1. Balancing Chemical Reactions

Introduction

Hydrometallurgy is a chemical process industry. At the heart of all hydrometallurgical extraction processes is a collection of chemical reactions. What the chemical reaction tells us is how much of the specified reagents will be needed to react with the substances of interest, and how much of the specified products will be produced. Reagents cost money and products either have value, or involve disposal costs. Flow rates (amounts) of materials, reagents and products in turn determine the size of the necessary equipment. Moreover, chemical processes involve enthalpy changes. Coping with the associated heat flows is another crucial aspect of engineering hydrometallurgical processes, and this in turn requires detailed knowledge of the chemical reactions involved. Hence the ability to balance chemical reactions is a fundamentally important skill. Finally, this skill is transferrable to any industry that involves chemical processing, for instance steel making, cement manufacture, petrochemicals (e.g. polymers used for composite manufacturing), ceramics manufacturing, etc. The central chemical reactions for a few hydrometallurgical processes are outlined below.

The Bayer process for leaching bauxite ores is at the heart of the aluminum industry. After steel aluminum is the biggest tonnage metal produced in the world.

AlO(OH) _(s) + H₂O _(l) + NaOH _(aq) = NaAl(OH)_{4 (aq)}          (1)

A significant fraction of the world’s copper is made electrolytically from a purified copper sulfate solution. Cu⁺² is reduced at the cathode and water is oxidized at the anode to form copper metal.

CuSO_{4 (aq)} + H₂O _(l) = Cu _(s) + O_{2 (g)} + H₂SO_{4 (aq)}          (2)

Zinc metal is obtained mainly from the mineral sphalerite, ZnS. There are two main routes for leaching (dissolving) ZnS into an aqueous solution. The first involves roasting ZnS to form ZnO and then dissolving it in acidic solution.

ZnS _(s) + O_{2 (g)} = ZnO _(s) + SO_{2 (g)}          (3)

ZnO _(s) + H₂SO_{4 (aq)} = ZnSO_{4 (aq)} + H₂O _(l)          (4)

The second involves reaction with oxygen at elevated temperature and pressure.

ZnS _(s) + 2 O_{2 (g)} = ZnSO_{4 (aq)}          (5)

1.1 Concept of mass balance

The conservation of matter principle states that matter is neither created nor destroyed. In other words, atoms don’t spontaneously appear or disappear (barring nuclear reactions). All the atoms that enter a process must still be present afterwards. The groupings of those atoms may change, forming new molecules or ions, but the number of atoms coming in equals the numbering going out. This is the basis for a balanced chemical reaction. A balanced chemical reaction is simply a statement of mass balance. As such we can use the “=” sign and remember that a balanced reaction meets the condition that all the atoms on the left are also present on the right.

1.2 Procedure for balancing reactions

There are a number of methods for balancing chemical reactions. One of the simplest and most foolproof is presented here. The best way to learn the method is through practice. In this method oxidation states need not be assigned.

Note: Reactions are conducted in acidic solution or in basic solution. This will be specified or be obvious from the context. The procedure for reactions in basic solution differs in only one additional step.

The procedures are provided below. Explanations and examples follow. The key is to perform each step, and in the order given. Hence it is imperative that you commit the procedures to memory!

In acid solution:

(a) Separate the half reactions

(b) Balance for the non-water elements (those other than H and O)

(c) Balance for oxygen atoms with H₂O

(d) Balance for hydrogen atoms with H⁺

(e) Balance for charge with electrons (denoted e^–)

(f) Recombine the balanced half reactions

(g) CHECK!

In basic solution:

(a) Separate the half reactions

(b) Balance for the non-water elements (those other than H and O)

(c) Balance for oxygen atoms with H₂O

(d) Balance for hydrogen atoms with H⁺

(e) Balance for charge with electrons (denoted e^–)

(f) Recombine the balanced half reactions

(g) Convert to base form by adding H₂O = H⁺ + OH^– or H⁺ + OH^– = H₂O

(h) CHECK!

