Volume 2
2.10. BALANCING SIMPLE REDOX EQUATIONS

Balancing Simple Redox Equations Using Oxidation States (Numbers) • Disproportionation and Comproportionation • Exercises
2.10.1. Balancing Simple Redox Equations Using Oxidation States (Numbers). There is a consistent pattern showing that those who excel at balancing redox reactions also excel at the basics of chemistry. As you practice the skill of balancing redox equations, you benefit from gradually learning new chemical transformations and trends in reactivity of various compounds while not even noticing that. We will start with some rather simple redox equations.

The very first thing we should do when there is a chemical equation to be balanced is to determine whether or not the reaction is a redox one. For that, we need to calculate the oxidation states of all of the elements involved. If the number(s) for the same element on the left differ from those on the right, this is a redox reaction. To avoid confusion, it is a good idea to write the oxidation states above the symbols of the elements in the formulas. Two simple examples are presented in Figure 2-111.
Figure 2-111. The reaction at the bottom is a redox reaction. The one at the top is not.

Although both transformations shown in Figure 2-111 produce NaCl, only the one at the bottom is a redox reaction, as follows from the analysis of the oxidation states of the elements involved. In the reaction of NaOH with HCl, none of the elements change their oxidation states. The oxidation state of sodium in NaOH and in NaCl is the same, +1. The oxidation state of hydrogen in NaOH and in H2O is the same, +1. The oxidation state of oxygen in NaOH and in H2O is also the same, -2. Like any other neutralization reaction between an acid and a base, this is not a redox reaction.

In contrast, the reaction of Na with HCl shown at the bottom of Figure 2-111 is a redox reaction. While the Cl remains in the same oxidation state (-1) throughout the transformation, both the Na and H change theirs. The oxidation state of Na increases from 0 to +1 and that of H lowers from +1 to 0. The Na is oxidized as it reduces the H of the HCl. Accordingly, the hydrogen of the HCl is reduced as it oxidizes the sodium.

To learn how to use the method of oxidation states for balancing redox reactions, we will start with a very simple transformation, the reaction of aluminum metal with oxygen.

Al + O2 = Al2O3 (not balanced)

One might doubt that aluminum is a flammable material. True, a chunk of aluminum metal would not burn if placed over a flame, aluminum pots and pans can be safely used on gas and electric stove burners, and the skin of modern aircraft is made of aluminum alloys. A fine aluminum powder, however, can be set on fire easily and burns energetically, as shown in Video 2-11. The product of this reaction is aluminum oxide, Al2O3, a white powder.
Video 2-11. Powdered aluminum metal is highly flammable (source).

As exemplified by the reaction of aluminum with oxygen, to balance a redox equation, we follow four steps outlined in Figure 2-112.
Figure 2-112. Balancing the equation for the redox reaction between Al and O2.

In step 1, we identify the elements that change their oxidations states in the reaction. Both the Al and O do. Before the reaction, the aluminum was a simple substance and so was the O2. The oxidation state of any simple substance is always 0. In the reaction, the oxidation state of the aluminum metal changes from 0 to +3 and that of the oxygen from 0 to -2. The aluminum is therefore oxidized and the oxygen is reduced.

In step 2, we write half-equations for the oxidation and reduction (Figure 2-112). One Al atom loses three electrons to get to the oxidation state of +3. One molecule of O2 gains four electrons, as each O atom gains two to attain the oxidation state of -2.

In step 3, we balance the number of electrons donated and the number of electrons accepted. For that, we multiply the oxidation half-equation by 4 and the reduction half-equation by 3. Having done that, we can see that the number of electrons in both half-equations is the same, 12, the least common multiple for 3 and 4.

In step 4, we use the multipliers found in step 3 (4 for Al and 3 for O2) as the correct coefficients for the reagents. These coefficients should be used as the guidance to balance the product. Placing the coefficient 2 before the Al2O3 formula on the right side concludes the balancing.

The next equation we will balance is for the reaction that is used for the recovery of refined gold from gold scraps. Watch Video 2-12 showing this fascinating process that employs aqua regia, a mixture of concentrated hydrochloric and nitric acids. Aqua regia, "royal water" or "king's water" in Latin, takes its name from its unique ability to dissolve gold, the "king of metals" that no other individual acid dissolves. An inexperienced person should keep away from extremely corrosive and hazardous aqua regia.
Video 2-12. Gold metal dissolves in aqua regia (source).

The reaction of the dissolution of gold in aqua regia is as follows.

