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22.7: Changes in Oxidation Number in Redox Reactions

Difficulty Level: At Grade Created by: CK-12
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Zinc is an important component of many kinds of batteries

Where does the zinc in batteries come from?

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:

\begin{align*}\text{ZnCO}_3(s) + (\text{heat}) \rightarrow \text{ZnO}(s) + \text{CO}_2(g) \end{align*}ZnCO3(s)+(heat)ZnO(s)+CO2(g)

2. Then the ZnO is treated with carbon:

\begin{align*}\text{ZnO}(s) +\text{C}(s) + (\text{heat}) & \rightarrow \text{Zn}(g) + \text{CO}(g)\\ \text{ZnO}(s)+\text{CO}(g)+ (\text{heat}) & \rightarrow \text{Zn}(g)+\text{CO}_2(g)\end{align*}ZnO(s)+C(s)+(heat)ZnO(s)+CO(g)+(heat)Zn(g)+CO(g)Zn(g)+CO2(g)

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:

\begin{align*}\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}\end{align*}Fe+CuSO4FeSO4+Cu

In the course of the reaction, the oxidation number of 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 Fe2+ while converting the Cu2+ to metallic copper (Cu0).

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. Table below summarizes the processes of oxidation and reduction.

Processes of Oxidation and 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

Sample Problem: Identifying Oxidation and Reduction

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.

\begin{align*}\text{Fe}_2\text{O}_3(s)+3\text{CO}(g) \rightarrow 2\text{Fe}(s)+3\text{CO}_2(g)\end{align*}

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.

\begin{align*}\overset{+3}{\text{Fe}_2} \overset{-2}{\text{O}_3} (s)+3 \overset{+2}{\text{C}} \overset{-2}{\text{O}} (g) \rightarrow 2 \overset{0}{\text{Fe}}(s)+3 \overset{+4}{\text{C}} \overset{-2}{\text{O}_2}(g)\end{align*}

The element carbon is oxidized because its oxidation number increases from +2 to +4. The iron(III) ion within the Fe2O3 is reduced because its oxidation number decreases from +3 to 0. The carbon monoxide (CO) is the reducing agent since it contains the element that is oxidized. The Fe3+ ion is the oxidizing agent since it is reduced in the reaction.




  1. In the example reaction above, list the ways the reduction of ferric oxide fits the definitions of oxidation and reduction.
  2. In the reaction below, what is oxidized and what is reduced?\begin{align*}\text{Cr}_2\text{O}_{3(s)} + 2\text{Al}_{(s)} \rightarrow 2\text{Cr}_{(s)} + \text{Al}_{2}\text{O}_{3(s)}\end{align*}
  3. List the ways that the conversion of aluminum in the reaction above fits the definitions in the table.

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Difficulty Level:
At Grade
Date Created:
Sep 06, 2013
Last Modified:
Sep 11, 2016
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