13.2 Writing redox and half-reactions | Electrochemical reactions

13.1 Revision of oxidation and reduction

13.2 Writing redox and half-reactions (ESCQY)

Redox reactions and half-reactions (ESCQZ)

Remember from Grade 11 that oxidation and reduction occur simultaneously in a redox reaction. The reactions taking place in electrochemical cells are redox reactions. Two questions should be asked to determine if a reaction is a redox reaction:

Is there a compound or atom being oxidised?
Is there a compound or atom being reduced?

If the answer to both of these questions is yes, then the reaction is a redox reaction. For example, the reaction given below is a redox reaction:

\(2\text{Fe}^{3+}(\text{aq}) + \text{Sn}^{2+}(\text{aq})\) \(\to\) \(2\text{Fe}^{2+}(\text{aq}) + \text{Sn}^{4+}(\text{aq})\)

You can write a redox reaction as two half-reactions, one showing the reduction process, and one showing the oxidation process. \(\text{Fe}^{3+}\) is gaining an electron to become \(\text{Fe}^{2+}\). \(\color{red}{\textbf{Iron}}\) is therefore being \(\color{red}{\textbf{reduced}}\) and \(\color{blue}{\textbf{tin}}\) is the \(\color{blue}{\textbf{reducing agent}}\) (causing iron to be reduced). The \(\color{red}{\textbf{reduction half-reaction}}\) is:

\(\color{red}{\textbf{Fe}^{3+}\textbf{(aq) + e}^{-} \to \textbf{Fe}^{2+}\textbf{(aq)}}\)

\(\text{Sn}^{2+}\) is losing two electrons to become \(\text{Sn}^{4+}\). \(\color{blue}{\textbf{Tin}}\) is therefore being \(\color{blue}{\textbf{oxidised}}\) and \(\color{red}{\textbf{iron}}\) is the \(\color{red}{\textbf{oxidising agent}}\) (causing tin to be oxidised). The \(\color{blue}{\textbf{oxidation half-reaction}}\) is:

\(\color{blue}{\textbf{Sn}^{2+}\textbf{(aq)} \to \textbf{Sn}^{4+}\textbf{(aq) + 2e}^{-}}\)

Notice that in the overall reaction the reduction half-reaction is multiplied by two. This is so that the number of electrons gained in the reduction half-reaction match the number of electrons lost in the oxidation half-reaction.

Balancing redox reactions (ESCR2)

Half-reactions can be used to balance redox reactions. We are going to use some worked examples to help explain the method.

Worked example 1: Balancing redox reactions

Chlorine gas oxidises \(\text{Fe}^{2+}\) ions to \(\text{Fe}^{3+}\) ions. In the process, chlorine is reduced to chloride ions. Write a balanced equation for this reaction.

Write down the unbalanced oxidation half-reaction

\(\text{Fe}^{2+}(\text{aq})\) \(\to\) \(\text{Fe}^{3+}(\text{aq})\)

Balance the number of atoms on both sides of the equation

There is one iron atom on the left and one on the right, so no additional atoms need to be added.

Once the atoms are balanced, check that the charges balance

The charge on the left of the equation is \(\text{+2}\), but the charge on the right is \(\text{+3}\). Therefore, one electron must be added to the right hand side so that the charges balance. The half-reaction is now:

\(\text{Fe}^{2+}(\text{aq})\) \(\to\) \(\text{Fe}^{3+}(\text{aq}) + \text{e}^{-}\)

Repeat steps 1 - 3 with the reduction half-reaction

The unbalanced reduction half-reaction is:

\(\text{Cl}_{2}(\text{g})\) \(\to\) \(\text{Cl}^{-}(\text{aq})\)

The atoms don't balance, so we need to multiply the right hand side by two to fix this.

\(\text{Cl}_{2}(\text{g})\) \(\to\) \(2\text{Cl}^{-}(\text{aq})\)

Two electrons must be added to the left hand side to balance the charges.

