Module 5

Lesson 1.2  Introducing the Half-Reaction



Key Concepts


Consider the reaction shown in the preceding video:


\( \mathrm { Cu(s) + 2AgNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2Ag(s) } \)


In one part of the reaction, an atom of copper is losing two electrons to become a Cu2+ ion. In the other part of the reactions, an Ag+ ion is gaining an electron to become a silver atom. These two parts are referred to as half-reactions.

Half-reaction: a balanced chemical equation that displays either oxidation or reduction

A half-reaction represents what is happening to one reactant in the reaction. As is the case with any other chemical reaction, each half-reaction must be balanced with respect to mass and charge.

To balance a half-reaction for mass, be sure that the number of atoms of each element present in the reactants is equivalent to the number of atoms of each element present in the products.

To balance for charge, add the charges on each side of the half-equation. If the net charges on each side are equal, the equation is balanced with respect to charge. If not, add electrons to balance the charges.



  1.   Writing Balanced Simple Half-Reactions


    Most metals and non-metals have relatively simple half-reaction equations. As an example, return to the example in which copper solid reacts with aqueous silver nitrate.


    \( \mathrm { Cu(s) + 2AgNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2Ag(s) } \)


    In the reaction of copper solid with silver nitrate, one half-reaction involves the copper solid becoming Cu2+.


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


    The half-reaction shown above is balanced with respect to mass. That is, one copper atom occurs on the left and one copper atom occurs on the right.


    In the reaction of copper solid with silver nitrate, the other half-reaction involves the silver ion becoming solid silver.


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


    This half-reaction is balanced with respect to mass. That is, one silver atom occurs on the left and one silver atom occurs on the right.


    Although the half-reactions shown above are both balanced for mass, they are not balanced for charge. How do we balance each for charge?


    First, consider the half-reaction involving copper. A zero charge occurs on the left-hand side of the reaction, but a charge of 2+ occurs on the right-hand side of the reaction. Therefore, the half-reaction is not balanced with respect to charge. However, if we add two electrons to the right-hand side of the half-reaction, the two electrons will cancel the two positive charges on the copper ion, resulting in zero charge on the right-hand side of the equation.


    \( \mathrm { Cu(s) \rightarrow Cu^{2+}(aq) + 2e^- } \)
    \( \mathrm { 0 = (2+) + (2-) } \)


    Because the copper atom is losing electrons in the above reaction, it is undergoing oxidation.


    Next, consider the half-reaction involving silver. A charge of 1+ occurs on the left-hand side of the reaction, however a charge of 0 occurs on the right-hand side. Thus, the half-reaction is not balanced with respect to charge. However, if we add one electron to the left-hand side of the half-reaction, the electron will cancel the positive charge on the silver ion, resulting in zero charge on the left-hand side of the equation.


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


    \( \mathrm { (1+) + (1-) = 0 } \)

Because the silver ion is gaining electrons, it is undergoing reduction.


The nitrate ion (NO3-) is neither gaining nor losing electrons; therefore, it is not shown in the half-reactions. The nitrate ion is said to be a "spectator ion".

Spectator ion - an ion that does not participate in the chemical reaction


Chemical reactions such as the reaction between the copper coil and silver nitrate are referred to as "redox" reactions (reduction-oxidation). In such reactions, reduction and oxidation are occurring simultaneously.


To summarize the process of oxidation:

  • Oxidation involves the loss of electrons (LEO).
  • Electrons are located on the product side of a half-reaction.
  • Solid metal atoms usually undergo oxidation.
  • Non-metallic ions usually undergo oxidation.
  • The entity undergoing oxidation will undergo an increase in oxidation number (to be discussed in more detail later).


Loss of Electrons is Oxidation (LEO)


Examples of oxidation half-reactions:

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

\( \mathrm { 2Br^-(aq) \rightarrow Br_2(l) + 2e^- } \)




To summarize the process of reduction:

  • Reduction involves the gain of electrons (GER).
  • Electrons are located on the reactant side of a half-reaction.
  • Non-metallic atoms usually undergo reduction.
  • Metallic ions usually undergo reduction.
  • The entity undergoing reduction will undergo a decrease in oxidation number (to be discussed in more detail later).


Gain of Electrons is Reduction (GER)


Examples of reduction half-reactions:

\( \mathrm { F_2(g) + 2e^- \rightarrow 2F^-(aq) } \)

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



A mnemonic device is a word or group of words that help you remember information.

"LEO the Lion says GER"


  • Loss of Electrons is Oxidation (LEO)
  • Gain of Electrons is Reduction(GER)

OIL RIG:     Oxidation Is Loss; Reduction Is Gain.





Watch the following video. Which atom is being reduced, and which substance is being oxidized?

Check Your Understanding


Complete Practice Questions 7 to 11 on page 564 of the textbook. Click on the banner below to check your work.

