Volume 3

The Ion Electron Method for Balancing Redox Reactions • Exercises
3.9.1. The Ion Electron Method for Balancing Redox Reactions. In Volume 2, we learned how to balance redox reactions. The balancing method was based on making the number of electrons lost by the reductant equal to the number of electrons gained by the oxidant in a given redox process. Although in many cases this method works quite well, try to use it to balance the following equation.

HNO3 + Zn = Zn(NO3)2 + NH4NO3 + H2O

Was it challenging? This equation is presented in its balanced form in Figure 3-56. I used another method to balance this reaction. This method, often referred to as the ion electron method, makes the balancing much easier, especially in the final stages. Let us balance this reaction of highly diluted HNO3 with Zn using the ion electron method.

In the first three steps, we write down the regular unbalanced equation (Step 1), convert it to the still unbalanced ionic equation (Step 2), and identify the elements that change their oxidation states (Step 3). These three steps are shown in Figure 3-109. We can see that the elements that change their oxidation states are Zn (going from 0 to +2) and N (going from +5 to -3).
Figure 3-109. The first three steps of balancing a redox reaction by the ion electron method.

In the next step (Step 4), we write half-reactions for the oxidation of the zinc atom and reduction of the nitrate anion (Figure 3-110). In the ion electron method, we write formulas of atoms, molecules, and ions present in the reaction mixture, not just the symbols of the elements and their oxidation states. In this particular case, a zinc atom is transformed into Zn2+ and a nitrate anion into NH4+. We must check if all of the elements except oxygen and hydrogen in both half-equations are balanced. In this case they are. In the top half-reaction of Figure 3-110, there is one Zn on the left and one Zn on the right. Likewise, there is one N on the left and one N on the right in the bottom half-reaction. If some balancing was required, we would do that.
Figure 3-110. Step 4: writing half-reactions and balancing them for all elements, except O and H.

The upper half-reaction (Zn → Zn2+) is balanced, but the bottom half-reaction (NO3- → NH4+) is balanced only for the nitrogen. We need to balance it for the oxygen and hydrogen, too. First, we balance the oxygen by adding the right number of molecules of water. There are three O atoms on the left and no oxygen on the right. We therefore add three molecules of H2O to the right part of the half-equation (Figure 3-111).
Figure 3-111. Step 5: balancing oxygen in half-reactions.

Now that the oxygen atoms are balanced, we balance the hydrogen atoms. There are no H atoms on the left. On the right, there are a total of ten H atoms (four in the NH4+ and six in the three molecules of H2O). To balance the hydrogen, we add 10 protons to the left part of the equation (Figure 3-112).
Figure 3-112. Step 6: balancing hydrogen in half-reactions.

All of the elements in both half-equations are now balanced. Now we need to balance the charges. For that (Figure 3-113), we calculate the total charges on the left part and on the right part for both half-reactions. In the upper half-equation, the charge on the left part is zero because a zinc atom is electroneutral. On the right part, we have a zinc cation that bears the charge of +2. By subtracting two electrons from the left we make the half-equation charge-balanced. This subtraction actually makes perfect chemical sense, as a zinc atom loses two electrons when oxidized.

In the bottom half-equation in Figure 3-113, the total charge on the left is +9 because the NO3- bears the charge of -1 and the total charge of ten H+ ions is +10. On the right, there is only one charged species, the ammonium cation that has a charge of +1. To balance the charges, we add 8 electrons to the left part of the half-reaction.
Figure 3-113. Step 7: adding electrons to charge-balance the half-reactions.

Now we need to balance the number of electrons donated by the Zn and the number of electrons accepted by the NO3- (Step 8). For that, we multiply the top half-equation by 4 (Figure 3-114). Then we add the two resultant half-equations (Step 9, Figure 3-115). By cancelling out the electrons, we derive the final, balanced net ionic equation for the redox reaction.
Figure 3-114. Step 8: equalizing the number of donated and accepted electrons.
Figure 3-115. Step 9: adding the half-reactions and simplifying the result produces the final balanced net ionic equation.

The resultant net ionic equation can be easily converted to total ionic and regular molecular equations (Figure 3-116). For that, we just need to add the right number of NO3- anions to both parts of the equation. How to calculate that number? Just think logically. There are ten positively charged H+ ions and one negatively charged NO3- in the left part of the equation, making the total charge of +10 + -1 = +9. To counterbalance the charge of +9, nine negative charges are needed, which are provided by nine NO3- anions.
Figure 3-116. Step 10: deriving total ionic and molecular equations from a net ionic equation.

Remember that a correctly balanced ionic equation has not only an equal number of atoms, but also the same total charge on the left and on the right. This total charge is not necessarily zero for a net ionic equation but is always zero for a total ionic equation. Always double check that your equation is correctly balanced. If the balancing is not working out, look for an error by reviewing every step right from the start.

Next we balance the following reaction.