Au + HNO3 + HCl = AuCl3 + NO + H2O (not balanced)

To balance this equation, first we identify the elements that change their oxidation states (Figure 2-113, step 1). These elements are Au, which is oxidized (the oxidation state changes from 0 to +3), and N, which is reduced (the oxidation state changes from +5 to +2). Next, we write the half-equations for the oxidation and reduction (step 2). As the number of electrons one Au atom loses equals the number of electrons one N atom gets, we do not need to use any multipliers in this case (step 3).
Figure 2-113. Balancing the equation for the reaction of gold metal with aqua regia (HNO3 + HCl).

The oxidation state method gives us the coefficients for Au, HNO3, AuCl3, and NO. That is all we need to finish the balancing. As three Cl atoms are needed to produce one molecule of AuCl3, we place 3 before the HCl formula to get the following equation.

Au + HNO3 + 3 HCl = AuCl3 + NO + H2O

Next we need to balance the H atoms. There are four H atoms on the left but only two on the right. Placing 2 in front of the formula of H2O on the right side of the equation makes it completely balanced.

Au + HNO3 + 3 HCl = AuCl3 + NO + 2 H2O

The beauty of the oxidation state method is that as long as the electrons lost and electrons gained are calculated correctly, the balancing is guaranteed to work.

Potassium permanganate (KMnO4) and potassium dichromate (K2Cr2O7) are widely used powerful oxidants (Figure 2-114). Both are salts and, consequently, strong electrolytes nearly completely dissociated in solution.
Figure 2-114. Structural formulas of KMnO4 and K2Cr2O7.

The very strong oxidizing power of KMnO4 and K2Cr2O7 stems from the high oxidation states of the Mn (+7) and Cr (+6) in the MnO4- (permanganate) and Cr2O72- (dichromate) anions. For example, permanganate can oxidize even chloride (Cl-) to chlorine (Cl2). Let us balance this reaction (Figure 2-115).
Figure 2-115. Balancing the reaction of KMnO4 with HCl.

As always, first we identify the elements that change their oxidations states in the reaction. These are Mn (+7) that is reduced to Mn (+2) and Cl (-1) that is oxidized to Cl (0). Next we write the half reactions and balance the number of electrons given and accepted. Finally, we balance the entire equation (Figure 2-115).

The final stage of the balancing is a bit tricky. The half-equation for the oxidation of the chloride anion suggests that the coefficient to be placed before the HCl formula should be 10. Why do we use 16, not 10 then? That is because in addition to the 10 Cl atoms that change their oxidation state to give 5 molecules of Cl2, we also need 6 Cl atoms that do not change their oxidation state of -1 to balance the equation. Of these six, two are needed for the formation of two molecules of KCl and four for two molecules of MnCl2. The total number of Cl atoms for the reaction is therefore 16 (10 + 6). When we use this coefficient for HCl in the equation, the balancing works like a charm.

There is a slightly different, more advanced technique to balance redox equations. While being more complex, that method makes the final balancing step much easier. We will learn this method in Volume 3 after we have studied the chemistry of selected elements.

2.10.2. Disproportionation and Comproportionation. In the 1780-90s, the famous French chemist Claude Louis Berthollet (1748-1822) experimented with chlorine gas in alkaline solutions. In one experiment, Berthollet passed Cl2 through a hot solution of KOH and noticed the formation of a white crystalline precipitate. He identified the precipitated crystals as KClO3, potassium chlorate, which since then has also been known as Berthollet's Salt. Curiously, it is the discovery of KClO3 that immortalized Berthollet's name, not his other, arguably much more significant contributions to science and technology, such as the determination of the composition of ammonia (NH3), the devising, in collaboration with Lavoisier, of a chemical nomenclature that provided the foundation for the modern one, and the first development of the modern commercial bleach.

In the Berthollet reaction, KClO3 is produced from Cl2 and a hot solution of KOH. The oxidation state of chlorine in Cl2, the reagent, is 0. In the product, KClO3, the oxidation state of Cl is +5. Clearly, the chlorine is oxidized in the reaction. If something is oxidized, something else should be reduced in the same reaction. What is it that is reduced during the formation of KClO3 from Cl2 and hot aqueous KOH?

Apart from chlorine, there are only three other elements involved in the reaction. These are K, O, and H. Of these three, the oxygen cannot be reduced, as its oxidation state is already -2. The oxidation state of both the K and H is +1, so, in principle, they could accept electrons. But would it not be strange if much more electronegative chlorine lost electrons to much less electronegative hydrogen or, especially, potassium? The reductant in this reaction is the same element as the oxidant: chlorine. Some of the Cl2 molecules are oxidized to KClO3 and some are reduced to KCl.

