redox

Oxidation and Reduction

Oxidation and Reduction In the early days of chemistry, a substance was said to be oxidized oxid ized af ter it it had had reacted reacted with oxyg en. The T he reaction its elf was called called o xid ati on. A familiar familiar examp le o f oxidation in this this sense sense is the rusting of iron. In this reaction, iron in the presence of moisture reacts with oxygen from the air to produce ordinary rust, iron oxide. Another example of oxidation is the burning of fuels. When wood, natural gas, or coal burns, the oxygen in the air combines with the carbon and hydrogen in the fuel to produce water and the oxides of carbon. Today, the term oxidation still is used in this sense, but it has a  bro  b roaa d er m eani ea ning ng,, as well. we ll. O x idat id atio ion n refe re fers rs tod to d ay to a cate ca teg g o ry o f  reactions with some characteristics similar to those of the reactions o f oxygen just descri described bed.. Originally, reduction referred to a chemical reaction in which a compound lost oxygen: 2Fe20 2Fe20 3(5) 3(5) + 3C (s) -► 4Fe (/) + 3C O, (g)

(Eq. 1)

In Equation 1, the iron (III) oxide is being reduced. This is  bec  b ecau ause se,, du d u rin ri n g the th e reac re actio tion, n, o x ygen yg en is remov rem oved ed from fro m the th e co c o m pou po u nd, nd , leaving elemental iron. Today, the term reduction has been given a second, broader meaning. To describe reactions in which oxidation and reduction occur  in the broader broad er sense of those terms, terms, chemists che mists use oxidation num bers. The next three sections explain their determination and use.

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 Inorganic Chemistry

118 Oxidation Numbers

The oxidatio n nu m b er , or oxidation state, of an atom is the apparent charge assigned to an atom o f an element. The charge is called  apparent because, while in some cases an oxidation number is the actual charge on an ion, in other cases there is no evidence for the presence o f  a true electric charge on an atom that has been assig ned an oxidation number. Oxidation num bers, therefore, are merely a convenient device for analyzing types o f reactions discussed in this chapter. The Rules fo r Determining O xidation Numbers

1. The oxidation nu mb er o f an atom in the unc om bined slate is zero. For  example, the oxidation numbers of the following atom s are all 0 : • An atom of phosp horus in the P4 molecule. • An atom of ch lorine in the C l2 molecule. • An atom o f sodium. Na. when no t  part o f a compound. 2. The oxidation nu mber o f a monatoinic ion is equal to the charge on the ion. ion: Mg2' in M gBr,; C l- ion in NH 4C1; 0 :~ ion in N a ,0 oxidation +2 -1 -2 number: 3. In most com pounds, the oxidation number o f hydrogen is +1. except in the case o f the hydride compounds o f the me tals from Grou ps 1 and 2. In hydrides, the oxidation number of hydrogen i s - l . For example, compound: H2S 0 4 NH 3 H ,0 NaH (sodium hydride) oxidation number of  hydrogen: +1 +1 +1 -1 4. In the vast majority o f compounds, the oxidation num ber o f oxygen i s -2 . In peroxides the oxidation num ber of oxygen is - 1 . In OF,, its oxidation number is +2. For example, compound: H 20 F e ,0 3 KClOj H20 2 O F, oxidation number o f  oxygen: -2 -2 -2 -I +2 5. In binary compo unds of nonmetals, the more electronega tive element is assigned the negative oxida tion num ber. For examp le, in PCI3, chlorine is more electrone gative than phosphorus. Each chlorine atom partially gains one electron, giving each chlorine atom an oxidation n umber o f- 1 and the  phosphor us atom an oxidation nu mbe r o f -3 . 6. The sum o f the oxida tion num bers for all the atoms in a neutral molecule or  formula is zero. The sum o f the oxidation numbers for nil atoms represented   by a polvatoniic ion is equal to the ch are e on the ion. For example. 11,0: ' (2 x + 1) + (- 2 ) = 0 oxidation num ber of each o f 2 II atoms H C O j ': ( + ! ) + (+ 4) - (3 * - 2 ) = - 1 oxidation number o f charge on ion each o f :3 oxygen a tom

Figure 4.1 : How oxidation number s should be determined.