1.3 Examples

The first step is to try to separate out the chemical species that might be oxidized and reduced. You do not need to know which, but you do need to determine what reactants form which products. Consider the following examples:

1. Pyrite (FeS₂) in itself is not valuable. However, when it contains finely disseminated gold the pyrite is recovered and oxidatively leached. The gold then is freed from the pyrite matrix and can be leached in subsequent steps. Oxygen gas is a suitable oxidant. The reaction is carried out in an acidic solution in an autoclave to form ferric sulfate. Translating this into a preliminary reaction form gives:

FeS₂ + O₂ = Fe⁺³ + SO₄^2-          (6)

The indicated reagents are FeS₂ and O₂. Acid is taken care of in the balancing sequence. The indicated products are Fe⁺³ and SO₄^2-. Note that you must be able to write chemical formulas for simple compounds, and recognize which salts are soluble in water, e.g. Fe₂(SO₄)₃. These are covered in later sections in this review. Mineral formulas will be provided.

Step (a). Looking at the reaction it is clear that FeS₂ reacts to form Fe⁺³ and SO₄^2-:

FeS₂ = Fe⁺³ + SO₄^2-          (7)

O₂ is also involved, but how? It is not clear from the information given. Therefore simply write:

O₂ = ?          (8)

Following the rules will take care of O₂. These are the crude, unbalanced half reactions.

Step (b). Balance all elements other than H and O in each half reaction. We need two S on the write to equal the two on the left in half reaction (7).

FeS₂ = Fe⁺³ + 2 SO₄^2-          (9)

And, note, S is present as SO₄^2-, so we need 2 SO₄^2-. Half reaction (8) involves only O, so nothing need be done to it at this stage.

Step (c). Balance the O atoms with H₂O. There are 8 O on the right of half reaction (9) and none on the left. Add 8 H₂O to the left side:

FeS₂ + 8 H₂O = Fe⁺³ + 2 SO₄^2-          (10)

Next we can start to deal with the O₂ = ? half reaction:

O₂ = 2 H₂O          (11)

Since no product for O₂ is specified, simply follow the steps outlined above.

Step (d). Balance for H atoms with H⁺. In half reaction (10) there are 16 H atoms as H₂O on the left; none on the right. Add 16 H⁺ to the right. For the O₂ half reaction (11) add 4 H⁺ to the left.

FeS₂ + 8 H₂O = Fe⁺³ + 2 SO₄^2- + 16 H⁺          (12)

O₂ + 4 H⁺ = 2 H₂O          (13)

Step (e). Balance for charge with e^–. The charge on the left side must equal the charge on the right. Just as mass is conserved, so charge is not created or destroyed either. Chemical reactions do not generate net charge. (The electrical energies associated with such a process on a bulk scale would be enormous.) For half reaction (12) the charge on the left is zero and that on the right is +15. If we are to balance charge with electrons, then we can only add e^– to the right side. On the other hand, e^– have to be added to the left side for half reaction (13).

FeS₂ + 8 H₂O = Fe⁺³ + 2 SO₄^2- + 16 H⁺ + 15 e^–          (14)

O₂ + 4 H⁺ + 4 e^– = 2 H₂O          (15)

Note: In an electron transfer reaction, something must donate the electrons (be oxidized), and something must  receive them. If you have two half reactions that have the electrons on the same side, you’ve made a mistake.

Step (f). Recombine the balanced half reactions. The key here is that in the final reaction, electrons do not appear. Electrons are transferred. They do not accumulate and they are not reactants in themselves. Again, bulk charge separation would involve huge energies. Recalling the mathematical equation analogy for a balanced chemical reaction, we can add them together and cancel any common terms on both sides.

In order for all the electrons given up in half reaction (14) to be used up by half reaction (15), they must be multiplied by factors such that the number of electrons is equal for both. Then add the two half reactions:

4 x {FeS₂ + 8 H₂O = Fe⁺³ + 2 SO₄^2- + 16 H₊ + 15e^–}          (16)

15 x {O₂ + 4 H⁺ + 4 e^– = 2 H₂O}          (17)

  4 FeS₂ + H₂O = 4 Fe⁺³ + 8 SO₄^2- + H⁺ + e^–          (18)

15 O₂ + H⁺ + e^– = H₂O          (19)

4 FeS₂ + 2 H₂O + 15 O₂ = 4 Fe⁺³ + 8 SO₄^2- + 4 H⁺          (20)

Step (g). CHECK! It is always advisable to check the final result. Does it truly balance? If so, then (a) all the atoms on one side are also present on the other and (b) the sum of the charges on the left equals that on the right. Both are required. If one or both are incorrect, then there is an error in the balancing process.

