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Chemistry LibreTexts

Half Reactions

Skills to Develop

  • Explain what half reactions are.
  • Write a chemical equation to represent a (half) reduction reaction.
  • Write a chemical equation to represent an oxidation reaction.
  • Combine reduction and oxidation equations to explain chemical reactions.
  • Write two half reactions for an unbalanced overall chemical reaction.

Half Reactions

A half reaction is a reduction or an oxidation reaction. For example, the following are half reactions.

\(\ce{2 H+ + 2 e- \rightarrow H2}\)
\(\ce{MnO4- + 5 e- + 8 H+ \rightarrow Mn^2+ + 4H2O}\)
\(\ce{Zn \rightarrow Zn^2+ + 2 e- }\)
\(\ce{Cu \rightarrow Cu^2+ + 2 e-}\)

A half reaction does not occur by itself; at least two such reactions must be coupled so that the electron released by one reactant is accepted by another in order to complete the reaction. Thus, oxidation and reduction reactions must take place simultaneously in a system, and this type of reaction is called oxidation reduction reaction or simply redox reaction. For example,

\(\ce{Zn + Cu^2+ \rightarrow Zn^2+ + Cu}\)

is such a redox reaction, \(\ce{Zn}\) being oxidized, and \(\ce{Cu^2+}\) being reduced. Redox reactions take place in battery operations.

Half-reaction equations are useful for balancing redox reaction equations.

Balancing Redox Reactions with Half Reactions

A redox reaction can take place even when the reactants are well separated in space, as long as the flow of electrons and ions are facilitated by electrical connections (salt bridge and wire). A simple redox reaction of the type:

\(\ce{Zn_{\large{(s)}} + Cu^2+_{\large{(aq)}} \rightarrow Cu_{\large{(s)}} + Zn^2+_{\large{(aq)}}}\)

may be carried out using a galvanic cell. This reaction may be written as two half-reactions and adding the two half reactions gives the overall equation representing a chemical process:

\(\begin{align}
& &\mathrm{Zn}\hspace{60px} & \rightarrow \mathrm{Zn^{2+}_{\large{(aq)}} + 2 e^-} \\
& &+\underline{\hspace{10px}\mathrm{Cu^{2+}_{\large{(aq)}} + 2 e^-}}  & \underline{\rightarrow \mathrm{Cu^{\:}_{\large{(s)}}}\hspace{60px}} \\
&   & \mathrm{Zn_{\large{(s)}} + Cu^{2+}_{\large{(aq)}}} & \rightarrow \mathrm{Cu_{\large{(s)}} + Zn^{2+}_{\large{(aq)}}}
\end{align}\)

In this case, the electrons travel from the \(\ce{Zn}\) electrode to the \(\ce{Cu}\) electrode via the wire connecting the two electrodes. The ions travel in a solution or through a salt bridge to balance the charge in the electrolyte solutions. You have already seen the diagram of a galvanic cell.

A redox reaction may be balanced by first writing two half-reactions, and then canceling the electrons by adding them algebraically. You will learn to balance half-reaction equations in this tutorial.

Guide for Writing and Balancing Half-Reaction Equations

  1. Identify the key element that undergoes an oxidation state change.
  2. Balance the number of atoms of the key element on both sides.
  3. Add the appropriate number of electrons to compensate for the change of oxidation state.
  4. Add \(\ce{H+}\) (in acid medium), or \(\ce{OH-}\) (in basic medium), to balance the charge on both sides of the half-reactions; and \(\ce{H2O}\), if necessary, to balance the equations.

Example 1

Balance the two half reactions for the reaction in an acid solution:

\(\ce{H2O2 + I- \rightarrow I2 + H2O}\)

Hint

  1. \(\ce{I-}\) is oxidized (oxidation state increases from -1 to 0).
    \(\ce{O}\) (oxygen) is reduced (oxidation state decreases from -1 to -2).
  2. The two half-reactions with balanced numbers of key atoms are:

    \(\ce{2 I- \rightarrow I2}\); (oxidation)
    \(\ce{H2O2 \rightarrow 2 H2O}\); (reduction of oxygen)

  3. Add electrons to compensate for the changes of oxidation state.

    \(\ce{2 I- \rightarrow I2 + 2 e-}\); (oxidation)
    \(\ce{H2O2 + 2 e- \rightarrow 2 H2O}\); (reduction)

  4. Obviously, \(\ce{H+}\) should be added to the reduction half-reaction, and the balanced equations are:

    \(\ce{2 I- \rightarrow I2}\); (oxidation)
    \(\ce{H2O2 + 2H+ \rightarrow 2 H2O}\); (reduction)

Example 2

Balance two half reactions for the reaction in a basic solution:

\(\ce{ClO2 + OH- \rightarrow ClO2- + ClO3-}\)

Hint

  1. In this reaction, \(\ce{Cl}\) from \(\ce{ClO2}\) is both oxidized and reduced.
  2. The two half-reactions are:

    \(\ce{ClO2 \rightarrow ClO2-}\); (reduction)
    \(\ce{ClO2 \rightarrow ClO3-}\); (oxidation)