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Oxidation num bers are assigned to atoms in chemical formulas. Rules have been developed for assigning oxidation num bers to atoms. These rules are given in Figure 4.1. Sample Problem 1 Using the rules for assigning oxidation nu mbers given in Figure 4.1, assign oxidation numbers to each element in the compound  calcium chloride C aCl2. Solution Calcium fluoride is an ionic compound composed o f the calcium ion, Ca2\ and the chloride ion. C l'. The calcium ion is a monatomic ion with a charge of 2+. Hence, by Rule 2, the oxidation number of  the calcium atom in calcium chloride is +2. The chloride ion is a monatomic ion with a charge of 1-. Hence by Rule 2, the oxidation number of an atom o f chlorine in calcium chloride is - 1. An example of oxidation-reduction. Oxidation is defined  simply as a change in which the oxidation number of an element increases. Such a change can take place only at the expense of a decrease in the oxidation num ber of some other element. The latter is termed  reduction. For example, consider the following reaction 0 0+3-2 2 A1 + 3 S -> AI,S3 in which the oxidation number of each element is written above the symbol. Aluminum and sulfur in the free state have, in accordance with our rules, an oxidation number of 0; whereas in the compound, aluminum sulfide, the oxidation number of aluminum is +3 and that o f sulfur is -2 . In this reaction, aluminum gains in oxidation number  from 0 to +3 and hence is oxidized; whereas sulfur loses in oxidation number from 0 to -2 and is reduced. Let us examine this reaction in another way. As it proceeds, each atom o f aluminium gives up 3 electrons and each atom o f sulfur  takes up 2 electrons. In the reaction as a whole, 6 electrons given up  by the 2 alum inum atom s are taken up by the 3 su lfur atom s. A transfer of electrons has taken place, and the resultant compound  A1,S3 is electrovalent; it is composed o f A P * and S ' ~ ions. In this  particular case, Al gains in oxidation number or loses electrons;

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likewise, sulfur loses in oxidation number or gains electrons. It is often convenient when dealing with electrovalent substances to discuss these changes in terms of a gain or loss in electrons as well as in terms o f changes in oxidation num ber. To summ arize: for  reactions involving a transfer o f electrons, Oxidation is gain in oxidation n umb er or loss o f electrons Reduction is loss in oxidation num ber or gain o f electrons.

Oxidation and reduction are mutually dependent processes; if  electrons are taken up by one substance, they must be given up by another. There can be no oxidation without reduction, nor reduction without oxidation. The union of aluminum and su lfur may be written as taking  place in tw o steps with e representing an electron: 2 AI -> 2 A I^ + + 6 e (oxidation) 6 e + 3 S - > 3 S “ ‘ (reduction)

(1) (2)

By adding the two steps and cancelling the electrons, w e obtain the net equation which was written above. Oxidizing an d red uc ing agents. Oxygen, sulfur, or other  nonm etals, as they combine with m etals, are recognized as oxidizing agents', therefore, the metals must be the reducing agents. In all oxidation-red uction chemical changes, the following relationship exists: Oxidizing agent = electron receiver = loser in oxidation number  Redu cing agent = electron giver = gaine r in oxidation number 

The receiver-loser and giver-gainer relationship is somewhat confusing, but it is helpful to consider that a particle giving away negative charge must itself become more positive. If it receives a negative charge, it becomes less positive. In the combination 2 Fe + 3 Cl2 -» 2 FeC l3 (2 Fei++, 6 Cl ) Cl2 is the oxidizing agent; each Cl receives 1 electron and loses 1 in oxidation number. Fe is the reducing agent: each Fe gives 3 electrons and gains 3 in oxidation number. In the replacement Zn + C u S 0 4 (solution) or

Z n S 0 4 (solution) + Cu

Zn + Cu+^ + S 0 4~ -> Zn'~ + S 0 4~~ + Cu

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Oxidation and Reduction

Cu " is the oxid izing agent: it receives 2 electrons and loses 2 in oxidation number. Zn is the reducing agent: it gives 2 electrons and  gains 2 in oxidation number. Balancing oxidation-reduction equations. To balance these equations, we make use of the rule that the total gain in oxidation numbers must equal the total loss in oxidation numbers o f the elements. Co nside r the reaction +J -I