   Left side   Right side

4 Fe   4 Fe

8 S   8 S

4 H   4 H

32 O   32 O

                                       0 charge   +12 + 4 – 16 = 0 net charge OK

Note that e- never appear in the final balanced chemical reaction!

The balanced reaction should also indicate the phases of the chemical species. The most stable form of the species is usually provided. These are indicated as subscripts:

aq = aqueous solution (for ions and other solutes)

s = solid (e.g. all minerals)

g = gas (e.g. O₂)

l = liquid (e.g. H₂O, the solvent)

Then the reaction is written:

4 FeS_{2 (s)} + 2 H₂O _(l) + 15 O_{2 (g)} = 4 Fe⁺³ _(aq) + 8 SO₄^2- _(aq) + 4 H⁺ _(aq)          (21)

Note: This is the reaction in ionic form. The following rules apply:

• • Soluble salts and strong acids are represented as ions.
  • Any ions not directly involved in the reaction are omitted.

The second rule can be illustrated better by another example. The balanced reaction for Al(O)OH leaching, reaction (1), was stated above to be:

AlO(OH) _(s) + H₂O _(l) + NaOH _(aq) = NaAl(OH)_{4 (aq)}          (22)

NaOH is present as Na⁺ and OH^– in solution. NaAl(OH)₄ is present as Na⁺ and Al(OH)^4-. The sodium ions are not involved in the reaction; they do not undergo chemical change. Hence we may write the reaction in ionic form as:

AlO(OH) _(s) + H₂O _(l) + OH^– _(aq) = Al(OH)^4- _(aq)           (23)

The other form for a chemical reaction is the “neutral form.” Here all cations are matched with anions such that all ions are represented as neutral salts. Once a reaction in ionic form is obtained, then the reaction in neutral form can be written. The following rules apply:

• Combine cations and anions that occur in the balanced reaction.
• Combine auxiliary ions (those not directly involved in the reaction, like Na⁺)
  as needed with ions present in the reaction.
• Combine cations with the anions, then move on to the next one, if required(or vice versa).
• More of the necessary counterions may be needed than are present in the reaction in ionic form. This is OK, so long as everything balances in the end.

Consider the pyrite oxidation example, reaction (21):

4 FeS_{2 (s)} + 2 H₂O _(l) + 15 O_{2 (g)} = 4 Fe⁺³ _(aq) + 8 SO₄^2- _(aq) + 4 H⁺ _(aq)          (24)

Start by combining the cations with sulfate. There is only one type of anion present; it can be used to combine with the two different cations.

4 Fe⁺³ _(aq) becomes 2 Fe₂(SO₄)_{3 (aq)}

4 H⁺ _(aq) becomes 2 H₂SO_{4 (aq)}

Note that by convention we will generally express sulfuric acid, H2SO4, as 2H+ and SO42- in ionic form, even though HSO4- exists. The reaction in neutral form is:

4 FeS_{2 (s)} + 2 H₂O _(l) + 15 O_{2 (g)} = 2 Fe₂(SO₄)_{3 (aq)} + 2 H₂SO_{4 (aq)}          (25)

Again, always check that it still balances:

Left side Right side

4 Fe   4 Fe

8 S   8 S

4 H   4 H

32 O   32 O

       0 charge   0 charge OK

Note on two common errors: Do not invoke chemical species that are not mentioned or that are not implicit from the context! This is a common error. For example, because pyrite is an iron sulfide (formally Fe⁺² and S₂^2-, actually), sometimes students decide that sulfide (S^2-) should be a product. This is not mentioned or implied in the problem, so don’t introduce it. There is no basis for it.