K2Cr2O7 + SO2 + H2SO4 = Cr2(SO4)3 + K2SO4 + H2O

In this reaction, potassium dichromate, a powerful oxidant, is reduced by SO2, a commonly used reducing agent. As the reaction rapidly occurs, the pungent odor of SO2 disappears and the orange color of K2Cr2O7 changes to the characteristic green color of Cr2(SO4)3 (and other salts of Cr3+). The first three steps of the balancing are presented in Figure 3-117.
Figure 3-117. Balancing the reaction of K2Cr2O7 with SO2 in the presence of H2SO4, steps 1-3.

The next three steps of the balancing are presented in Figure 3-118. Note that before we balance the oxygen and hydrogen in Steps 5 and 6, we must balance the atoms of the other elements. Since Cr2O72- contains two Cr atoms, we have to put the coefficient 2 in front of the Cr3+ on the right of the half-equation.
Figure 3-118. Balancing the reaction of K2Cr2O7 with SO2 in the presence of H2SO4, steps 4-6.

Next, we balance the charges and equalize the donated and accepted electrons (Figure 3-119).
Figure 3-119. Balancing the reaction of K2Cr2O7 with SO2 in the presence of H2SO4, steps 7 and 8.

Finally, we add the two half-equations, simplify the result, and derive the net ionic, total ionic, and molecular equations (Figure 3-120). Done.
Figure 3-120. Balancing the reaction of K2Cr2O7 with SO2 in the presence of H2SO4, steps 9 and 10.

In the two examples above, we used H2O and H+ for balancing the oxygen (Step 5) and hydrogen (Step 6) in the half-equations. That was appropriate because both reactions occur in the presence of an acid. However, some redox reactions occur under basic conditions where the concentration of H+ is negligible because of the presence of hydroxide ions, OH-. Therefore, when balancing oxygen and hydrogen atoms for such reactions we cannot use hydrogen ions, H+. We should use H2O and OH- instead. Let me show you how to do that by balancing the following reaction.

K2SO3 + KMnO4 + KOH = K2SO4 + K2MnO4 + H2O

In this redox reaction, dark purple potassium permanganate (KMnO4; Mn oxidation state = +7) is reduced to dark green potassium manganate (K2MnO4; Mn oxidation state = +6). The reductant is potassium sulfite (K2SO3; S oxidation state = +4), which undergoes oxidation to potassium sulfate (K2SO4; S oxidation state = +6).

The first three steps are, as always, the conversion of the unbalanced molecular equation to an unbalanced ionic equation and identifying the atoms that change their oxidation states in the reaction (Figure 3-121).
Figure 3-121. Balancing the reaction of KMnO4 with K2SO3 in the presence of KOH, steps 1-3.

Next, we come up with half-reactions for the overall redox transformation, also as before (Figure 3-122).
Figure 3-122. Balancing the reaction of KMnO4 with K2SO3 in the presence of KOH, step 4.

Now we have to balance the half-reactions. This could not be easier for the upper one, since atom-wise it is already balanced. As for the bottom half-reaction, there is a deficiency of two oxygen atoms on the left side of the half-equation. Somehow, we need to add two O atoms to the left. Although using H2O and H+ would be easy (SO32- + 2 H2O → SO42- + 4 H+), we cannot do it in this case because there are virtually no H+ ions in the alkaline solution. We need to get by with just H2O and OH-, and this is actually doable (Figure 3-123). For each one oxygen atom to be added, add two OH- anions to the oxygen-deficient part of the equation and one molecule of H2O to the other part. That takes care of the issue.
Figure 3-123. Balancing both the oxygen and hydrogen atoms using H2O and OH-.

The rest of the balancing is done (Figure 3-124) in the same way as for redox reactions occurring under acidic conditions.
Figure 3-124. Balancing the reaction of KMnO4 with K2SO3 in the presence of KOH: final steps.

So, to balance half-reactions under acidic conditions, H2O and H+ are used. The balancing of half-reactions under basic conditions is done by using H2O and OH-. What do we do if a redox reaction occurs under neutral conditions? In such cases, use H2O on the left and either H+ or OH- on the right of a half-reaction.

The ion electron balancing method is a powerful tool. Master it and you will be amazed how much you have improved not only your redox reaction balancing skills, but also your overall knowledge and understanding of chemistry.

There are some redox reactions, however, that should not be balanced by the ion electron method. Those are reactions where oxidation and reduction processes occur not in solution, but in the gas phase or at the gas-solid or solid-solid interface. Examples of such transformations include oxidations with O2, the synthesis of ammonia from N2 and H2, and reactions of halogens with metals or hydrogen. To balance such reactions we may use the simpler oxidation state method described in Volume 2. In many cases, however, the balancing can be done easily without even thinking about oxidation states and electrons moving from one species to another. As we have learned from Volume 1, the balancing of oxidation reactions with O2 is made easy by just remembering that the number of O atoms provided by any number of O2 molecules is always even. Let me show you how to balance one of the trickiest oxidation reactions involving O2.