+1 -2

+2 -1

+1 -1

o

FeC lj + H,S -> FeClj + HC1 + S

I

 —

_

i

1

----------------

in which the oxidation number of each element is placed above the symbol. Two o f the elements undergo a change in oxidation number; each iron atom loses 1 (from +3 to +2) and each sulfur atom gains 2 (from -2 to 0) in oxidation number. The loss o f 1 in oxidation num ber o f Fe is indicated by - I , and the gain o f 2 in oxidation number of S is indicated by +2. To balance the gain and loss, 2 iron atoms will be required for each sulfur atom, and this will call for a ratio o f 2 FeCI3 to 1 H,S. This must be the ratio in which these two substances react: 2 FeClj + 1 H2S -> 2 FeCl2 + 2 HC1 + S Having obtained the ratio of the oxidizing agent, FeCl3, to the reducing agent H2S, it is easy to complete the balancing of  the equation on the right side. In this equation, the change in oxidation number of sulphur (2) is made the coefficient of FeCl3, while the chang e in oxidation numb er o f iron ( 1) is the coefficient Of H,S. In general, we may employ the following steps in balancing oxidation-reduction equations: 1. Write the oxidation num ber of each o f the elements in each compound above the symbol o f the element in the equation. 2. Determine which elements have changed in oxidation number  and the amount o f change. Usually two elements only will have changed; one gains in oxidation num ber while the other  loses.

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3. Balance the gain and loss in oxidation numbers. This may be accomplished automatically by placing the number  representing the change in the oxidizing agent in front o f the formula for the reducing agent, and vice versa. 4. Com plete the balancing o f the two elemen ts which have changed oxidation nu mber on the right side o f the equation, keeping in mind that the ratio of oxidizing agent to reducing agent as determined from the change in oxidation numbers must be maintained. Balance hydrogen and oxygen atoms after all other elements have been balanced. Let us illustrate the outlined steps by balancing an equation: (1 )

0 1+ 5-2

+1+5-2

+2 -2

+1-2

P + UNO, -> HPQj + NO + HjO

0

(2)

+5

+5

+2

P + HN O 3 -► HPOj + NO + HjO

5 (3)

3 P + T H N 0 3 - » H PO j

(4)

3 P + 5 MNOj

t

NO + 11,0

3 HP Oj + 5 NO + H20

(balan ced)

Oxidation-reduction equations are m ore difficult to balance by ordinary trial and error methods than those which involve no change in oxidation num ber. The “change in oxidation num ber” method for   bala ncin g oxid ation-reduction equations, however, is ra pid and  accurate. BALANCING

EQUATIONS

FOR

REDOX

REACTIONS—NO

In general when you are faced with balancing an equation for a nonaqueo us redox reaction, you will be provided with formulas for all reactants and produc ts. The m ethod used, the oxidalion-number m ethod is outlined below :

so l v e n t

pr e s e n t

.

BALANCING REDOX EQUATIONS : OXIDATION-NUMBER METHOD

Step 1 Assign oxidation number to all atoms. Step 2 N otewhichatom sappeartoloseandw hichappeartogainelectrons, and determine how m any electrons are lost and gained. Step 3 If there is more than one atom losing or gaining electrons in a formula unit, determine the total loss or gain of electrons  per form ula unit.

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Oxidation and Reduction

Step 4 Make the gain of electrons by the oxidizing agent equal to the loss by the reducing agent by inserting the appropriate coefficient before the formula of each on the left-hand side of the equation. (See the following examples.) Step 5 Complete the balancing o f the equation by inspection. First  balance atoms which have gained or lost electrons; second, atoms other than O or H; third, O atoms; and last, H atoms. Example 1 Problem : Balance the following equation MnO, + KCI06 + KOH -> K ,M n04 + KCI + H20 Solution : Step 1: M n02 + K CIO- + K O H -> K,M n04 + K Cl + H, O +4-2

+ 1 + 5 - 2 + 1 -2 + 1

+1+ 6-2

+ 1 —1 + 1 —2

Step 2: M anganese, Mn, changes oxidation num ber from +4 to +6 and so appears to lose two electrons. Cl changes from +5 to -1 and so app^rs to gain s,ix electrons. MnO, + KCIO, + KOH ---- ► K2Mn02 + KCI + H2 +4

"

+5

+6

'

-1

A

 Reduction: Each Cl atom :ains 6e-

Each KC103 formula unii gains 6eOxidation: Each Mn atom loses 2e-

Each MnO , formula unit loses 2e-

Step 3: As is shown above, since there is only one Mn atom in each MnO, formula unit, the electron loss per formula unit is two. Also, since there is but one Cl atom in each KCIO, formula unit, the electron gain per formula unit is six. Step 4: 3M nO, + K C I0 3 + KOH -» K2M n 0 4 + KCI + H20 Just look at the left-hand side for now. By placing a 3 before M nO,, we are saying that if one M n0 2 formu la unit loses two electrons, then three will lose 3 * 2, or six, electrons. This will balance the electrons lost (six) against the electrons gained (six) by one KC103 form ula unit. Now that this 3 : 1 ratio (M nO ,/KC I3) has been established, it must not be changed. Step 5 : Add coefficient to the right-hand side to balance the Mn and Cl atoms, b ecause they were involved in the electron transfer.