\(\text{Cl}_{2}(\text{g}) + 2\text{e}^{-}\) \(\to\) \(2\text{Cl}^{-}(\text{aq})\)

Compare the number of electrons in each equation

Multiply each half-reaction by a suitable number so that the number of electrons released (oxidation) is equal to the number of electrons accepted (reduction).

oxidation half-reaction: \(\color{red}{\times \textbf{2}}\): \(\color{red}{2}\)\(\text{Fe}^{2+}(\text{aq})\) \(\to\) \(\color{red}{2}\)\(\text{Fe}^{3+}(\text{aq}) +\)\(\color{red}{\textbf{2}}{\textbf{e}^{-}}\)

reduction half-reaction: \(\color{red}{\times \textbf{1}}\): \(\text{Cl}_{2}(\text{g}) +\) \(\textbf{2e}^{-} \to\) \(2\text{Cl}^{-}(\text{aq})\)

Combine the two half-reactions to get a final equation for the overall reaction

\(2\text{Fe}^{2+}(\text{aq}) + \text{Cl}_{2}(\text{g})\) \(\to\) \(2\text{Fe}^{3+}(\text{aq}) + 2\text{Cl}^{-}(\text{aq})\)

Do a final check to make sure that the equation is balanced

We check the number of atoms and the charges and find that the equation is balanced.

Worked example 2: Balancing redox reactions in an acid medium

The following reaction takes place in an acid medium:

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + \text{H}_{2}\text{S}(\text{g})\) \(\to\) \(\text{Cr}^{3+}(\text{aq}) + \text{S}(\text{s})\)

Write a balanced equation for this reaction.

Write down the unbalanced reduction half-reaction

In \(\text{Cr}_{2}\text{O}_{7}^{2-}\) chromium exists as \(\text{Cr}^{6+}\). It becomes \(\text{Cr}^{3+}\). Therefore electrons are gained, this is the reduction half-reaction:

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq})\) \(\to\) \(\text{Cr}^{3+}(\text{aq})\)

Balance the number of atoms on both sides of the equation

We need to multiply the right side by two so that the number of Cr atoms will balance. In an acid medium there are water molecules and \(\text{H}^{+}\) ions in the solution, so these can be used to balance the equation.

To balance the oxygen atoms, we will need to add water molecules to the right hand side:

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq})\) \(\to\) \(2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{ℓ})\)

Now the oxygen atoms balance but the hydrogens don't. Because the reaction takes place in an acid medium, we can add hydrogen ions to the left side.

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq})\) \(\to\) \(2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{ℓ})\)

Once the atoms are balanced, check that the charges balance

The charge on the left of the equation is (\(-\text{2}\) + \(\text{14}\)) = \(\text{+12}\), but the charge on the right is \(\text{+6}\). Therefore, six electrons must be added to the left hand side so that the charges balance. This makes sense as electrons are gained in the reduction half-reaction. The half-reaction is now:

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq}) + 6\text{e}^{-}\) \(\to\) \(2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{ℓ})\)

Repeat steps \(\text{1}\) - \(\text{3}\) with the oxidation half-reaction

The unbalanced oxidation half-reaction is:

\(\text{H}_{2}\text{S}(\text{g})\) \(\to\) \(\text{S}(\text{s})\)

However, you can ignore the \(\text{H}^{+}\) in \(\text{H}_{2}\text{S}\) as they are accounted for by the acid medium and the reduction half-reaction.

\(\text{S}^{2-}(\text{aq})\) \(\to\) \(\text{S}(\text{s})\)

The atoms balance, however the charges do not. Two electrons must be added to the right hand side of the equation.

\(\text{S}^{2-}(\text{aq})\) \(\to\) \(\text{S}(\text{s}) + 2\text{e}^{-}\)

Compare the number of electrons in each equation

Multiply each half-reaction by a suitable number so that the number of electrons released (oxidation) is equal to the number of electrons accepted (reduction).

oxidation half-reaction: \(\color{red}{\times \textbf{3}}\): \(\color{red}{\text{3}}\)\(\text{S}^{2-}(\text{aq})\) \(\to\) \(\color{red}{\text{3}}\)\(\text{S}(\text{s}) +\) \(\color{red}{\textbf{6}}{\textbf{e}^{-}}\)

reduction half-reaction: \(\color{red}{\times \textbf{1}}\): \(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq}) +\) \(\textbf{6e}^{-} \to\) \(2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{ℓ})\)