Page 564 Practice Question 7
  1. redox reaction: a reaction in which electrons are transferred between entities
  2. reduction: a chemical process involving the gain of electrons by an entity
  3. oxidation: a chemical process involving the loss of electrons by an entity


Page 564 Practice Question 8
  1. \( \mathrm { Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) } \)
    (gain of electrons)
  2. \( \mathrm { Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- } \)
    (loss of electrons)
  3. \( \mathrm { 2 H^+(aq) + 2e^- \rightarrow H_2(g) } \)
    (gain of electrons)
  4. \( \mathrm { Mg(s) \rightarrow Mg^{2+}(aq) + 2e^- } \)
    (loss of electrons)


Page 564 Practice Question 9

  1. \( \mathrm { Ni(s) \rightarrow Ni^{2+}(aq) + 2e^- } \)
    (oxidation half-reaction)

    \( \mathrm { Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) } \)
    (reduction half-reaction)

  2. \( \mathrm { Pb(s) \rightarrow Pb^{2+}(aq) + 2e^- } \)
    (oxidation half-reaction)

    \( \mathrm { Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) } \)
    (reduction half-reaction)

  3. \( \mathrm { Ca(s) \rightarrow Ca^{2+}(aq) + 2e^- } \)
    (oxidation half-reaction)

    \( \mathrm { 2H^+(aq) + 2e^- \rightarrow H_2(g) } \)
    (reduction half-reaction)

  4. \( \mathrm { Al(s) \rightarrow Al^{3+}(s) + 3e^- } \)
    (oxidation half-reaction)

    \( \mathrm { Fe^{3+}(s) + 3e^- \rightarrow Fe(l) } \)
    (reduction half-reaction)

You may have noticed that nitrate is a spectator in a), b), and c).

Oxide is a spectator in d).



Page 564 Practice Question 10

\( \mathrm { Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) } \)

(gain of electrons)


\( \mathrm { 2 I^-(aq) \rightarrow I_2(s) + 2e^- } \)

(loss of electrons)



Page 564 Practice Question 11

A redox reaction has not occurred. The entities present before the reaction are Fe3+, Cl-, Na+, and OH-. Exactly the same entities are present after the reaction; therefore, no transfer of electrons has occurred from one entity to another.




  1.  
  2.   Writing Balanced Complex Half-Reactions


    Although most metals and non-metals have relatively simple half-reaction equations, polyatomic ions and molecular compounds undergo more complicated reduction and oxidation processes.


    You are expected to write balanced complex half-reactions for reactions occurring in neutral or acidic solutions. In these cases, water and hydrogen ions are present and have important roles.


    To write balanced complex half-reactions in a neutral or acidic solution, follow these steps:


    1. Write the chemical formulae for the reactants and the products.
    2. Balance atoms other than O and H.
    3. Balance O atoms by adding water.
    4. If the conditions are acidic, balance H atoms by adding H+(aq).
    5. Balance electrical charges by adding e-.


    Note: Although the textbook covers how to write complex half-reactions in a basic solution, you are not required to know this. You are required only to write complex half-reactions in neutral and acidic solutions.


Watch



Read pages 564 to 567 in the textbook. Work through Sample Problem 13.2 and Communication Example 2.

Check Your Understanding

To gain practice writing complex half-reactions in acidic solutions, complete Practice Questions 12 (a and d only) on page 566.

Page 566 Practice Question 12 (a and d)


  1. Step 1: \( \mathrm { N_2O(g) } \) \( \mathrm { \rightarrow } \) \( \mathrm { N_2(g) } \)
    Step 2: \( \mathrm { N_2O(g) } \) \( \mathrm { \rightarrow } \) \( \mathrm { N_2(g) } \) Nitrogen is balanced
    Step 3: \( \mathrm { N_2O(g) } \) \( \mathrm { \rightarrow } \) \( \mathrm { N_2(g) } \) \( \mathrm { + } \) \( \mathrm { H_2O(l) } \)
    Step 4: \( \mathrm { 2 H^+(aq) } \) \( \mathrm { + } \) \( \mathrm { N_2O(g) } \) \( \mathrm { \rightarrow } \) \( \mathrm { N_2(g) } \) \( \mathrm { + } \) \( \mathrm { H_2O(l) } \)
    Step 5: \( \mathrm { 2^+ } \) \( \mathrm { 0 } \) \( \mathrm { = } \) \( \mathrm { 0 } \) \( \mathrm { 0 } \)
    \( \mathrm { 2^+ } \) \( \mathrm { = } \) \( \mathrm { 0 } \)

    Add e- to the most positive side.

    \( \mathrm { 2e^- + 2H^+(aq) + N_2O(g) \rightarrow N_2(g) + H_2O(l) } \)    (reduction)

  2.  
  3.  

  4. Step 1 and 2: \( \mathrm { NO_3^-(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { HNO_2(aq) } \)
    Step 3: \( \mathrm { NO_3^-(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { HNO_2(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(l) } \)
    Step 4: \( \mathrm { 3H^+(aq) } \) \( \mathrm { + } \) \( \mathrm { NO_3^-(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { HNO_2(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(l) } \)
    Step 5: \( \mathrm { 2e^- } \) \( \mathrm { + } \) \( \mathrm { 3H^+(aq) } \) \( \mathrm { + } \) \( \mathrm { NO_3^-(aq) } \) \( \mathrm { \rightarrow } \) \( \mathrm { HNO_2(aq) } \) \( \mathrm { + } \) \( \mathrm { H_2O(l) } \) (reduction)