  3. Add electrons to compensate for the oxidation changes:

    \(\ce{ClO2 + e- \rightarrow ClO2-}\); (reduced, 4 \(\rightarrow\) 3 for \(\ce{Cl}\))
    \(\ce{ClO2 \rightarrow ClO3- + e-}\); (oxidized, 4 \(\rightarrow\) 5)

  4. Add \(\ce{H+}\), \(\ce{OH-}\), or \(\ce{H2O}\) to balance both equations which results in

    \(\ce{ClO2 + e- \rightarrow ClO2-}\)
    \(\ce{ClO2 + 2 OH- \rightarrow ClO3- + e- + H2O}\)

    Now add the two half reactions together to give the overall reaction:

    \(\ce{2 ClO2 + 2 OH- \rightarrow ClO2- + ClO3- + H2O}\)

Example 3

Balance two half reactions for the reaction in an acidic solution:

\(\ce{HS2O3- \rightarrow S + HSO4-}\)

Hint

You may think that the two sulfur atoms in the formula are identical, but they are different. You have to understand the chemistry of these ions and then start to investigate their chemical reaction. The structure of \(\ce{HS2O3-}\) may be compared to that of \(\ce{HSO4-}\):

       S     O-       O     O-
        \\ /           \\ /
          S              S
        // \           // \
       O    OH        O    OH

Thus, one of the two \(\ce{S}\) atoms has an oxidation state of -2, and we represent this \(\ce{S}\) atom by (\(\ce{=S}\)) to indicate that it is attached to another \(\ce{S}\) atom by a double bond (=).

  1. In this reaction, one \(\ce{S}\) atom goes from -2 to 0, whereas the oxidation state of the other \(\ce{S}\) atom does not change. You have to assume that the \(\ce{S}\) atom is oxidized by a reducing agent, \(\ce{H2O}\).
  2. Only the key elements are given on the left in the half-reactions:

    \(\ce{HS(=S)O_3^- \rightarrow S +} \textrm{...}\ce{(HSO_4^- )}\)
    \(\ce{H_2O \rightarrow H_{2\large{(g)}}} + \textrm{...}\ce{(HSO4- )}\)

  3. Add electrons to compensate for the oxidation changes:

    \(\ce{HS(=S)O3- \rightarrow S + 2 e- + HSO4-}\)
    \(\ce{H2O + 2 e- \rightarrow H2(g) + HSO4-}\)

  4. Combining the two half-reactions gives the following balanced chemical equation:

    \(\ce{HS(=S)O3- + H2O \rightarrow S + H_{2\large{(g)}} + HSO4-}\)

Example 4

Balance the two half reaction equations for the following reaction in acidic solution:

\(\ce{S(=S)O3^2- + I2 \rightarrow I- + S4O6^2-}\)

Hint

The chemistry of the above reaction is complicated, but you don't have to worry about that at this time. You may use the above method even if you do not know the structure of these species.

  1. The \(\ce{S}\) atoms are oxidized. For convenience, the best way is to assume the average oxidation state for both \(\ce{S}\) atoms.

    \(\ce{S}\) oxidized (oxidation state (+2) \(\rightarrow\) (+10/4 or +2.5)
    \(\ce{I}\) reduced ( 0 \(\rightarrow\) -1)

  2. Write the half-reactions and balance the key elements:

    \(\ce{2 S2O3^2- \rightarrow S4O6^2-}\)
    \(\ce{I2 \rightarrow 2 I^-}\)

  3. Add electrons to compensate the oxidation state changes:

    \(\ce{2 S2O3^2- \rightarrow S4O6^2- + 2 e-}\)
    \(\ce{I2 + 2 e- \rightarrow 2 I-}\)

  4. Since the half-reactions are balanced with respect to charges and number of atoms, no further work is required. Just add the two equations and get a balanced equation.

    \(\ce{2 S2O3^2- + I2 \rightarrow 2 I- + S4O6^2-}\)

The four (4) examples above illustrate how to break down a task in steps. You will need the practice in order to master the skills. Take any redox chemical reaction equation and try to balance the two half reactions. The Confidence Building Questions below have redox reaction equations for you to practice.

The following questions require one step at a time, but you may take any question and follow the four steps as illustrated in the above examples.

Confidence Building Questions

  1. In the reaction

    \(\ce{I2 + H2S \rightarrow H+ + I- + S_{\large{(s)}}}\)

    what element is oxidized?

    Hint: Sulfur, \(\ce{S}\), is oxidized.

  2. In the reaction

    \(\ce{I2 + H2S \rightarrow H+ + I- + S_{\large{(s)}}}\)

    what element is reduced?

    Hint: Iodine, \(\ce{I}\), is reduced.

    Skill - Identify elements oxidized and reduced.

    The oxidation state for \(\ce{S}\) goes from -2 in \(\ce{H2S}\) to 0 in \(\ce{S}\), an element.
    The oxidation state of \(\ce{I}\) goes from 0 to -1.

  3. In the reaction

    \(\ce{Zn + NO3- \rightarrow Zn^2+ + NH4+}\)

    what element is reduced?

    Hint: Nitrogen in \(\ce{NO3-}\) is reduced.