Combine the two half-reactions to get a final equation for the overall reaction

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq}) + 3\text{S}^{2-}(\text{aq})\) \(\to\) \(3\text{S}(\text{s}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{ℓ})\)

However, \(\text{H}_{2}\text{S}(\text{g})\) is the reactant, so it would be better to write:

\(\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 3\text{H}_{2}\text{S}(\text{g})\) \(\to\) \(3\text{S}(\text{s}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{ℓ})\)

Do a final check to make sure that the equation is balanced

We check the number of atoms and the charges and find that the equation is balanced.

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Worked example 3: Balancing redox reactions in an alkaline medium

The complex ion hexaamminecobalt(II) (\(\text{Co}(\text{NH}_{3})_{6}^{2+}\)) is oxidised by hydrogen peroxide to form the hexaamminecobalt(III) ion (\(\text{Co}(\text{NH}_{3})_{6}^{3+}\)). Write a balanced equation for this reaction.

Write down the unbalanced oxidation half-reaction

Ammonia (\(\text{NH}_{3}\)) has an oxidation number of \(\text{0}\). Therefore, in \(\text{Co}(\text{NH}_{3})_{6}^{2+}\) cobalt exists as \(\text{Co}^{2+}\). In \(\text{Co}(\text{NH}_{3})_{6}^{3+}\) cobalt exists as \(\text{Co}^{3+}\). Electrons are lost and this is the oxidation half-reaction:

\(\text{Co}^{2+}(\text{aq})\) \(\to\) \(\text{Co}^{3+}(\text{aq})\)

Balance the number of atoms on both sides of the equation

The number of atoms are the same on both sides.

Once the atoms are balanced, check that the charges balance

The charge on the left of the equation is \(\text{+2}\), but the charge on the right is \(\text{+3}\). One electron must be added to the right hand side to balance the charges in the equation:

\(\text{Co}^{2+}\) \(\to\) \(\text{Co}^{3+} + \text{e}^{-}\)

Repeat steps \(\text{1}\) - \(\text{3}\) with the reduction half-reaction

Cobalt is oxidised by hydrogen peroxide, therefore hydrogen peroxide is reduced. Reduction means a gain of electrons. The product of the reduction of \(\text{H}_{2}\text{O}_{2}\) in an alkaline medium is \(\text{OH}^{-}\):

\(\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(\text{OH}^{-}(\text{aq})\)

Next you need to balance the atoms:

\(\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(2\text{OH}^{-}(\text{aq})\)

Then you need to balance the charges:

\(\text{H}_{2}\text{O}_{2}(\text{ℓ}) + 2\text{e}^{-}\) \(\to\) \(2\text{OH}^{-}(\text{aq})\)

Compare the number of electrons in each equation

Multiply each half-reaction by a suitable number so that the number of electrons released (oxidation) is equal to the number of electrons accepted (reduction):

oxidation half-reaction: \(\color{red}{\times \textbf{2}}\): \(\color{red}{2}\)\(\text{Co}^{2+}(\text{aq})\) \(\to \color{red}{2}\)\(\text{Co}^{3+}(\text{aq}) +\) \(\color{red}{\textbf{2}}{\textbf{e}}^{-}\)

reduction half-reaction: \(\color{red}{\times \textbf{1}}\): \(\text{H}_{2}\text{O}_{2}(\text{ℓ}) +\) \(\textbf{2e}^{-} \to\) \(2\text{OH}^{-}(\text{aq})\)

Combine the two half-reactions, and add in the spectator ions, to get a final equation for the overall reaction

\(2\text{Co}(\text{NH}_{3})_{6}^{2+}(\text{aq}) + \text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(2\text{Co}(\text{NH}_{3})_{6}^{3+}(\text{aq}) + 2\text{OH}^{-}(\text{aq})\)

Do a final check to make sure that the equation is balanced

We check the number of atoms and the charges and find that the equation is balanced.