Another common error that can sometimes follow from the first error is to inadvertently generate two oxidation or two reduction half reactions. For example, suppose someone mistakenly writes that elemental sulfur and sulfate are both products of pyrite oxidation, i.e.,

FeS₂ + O₂ = Fe⁺³ + S + SO₄^2-          (26)

(Again, sulfur is not mentioned, so don’t invoke it.) Starting with (26) as written, the balanced half reaction would be:

FeS₂ + 4 H₂O = Fe⁺³ + S + SO₄^2- + 8 H⁺ + 9 e^–          (27)

But, why does the S:SO₄^2- ratio have to be 1:1? A priori it could be anything:

FeS₂ + 6 H₂O = Fe⁺³ + S + SO₄^2- + 12 H⁺ + 12 e^–          (28)

FeS₂ + 2 H₂O = Fe⁺³ + S + SO₄^2- + 4 H⁺ + 6 e^–          (29)

etc., etc.

In fact, what has been done here is to generate a half reaction that is actually two oxidation half reactions:

FeS₂ = Fe⁺³ + 2 S + 3 e^–          (30)

FeS₂ + 8 H₂O = Fe⁺³ + 2 SO₄^2- + 16 H⁺ + 15 e^–          (31)

And there are an infinite number of ways that these can be combined. Hence no one unique half reaction, nor balanced reaction can be written. (It may be noted that in practice pyrite oxidation under some conditions can indeed produce both elemental sulfur and sulfate. However, specific information as to the molar ratio of S and SO₄^2- is required to write an overall reaction, if that is desired.)

2. Pyrolusite is a manganese mineral, MnO₂. Pyrolusite is a strong oxidizing agent and one proposed way to leach it is using sulfur dioxide, a good reducing agent (also one of the “acid rain” gases), to form Mn⁺² and sulfate. The process is conducted in acidic solution. Translating this into a crude reaction gives,

MnO₂ + SO₂ = Mn⁺² + SO₄^2-          (32)

Note: A common error is to invoke O^2- as a product. The mistaken rationale is that since MnO₂ contains (formally) oxide ions, leaching to form Mn⁺² must then release oxide anions. But this is wrong; O^2- is such a strong base that the reaction,

O^2- + H₂O → 2 OH^–          (33)

goes to completion and no O^2- can be detected in solution. Thus O^2- ion never appears in a balanced reaction in water!

In acidic solution,

O^2- + 2 H⁺ = H₂O          (34)

likewise goes to completion.

Step (a). Separate the half reactions. From the crude reaction we see that:

MnO₂ = Mn⁺²          (35)

SO₂ = SO₄^2-          (36)

Note that at this stage we are not concerned about the oxygens; they will be taken care of in the balancing process. Usually this proves to be the case, though there are exceptions (see example 5).

Step (b). Balance for non-water elements. Both crude half reactions are already balanced in this regard.

Step (c). Balance for oxygens with water:

MnO₂ = Mn⁺² + 2 H₂O          (37)

SO₂ + 2 H₂O = SO₄^2-          (38)

Step (d). Balance for H atoms with H⁺:

MnO₂ + 4 H⁺ = Mn⁺² + 2 H₂O          (39)

SO₂ + 2 H₂O = SO₄^2- + 4 H⁺          (40)

Step (e). Balance for charge with e^–:

MnO₂ + 4 H⁺ + 2 e^– = Mn⁺² + 2 H₂O          (41)

SO₂ + 2 H₂O = SO₄^2- + 4 H⁺ + 2 e^–          (42)

Step (f). Recombine the half reactions so that the e- cancel:

MnO₂ + 4 H⁺ + 2 e^– = Mn⁺² + 2 H₂O          (43)

SO₂ + 2 H₂O = SO₄^2- + 4 H⁺ + 2 e^–          (44)

MnO_{2 (s)} + SO_{2 (g)} = Mn⁺²_aq + SO₄^2-_(aq) (including phases)          (45)

This is a simple reaction in the end. Note that from the involvement of H⁺ in the half reactions, the reaction as written would occur in acid solution, however, acid is not involved in the final balanced reaction, i.e. it does not appear because there is no net consumption or production.

