Oxidation Numbers and Balancing Redox Equations

The oxidation number, or oxidation state, is a bookkeeping device used to keep track of the number of electrons formally associated with a particular element. The oxidation number is meant to tell how many electrons have been lost or gained by a neutral atom when it forms a compound. Because oxidation numbers have no real physical meaning, they are somewhat arbitrary, and not all chemists will assign the same oxidation number to a given element in an unusual compound. However, there are some ground rules that provide a useful start. 

The oxidation number of H is almost always +1, except in metal hydrides—e.g., NaH, in which H is—1. 

The oxidation number of oxygen is almost always -2. The only common exceptions are peroxides, in which two oxygen atoms are connected and each has an oxidation number of -1. Two examples are hydrogen peroxide (H—O—O—H) and its anion (H—O—O ). The oxidation number of oxygen in gaseous O2 is 0. 

The alkali metals (Li, Na, K, Rb, Cs, Fr) almost always have an oxidation number of +1. The alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra) are almost always in the +2 oxidation state. 

The sum of the oxidation numbers of each atom in a molecule must equal the charge of the molecule. In H2O, for example, we have 

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APPENDIX D Oxidation Numbers and Balancing Redox Equations... 

Use oxidation numbers to balance redox equations in which the reactants and products are known. (Questions 2.3 and 2.4)... 

Fe3+ is the oxidizing agent because it takes an electron from V2+. V2+ is the reducing agent because it gives an electron to Fe3+. Fe3+ is reduced, and V2+ is oxidized as the reaction proceeds from left to right. Appendix D reviews oxidation numbers and balancing of redox equations. 

Most redox equations can be balanced and (2) the change-in-oxidation-number method. Many redox equations can be balanced... 

In earlier sections of this chapter, we showed how to write and balance equations for precipitation reactions (Section 4.2) and acid-base reactions (Section 4.3). In this section we will concentrate on balancing redox equations, given the identity of reactants and products. To do that, it is convenient to introduce a new concept, oxidation number. 

When balancing redox equations, we consider the gain of electrons (reduction) separately from the loss of electrons (oxidation), express each of these processes as a halfreaction, and then balance both atoms and charge in each of the two half-reactions. When we combine the halfreactions, the number of electrons released in the oxidation must equal the number used in the reduction. 

The formula you need for this problem is AG° = -ncSSE°. The Faraday constant, <3, is equal to 9.65 x 104 joules volt-1 mole n is the number of electrons transferred between oxidizing and reducing agents in a balanced redox equation. 

Sometimes the two different oxidation states are indicated by the use of Roman numerals after the formula, e.g. Fe(II)2+ and Fe(III)3+. These Roman numerals are called oxidation number . When the oxidation number of an element increases, it has been oxidized. A decrease in the oxidation number shows the element has been reduced. A further method of balancing redox equations is based on these oxidation numbers, but the concepts will not be pursued further in this book ... 

Equations for redox reactions are sometimes difficult to balance. Use the steps in Skills Toolkit 2 below to balance redox equations for reactions in acidic aqueous solution. An important step is to identify the key ions or molecules that contain atoms whose oxidation numbers change. These atoms are the starting points of the unbalanced half-reactions. For the reaction of zinc and hydrochloric acid, the unbalanced oxidation and reduction half-reactions would be as follows ... 

Two methods are commonly used to balance redox equations the half-reaction method and the oxidation number method. 

M Note that the product ab is the total number of electrons gained in the reduction (and lost in the oxidation) represented by the balanced redox equation. Thus, if a = h, it is not necessary to multiply the half-reactions by a and b.lia = b = n, the equilibrium constant is determined from... 

One important use of oxidation numbers is in balancing redox equations. There are essentially two methods to balance redox reactions the oxidation number change method and the ion-electron method. In the former method, the changes in oxidation number are used to balance the species in which the elements that are oxidized and reduced appear. The numbers of atoms of each of these elements is used to give equal numbers of electrons gained and lost. If necessary, first balance the number of atoms of the element oxidized and/or the number of atoms of the element reduced. Then, balance by inspection, as was done in Chapter 7. 

We know that in an oxidation-reduction reaction we must ultimately have equal numbers of electrons gained and lost, and we can use this principle to balance redox equations. For example, in this case, 2 Ag ions must be reduced for every Cu atom oxidized ... 

Q Deduce oxidation numbers and use them to balance redox equations. Recall the principles of manganate(Vll) and thiosulphate titrations. Recall the corrosion of iron and its prevention. 

The main difference between balancing ionic redox equations and molecular redox equations is in how we handle the ions. In the ionic redox equations, besides having the same number of atoms of each element on both sides of the final equation, we must also have equal net charges. In assigning oxidation numbers, we must therefore remember to consider the ionic charge. 