Balancing redox reactions

Textbook Exercise 13.2

\(\text{HNO}_{3}(\text{ℓ}) + \text{PbS}(\text{s})\) \(\to\) \(\text{PbSO}_{4}(\text{s}) + \text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{ℓ})\)

\(\text{Pb}^{2+}\) is a spectator ion.

The unbalanced reduction half-reaction is:

\(\text{NO}_{3}^{-}(\text{aq})\) \(\to\) \(\text{NO}_{2}(\text{g})\)

Add water molecules to the right and \(\text{H}^{+}\) ions to the left (acid medium) to balance the oxygen and hydrogen atoms:

\(\text{NO}_{3}^{-}(\text{aq}) + 2\text{H}^{+}(\text{aq})\) \(\to\) \(\text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{ℓ})\)

Balance the charge by adding an electron to the left (this makes sense as this is the reduction half-reaction, and \(\text{N}^{5+}\) \(\to\) \(\text{N}^{4+}\)):

\(\text{NO}_{3}^{-}(\text{aq}) + 2\text{H}^{+}(\text{aq}) + \text{e}^{-}\) \(\to\) \(\text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{ℓ})\)

The unbalanced oxidation half-reaction is:

\(\text{S}^{2-}(\text{aq})\) \(\to\) \(\text{SO}_{4}^{2-}(\text{aq})\)

Add water molecules to the left and \(\text{H}^{+}\) ions to the right to balance the oxygen and hydrogen atoms:

\(\text{S}^{2-}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{ℓ})\) \(\to\) \(\text{SO}_{4}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq})\)

Balance the charge by adding eight electrons to the right (this makes sense as this is the oxidation half-reaction, and \(\text{S}^{2-}\) \(\to\) \(\text{S}^{6+}\)):

\(\text{S}^{2-}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{ℓ})\) \(\to\) \(\text{SO}_{4}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 8\text{e}^{-}\)

We multiply the reduction half-reaction reaction by \(\text{8}\) to balance the number of electrons in both equations:

\(8\text{NO}_{3}^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + 8\text{e}^{-}\) \(\to\) \(8\text{NO}_{2}(\text{g}) + 8\text{H}_{2}\text{O}(\text{ℓ})\)

Adding the two equations together gives the balanced equation (electrons are equal on both sides and can be removed):

\(8\text{NO}_{3}^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + \text{S}^{2-}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{ℓ})\) \(\to\) \(\text{SO}_{4}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 8\text{NO}_{2}(\text{g}) + 8\text{H}_{2}\text{O}(\text{ℓ})\)

Putting the spectator ions back into the equation and removing any extra \(\text{H}^{+}\) ions and water molecules we get:

\(8\text{HNO}_{3}(\text{ℓ}) + \text{PbS}(\text{s})\) \(\to\) \(\text{PbSO}_{4}(\text{s}) + 8\text{NO}_{2}(\text{g}) + 4\text{H}_{2}\text{O}(\text{ℓ})\)

\(\text{NaI}(\text{aq}) + \text{Fe}_{2}(\text{SO}_{4})_{3}(\text{aq})\) \(\to\) \(\text{I}_{2}(\text{s}) + \text{FeSO}_{4}(\text{aq}) + \text{Na}_{2}\text{SO}_{4}(\text{aq})\)

\(\text{Na}^{+}\) and \(\text{SO}_{4}^{2-}\) are spectator ions.

\(\text{Fe}\) has the oxidation number \(\text{+3}\) in \(\text{Fe}_{2}(\text{SO}_{4})_{3}\), and \(\text{+2}\) in \(\text{FeSO}_{4}\). This must be the reduction half-reaction:

\(\text{Fe}^{3+}(\text{aq})\) \(\to\) \(\text{Fe}^{2+}(\text{aq})\)

The atoms are balanced. Balance the charge by adding an electron to the left:

\(\text{Fe}^{3+}(\text{aq}) + \text{e}^{-}\) \(\to\) \(\text{Fe}^{2+}(\text{aq})\)

The unbalanced oxidation half-reaction is:

\(\text{I}^{-}(\text{aq})\) \(\to\) \(\text{I}_{2}(\text{s})\)

There must be \(\text{2}\) \(\text{I}^{-}\) to balance the atoms:

