Module 5

Lesson 2.4  Balancing Redox Reactions


Key Concepts

A redox reaction includes both oxidation and reduction. One substance is losing electrons and the other substance is gaining electrons. The theory of electron transfer requires that the total number of electrons gained in a reaction must equal the total number of electrons lost.


You are expected to know how to balance a skeleton equation. A skeleton equation, sometimes called an unbalanced equation, shows only the main reactants and products of the reaction. A skeleton equation does not indicate quantities. In other words, a skeleton equation is not balanced.


Following are two methods used to balance skeleton equations - the half-reaction method and the oxidation number method.


  1. Half-Reaction Method
    1. Split the skeleton equation into two half-reactions.
    2. Find the half-reactions on the redox table. If the redox table does not provide the half-reaction, you can construct your own half-reactions using the method you learned in Lesson 1. Balance each half-reaction.
    3. To balance the electrons, multiply one or both half-reactions by the simplest whole numbers.
    4. Cancel anything that appears on both sides of the arrow, including the electrons.
    5. Add the two balanced half-reactions to form an equation, which we refer to as a net ionic equation.

Example 

Nickel metal is reacted with a silver ion solution.

Write the skeleton equation:

\( \mathrm { Ni(s) + Ag^+(aq) \rightarrow Ni^{2+}(aq) + Ag(s) } \)


Split equation into two half-reactions. In this case, both half-reactions appear on the redox table in your data booklet.

\( \mathrm { Ag^+(aq) + 1e^- \rightarrow Ag(s) } \)

\( \mathrm { Ni(s) \rightarrow Ni^{2+}(aq) + 2e^- } \)


Multiply by coefficients to balance the electrons.

\( \mathrm { Ag^+(aq) + 1e^- \rightarrow Ag(s) \bf{ ~\times~2 } } \)

\( \mathrm { Ni(s) \rightarrow Ni^{2+}(aq) + 2e^- } \)


Add the half-reactions to form a "net ionic equation", cancelling any terms that appear on both sides of the equation.


Fig. 1


Although most metals and non-metals have relatively simple half-reaction equations, polyatomic ions and molecular compounds undergo more complex redox processes.  In these cases, the reactions often occur in an acidic or basic solution.  You are expected to devise balanced half-reactions in neutral or acidic conditions, but not basic.

Watch the video below for examples of how to balance redox reactions using the half-reaction method.

Watch




Check Your Understanding


Go to the textbook and complete Practice Question 31 on page 581. Click on the banner below to check your work.

Page 581 Practice Question 31

  1. \( \mathrm { 4 [Zn(s) \rightarrow Zn^{2+}(aq) + 2 e^-] } \)
    \( \mathrm { NO_3^-(aq) + 10 H^+(aq) + 8 e^- \rightarrow NH_4^+(aq) + 3 H_2O(l) } \)
    \( \mathrm { 4 Zn(s) + NO_3^-(aq) + 10 H^+(aq) \rightarrow 4 Zn^{2+}(aq) + NH_4^{+}(aq) + 3 H_2O(l) } \)


  2. \( \mathrm { Cl_2(g) + 2 e^- \rightarrow 2 Cl^-(aq) } \)
    \( \mathrm { SO_2(g) + 2 H_2O(l) \rightarrow SO_4^{2-}(aq) + 4 H^+(aq) + 2 e^- } \)
    \( \mathrm { Cl_2(g) + SO_2(g) + 2 H_2O(l) \rightarrow 2 Cl^-(aq) + SO_4^{2-}(aq) + 4 H^+(aq) } \)


  1.  
  2. Oxidation Number Method


    An alternate method of balancing redox reactions does not require half-reactions. This method uses oxidation numbers, which you learned about earlier in this module.

    1. Assign oxidation numbers and identify the entities whose oxidation numbers change.
    2. Use the change in oxidation number to identify the number of electrons transferred per atom.
    3. Use the chemical formula subscripts to identify the number of electrons transferred per reactant.
    4. Calculate the simplest whole number coefficients that will balance the number of electrons transferred for each reactant. Use these coefficients to balance the species that are oxidized and reduced.
    5. Balance the rest of the elements by inspection. Leave H and O until the end.
    6. Balance the O atoms using H20(l).
      • In acidic solutions, add H+ to the side deficient in hydrogen atoms.
      • In basic solutions, there is one additional step. Add hydroxide ions to both sides equal in number to the number of H+ present. Then combine the hydroxide ions and the H+ ions to form water molecules.  Cancel the same number of water molecules on both sides. 



Example 1

Balance the following reaction using oxidation numbers.


First, assign oxidation numbers.



Fig. 2


Determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. These numbers become the coefficients for the reactants.