  4. In the half-reaction

    \(\ce{NO3- \rightarrow NH4+}\)

    should electrons be added to the side on which \(\ce{NO3-}\) is found or that on which \(\ce{NH4+}\) is found?
    Hint: Electrons should be added to the nitrate side.

    Skill - Add electrons to balance half reactions.
    \(\ce{Zn}\) is oxidized, \(\ce{NO3-}\) is reduced. You are probably thinking that \(\ce{NH4+}\) carries a positive charge, and \(\ce{NO3-}\) carries a negative charge. But you should add electrons to compensate the oxidation state change. See next question.

  5. In the half-reaction

    \(\ce{NO3- + 8 e- \rightarrow NH4+}\)

    how many \(\ce{H+}\) ions should be added to balance the charges?
    Hint: 10 \(\ce{H+}\) ions

    Skill - Add \(\ce{H+}\) to balance the charge as a step to balance equations.

  6. How should you balance the equation?

    \(\ce{10 H+ + NO3- + 8 e- \rightarrow NH4+}\)

    Hint: Add 3 \(\ce{H2O}\) on the right-hand side.

    Skill - Add water to balance \(\ce{H}\) and \(\ce{O}\) in the equation.
    The balanced equation is

    \(\ce{10 H+ + NO3- + 8 e- \rightarrow NH4+ + 3H2O}\).

  7. In the reaction

    \(\ce{ClO3- + As2S3 \rightarrow Cl- + H2AsO4- + SO4^2-}\)

    what is the net oxidation state change of \(\ce{Cl}\)?
    Hint: It changes from +5 to -1, a decrease of 6.

    Skill - Identify element oxidized or reduced and changes of oxidation state.

  8. In the reaction

    \(\ce{ClO3- + As2S3 \rightarrow Cl- + H2AsO4- + SO4^2-}\)

    what is the net oxidation state change of \(\ce{As}\)?
    Hint: The oxidation state increases by 2.

    Discussion - \(\ce{As}\) in \(\ce{As2S3}\) has an oxidation state of +3.
    \(\ce{As}\) in \(\ce{H2As4-}\) has an oxidation state of +5.

  9. In the reaction

    \(\ce{ClO3- + As2S3 \rightarrow Cl- + H2AsO4- + SO4^2-}\)

    which two elements are oxidized?
    Hint: Sulfur, \(\ce{S}\), and arsenic, \(\ce{As}\).

    Discussion -
    The oxidation states for \(\ce{S}\) are -2 in \(\ce{As2S3}\), and +6 in \(\ce{SO4^2-}\). The oxidation state of \(\ce{S}\) changes from -2 to +6, an increase of 8.

  10. In the half-reaction

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + SO4^2-}\)

    which key element is not balanced before you calculate the number of electrons to be added?
    Hint: The element \(\ce{S}\) is not balanced yet.

    Skill - Balance all key elements if more than one is involved before adding electrons.
    There are 3 \(\ce{S}\) on the left, and only one on the right.

  11. In the half-reaction

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2-}\)

    to which side should electrons be added, the left (\(\ce{As2S3}\)) or the right (\(\ce{SO4^2-}\))?
    Hint: Both \(\ce{As}\) and \(\ce{S}\) are oxidized, add electrons on the right side.

    Skill - Add electrons as a step in balancing half-reaction equations.
    Since both \(\ce{As}\) and \(\ce{S}\) are oxidized, both give off electrons.

  12. In the half-reaction

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2-}\)

    how many electrons should be added to the right?
    Hint: Add 28 e- to the right.

    Skill - Add the correct number electrons as a step in balancing half-reaction equations.

    \(\mathrm{As\textrm : \: +3 \rightarrow +5, \: (2\: e^-)\times2}\)
    \(\mathrm{S\textrm : \: -2 \rightarrow +6, \: (8\: e^-)\times3}\)

    Total number of electrons: \(\mathrm{(2\times2 + 3\times8)\: e^- = 28\: e^-}\)

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2- + 28 e-}\)

  13. The half-reaction is carried out in an acidic solution:

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2- + 28 e-}\)

    What should you add next in order to balance it?
    Hint: Add \(\ce{H+}\) ion to balance the charges.

    Skill - Balance the charge using \(\ce{H+}\) or \(\ce{OH-}\).
    Balance the charge on the right using \(\ce{H+}\), which is abundant in an acidic solution.

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2- + 28 e- + 36 H+}\)

  14. The half-reaction is carried out in an acidic solution:

    \(\ce{As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2- + 28 e- + 36 H+}\)

    How many water molecules should be added in order to balance the half-reaction?
    Hint: Add formula for water, \(\ce{H2O}\), to balance.

    Skill - Balance half-reaction equation.
    The balanced half-reaction equation is

    \(\ce{20 H2O + As2S3 \rightarrow 2 H2AsO4- + 3 SO4^2- + 28 e- + 36 H+}\)

The skill to balance redox equations can be broken down in several steps. Each question above involves a small step. You should review the steps and acquire all the skills needed. These questions also show that you can be tested for skills to balance equations by one of these questions.

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