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3 M n 0 2 + KCIOj + KOH -> 3K2M n 0 4 + KC1 + H20 Then, balance the K atoms (note : the coefficient before K C I0 3 must be changed) 3M nO, + KCIO , + 6KOH -> 3K 2M n 0 4 + KC1 + 3H20 then, O atoms 3 M n0 2 + KCIO3 + 6KOH -*• 3 K ,M n 0 4 + KC1 + 3H 20 and last, H atoms. But note that they are already balanced. Example 2 Problem: Balance the following equation H2C20 4 + K M n04 -► C 0 2+ MnO + K20 + H20 Solution: Step 1 : H2C20 4 + K M n 0 4 -> C 0 2+ MnO + K ,0 + H ,0 +1 +3 - 2

+1 +7 -2

+4 - 2 +2 -2

+1 - 2 +1 -2

+

K ,0

Step 2 : H 2C 20 4

+

-3

KM nO , ->

*7

M

H ,C ,04 + KM n04 +3

C 02

MnO

+

+

H:0

O ----- ►

CO ,

+7

+

+4 i‘

MnO

+

K ,0

+

HzO

+2

^

 Reduction: Each Mn  jains 5e-

Oxidation: Hacli C' loses 1e-

Step 3: H ,C ,04 + K M n0 4 -> CO, + +4 -3 +7 H2C20 4 + K M n0 4 ----- *• +3

•-

MnO + K20 + H ,0 C 02+ MnO+ K ,0 +H ,0

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Step 5 : 5H,CjO< + 2KMnO, -* 10C02+ 2MnO + K,0 + 5H,0 BALANCING EQUATIONS FOR REDOX REACTIONS—AQUEOUS

For reactions which occur in aqueous solution you may  be given only a skeletal equation, one in which only the principal reactants and products are given. You m ust complete the equation, as well as balance it. There are two methods for doing this. These methods, the oxidation-number method and the half-reaction method, are different and should not be confused with each other. Each has its advantages and disadvantages. so l u t io n s

.

BALANCING REDOX EQUATIONS (AQUEOUS SOLUTIONS) :

OXIDATION-NUMBER METHOD Step 1 Assign oxidation numbers to all atoms. Step 2 Note w hich atoms appear to lose and which ap pear to gain electrons, and determine how many electrons are lost and gained. Step 3 If more than one atom in a formula unit loses or gains electrons, determine the total loss or gain per formula unit. Step 4 Make the gain of electrons by the oxidizing agent equal to the loss by the reducing agent by inserting an appro priate coefficient before the formula of each on the left-hand  side o f the equation. (See the following examples.) Step 5 Balance the atoms which have gained or lost electrons by adding appropriate coefficients on the right. Step 6 Balance all other atoms except for O and H. Step 7 Balance the charge (the sum o f all the ionic charges) so tha t it is the sam e on both sides, by a dd ing either H+ or 

OH". (a) If the reaction takes place in acidic solution, add H+ ions to the side deficient in positive charges. (b) If the reaction takes place in basic solution, add OH' ions to the side deficient in negative charges. Step 8 Balance O atoms by adding H ,0 to the appropriate side. Check to see that the H atoms are now balanced. (They will be, if  you have made no mistakes.)

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Example 3 Problem : Complete and balance the following equation for a reaction which takes place in acidic solution : Cr20 72' + Fe2* -» Cr34 + Fe3+ (acidic) Solution : Step 1: Cr20 72" + Fe2" -> Cr34 + Fe3+ +6 —2 +2 +3 +3 Step 2 : C r ,0 72-

+

+6

Fe2+

-------►

+2

Cr3"

+

Fe3*

+3

+3

ii

A

Oxidation: Loss of  i 5-pcrFe

 Reduction: Gain of 3e- per Cr 

Step 3 : Cr20 72+6

+

Fe2+ +2

----- *

Cr-3+ +

Fe3*

+3

+3

 j \ 

A

Loss o f Ie  per Fc^*

Total Gain of 2 X 3, or le per Cr20 72'

Step 4: Step 5: Step 6 : Step 7:

C r,0 72" + 6 Fe2+ -> C r3+ + Fe3, Cr20 72' + 6F e2+ ^ 2C r34 + 6 Fe34 Done! Total charge on left = -2 + 6 (+2) = +10. Total charge on right = 2 (+3) + 6 (+3) = +24. Added positive charge needed on left  = +14. 14H4 + C r,0 72 + 6Fe2" -» 2Cr,+ + 6 Fe3’

Step 8 :14 H + Cr20 72' + 6Fe2' -► 2C rv + 6Fe3* + 7H20 .

