In the Classroom
A New Inspection Method for Balancing Redox Equations Chunshi Guo Department of Chemistry, Siping Teachers’ College, Siping 13600, Jilin Province, PRC
Although simple chemical equations are generally balanced by inspection, many chemists seem to believe that complex redox equations require special balancing techniques. Chemistry teachers often tell their students that they must learn how to use either the “change in oxidation number” or the “ion–electron half-reaction” method because it is next to impossible to balance a complicated redox equation simply by inspection. Actually they are quite mistaken about that! Not only can all chemical equations be balanced by inspection, but inspection is often the quickest and easiest way to balance a particular redox equation. However, in order to balance a complex chemical equation by inspection, you must follow some kind of logical procedure. The one described here involves the use of “linked sets”. Consider the following equation:
5 and the second set by 22. 15Pb(N3)2 + 44Cr(MnO4 ) 2 → 5Pb3O4 + 90 NO + 22Cr2 O3 + 88 MnO2 The equation is now balanced. In spite of the fact that there are 4 different elements being oxidized or reduced in this equation, it is readily balanced by inspection, without any need for oxidation numbers or half-reactions. In this next reaction, phosphorus is the only element being oxidized or reduced. The equation is difficult to balance using conventional balancing techniques, but it is easy to balance by inspection. P2I4 + P 4 + H 2O → PH 4I + H3PO4 First balance I and O and indicate their linked sets. +16 _______________________
Pb(N3)2 + Cr(MnO4)2 → Pb3O4 + NO + Cr2O3 + MnO 2 This equation is not easy to balance by any of the standard methods described in textbooks, but it is easily balanced by inspection. First Pb is balanced by placing a 3 in front of the first term, and N can then be balanced with an 18 in front of NO: ___________________________________ |
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3 Pb(N 3)2 + Cr(MnO4)2 → Pb3O4 + 18NO + Cr2O3 + MnO2 Two elements (Pb and N), which occur in three terms, have now been balanced with respect to each other. The three terms in which they occur constitute a “linked set”. (A connecting line above the equation has been used to identify the set.) Meanwhile, Cr can be balanced with a 2 in front of the second term, and then Mn is balanced with a 4 in front of MnO2. ___________________________________ |
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3 Pb(N3)2 + 2 Cr(MnO4)2 → Pb3O4 + 18NO + Cr2O3 + 4MnO2 |________________________________________ | |
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P2I4 + P4 + 4 H2 O → 4 PH4 I + H3PO4 |________________| {5
Then balance the H to tie the linked sets together. (In the first set the number of H atoms is increasing by 16, while in the second set the number is decreasing by 5.) Multiply the first set by 5 and the second by 16. 5P 2I4 + P4 + 64H2 O → 20 PH4I + 16H 3PO4 All elements have now been balanced except P, which can be balanced by choosing the right coefficient for P4. Other terms in the equation show 10 P atoms on the left and 36 on the right, indicating a need for 26 more P atoms on the left side of the equation. That means that we need 26 /4 (or 13 /2) molecules of P4. 5P 2I4 + 13/2P4 + 64 H2O → 20 PH4I + 16H 3PO4 To get rid of the fraction, simply multiply through by 2.
The Cr and Mn in Cr(MnO4)2, Cr2O3, and MnO2 are now balanced with respect to each other, and so these three terms become a second “linked set”. (Note the connecting line below the equation.) Each of the two sets contains balanced elements within the set, but the two sets are not balanced with respect to each other. At this point, all of the elements in the equation have been balanced except O, which occurs in both linked sets. We can now use the balancing of O to tie the two sets together. Notice that in the first set there are no O atoms on the left side of the equation, but there are 22 on the right. (We might write “+22” above the connecting line to indicate the increase in O atoms.) In the second set, there are 16 O atoms on the left, but only 11 on the right. (We write “-5” to show the decrease in O atoms.) |
The equation is now balanced. In this next example, the use of a “twin” term is helpful in balancing the equation. (See Rule 3.) C 6H12O6 + KMnO4 + H 2SO4 → CO2 + K 2SO4 + MnSO4 + H2 O First the C atoms are balanced by placing a 6 in front of CO2. Then K is balanced by placing a 2 in front of KMnO4, Mn is balanced with a 2 in front of MnSO4, and S is balanced with a 3 in front of H2 SO4. There are now two linked sets, indicated by connecting lines, plus the H 2O term. ____________________________________ |
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C6H12O6 + 2KMnO4 + 3H2SO4 → 6CO2 + K2SO4 + 2MnSO4 + H2O |___________________________|
+22 _________________________________
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10 P2 I4 + 13 P4 + 128 H 2O → 40 PH4 I + 32H3 PO4
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3 Pb(N3)2 + 2 Cr(MnO4)2 → Pb3O4 + 18NO + Cr2O3 + 4MnO2 |_______________________________________ | | {5
To balance the two linked sets with respect to each other, simply balance the O atoms by multiplying the first set by
