced equation, in ionic form. The rules for halancing redox equations by the ion-electron method may he summarized as follows: 1) Write the equation in ionic form, and separate it into two half-
reactions. 2) Balance each half-reaction separately, adding HEOwhere needed
to balance oxygen atoms and H+where needed to balance hydroeen atoms. (For alkaline solutions use OH- to halanee oxveen ." and ifn0to balance hydrogen.) 3) Add electrons to one side of each half-reactionso as to halance the charges. 4) Make the electron changes equal by multiplying one or both half-reactionsby appropriate numbers. 5) Add the two half-reactions together, eliminating all dnplications. Algebraic Method -
Other methods for halancing redox equations have included the anhvdride method, the hvdride-hvdroxide method, the partial equation method, the method of positive and negative bonds, and the algebraic method. Of these older methods, only the algebraic method will be discussed here. Students often find it interesting to use this purely mathematical approach to balancing redox equations. In using the algehraic method we first write the equation as follows: xS+yHNO,-zS02+pNO+qHxO The x , y, z, p, and q are arbitrary symbols used to represent the five unknown coefficients needed to balance the equation. Then in order to balance sulfur we set x equal to z , and each of the other elements is treated accordingly.
+
Oxidation: 2 Hz0 S SO2 4 H+ + 4 e- (balanced) The reduction step is treated similarly. The nitrogen atoms are alreadv balanced. but there are 3 oxvaen atoms on the left that musthe balanced on the right, so add 2 molecules of Hz0 to balance the oxygen.
-
Reduction: NOa- NO + 2 Hz0 Now the left side needs 4 hydrogen atoms, which we supply as H+ions.
-
+
+
NO 2 Hz0 Reduction: 4 H+ NO,The net charge on the left is +3, while there is no charge on the right. Addition of 3 electrons to the left side balances the charges.
-
Reduction: 3 e- + 4 H+ + NOS- NO + 2 Hz0 (balanced) The two balanced half-reactions are:
-
Oxidation: 2 Hz0 + S SO2 + 4 H+ + 4eReduction: 3ec + 4 H+ + Nos- -NO + 2 Hz0 Before we add them together, we must make the electron loss in the oxidation step equal to the electron gain in the reduction step. (The electron terms will then cancel each other when we add the half-reactions.) We can make the electron change equal 12 electrons in both cases if we multiply the oxidation step by 3 and the reduction step by 4.
(Sulfur) x = r (Nitrogen)y = p (Hydrogen)y = 2q (Oxygen)3y = 2z + p + q I t follows that: 3 y = 2 z + y + y / 2 or 3 y - y - y / 2 = 2 i 3y/2 = 22 and 3y = 42 If we let y = 4, then z = 3, x = 3, p = 4, and q = 2. Inserting these coefficients into the equation, we obtain the balanced chemical equation: ~S+~HNO~-~SOS+~NO+~H~O With the algehraic method it is not necessary to locate the elements being oxidized and reduced or to determine oxidation numhers, but the algebraic method does have one distinct drawback. Algebraic equations can have more than one solution, and sometimes several different sets of coefficients can be obtained for the same equation, although only one set represents the correct stoichiometric coefficients. Some Special Examples
Here are a few additional equations that represent some common problems that students have in balancing redox equations. Volume 55. Number 5, May 1978 / 329
1) Subscript Numbers. In the following equation the oxidation step involves an increase of 5 units in the oxidation number of iodine, hut the subscript in I2 tells us that there must he at least two iodine atoms in the equation, so we immediately add the parenthetical (2).
Let us first balance the changes in oxidation number
+5 (2) +5 HNOa 12 HI03 + NOz + Hz0 Since nitrogen goes down 1in oxidation numher, we equalize the changes by multiplying the nitrogen reduction step by 10.
+
-
+5 (2) = +10 p 7 + 5 10 HN03 + I2 2 HIOs + 10 NOz + Hz0 +4 +5 -1 (10) = -10 The numhers in parentheses are the numhers of iodine and nitrogen atoms needed in the equation. The hydrogen and 3xygen are balanced by inspection.
-
-
10 HN03 + I2 2 HI03 + 10 NOa + 4 Hz0 2) Dual Function Reagents. The following equation can :ause problems because the HC1 in this reaction has two different functions. It is a reducing agent, but it is also an acid mpplying chloride ions.
The 7 oxygen atoms on the left will require 7 H 2 0 molecules on the right, which in turn will need 14 H+ on the left to halance the hydrogen. Cr2072-+3S2-+14H+-2Cr3++3S+IH20
+
-
- -
Oxidation: Mn2+ MnO44 Hz0 + Mn2+ MnOl- 8 Hf 5eReduction: PbO2 Pb2+ 2e- + 4 HC PbOz Pb2++ 2 Hz0 Before adding the half-reactions together, we must multiply the oxidation step by 2 and the reduction step by 5 so that the electron terms will cancel out.
+
'Jotice that some chlorine ends up as MnCl2 or KC1 (not oxilized), but some ends up as Cl2 (oxidized). In balancing the :hanges in oxidation numher, the (2) in the oxidation step is ~eededbecause of the subscript 2 in Clz. The oxidation lumber changes are then +2 and -5, so we make them hoth !qua1 to 10. The numbers in parentheses tell us what the :oefficients must be to balance the oxidation and reduc.ion. +1(2) (5) = + l o ' 02 KMn04+ 10 HCI 2 MnC12 + 5 C12 + KC1 + Hz0 +2 +I -5 (2) = -10 rhe rest of the balancing is done by inspection. Since there Ire 2 atoms of potassium on the left, we need 2 KC1 on the .ight.