\( \mathrm { 1 } \)\( \mathrm { I_2(s) } \) \( \mathrm { + } \) \( \mathrm { 5 } \)\( \mathrm { NaOCl(aq) } \) \( \mathrm { + } \) \( \mathrm { NaOH(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { NaCl(aq) } \) \( \mathrm { + } \) \( \mathrm { NaIO_3(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(aq) } \)
\( \mathrm { 5e^- / I } \) \( \mathrm { 2e^-/Cl } \)
\( \mathrm { 1(10e^-/I_2) } \) \( \mathrm { 5(2e^-/NaOCl) } \)

The coefficients for the reduced and oxidized products are determined next

\( \mathrm { 1 } \)\( \mathrm { I_2(s) } \) \( \mathrm { + } \) \( \mathrm { 5 } \)\( \mathrm { NaOCl(aq) } \) \( \mathrm { + } \) \( \mathrm { NaOH(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { 5 } \)\( \mathrm { NaCl(aq) } \) \( \mathrm { + } \) \( \mathrm { 2 } \)\( \mathrm { NaIO_3(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(aq) } \)

Balance other atoms (such as Na).

\( \mathrm { 1 } \)\( \mathrm { I_2(s) } \) \( \mathrm { + } \) \( \mathrm { 5 } \)\( \mathrm { NaOCl(aq) } \) \( \mathrm { + } \) \( \mathrm { 2 } \)\( \mathrm { NaOH(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { 5 } \)\( \mathrm { NaCl(aq) } \) \( \mathrm { + } \) \( \mathrm { 2 } \)\( \mathrm { NaIO_3(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(aq) } \)

Balance H and O last.
Remember that you are not told that this is an acidic solution; therefore, you cannot add H+ to balance the hydrogen. As it turns out, the hydrogen and oxygen atoms are balanced already.
Final balanced equation:

\( \mathrm { 1 } \) \( \mathrm { I_2(s) } \) \( \mathrm { + } \) \( \mathrm { 5 } \) \( \mathrm { NaOCl(aq) } \) \( \mathrm { + } \) \( \mathrm { 2 } \) \( \mathrm { NaOH(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { 5 } \) \( \mathrm { NaCl(aq) } \) \( \mathrm { + } \) \( \mathrm { 2 } \) \( \mathrm { NaIO_3(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(aq) } \)


Example 2

Balance the following skeletal equation. Assume acidic conditions.


Assign oxidation numbers



Fig. 3


Determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. These numbers become the coefficients for the reactants.


\( \mathrm { 3 } \)\( \mathrm { As_2O_3(s) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO_3^-(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { H_3AsO_4(aq) } \) \( \mathrm { + } \) \( \mathrm { NO(g) } \)
\( \mathrm { 2e^- / As } \) \( \mathrm { 3e^-/N } \)
\( \mathrm { 3(4e^-/ As_2O_3) } \) \( \mathrm { 4(3e^-/ NO_3^-) } \)

Determine the coefficients for the reduced and oxidized products. Leave oxygen and hydrogen until the next step.

\( \mathrm { 3 } \)\( \mathrm { As_2O_3(s) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO_3^-(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { 6 } \)\( \mathrm { H_3AsO_4(aq) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO(g) } \)

Balance oxygen atoms using H20(l).
The reactant side has 21 oxygen atoms, but the product side has 28 oxygen atoms. Therefore, add 7 H2O(l) to the reactant side

\( \mathrm { 3 } \)\( \mathrm { As_2O_3(s) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO_3^-(aq) } \) \( \mathrm { + } \) \( \mathrm { 7 } \)\( \mathrm { H_2O(l) } \) \( \mathrm { \rightarrow } \) \( \mathrm { 6 } \)\( \mathrm { H_3AsO_4(aq) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO(g) } \)

Finally, balance the H atoms.
The reactant side has 14 hydrogen atoms and the product side has 18 hydrogen atoms. To balance, add 4 H+(aq) on the reactant side. (You are allowed to do this because the reaction is occurring in an acidic solution.)

\( \mathrm { 3 } \)\( \mathrm { As_2O_3(s) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO_3^-(aq) } \) \( \mathrm { + } \) \( \mathrm { 7H_2O(l) } \) \( \mathrm { + } \) \( \mathrm { 4H^+(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { 6 } \)\( \mathrm { H_3AsO_4(aq) } \) \( \mathrm { + } \) \( \mathrm { 4 } \)\( \mathrm { NO(g) } \)

Note: The reaction is electrically balanced. The reactant side has 4- and 4+, which equals 0. The product side has no charges; therefore, it equals 0 also.


Watch


 Read pages 589 to 593 in the textbook


Check Your Understanding


Complete "Practice" question 12 on page 593 of the textbook.

Hint: After balancing a net ionic equation using oxidation numbers, check your work. Count up the total numbers of each atom/ion on each side and check that the charges on both sides balance.

Click on the link below to check your work.

Page 593 Practice Question 12