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22.7: Changes in Oxidation Number in Redox Reactions

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    53961
  • Zinc is an important component of many kinds of batteries. This metal is mined as zinc compounds, one of which is zinc carbonate. To obtain the pure metal, the ore must go through the following chemical processes:

    1. High temperatures and hot air blasts are used to roast the ore:
      \[\ce{ZnCO_3} \left( s \right) + \text{heat} \rightarrow \ce{ZnO} \left( s \right) + \ce{CO_2} \left( g \right)\]
    2. Then the \(\ce{ZnO}\) is treated with carbon.
      \[\begin{align} \ce{ZnO} \left( s \right) + \ce{C} \left( s \right) + \text{heat} &\rightarrow \ce{Zn} \left( g \right) + \ce{CO} \left( g \right) \\ \ce{ZnO} \left( s \right) + \ce{CO} \left( g \right) + \text{heat} &\rightarrow \ce{Zn} \left( g \right) + \ce{CO_2} \left( g \right) \end{align}\]

    The result is the pure metal which can then be fabricated into a variety of products.

    Changes in Oxidation Number in Redox Reactions

    Consider the reaction below between elemental iron and copper sulfate:

    \[\ce{Fe} + \ce{CuSO_4} \rightarrow \ce{FeSO_4} + \ce{Cu}\]

    In the course of the reaction, the oxidation number of \(\ce{Fe}\) increases from zero to \(+2\). The oxidation number of copper decreases from \(+2\) to \(0\). This result is in accordance with the activity series. Iron is above copper in the series, so will be more likely to form \(\ce{Fe^{2+}}\) while converting the \(\ce{Cu^{2+}}\) to metallic copper \(\left( \ce{Cu^0} \right)\).

    A loss of negatively-charged electrons corresponds to an increase in oxidation number, while a gain of electrons corresponds to a decrease in oxidation number. Therefore, the element or ion that is oxidized undergoes an increase in oxidation number. The element or ion that is reduced undergoes a decrease in oxidation number. The table below summarizes the processes of oxidation and reduction.

    Table 22.7.1: Processes of Oxidation and Reduction
    Oxidation Reduction
    Complete loss of electrons (ionic reaction) Complete gain of electrons (ionic reaction)
    Gain of oxygen Loss of oxygen
    Loss of hydrogen in a molecular compound Gain of hydrogen in a molecular compound
    Increase in oxidation number Decrease in oxidation number

    Example 22.7.1

    Use changes in oxidation number to determine which atoms are oxidized and which atoms are reduced in the following reaction. Identify the oxidizing and reducing agent.

    \[\ce{Fe_2O_3} \left( s \right) + 3 \ce{CO} \left( g \right) \rightarrow 2 \ce{Fe} \left( s \right) + 3 \ce{CO_2} \left( g \right)\]

    Solution:

    Step 1: Plan the problem.

    Use the oxidation number rules to assign oxidation numbers to each atom in the balanced equation. Coefficients do not affect oxidation numbers. The oxidized atom increases in oxidation number and the reduced atom decreases in oxidation number.

    Step 2: Solve.

    \[\overset{+3}{\ce{Fe_2}} \overset{-2}{\ce{O_3}} \left( s \right) + 3 \overset{+2}{\ce{C}} \overset{-2}{\ce{O}} \left( g \right) \rightarrow 2 \overset{0}{\ce{Fe}} \left( s \right) + 3 \overset{+4}{\ce{C}} \overset{-2}{\ce{O_2}} \left( g \right)\]

    The element carbon is oxidized because its oxidation number increases from \(+2\) to \(+4\). The iron (III) ion within the \(\ce{Fe_2O_3}\) is reduced because its oxidation number decreases from \(+3\) to \(0\). The carbon monoxide \(\left( \ce{CO} \right)\) is the reducing agent since it contains the element that is oxidized. The \(\ce{Fe^{3+}}\) ion is the oxidizing agent since it is reduced in the reaction.

    Summary

    • Processes for determining which atoms are oxidized and which are reduced in a chemical reaction are described.

    Contributors

    • CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.