-
rhe Mn atoms are alreadv balanced (2 on each side), but let 1s now check the chlorine. (Hydrogen and oxvgen halancing ire u s ~ ~ a lhest l \ ~ left until last., The chlorine atoms are not danced. o n the left we have 10 chlorine atoms to balance the iC12on the right, hut there are 6 more chlorine atoms on the ,ight that were not involved in the oxidation. We need to add imore HCl molecules on the left, making a total of 10 6 or 16.
+
2 KMnOl
+ 16 HC1-
2 MnC12
+ 5 Clz + 2 KC1 + 8 Hz0
+
HCI can also he listed twice in the equation, i.e., 10 HCI 6 ICI.) The 8 Hz0 molecules on the right are needed to balance .he 16 hydrogen atoms on the left, and a check of the oxygen Itoms shows 8 on each side. 3) Ionic Equations. When equations are ionic, i t is imIortant that the charges he balanced as well as the atoms. The 'ollowing equation might at first appear to he balanced if we ;imply use 14 H+, 2 Cr3+,and 7 H20, hut the charges will not lalance. 330 / Journal of Chemical Education
+
Now let us check the charges: on the left -2 3(-2) 14(+1) = -2 -6 +14 = +6, andon the right 2(+3) = +6. The check of the charees on hoth sides confirms that the eauation is balanced. 4) Incom~leteEauations. Sometimes redox reactions are not given in complete form. Suppose we want to write a halanced equation for a reaction in which Mn2+ is oxidized to Mn04- and P h 0 2 is reduced to Pb2+. There may he other reactants or products (such as water), hut they are not given. The easiest method to use in this case is the ion-electron or half-reaction method.
Reduction: W e
+
4
~e H+
+
+
-
+ 5 Pb02
-
5 Pbz+
+ fe2 Hz0
The equation is now complete, and it is balanced with respect to charge as well as numhers of atoms. 5) Same Element Oxidized and Reduced. Sometimes there is only one element involved in the oxidation and reduction. When hypochlorous acid decomposes, for example, chlorine is hoth oxidized and reduced. +4 (1)= +4 1+1+5 3 HClO HClOs 2 HCI )+I 1-1 -2 (2) = -4 The coefficient 3 placed in front of HClO is the sum of 1(for the oxidation step) and 2 (for the reduction step). 6) More Than Two Redox Elements. In the following equation silver and arsenic are both heing reduced and zinc is heing oxidized.
-
+
-
+
Ag3As04+ Zn + H2SO4 AsH3 Ag + ZnSO4 + Hz0 Remember that the total increase in oxidation numher must equal the total decrease. +2 (11)= +22
0 2 Ag3As04+ 11Zn + HzSOl 1+11+5 -1 (3) (2) = -6 -8 (2) = -16
-
2 AsH3
1-3
+ 6 Ag +
+2 11ZnSOl + Hz0
0 -6 -16 = -22
The H2S04and H20 are balanced by inspection to give the completely balanced equation. 2 Ag&04
+ 11Zn + 11H 2 S O a -2AsH3+6Ag+llZnSO4+8H20
Importance of Oxidation-Reduction
Oxidation in the form of burning of fuels (coal in a power plant, oil in a furnace, gasoline in a car engine, natural gas in a stove, wood in a fireplace,etc.) is the world's primary source of energy. It probably will remain so in the foreseeable future, even though our supplies of fossil fuels are rapidly dwindling. The energy derived from "batteries" (voltaic cells) is generated by allowing spontaneous redox reactions to wcur under such conditions that the site of oxidation is physically separated from the site of reduction. The electrons given up at the anode (the electrode where oxidation is occurring) must pass though awire toget to the cathode (where reduction occurs), and the flow of electrons through the wire constitutes an electric current. The two most widely used voltaic cells are the automotive lead storage battery (in which lead is oxidized and lead dioxide reduced) and the "dry cell" (in which zinc metal is oxidized and M n O 2 reduced). These and other examples of important oxidation-reduction processes encountered in ev-
The word REDOX is more than just a neat abbreviation; It integrates two words into a simple combination. The two are really one-that is its subtle connotation, And so they are united in one name. Each needs the other; neither con occur in isolation. One substance takes electrons, while the other makes donation. Whenever there? reduction, there is always oxidation, And the amounts of each must be the same. References Bottomley, J.."Noteana method for determining tbs cmffleienbin ehemidquationa", Chsm. NWJS,37.110 11373). Mming, H. G.,'"Balancing Equation, Algobrsiely", J. CHEM.EDUC., 11,125 (1934). Gmdstein, M. P., "Intc~pntatianof Oxidation-Redunion'; J. CHEM. EDUC., 47. 452 11970). Karslake, W. d., "Balancing of Ionic Chemical Equations", Chrm. Noma, 66.41 11907). Loekwood, K. L., "Redox Rmiaited", J. CHEM. EDUC., 38.323 (1961). Morris, K. B.."The Balancing of Oxidation-Reduction Equations". J. CHEM. EDUC.. 15. 533 (1938). Steinbsch, 0. F.,"Non-Staichiometric Equations". J. CHEM. EDUC.. 21.66 11944). Vanderwerf, C. A,. Davidson, A. W.. and Sialer, H. R.,"Oxidation-Reduerim: ARe-evaluatian",J. CHEM.EDUC.,22,450 11945). Vanderwed C. A., "A Conaiatent Treatmentof Oxidation-bdudion", J. CHEM EDUC., 25,547 11948). Yalman. R. G., "Writing Oxidation-Reduction Equations", J. CHEM. EDUC.. 36. 215 (1959).
Volume 55, Number 5, May 1978 1 331
0
