7.1 Redox reactions

Redox reactions

A second type of reaction is the redox reaction, in which both oxidation and reduction take place.

Oxidation and reduction

If you look back to chapter [link] , you will remember that we discussed how, during a chemical reaction, an exchange of electrons takes place between the elements that are involved. Using oxidation numbers is one way of tracking what is happening to these electrons in a reaction. Refer back to [link] if you can't remember the rules that are used to give an oxidation number to an element. Below are some examples to refresh your memory before we carry on with this section!

Examples:

1. CO ${}_2$ Each oxygen atom has an oxidation number of -2. This means that the charge on two oxygen atoms is -4. We know that the molecule of CO ${}_2$ is neutral, therefore the carbon atom must have an oxidation number of +4.
2. KMnO ${}_4$ Overall, this molecule has a neutral charge, meaning that the sum of the oxidation numbers of the elements in the molecule must equal zero. Potassium (K) has an oxidation number of +1, while oxygen (O) has an oxidation number of -2. If we exclude the atom of manganese (Mn), then the sum of the oxidation numbers equals +1+(-2x4)= -7. The atom of manganese must therefore have an oxidation number of +7 in order to make the molecule neutral.

By looking at how the oxidation number of an element changes during a reaction, we can easily see whether that element is being oxidised or reduced .

Oxidation and reduction

Oxidation is the loses of an electron by a molecule, atom or ion. Reduction is the gain of an electron by a molecule, atom or ion.

Example:

$\mathrm{Mg}+{\mathrm{Cl}}_2\to {\mathrm{MgCl}}_2$

As a reactant , magnesium has an oxidation number of zero, but as part of the product magnesium chloride, the element has an oxidation number of +2. Magnesium has lost two electrons and has therefore been oxidised . This can be written as a half-reaction . The half-reaction for this change is:

$\mathrm{Mg}\to {\mathrm{Mg}}^{2+}+2{\mathrm{e}}^{-}$

As a reactant , chlorine has an oxidation number of zero, but as part of the product magnesium chloride, the element has an oxidation number of -1. Each chlorine atom has gained an electron and the element has therefore been reduced . The half-reaction for this change is:

${\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}$

Half-reaction

A half reaction is either the oxidation or reduction reaction part of a redox reaction. A half reaction is obtained by considering the change in oxidation states of the individual substances that are involved in the redox reaction.

An element that is oxidised is called a reducing agent , while an element that is reduced is called an oxidising agent .

Redox reactions

Redox reaction

A redox reaction is one involving oxidation and reduction, where there is always a change in the oxidation numbers of the elements involved.

Demonstration : redox reactions

Materials:

A few granules of zinc; 15 ml copper (II) sulphate solution (blue colour), glass beaker.

Method:

Add the zinc granules to the copper sulfate solution and observe what happens. What happens to the zinc granules? What happens to the colour of the solution?

Results:

• Zinc becomes covered in a layer that looks like copper.
• The blue copper sulfate solution becomes clearer.

Cu ${}^{2+}$ ions from the CuSO ${}_4$ solution are reduced to form copper metal. This is what you saw on the zinc crystals. The reduction of the copper ions (in other words, their removal from the copper sulphate solution), also explains the change in colour of the solution (copper ions in solution are blue). The equation for this reaction is:

${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$

Zinc is oxidised to form Zn ${}^{2+}$ ions which are clear in the solution. The equation for this reaction is:

$\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}$

The overall reaction is:

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Zn}\left(\mathrm{s}\right)\to \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)$

Conclusion:

A redox reaction has taken place. Cu ${}^{2+}$ ions are reduced and the zinc is oxidised.

Below are some further examples of redox reactions:

• ${\mathrm{H}}_2+{\mathrm{F}}_2\to 2\mathrm{HF}$ can be re-written as two half-reactions: ${\mathrm{H}}_2\to 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ (oxidation) and ${\mathrm{F}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{F}}^{-}$ (reduction)
• ${\mathrm{Cl}}_2+2\mathrm{KI}\to 2\mathrm{KCl}+{\mathrm{I}}_2$ or ${\mathrm{Cl}}_2+2{\mathrm{I}}^{-}\to 2{\mathrm{Cl}}^{-}+{\mathrm{I}}_2$ , can be written as two half-reactions: ${\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}$ (reduction) and $2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+2{\mathrm{e}}^{-}$ (oxidation)

In Grade 12, you will go on to look at electrochemical reactions, and the role that electron transfer plays in this type of reaction.

Redox reactions

1. Look at the following reaction: $2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$
   1. What is the oxidation number of the oxygen atom in each of the following compounds?
      1. H ${}_2$ O ${}_2$
      2. H ${}_2$ O
      3. O ${}_2$
   2. Does the hydrogen peroxide (H ${}_2$ O ${}_2$ ) act as an oxidising agent or a reducing agent or both, in the above reaction? Give a reason for your answer.
2. Consider the following chemical equations: 1. $\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}$ 2. $4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ Which one of the following statements is correct?
   1. Fe is oxidised and H ${}^{+}$ is reduced
   2. Fe is reduced and O ${}_2$ is oxidised
   3. Fe is oxidised and O ${}_2$ is reduced
   4. Fe is reduced and H ${}^{+}$ is oxidised
   (DoE Grade 11 Paper 2, 2007)
3. Which one of the following reactions is a redox reaction?
   1. $\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$
   2. ${\mathrm{AgNO}}_3+\mathrm{NaI}\to \mathrm{AgI}+{\mathrm{NaNO}}_3$
   3. $2{\mathrm{FeCl}}_3+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2\to {\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{HCl}+2{\mathrm{FeCl}}_2$
   4. ${\mathrm{BaCl}}_2+{\mathrm{MgSO}}_4\to {\mathrm{MgCl}}_2+{\mathrm{BaSO}}_4$