Step (g). CHECK:

Left side   Right side

1 Mn   1 Mn

4 O   4 O

1 S   1 S

              0 charge   0 net charge OK

Finally, the reaction can be easily converted to neutral form:

MnO_{2 (s)} + SO_{2 (g)} = MnSO_{4 (aq)}          (46)

CHECK:

  Left side   Right side

1 Mn   1 Mn

4 O   4 O

1 S   1 S

              0 charge   0 net charge OK

3. Thiosulfate is a prospective leaching agent for gold and silver ores. It offers the possibility of replacing the more toxic sodium cyanide. Thiosulfate undergoes a range of decomposition reactions, one of which is hydrolysis. The nature of the reaction depends critically on pH (whether the solution is acidic or basic). In basic solution the bisulfide and sulfate may form. Anionic species are present as sodium salts. Write the reaction for thiosulfate hydrolysis in basic solution.

Again, the first thing to do is to translate the information into a crude reaction. You would be expected to know that thiosulfate is S₂O₃^2-, bisulfide is HS- and sulfate is SO₄^2-. This example involves basic solution. Further, only one reactant is mentioned, i.e. S₂O₃^2-, which forms two products, i.e. HS^– and SO₄^2-. This is an example of a disproportionation reaction; a reaction in which one species forms two new products. Hence we must write:

S₂O₃^2- = SO₄^2- + HS^–          (47)

Step (a). Separate the half reactions. The only possibility is:

S₂O₃^2- = SO₄^2-           (48)

S₂O₃^2- = HS^–           (49)

Step (b). Balance the non-H₂O elements; in this case only S.

S₂O₃^2- = 2 SO₄^2-           (50)

S₂O₃^2- = 2 HS^–           (51)

Step (c). Balance for O with H₂O:

S₂O₃^2- + 5 H₂O = 2 SO₄^2-          (52)

S₂O₃^2- = 2 HS^– + 3 H₂O           (53)

Step (d). Balance for H with H⁺:

S₂O₃^2- + 5 H₂O = 2 SO₄^2- + 10 H⁺          (54)

S₂O₃^2- + 8 H⁺ = 2 HS^– + 3 H₂O           (55)

Step (e). Balance for charge with e^–:

S₂O₃^2- + 5 H₂O = 2 SO₄^2- + 10 H⁺ + 8 e^–           (56)

S₂O₃^2- + 8 H⁺ + 8 e^– = 2 HS^– + 3 H₂O           (57)

Step (f). Recombine the half reactions so that all e^– cancel:

S₂O₃^2- + 5 H₂O = 2 SO₄^2- + 10 H⁺ + 8 e^–           (58)

S₂O₃^2- + 8 H⁺ + 8 e^– = 2 HS^– + 3 H₂O           (59)

2 S₂O₃^2- + 2 H₂O = 2 SO₄^2- + 2 HS^– + 2 H+           (60)

Note that there is one thiosulfate in each half reaction, so two result when the half reactions are added.

Step (g). For basic solutions only, convert from H+ form to basic form. This is done by using suitable multiples of the water dissociation reaction, either:

H₂O = H⁺ + OH^–          (61)

or,

H⁺ + OH^– = H₂O          (62)

in order to cancel out all H⁺, while still keeping a balanced reaction.

2 S₂O₃^2- + 2 H₂O = 2 SO₄^2- + 2 HS^– + 2 H⁺          (63)

2 H⁺ + 2 OH^– = 2 H₂O          (64)

2 S₂O₃^2- _(aq) + 2 OH^– _(aq) = 2 SO₄^2- _(aq) + 2 HS^– _(aq)          (65)

Step (h). CHECK:

  Left side   Right side

4 S   4 S

8 O   8 O

2 H   2 H

       -6 charge   -6 charge OK

Finally, convert this to neutral form by balancing negative charges with Na⁺:

2 Na₂S₂O_{3 (aq)} + 2 NaOH _(aq) = 2 Na₂SO_{4 (aq)} + 2 NaHS _(aq)          (66)

Step (h). CHECK:

  Left side   Right side

4 S   4 S

8 O   8 O

2 H   2 H

6 Na   6 Na

       0 charge   0 charge OK

4. Digenite is a copper mineral, Cu_1.8S, or Cu₉S₅. It can be leached by ferric sulfate in acid solution to form cupric sulfate and ferrous sulfate and elemental sulfur.