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Example 4 Problem: Complete and balance the following equation for a reaction in basic solution : C r0 42 + Fe(OH ), -> C r0 2 + Fe(O H)3 (basic) Solution: Cr042Step 1:

+

+6

C r0 2-

Fe(O H)2

♦2

+3 ii

+ Fe(O H)3

•

+3

A

Oxidation: Loss ot le~ oerFe

 Reduction Gain of 3e perCr 

Gain of 3c” pcrCr04-’

Step 4:

C r0 42" + 3Fe(OH)j -► C r0 2~ + Fe(OH )3 Step 5:

C r0 42' + 3Fe(OH),

C r 0 2" + 3Fe(OH)3

Step 6 : Done!

Step 7 : Total charge on left = -2 + 3 (0) = -2 . Total charge on right = -1 + 3 (0) = - I . Added negative charge needed on right = - I. C r0 42" + 3Fe(OH),

C rO ,' + 3F e(O H ),O H '

Step 8 : 2H20 + C r0 4- + 3Fe(O H),

C r0 2 + Fe(OH )3+ OH

The second method for balancing redox equations for reactions which occur in aqueous solution is the half-reaction method, also known as the ion-electron method. Note that in it you do not  assign any oxidation numbers. BALANCING REDOX EQUATIONS (AQUEOUS SO LU TIO N S):

11AI-F-ftKACTION METHOD Step I Separate the skeletal equation into two half-reactions, one an oxidation and the other a reduction. Step 2 Balance each h alf-reaction sep arately ac cord ing to this sequence :

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^

 Inorganic Chemistry

a Balance all atoms other than H and O by inspection,  b Balance O atoms by adding H20 to the appropriate side, c Balance the H atoms. The way this is done depends on whether the solution is acidic or basic. i In acidic solution, add the appro priate number o f H+ to the side deficient in H. n In basic solution, instead o f add ing H4 ions, add one  H20 molecule to the side deficient in H plus one OH  ion to the opposite side fo r each needed H atom. You may wish to cancel out HzO molecules duplicated on each side at this point. d Balance the charge by adding electrons (e~) to the side deficient in negative charge. Step 3 Multiply each balanced half-reaction by an appropriate number  in order to balance the electron loss against the electron gain Then add the two half-reactions. Step 4 Subtract out (cancel) anything which appears on both sides. (The electrons should all disappear in this step.) Example 5 Problem : Using the half-reaction metho d com plete and balance the following equation for a reaction occurring in acidic solution. Cr20 72' + Fe2* -> Cr3* + Fe3* Solution : Step I : Cr20 72" -> Cr3* Step 2 : a Cr20 72- ->  ICr3*

Step 3 :

(acidic) Fe2* -> Fe3< (done!)

 b Cr20 72' -+  IC i3* + 7H20

(done!)

c 14H* + Cr20 72" -> 2 0 ’* + 7H20

(done!)

d be- + 14H' Cr20 72' ->  IC t3* + 7H20

Fe2*-» Fe3* + e~

6 e-

I4H* + Cr20 , 2' -» 2C r* + 7H20

[Fe2* -» Fe3* e ] x 6

be- I4H’ + Cr20 72- -> 2Cr3* + 7H20

6 Fe2+^ 6 Fe3*+ 6e‘ be- I4H* + Cr20 , 2' + 6Fe2* -> 2Cru 6 Fe5* + 7H20 + be

Step 4 : I4H* + Cr20 72‘ + 6Fe2< -» 2Cr‘* 6 Fe3* + 7H,0

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Example 6 Problem : Complete and balance the following equation for a reaction taking place in basic solution using the half-reaction method : C r0 42' + Fe(OH )2 —* C r0 2~+ Fe(OH)3 (basic) Solution: Step I : C r 0 42 - > C r 0 2~

Fe(OH), -> Fe(OH)j

Step 2 : a (done)

(done!)

 b C r 0 42" -> C r0 2“ + 2 H: 0

H20 + Fe(OH), -+ Fe(OH )3

c4 H 20 + C r0 42 -

OH + H20 + Fe(OH ), -> Fe(O H)3+ H.O

C r 0 , + 2H 20 + 4 0 H “ 2 H ,0 + C r0 42~ -

OH' + Fe(OH), -> Fe(OH)3

CKV+40H' Step 3 : 3 tf + 2H;0 + C r 0 42~ —»

[O ir + Fe(OH), -> Fe(OH)j + c‘) * 3

CK V + 40H-

3