{"id":1305,"date":"2025-11-26T16:42:18","date_gmt":"2025-11-26T21:42:18","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/?post_type=chapter&p=1305"},"modified":"2026-03-23T15:08:14","modified_gmt":"2026-03-23T19:08:14","slug":"1305","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/chapter\/1305\/","title":{"raw":"1. Balancing Chemical Reactions","rendered":"1. Balancing Chemical Reactions"},"content":{"raw":"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.\r\n\r\nThe 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.\r\n\r\n\\[\\ce{AlO(OH)_{(s)} + H2O_{(l)} + NaOH_{(aq)} = NaAl(OH)4_{(aq)}} \\tag{1}\\]\r\n\r\nA significant fraction of the world\u2019s 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.\r\n\r\n\\[\\ce{CuSO4_{(aq)} + H2O_{(l)} = Cu_{(s)} + 1\/2O2_{(g)} + H2SO4_{(aq)}} \\tag{2}\\]\r\n\r\nZinc 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 [pb_glossary id=\"2083\"]roasting[\/pb_glossary] ZnS to form ZnO and then dissolving it in acidic solution.\r\n\r\n\\[\\ce{ZnS_{(s)} + 3\/2O2_{(g)} = ZnO_{(s)} + SO2_{(g)}} \\tag{3}\\]\r\n\r\n\\[\\ce{ZnO_{(s)} + H2SO4_{(aq)} = ZnSO4_{(aq)} + H2O_{(l)}} \\tag{4}\\]\r\n\r\nThe second involves reaction with oxygen at elevated temperature and pressure.\r\n\r\n\\[\\ce{ZnS_{(s)} + 2O2_{(g)} = ZnSO4_{(aq)}} \\tag{5}\\]\r\n

1.1 Concept of Mass Balance<\/h2>\r\nThe conservation of matter principle states that matter is neither created nor destroyed. In other words, atoms don\u2019t 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 \u201c=\u201d sign and remember that a balanced reaction meets the condition that all the atoms on the left are also present on the right.\r\n

1.2 Procedure for Balancing Reactions<\/h2>\r\nThere 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.\r\n
Note:<\/strong> 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.<\/blockquote>\r\nThe 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!\r\n\r\nIn acid solution:\r\n
    \r\n \t
  1. Separate the half reactions<\/li>\r\n \t
  2. Balance for the non-water elements (those other than H and O)<\/li>\r\n \t
  3. Balance for oxygen atoms with H2O<\/li>\r\n \t
  4. Balance for hydrogen atoms with H+.<\/li>\r\n \t
  5. Balance for charge with electrons (denoted e-)<\/li>\r\n \t
  6. Recombine the balanced half reactions<\/li>\r\n \t
  7. CHECK!<\/li>\r\n<\/ol>\r\nIn basic solution:\r\n
      \r\n \t
    1. Separate the half reactions<\/li>\r\n \t
    2. Balance for the non-water elements (those other than H and O)<\/li>\r\n \t
    3. Balance for oxygen atoms with H2<\/sub>O<\/li>\r\n \t
    4. Balance for hydrogen atoms with H+.<\/li>\r\n \t
    5. Balance for charge with electrons (denoted e-)<\/li>\r\n \t
    6. Recombine the balanced half reactions<\/li>\r\n \t
    7. Convert to base form by adding H2<\/sub>O = H+<\/sup> + OH-<\/sup>\u00a0 or\u00a0 H+<\/sup> + OH-<\/sup> = H2<\/sub>O<\/li>\r\n \t
    8. Check!<\/li>\r\n<\/ol>\r\n