First translate the information into a crude reaction. Cu_1.8S and Fe⁺³ are reactants. (Ferric = Fe⁺³; sulfates are generally soluble in water. You will need to know this.) Products are cupric ion (Cu⁺²) and ferrous ion (Fe⁺²). Here one must be very careful to realize that the only sulfur reaction product is sulfur, not sulfate and not both. From the first example we saw that assuming both S and SO₄^2- to be reaction products generates two oxidation half reactions, which results in an infinite number of possibly balanced reactions. First, sulfate does not undergo any chemical change. Second, elemental sulfur is a product, as stated. Third, the reaction is done in sulfate medium (as indicated by the sulfate counterion for ferric, ferrous and cupric.) This leads to:

Cu_1.8S + Fe⁺³ = Cu⁺² + Fe⁺² + S          (67)

Step (a). Separate the half reactions:

Cu_1.8S = Cu⁺² + S          (68)

Fe⁺³ = Fe⁺²          (69)

Step (b). Balance the non-water elements:

Cu_1.8S = 1.8 Cu⁺² + S          (70)

Fe⁺³ = Fe⁺²          (71)

Step (c). Balance for O with H₂O. No oxygens involved.

Step (d). Balance for H with H⁺. No H is involved in either half reaction.

Step (e). Balance for charge with e^–:

Cu_1.8S = 1.8 Cu⁺² + S + 3.6 e^–          (72)

Fe⁺³ + e^– = Fe⁺²          (73)

Step (f). Recombine the half reactions:

Cu_1.8S = 1.8 Cu⁺² + S + 3.6 e^–          (74)

3.6 Fe⁺³ + 3.6 e^– = 3.6 Fe⁺²          (75)

Cu_1.8S _(s) + 3.6 Fe⁺³ _(aq) = 1.8 Cu⁺² _(aq) + S _(s) + 3.6 Fe⁺² _(aq)          (76)

Step (g). CHECK:

Left side   Right side

1.8 Cu   1.8 Cu

1 S   1 S

1.8 Fe   1.8 Fe

      10.8+   10.8+ OK

To put the reaction in neutral form, combine cations with SO₄^2-.

1.8 Cu⁺² = 1.8 CuSO₄

3.6 Fe⁺³ = 1.8 Fe₂(SO₄)₃

3.6 Fe⁺² = 3.6 FeSO₄

Cu_1.8S _(s) + 1.8 Fe₂(SO₄)_{3 (aq)} = 1.8 CuSO_{4 (aq)} + S _(s) + 3.6 FeSO_{4 (aq)}          (77)

CHECK:

   Left side   Right side

1.8 Cu   1.8 Cu

3.6 Fe   3.6 Fe

6.4 S   6.4 S

21.6 O   21.6 O

       0 charge   0 charge OK

5. Hydrogen peroxide, H₂O₂, may be used to destroy cyanide after gold leaching. It is unstable, however, and undergoes disproportionation to form oxygen gas.

Very little information is given here, but it is enough. The key is to follow the rules and in the order given. Also, nothing is said about the solution being acidic or basic.

Step (a). Separate the half reactions. First write the one half reaction that is indicated:

H₂O₂ = O₂         (78)

Can we write another? If this is an electron transfer reaction, then there must be a second. But it is not immediately obvious. Therefore write:

H₂O₂ = ?         (79)

Step (b). Balance the non-water elements. There are none.