3 Cl2 + 6 KOH = KClO3 + 5 KCl + 3 H2O

A redox reaction in which the same element is both reduced and oxidized to give two different products is called disproportionation. In the reaction with KOH, chlorine (Cl2) disproportionates. This disproportionation reaction can be balanced as shown in Figure 2-116. At the very end of the balancing process there is one thing that we should never forget about, especially when dealing with disproportionation reactions. The coefficients derived directly from the half-equations are not always the smallest possible whole numbers. To follow the common practice to use the smallest possible whole-number coefficients in a chemical equation, we need to divide the originally derived numbers by 2. After that, the equation is in perfect shape.
Figure 2-116. Balancing the equation for the disproportionation reaction of Cl2 with hot KOH.

Disproportionation is possible only for an element in an intermediate oxidation state, so that some of it can be reduced and some oxidized. For instance, compounds of chlorine in the oxidation state of -1 cannot undergo disproportionation because they can be only oxidized and cannot be reduced. In contrast, Cl2 can be both oxidized and reduced, and therefore can undergo disproportionation. Reaction conditions can alter the course and outcome of a disproportionation reaction. While the reaction of a hot solution of KOH with Cl2 gives rise to KCl and KClO3 (Figure 2-116), a cold solution of KOH reacts with Cl2 to give potassium hypochlorite (KOCl). A cold solution of NaOH reacts with Cl2 similarly. This reaction is run on a large industrial scale to make household bleach, a 3-6% solution of NaOCl.

Cl2 + NaOH = NaCl + NaOCl + H2O
(balance this reaction)

The oxidation state of oxygen in hydrogen peroxide, H2O2, is -1 (Figure 2-110). As this is an intermediate oxidation state between the two much more common ones, -2 and 0, one would expect H2O2 to disproportionate. It does. In Volume 1, we learned about the decomposition of H2O2, catalyzed by Baker's yeast or MnO2.

2 H2O2 = 2 H2O + O2

In this decomposition reaction, one of the two O atoms of H2O2 is reduced to the oxidation state of -2 (water) and the other oxidized to the oxidation state of 0 (oxygen gas).

Opposite to disproportionation is comproportionation, a redox process where two compounds of the same element in a higher and lower oxidation state react to give a product containing the element in an intermediate oxidation state. An example of comproportionation is the formation of sulfur (S) from hydrogen sulfide (H2S) and sulfur dioxide (SO2). The balancing of this reaction is straightforward (Figure 2-117). Two of the three S atoms produced in the reaction come from the oxidation of H2S and one from the reduction of SO2.
Figure 2-117. Balancing the comproportionation reaction of SO2 with H2S.

2.10.3. Exercises.

1. Identify redox reactions among the following ones.

(a) Cu(OH)2 = CuO + H2O

(b) AgNO3 + KCl = AgCl + KNO3

(c) 2 AgNO3 + Cu = 2 Ag + Cu(NO3)2

(d) NH4Cl + NaOH = NaCl + NH3 + H2O

(e) 2 CuSO4 + 4 KI = 2 CuI + 2 K2SO4 + I2

(f) 2 Na + S = Na2S

(g) 2 K + 2 H2O = 2 KOH + H2

(h) K2O + H2O = 2 KOH

(i) 2 FeCl2 + Cl2 = 2 FeCl3

(j) 2 NaHCO3 = Na2CO3 + CO2 + H2O

2. Balance the following reactions using the oxidation state method. Follow the 4-step procedure as exemplified by Figures 2-112, 2-113, 2-115, and 2-117.

(a) Ag + HNO3 = AgNO3 + NO2 + H2O

(b) F2 + H2O = HF + O2

(c) NH3 + O2 = NO + H2O

(d) N2 + H2 = NH3

(e) KClO3 + FeCl2 + HCl = KCl + FeCl3 + H2O

(f) KMnO4 + H2O2 = MnO2 + KOH + O2 + H2O

(g) BrF = BrF3 + Br2

(h) P + HNO3 = H3PO4 + NO2 + H2O

(i) Fe2O3 + C = Fe + CO

(j) HIO3 + HI = I2 + H2O

3. Define disproportionation and comproportionation in chemistry. Are there disproportionation and/or comproportionation reactions among those in Problem 2 above? Answer

4. Calcium in its salts can disproportionate. True or false? Answer

5. Would you expect any disproportionation and/or comproportionation reactions for compounds of Cu and Fe? Explain. Answer