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++ZO2-

becomes 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 C1-

le 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 i i that the oxidizing agent the second reaction is chlorine rather than oxygen. Oxidation is an extremely common and also a highly im- xtant kind of chemical reaction. Probably the most familiar rm of oxidation is combustion (burning), a very rapid oxi- tio on process that occurs with much evolution of heat and ght. I t is a chemical reaction much older than recorded his- fry. 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 Com- pany and as a television lecturer in a series "Sootlieht on Research." She hna a&ve$on the staffs of --~. ~~-. .~- ~~~ - ~ - ~-~~ - ~ ~ ~ ~ ~ - ~-

Corning Community College and Doris Kolb Bradley University. Since 1967, Illinois Central College she has been Professor of Chem- East Peoria, Illinois 61635 istry at Illinois Central College.

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 to as calcination. since the ash was called calx). ~ - ~ - ~ ~ ~ ~ ~ ~~ ~

Hoyle's assistant. ~ o h e i t Houke, 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 combus- tildity.

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 be- lieved 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 the ohloeiston theorv. which was introduced about this .- ~ - ~ - A ~ ~ ~ ~ ~ - ~~~ . . same time and which explained combustion in a quite dif- ferent wav. ~ ~

dohann'~oachirn Becher 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 "phlogis- ton" (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.

Wood - Ash (Calx) + Phlogiston f

When wwd hurned, it lost weight because of its loss of phlo- giston; 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 in- crease.)

Whenever something hurned, i t was said to he "dephlog- isticated". 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

326 / Journal of Chemical Education

until 1777, so the credit for the discovery went to Priestley. In soite of their im~ortant discovew. neither of these men ever re& at~andoned his belief in the phi~gist*,n theory. 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 flam- mable materials was really just a combination of these suh- stances 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" (combi- e. -- nation with oxygen) to replace dephlogistication and "re- duction" (removal of oxygen) as a substitute for phlogistica- tion. Not only did Lavoisier look upon combustion as a com- bination with oxveen. hut he also recoenized the calcination .- . "

of metals and the respiration of animals to be oxidation re- actions.

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 im~lv 0 - . . .

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~oreciahlv until 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 elec- trons, and reduction a gain of electrons.

Defining oxidation and reduction as electron loss or gain eliminates the need for any oxygen involvement at all.

Ca t Clz - Ca2+ t 2 CI- element oxidized (oxidizing agent) (reducing agent) element reduced)

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 re- duction 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 va- lence electrons are not transferred to oxygen.

c toz-CO2

In Con 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 con- fused. since reduction is loeicallv defined in terms of de- creasing 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 as- siened to individual atoms in substances for ourooses of ecctron "bookkeeping". Although the assignment is some- what arbitrary, it provides a consistent method for deter- mining 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.

+I -1 t 2 -2 +2 -1(2) 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 electroneg- ative partner in each case. Let us take nitrous acid, for ex- ample, 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 (except in

peroxides, in which it is -1). (In OF2 oxygen has an oxidation numher of t 2 . since fluorine is the single element more electro- negative than oxygen.)

71 In rhtir compounds Group I A elementsare always +I, Group IIA elements arc nlnn).s +2, and Group llIA elements are almost al- ways +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).

8 + l ? -2(3) (2)+1 ? -2(4) ? 4 3 ) K ~ 1 0 3 ' Hz SO4 C 0z2-

In 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 car- bon.

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 hasanox- dation 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 man- pnese goes from +4 to +I. Its oxidation number is increasing, lo Mn is heing oxidized. When NaHSOa is changed to ele- nentary 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 re- hction.

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 chem- ical 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 ox- idizing agents (e.g. Fe3+, NOa-, Mn04-), although they need not he good ones (Nat and P O P , for example, are difficult to reduce); ele- ments 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 oxi- dation 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

agent must be exactly the same as the number lost by the re- ducing 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 meth- ods. 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 tech- nique.

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 dif- ferent 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 hy- drogen is +1 and oxygen is -2). The remaining element, ni- trogen, is in a +5 oxidation state in HN08 and +2 in NO. Since its oxidation numher is decreasing by 3, nitrogen is being re- duced. 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 ni- trogen 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 coeffi- cient, 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

328 1 Journal of Chemical Education

are basically the same. They may be summarized as fol- lows:

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.