Step (c). Balance for O with H₂O:

H₂O₂ = O₂ (unchanged)         (80)

H₂O₂ = 2 H₂O         (81)

Step (d). Balance for H with H⁺:

H₂O₂ = O₂ + 2 H⁺         (82)

H₂O₂ + 2 H⁺ = 2 H₂O         (83)

Step (e). Balance for charge with e^–:

H₂O₂ = O₂ + 2 H⁺ + 2 e^–         (84)

H₂O₂ + 2 H⁺ + 2 e^– = 2 H₂O         (85)

Step (f). Recombine the half reactions:

H₂O₂ = O₂ + 2 H⁺ + 2 e^–         (86)

H₂O₂ + 2 H⁺ + 2 e^– = 2 H₂O         (87)

2 H₂O_{2 (aq)} = O_{2 (g)} + 2 H₂O _(l)          (88)

Step (g). CHECK:

   Left side   Right side

4 H   4 H

4 O   4 O

 0 charge   0 charge

OK

The reaction is already in neutral form – no ions are involved. Note that despite not having water specified as a product, we could still write the balanced reaction, simply by following the rules.

6. Gold ores are usually processed using sodium cyanide, which dissolves the gold as a cyanide complex. Many other minerals also react with cyanide. A good example is copper minerals. Cupric oxide (tenorite) reacts with cyanide in basic solution to form [Cu(CN)₃]^2- and cyanate.

Cyanate is CNO^–, whereas cyanide is CN^–. The oxygen added to cyanate suggests that electron transfer has occurred, i.e. CN^– is oxidized to CNO^–. However, neither oxygen nor any other oxidant is mentioned. (Therefore, do not invoke another oxidant!) Most sodium salts are soluble, and most oxides are insoluble. Translating the text into a crude reaction yields:

CuO + CN^– = [Cu(CN)₃]^2- + CNO^–          (89)

It would appear that CuO is acting as an oxidant toward CN^–.

Step (a). Separate the half reactions. The first is fairly obvious:

CuO + CN^– = [Cu(CN)₃]^2-          (90)

We need a copper reactant and product, so CuO and [Cu(CN)₃]^2- are implicated. If [Cu(CN)₃]^2- is involved, then CN^– must be also, since it appears on the right in the complex.

For the second we have CNO^– as a product. The only reagent we have available that would work is CN^–.

CN^– = CNO^–          (91)

Step (b). Balance the non-water elements:

CuO + 3 CN^– = [Cu(CN)₃]^2-          (92)

CN^– = CNO^–          (93)

Step (c). Balance O with H₂O:

CuO + 3 CN^– = [Cu(CN)₃]^2- + H₂O          (94)

CN^– + H₂O = CNO^–          (95)

Step (d). Balance for H with H⁺:

CuO + 3 CN^– + 2 H⁺ = [Cu(CN)₃]^2- + H₂O          (96)

CN^– + H₂O = CNO^– + 2 H⁺          (97)

Step (e) Balance for charge with e^–:

CuO + 3 CN^– + 2 H⁺ + e^– = [Cu(CN)₃]^2- + H₂O          (98)

CN^– + H₂O = CNO^– + 2 H⁺ + 2 e^–          (99)

Step (f). Recombine the half reactions:

2 CuO + 6 CN^– + 4 H⁺ + 2 e^– = 2 [Cu(CN)₃]^2- + 2 H₂O          (100)

CN^– + H₂O = CNO^– + 2 H⁺ + 2 e^–          (101)

2 CuO + 7 CN^– + 2 H⁺ = 2 [Cu(CN)₃]^2- + CNO^– + H₂O           (102)

Step (g). Convert to basic form:

2 CuO + 7 CN^– + 2 H⁺ = 2 [Cu(CN)₃]^2- + CNO^– + H₂O          (103)

2 H₂O = 2 H⁺ + 2 OH^–          (104)

2 CuO _(s) + 7 CN^– _(aq) + H₂O _(l) = 2 [Cu(CN)₃]^2- _(aq) + CNO^– _(aq) + 2 OH^– _(aq)          (105)

Step (h). CHECK:

  Left side   Right side

2 Cu   2 Cu

7 CN   7 CN

2 H   2 H

3 O   3 O

 -7    -7

OK

To convert to neutral form: The reaction involves sodium cyanide, so use sodium (Na⁺) counterions:

CN^– = NaCN

[Cu(CN)₃]^2- = Na₂[Cu(CN)₃]