Fe is reduced (its oxidation state goes down from +3 to +2), and V is oxidized (its oxidation state goes up from +2 to +3). Appendix D discusses oxidation numbers and how to balance redox equations. You should be able to write a balanced redox reaction as the sum of two half-reactions, one an oxidation and the other a reduction. 

The half-reaction method, or ion-electron method, for balancing redox equations consists of seven steps. Oxidation numbers are assigned to all atoms and polyatomic ions to determine which species are part of the redox process. The oxidation and reduction equations are balanced separately for mass and charge. They are then added together to produce a complete balanced equation. These seven steps are applied to balance the reaction of hydrogen sulfide and nitric acid. Sulfuric acid, nitrogen dioxide, and water are the products of the reaction. 

Oxides. Chlorine oxides in which Cl has an even oxidation number undergo self-redox. In basic solution, the products are oxyanions such as C102, CIOs", and CIO4". Write the balanced ionic equations for the self-redox of (a) CIO2, (b) CI2O3. 

The term displaces means that aluminum goes into solution as Al (aq), forcing Zn (aq) out of solution as zinc metal. A1 is oxidized to Al +, and Zn + is reduced to Zn. In combining the half-cell equations to produce the overall equation, we must take care to ensure that the number of electrons involved in reduction equals the number involved in oxidation. (This is the half-reaction method of balancing redox equations discussed in Section 5. ) The cell diagram is written with the reduction half-cell equations as the right-hand electrode. 

But in the covalent compound SCI2, you have seen that the sulfur oxidation number of +2 was obtained by an imaginary distortion of the distribution of electrons, so that the shared electron pairs in the covalent bonds are transferred completely to the more electronegative chlorine. The charge carried by sulfur in the real SCI2 molecule must therefore be less than the oxidation number of +2. To emphasize this, oxidation numbers are often written as Roman numerals in parentheses after the element. Thus, SCI2 is written as sulfur(II) chloride, and SO2 as sulfur(IV) oxide. Oxidation numbers can be used to balance redox equations, which we consider next. 

Balancing the chemical equation for a redox reaction by inspection can be a real challenge, especially for one taking place in aqueous solution, when water may participate and we must include HzO and either H+ or OH. In such cases, it is easier to simplify the equation by separating it into its reduction and oxidation half-reactions, balance the half-reactions separately, and then add them together to obtain the balanced equation for the overall reaction. When adding the equations for half-reactions, we match the number of electrons released by oxidation with the number used in reduction, because electrons are neither created nor destroyed in chemical reactions. The procedure is outlined in Toolbox 12.1 and illustrated in Examples 12.1 and 12.2. 

The key to balancing complicated redox equations is to balance electrons as well as atoms. Because electrons do not appear in chemical formulas or balanced net reactions, however, the number of electrons transferred in a redox reaction often is not obvious. To balance complicated redox reactions, therefore, we need a procedure that shows the electrons involved in the oxidation and the reduction. One such procedure separates redox reactions into two parts, an oxidation and a reduction. Each part is a half-reaction that describes half of the overall redox process. 

After oxidation and reduction half-reactions are balanced, they can be combined to give the balanced chemical equation for the overall redox process. Although electrons are reactants in reduction half-reactions and products in oxidation half-reactions, they must cancel in the overall redox equation. To accomplish this, multiply each half-reaction by an appropriate integer that makes the number of electrons in the reduction half-reaction equal to the number of electrons in the oxidation half-reaction. The entire half-reaction must be multiplied by the integer to maintain charge balance. Example illustrates this procedure. 

One of the most important uses of oxidation numbers is in balancing redox (oxidation-reduction) equations. These equations can get very complicated, and a systematic method of balancing them is essential. There are many such methods, however, and each textbook seems to use its own. There are many similarities among the methods, however, and the following discussion will help no matter what method your instructor and your textbook use. 

In this section, you learned the half-reaction method for balancing equations for redox reactions. You investigated the redox reactions of metals with acids, and the combustion of two hydrocarbons. After applying the half-reaction method in the following review problems, you will learn a different method in section 10.4. This method will make greater use of oxidation numbers. 

In section 10.2, you learned that a redox reaction involves changes in oxidation numbers. If an element undergoes oxidation, its oxidation number increases. If an element undergoes reduction, its oxidation number decreases. When balancing equations by the half-reaction method in section 10.3, you sometimes used oxidation numbers to determine the reactant(s) and product(s) in each half-reaction. 

An alternative to the oxidation-number method for balancing redox reactions is the half-reaction method. The key to this method is to realize that the overall reaction can be broken into two parts, or half-reactions. One half-reaction describes the oxidation part of the process, and the other half-reaction describes the reduction part. Each half is balanced separately, and the two halves are then added to obtain the final equation. Let s look at the reaction of aqueous potassium dichromate (K2Cr2C>7) with aqueous NaCl to see how the method works. The reaction occurs in acidic solution according to the unbalanced net ionic equation... 

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