\(2\text{I}^{-}(\text{aq})\) \(\to\) \(\text{I}_{2}(\text{s})\)

Balance the charge by adding two electrons to the right:

\(2\text{I}^{-}(\text{aq})\) \(\to\) \(\text{I}_{2}(\text{s}) + 2\text{e}^{-}\)

We multiply the reduction half-reaction by \(\text{2}\) to balance the number of electrons in both equations:

\(2\text{Fe}^{3+}(\text{aq}) + 2\text{e}^{-}\) \(\to\) \(2\text{Fe}^{2+}(\text{aq})\)

Adding the two equations together and removing the electrons gives:

\(2\text{I}^{-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})\) \(\to\) \(\text{I}_{2}(\text{s}) + 2\text{Fe}^{2+}(\text{aq})\)

Putting the spectator ions back into the equation we get:

\(2\text{NaI}(\text{aq}) + \text{Fe}_{2}(\text{SO}_{4})_{3}(\text{aq})\) \(\to\) \(\text{I}_{2}(\text{s}) + 2\text{FeSO}_{4}(\text{aq}) + \text{Na}_{2}\text{SO}_{4}\)

Permanganate(VII) ions ( \(\text{MnO}_{4}^{-}\) ) oxidise hydrogen peroxide ( \(\text{H}_{2}\text{O}_{2}\) ) to oxygen gas. The reaction is done in an acid medium. During the reaction, the permanganate(VII) ions are reduced to manganese(II) ions (\(\text{Mn}^{2+}\) ). Write a balanced equation for the reaction.

The unbalanced reduction half-reaction is:

\(\text{MnO}_{4}^{-}(\text{aq})\) \(\to\) \(\text{Mn}^{2+}(\text{aq})\)

As this is in an acid medium, we can add water molecules to the right and \(\text{H}^{+}\) ions to the left to balance the oxygen and hydrogen atoms:

\(\text{MnO}_{4}^{-}(\text{aq}) + 8\text{H}^{+}\) \(\to\) \(\text{Mn}^{2+}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{ℓ})\)

Balance the charge by adding five electrons to the left (this makes sense as this is the reduction half-reaction, and \(\text{Mn}^{7+}\) \(\to\) \(\text{Mn}^{2+}\)):

\(\text{MnO}_{4}^{-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 5\text{e}^{-}\) \(\to\) \(\text{Mn}^{2+}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{ℓ})\)

The unbalanced oxidation half-reaction is:

\(\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(\text{O}_{2}(\text{g})\)

Add \(\text{H}^{+}\) ions to the right to balance the hydrogen atoms:

\(\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(\text{O}_{2}(\text{g}) + 2\text{H}^{+}(\text{aq})\)

Balance the charge by adding two electrons to the right (this makes sense as this is the oxidation half-reaction, and \(2\text{O}^{-}\) \(\to\) \(\text{O}_{2}\)):

\(\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(\text{O}_{2}(\text{g}) + 2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}\)

We multiply the reduction half-reaction by \(\text{2}\) and the oxidation half-reaction by \(\text{5}\) to balance the number of electrons in both equations:

\(2\text{MnO}_{4}^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + 10\text{e}^{-}\) \(\to\) \(2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_{2}\text{O}(\text{ℓ})\)

\(5\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(5\text{O}_{2}(\text{g}) + 10\text{H}^{+}(\text{aq}) + 10\text{e}^{-}\)

Adding the two equations together gives the balanced equation (electrons are equal on both sides and can be removed):

\(2\text{MnO}_{4}^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + 5\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(5\text{O}_{2}(\text{g}) + 10\text{H}^{+}(\text{aq}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_{2}\text{O}(\text{ℓ})\)

Removing any extra \(\text{H}^{+}\) ions we get:

\(2\text{MnO}_{4}^{-}(\text{aq}) + 6\text{H}^{+}(\text{aq}) + 5\text{H}_{2}\text{O}_{2}(\text{ℓ})\) \(\to\) \(5\text{O}_{2}(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_{2}\text{O}(\text{ℓ})\)

13.1 Revision of oxidation and reduction

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13.3 Galvanic and electrolytic cells