      1.3 Examples<\/h2>\r\nThe 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:\r\n

      Example 1<\/h1>\r\nPyrite (FeS2<\/sub>) 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:\r\n\r\n\\[\\ce{FeS2 + O2 = Fe^{3+} + SO4^{2-}} \\tag{6}\\]\r\n\r\nThe indicated reagents are FeS2<\/sub> and O2<\/sub>. Acid is taken care of in the balancing sequence. The indicated products are Fe+3<\/sup> and SO4<\/sub>2-<\/sup>.\r\n
      Note<\/strong> that you must be able to write chemical formulas for simple compounds, and recognize which salts are soluble in water, e.g. Fe2<\/sub>(SO4<\/sub>)3<\/sub>. These are covered in later sections in this review. Mineral formulas will be provided.<\/blockquote>\r\n
      Step (a):<\/h5>\r\nLooking at the reaction it is clear that FeS2<\/sub> reacts to form Fe+3<\/sup> and SO4<\/sub>2-<\/sup>:\r\n\r\n\\[\\ce{FeS2 = Fe^{3+} + SO4^{2-}} \\tag{7}\\]\r\n\r\nO2<\/sub> is also involved, but how? It is not clear from the information given. Therefore simply write:\r\n\r\n\\[\\ce{O2 = ?} \\tag{8}\\]\r\n
      Step (b):<\/h5>\r\nBalance 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).\r\n\r\n\\[\\ce{FeS2 = Fe^{3+} + 2SO4^{2-}} \\tag{9}\\]\r\n\r\nAnd, note, S is present as SO4<\/sub>2-<\/sup>, so we need 2 SO4<\/sub>2-<\/sup>. Half reaction (8) involves only O, so nothing need be done to it at this stage.\r\n
      Step (c):<\/h5>\r\nBalance the O atoms with H2<\/sub>O. There are 8 O on the right of half reaction (9) and none on the left. Add 8 H2<\/sub>O to the left side:\r\n\r\n\\[\\ce{FeS2 + 8H2O = Fe^{3+} + 2SO4^{2-}} \\tag{10}\\]\r\n\r\nNext we can start to deal with the O2<\/sub> = ? half reaction:\r\n\r\n\\[\\ce{O2 = 2H2O} \\tag{11}\\]\r\n\r\nSince no product for O2<\/sub> is specified, simply follow the steps outlined above.\r\n
      Step (d)<\/h5>\r\nBalance for H atoms with H+<\/sup>. In half reaction (10) there are 16 H atoms as H2<\/sub>O on the left; none on the right. Add 16 H+<\/sup> to the right. For the O2<\/sub> half reaction (11) add 4 H+<\/sup> to the left.\r\n\r\n\\[\\ce{FeS2 + 8H2O = Fe^{3+} + 2SO4^{2-} + 16H+} \\tag{12}\\]\r\n\r\n\\[\\ce{O2 + 4H+ = 2H2O} \\tag{13}\\]\r\n
      Step (e)<\/h5>\r\nBalance 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).\r\n\r\n\\[\\ce{FeS2 + 8H2O = Fe^{3+} + 2SO4^{2-} + 16H+ + 15e-} \\tag{14}\\]\r\n\r\n\\[\\ce{O2 + 4H+ + 4e- = 2H2O} \\tag{15}\\]\r\n
      Note<\/strong> that 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\u2019ve made a mistake.<\/blockquote>\r\n
      Step (f)<\/h5>\r\nRecombine the balanced half reactions. The key here is that in the final reaction, electrons do not appear<\/em>. 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.\r\n\r\nIn 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:[latex]\\begin{align*}\r\n\\ce{4FeS2}&+\\ce{\\cancel{32}2H2O}=\\ce{4Fe^{+3}}+\\ce{8SO4^{2-}}+\\ce{\\cancel{64}4H+}+\\ce{\\cancel{60}e-}\\tag{16} \\\\\r\n\\ce{15O2}&+\\ce{\\cancel{60}H+}+\\ce{\\cancel{60}e-}=\\ce{\\cancel{30}H2O}\\tag{17} \\\\\r\n\\hline\r\n\\ce{4FeS2}&+\\ce{2H2O}+\\ce{15O2}=\\ce{4Fe^{+3}}+\\ce{8SO4^{2-}}+\\ce{4H+}\\tag{18}\r\n\\end{align*}[\/latex]\r\n

      [latex]\r\n4FeS_{2} + \\overset{2}{\\cancel{32}}H_2O = 4Fe^{+3} + 8SO_{4\\,2-} + \\overset{4}{\\cancel{64}}H^{+} + \\cancel{60}e^{-}\r\n[\/latex] \\[\\ce{} \\tag{19}\\]<\/p>\r\n[latex]\\begin{align*}\r\n\\ce{15O2}&+\\ce{\\cancel{60}H+}+\\ce{\\cancel{60}e-}=\\ce{\\cancel{30}H2O}\\tag{20} \\\\\r\n\\hline\r\n\\ce{4FeS2}&+\\ce{2H2O}+\\ce{15O2}=\\ce{4Fe^{3+}}+\\ce{8SO4^{2-}}+\\ce{4H+}\\tag{21}\r\n\\end{align*}[\/latex]\r\n\r\n \r\n

      Step (g)<\/h5>\r\nCHECK! 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.\r\n\r\n[table id=20 \/]\r\n
      Note<\/strong> that e- never<\/em> appear in the final balanced chemical reaction!\r\n\r\n <\/blockquote>\r\nThe 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:\r\n\r\naq = aqueous solution (for ions and other solutes)\r\ns = solid (e.g. all minerals)\r\ng = gas (e.g. O2<\/sub>)\r\nl = liquid (e.g. H2<\/sub>O, the solvent)\r\n\r\nThen the reaction is written:\r\n\r\n\\[ \\ce{4FeS2_{(s)} + 2H2O_{(l)} + 15O2_{(g)} -> 4Fe^{3+}_{(aq)} + 8SO4^{2-}_{(aq)} + 4H+_{(aq)}} \\tag{22}\\]\r\n\r\nNote<\/strong>: This is the reaction in ionic form<\/em>. The following rules apply:\r\n