Notice that the element H appears in both sets on the left side of the equation, but it appears only in H 2O on the right side. In a case like this, it is useful to add a “twin” H2 O term, so that the H2 O can be included in both linked sets. In one case it takes 6 molecules of H2 O to balance the
Vol. 74 No. 11 November 1997 • Journal of Chemical Education
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In the Classroom H in C6 H12O 6 , while in the other case it takes 3 (H2 O) molecules to balance the H in 3 H2 SO4. (The H2 O term has now been added to each set, and all the elements have been balanced within the two sets except O.) {5
C6H12O6 + 2 KMnO4 + 3 H2SO4 → 6CO2 + K2SO4 + 2MnSO4 + 6 H2O + 3(H2O) +12
Once again, balancing O can be used to tie the two linked sets together. In the first set there are 6 O atoms on the left and 18 on the right, so the set is marked “+12”. In the second set there are 20 O atoms on the left and 15 on the right, so the set is marked “{5”. To balance O, the first set must be multiplied by 5 and the second by 12. 5 C 6H12O6 + 24 KMnO 4 + 36 H2SO4 → 30 CO2 + 12K2 SO4 + 24MnSO 4 + 30 H2O + 36 (H2O) Combining the last two terms yields the balanced equation. 5C 6H12O6 + 24KMnO4 + 36H2 SO4 → 30 CO2 + 12 K2SO 4 + 24MnSO4 +66H2 O Once again, this is a fairly complicated redox equation, and it has been balanced strictly by inspection. What about ionic equations? This method works for them, too. Consider the following equation: MnO4{ + Cl { + H+ → Mn2+ + Cl2 + H2 O
+8. For the N set, the charges go from +3 on the left to 0 on the right, for a change of {3. {3
4H2 O + 4 H+ + S2{ + NO3{ → NO + SO4 2{ + 2 H2O + 8 H+ +8
To balance the charges, simply multiply the S set by 3 and the N set by 8. 12 H2 O + 32 H+ + 3 S2{ + 8NO 3{ → 8 NO + 3 SO42 { + 16 H 2O + 24 H+ Combining the H2O and H+ terms results in a fully balanced ionic equation. 8H + + 3S2 { + 8 NO3 { → 8NO + 3 SO 42{ + 4 H2O The rules for balancing complex equations by inspection are summarized below. This method is generally useful for balancing all kinds of chemical equations, and it does not require the assignment of oxidation numbers or the splitting of redox reactions into “half-reactions”. The method can be especially helpful when the equation is quite complicated and difficult to balance by more traditional techniques. Acknowledgment I would like to thank Doris Kolb at Bradley University for her collaboration in the writing of this article.
All the elements, Mn, O, Cl, and H, are readily balanced, and the terms are easily placed into two linked sets. Rules for Balancing Complex Equations by Inspection
MnO4{ + 2 Cl{ + 8 H+ → Mn2+ + Cl2 + 4 H2 O Then, examine the charges. The amount of charge increase in one set must equal the charge decrease in the other. The charge increase in the Cl set is +2 ({ 2 → 0) and the charge decrease in the Mn set is { 5 ({ 1 + 8 → + 2). {5
MnO4{ + 2 Cl{ + 8 H+ → Mn2+ + Cl2 + 4 H2 O +2
Multiplying the Mn set by 2 and the Cl set by 5 results in a completely balanced ionic equation. 2MnO4{ + 10 Cl { + 16H+ → 2Mn2+ + 5 Cl2 + 8 H2 O The method also works with ionic equations that are incomplete, such as this one: S 2{ + NO3{ → NO + SO42{
(acid solution)
The S and N are already balanced, but they constitute two different linked sets. The O atoms can now be balanced by adding H2 O and H+. For the S set, 4 O atoms are needed on the left side of the equation, so they are added as 4 H2 O molecules; 8 H+ ions must then be added on the right to balance H. For the N set, 2 O atoms are needed on the right side of the equation, so they are added as 2 H2O, with 4 H+ then being added on the left side (as “twin” terms). 4H2 O + 4 H+ + S2{ + NO3{ → NO + SO4 2{ + 2 H2O + 8 H+ Now examine the charges. For the S set, the charge goes from {2 on the left to +6 on the right, for an increase of
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1. Any elements that appear only once on each side of the equation are balanced first. 2. Terms containing elements that have been balanced with respect to each other become part of a “linked set”. Usually there will be two different linked sets in an equation. 3. Sometimes compounds that contain elements common to both linked sets are “twinned”; that is, the same compound is written twice in the same equation, so that it can become part of both linked sets. (The twin terms are later combined.) 4. An element that occurs in both linked sets can be used to tie them together. Often this will be the last element in the equation remaining unbalanced. Determine the increase or decrease in number of atoms of that element as you go from the left to the right side of the equation in each linked set. Then use these numbers as factors for balancing the two sets with respect to each other. 5. For ionic equations, first split the equation into linked sets and balance all the elements within each set (using twin terms when needed). Then determine the increase or decrease in charge for each linked set, and use these numbers as factors for balancing the two sets with respect to each other. (On occasion it may be simpler to balance the charges before balancing the last several elements.)
Journal of Chemical Education • Vol. 74 No. 11 November 1997