The air-oxidation of the mineral pyrite (FeS2) is the first step of the production of sulfuric acid (Figure 3-36). Let us balance this redox reaction.

FeS2 + O2 = Fe2O3 + SO2

First, we notice that while the number of oxygen atoms in SO2 is even, in Fe2O3 it is odd. To make it even, we place 2 in front of the Fe2O3:

FeS2 + O2 = 2 Fe2O3 + SO2

Now we have four Fe atoms on the right but only one on the left. To balance the iron, we put 4 before FeS2:

4 FeS2 + O2 = 2 Fe2O3 + SO2

The iron is now balanced, but the sulfur is not. There are 8 S atoms on the left, but only one on the right. So, we place 8 before SO2:

4 FeS2 + O2 = 2 Fe2O3 + 8 SO

Now that both the Fe and S are balanced, we need to balance the oxygen. That should be doable because we have already ensured that the number of O atoms on the right is even. On the right, there are 6 O atoms in the two molecules of Fe2O3 and 16 O atoms in the 8 molecules of SO2. The total is 6 + 16 = 22, meaning that we need 22 atoms of oxygen in the left part of the equation. By putting 11 in front of the O2 we get our equation balanced:

4 FeS2 + 11 O2 = 2 Fe2O3 + 8 SO2

3.9.2. Exercises.

1. Balance both the atoms and charges in the following half-reactions.

(a) HSO3- → SO42- (acidic conditions)

(b) NO3- → N2 (acidic conditions)

(c) NO3- → N2O (acidic conditions)

(d) NO3- → NO2 (acidic conditions)

(e) H2O2 → O2 (basic conditions)

(f) P → PO43- (acidic conditions)

(g) Cl2 → ClO3- (basic conditions)

(h) NO3- → NH3 (basic conditions)

(i) MnO4- → MnO2 (neutral conditions)

(j) H2 → H+ (acidic conditions)

(k) Cl2 → ClO- (basic conditions)

(l) NO2- → NO3- (acidic conditions)

(m) NO2 → NO3- (basic conditions)

(n) NH3 → NH4+ (acidic conditions)

(o) H2S → SO42- (acidic conditions)

(p) Cr2O72- → Cr3+ (acidic conditions)

(q) MnO4- → Mn2+ (acidic conditions)

(r) I- → IO3- (acidic conditions)

(s) Zn → ZnO22- (basic conditions)

(t) Cu2+ → Cu2O (basic conditions)

(u) Cu2O → Cu2+ (acidic conditions)

(v) ClO3- → ClO2 (acidic conditions)

(w) Mn2O7 → Mn2+ (acidic conditions)

(x) CrO42- → Cr(OH)3 (basic conditions)

(y) Al → AlO2- (basic conditions)

(z) FeO → Fe3+ (acidic conditions)


2. Balance the following redox reactions using the ion electron method.

(a) Cl2 + NaOH = NaCl + NaOCl + H2O

(b) Cu + HNO3 (concentrated) = Cu(NO3)2 + NO2 + H2O

(c) Hg + HNO3 (dilute) = Hg(NO3)2 + NO + H2O

(d) KMnO4 + HCl = KCl + MnCl2 + Cl2 + H2O

(e) AgNO3 + KOH + H2O2 = Ag + KNO3 + O2

(f) Cl2 + KOH = KClO3 + KCl + H2O

(g) CrO3 + SO2 = Cr2(SO4)3 (acidic conditions)

(h) Cl2 + H2S + H2O = HCl + H2SO4 (acidic conditions)

(i) FeSO4 + O2 + H2SO4 = Fe2(SO4)3 + H2O

(j) K2Cr2O7 + NO2 + HNO3 = KNO3 + Cr(NO3)3 + H2O

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

(l) HIO3 + HI = I2 + H2O

(m) I2 + Cl2 + H2O = HIO3 + HCl (acidic conditions)

(n) Zn + NaOH = Na2ZnO2 + H2

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

(p) C + HNO3 = CO2 + NO2 + H2O

(q) H2SO3 + H2S = S + H2O

(r) Zn + HNO3 (dilute) = Zn(NO3)2 + N2O + H2O

(s) Zn + HNO3 (dilute) = Zn(NO3)2 + N2 + H2O

(t) SO2 + Br2 + H2O = H2SO4 + HBr (acidic conditions)

(u) CuSO4 + KI = K2SO4 + CuI + I2

(v) Si + H2O + KOH = K2SiO3 + H2

(w) K2Cr2O7 + CO + H2SO4 = Cr2(SO4)3 + K2SO4 + CO2 + H2O

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

(y) HIO3 + HI = I2 + H2O

(z) KClO3 + H2O2 = KCl + O2 + H2O (neutral conditions)

[Answer: No answer. You should be able to balance all of these equations yourself. If you have problems, reread this section carefully and practice more]