CNO^– = NaCNO

OH^– = NaOH

2 CuO _(s) + 7 NaCN _(aq) + H₂O _(l) = 2 Na₂[Cu(CN)₃] _(aq) + NaCNO _(aq) + 2 NaOH _(aq) (106)

CHECK:

  Left side   Right side

2 Cu   2 Cu

3 O   3 O

7 Na   7 Na

7 CN   7 CN

2 H   2 H

0 charge   0 charge

OK

7. Aluminum(III) can precipitate from a sulfuric acid solution as sodium aluminite, also known as basic aluminum sulfate, i.e. NaAl₃(SO₄)₂(OH)₆. It can be used as a means of disposing of unwanted aluminum in solution, but it can cause scaling problems as well. Write a balanced reaction for its formation.

We have a sulfate solution and Al⁺³, forming NaAl₃(SO₄)₂(OH)₆. In this case a sodium salt is not soluble, which is rare, and Na⁺ must be a reagent.

Step (a). Separate the half reactions:

We can write:

Al⁺³ + SO₄^2- + Na⁺ = NaAl₃(SO₄)₂(OH)₆          (107)

But this uses up everything mentioned in the question. We are not able to write a second “half” reaction. Therefore proceed with the one you have and see where that gets you.

Step (b). Balance non-water elements:

3 Al⁺³ + 2 SO₄^2- + Na⁺ = NaAl₃(SO₄)₂(OH)₆          (108)

Step (c). Balance for O with H₂O:

3 Al⁺³ + 2 SO₄^2- + Na⁺ + 6 H₂O = NaAl₃(SO₄)₂(OH)₆          (109)

Step (d). Balance for H with H⁺:

3 Al⁺³ + 2 SO₄^2- + Na⁺ + 6 H₂O = NaAl₃(SO₄)₂(OH)₆ + 6 H⁺         (110)

Step (e) Balance for charge with e^–:

Left side: +6   Right side: +6

No e^– are needed. The reaction does not involve electron transfer. Hence, no second “half” reaction was needed. This confirms the decision to proceed with only one “half” reaction. If e^– were to be involved, then something would be missing.

3 Al⁺³ _(aq) + 2 SO₄^2- _(aq) + Na⁺ _(aq) + 6 H₂O _(l) = NaAl₃(SO₄)₂(OH)_{6 (s)} + 6 H⁺ _(aq)          (111)

Step (f). Not needed.

Step (g). CHECK:

  Left side   Right side

3 Al   3 Al

2 S   2 S

14 O   14 O

12 H   12 H

To convert to neutral form, use sulfate anions to combine with cations, since the reaction occurs in sulfate medium.

2 Al⁺³ = Al₂(SO₄)₃

2 Na⁺ = Na₂SO₄

2 H⁺ = H₂SO₄

But this seems to cause problems. First there aren’t enough sulfates in reaction (111) to combine with all the cations. That problem will sort itself out. We can add sulfates to both sides. Second, if we use sodium counterions for the sulfate in reaction (111) we get 2 Na₂SO₄ for the 2 SO₄^2-, and we need SO₄^2- as a counterion for the Na⁺. This yields:

 Al₂(SO₄)₃ + 2 Na₂SO₄ + Na₂SO₄ + 6 H₂O = NaAl₃(SO₄)₂(OH)₆ + 3 H₂SO₄ (112)

And the reaction is not balanced; too many Na₂SO₄. In fact, what we have done is use Na⁺ as counterion for SO₄^2- and SO₄^2- as counterion for Na⁺. In doing so, we have introduced too much Na₂SO₄. We need 1 Na⁺ on the let to balance the one on the right. This gives:

Al₂(SO₄)_{3 (aq)} + Na₂SO_{4 (aq)} + 6 H₂O _(l) = NaAl₃(SO₄)₂(OH)_{6 (s)} + 3 H₂SO_{4 (aq)} (113)

This will be seen to be balanced. CHECK:

  Left side   Right side

3 Al   3 Al

5 S   5 S

26 O   26 O

1 Na   1 Na

0 charge   0 charge

OK