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.

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.

Reduction: 4 H+ + NO,- - NO + 2 Hz0

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+ + 4e-

Reduction: 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.

Oxidation: 6 H20 + 3 S - 3 SOz + 12 HC + 12 e-

Reduction: 12 e- + 16 H+ + 4 Nos- - 4 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

3 S + 4 H f + 4 N O s - - 3 S O z + 4 N O + 2 H z 0 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 HEO where needed to balance oxygen atoms and H+ where needed to balance hydro- een atoms. (For alkaline solutions use OH- to halanee oxveen and ." ifn0 to balance hydrogen.)

3) Add electrons to one side of each half-reaction so as to halance the charges.

4) Make the electron changes equal by multiplying one or both half-reactions by appropriate numbers.

5) Add the two half-reactions together, eliminating all dnplica- tions.

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.

(Sulfur) x = r

(Nitrogen) y = p

(Hydrogen) y = 2q

(Oxygen) 3y = 2z + p + q

I t follows that:

3y=2z+y+y/2 or 3y-y-y/2=2i

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 + ~ H N O ~ - ~ S O S + ~ N O + ~ H ~ O

With the algehraic method it is not necessary to locate the elements being oxidized and reduced or to determine oxida- tion numhers, but the algebraic method does have one distinct drawback. Algebraic equations can have more than one solu- tion, 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 oxi- dation 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 im- mediately add the parenthetical (2).

+5 (2)

+5 HNOa + 12 - HI03 + NOz + Hz0

Since nitrogen goes down 1 in oxidation numher, we equalize the changes by multiplying the nitrogen reduction step by 10.

'Jotice that some chlorine ends up as MnCl2 or KC1 (not oxi- lized), but some ends up as Cl2 (oxidized). In balancing the :hanges in oxidation numher, the (2) in the oxidation step is ~eeded because 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.

+5 (2) = +10

p 7 + 5 10 HN03 + I2 - 2 HIOs + 10 NOz + Hz0

+1(2) (5) = + l o ' -0

2 KMn04 + 10 HCI - 2 MnC12 + 5 C12 + KC1 + Hz0

+5 +4

rhe Mn atoms are alreadv balanced (2 on each side), but let

-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 dif- ferent functions. It is a reducing agent, but it is also an acid mpplying chloride ions.

+I

1s now check the chlorine. (Hydrogen and oxvgen halancing ire u s ~ ~ a l l \ ~ hest left until last., The chlorine atoms are not

+2

danced. o n the left we have 10 chlorine atoms to balance the i C12 on the right, hut there are 6 more chlorine atoms on the ,ight that were not involved in the oxidation. We need to add i more 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 im- Iortant 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.

-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.

Let us first balance the changes in oxidation number

The 7 oxygen atoms on the left will require 7 H20 molecules on the right, which in turn will need 14 H+ on the left to hal- ance the hydrogen.

Cr2072-+3S2-+14H+-2Cr3++3S+IH20

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~lete Eauations. Sometimes redox reactions are not given in complete form. Suppose we want to write a hal- anced equation for a reaction in which Mn2+ is oxidized to Mn04- and Ph02 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.

Oxidation: Mn2+ - MnO4-

4 Hz0 + Mn2+ - MnOl- + 8 Hf + 5e-

Reduction: 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.

4 2 Reduction: W e + ~e H+ + 5 Pb02 - 5 Pbz+ + fe 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

330 / Journal of Chemical Education

Remember that the total increase in oxidation numher must equal the total decrease.

+2 (11) = +22

0 +2

2 Ag3As04 + 11 Zn + HzSOl - 2 AsH3 + 6 Ag + 11 ZnSOl + Hz0

1+11+5 1-3 0

-1 (3) (2) = -6

-8 (2) = -16 -6 -16 = -22

The H2S04 and H20 are balanced by inspection to give the completely balanced equation.

2 Ag&04 + 11 Zn + 11 H 2 S O a

- 2 A s H 3 + 6 A g + l l Z n S O 4 + 8 H 2 0

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 dwin- dling.

The energy derived from "batteries" (voltaic cells) is gen- erated by allowing spontaneous redox reactions to wcur under such conditions that the site of oxidation is physically sepa- rated 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

Volume 55, Number 5, May 1978 1 331

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