Balancing Redox Reactions

Using what you have learnt about oxidation numbers and redox reactions, we can balance redox reactions in the same way that you have learnt to balance other reactions. The following worked examples will show you how.

Example 1: Balancing Redox Reactions

Question

Balance the following redox reaction: \[\text{Fe}^{2+}\text{(aq)} + \text{Cl}_{2}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + \text{Cl}^{−}\text{(aq)}\]

Step 1: Write a reaction for each compound

\[\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}\] \[\text{Cl}_{2} \rightarrow \text{Cl}^{−}\]

Step 2: Balance the atoms on either side of the arrow

We check that the atoms on both sides of the arrow are balanced:

\[\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}\] \[\text{Cl}_{2} \rightarrow 2\text{Cl}^{−}\]

Step 3: Add electrons to balance the charges

We now add electrons to each reaction so that the charges balance.

We add the electrons to the side with the greater positive charge.

\[\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-}\] \[\text{Cl}_{2} + 2e^{-} \rightarrow 2\text{Cl}^{−}\]

Step 4: Balance the number of electrons

We now make sure that the number of electrons in both reactions is the same.

The reaction for iron has one electron, while the reaction for chlorine has two electrons. So we must multiply the reaction for iron by \(\text{2}\) to ensure that the charges balance.

\[2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^{-}\] \[\text{Cl}_{2} + 2e^{-} \rightarrow 2\text{Cl}^{−}\]

We now have the two half-reactions for this redox reaction.

The reaction for iron is the oxidation half-reaction as iron became more positive (lost electrons). The reaction for chlorine is the reduction half-reaction as chlorine has become more negative (gained electrons).

Step 5: Combine the two half-reactions

 \(2\text{Fe}^{2+}\)\(\rightarrow\)\(2\text{Fe}^{3+} + 2e^{-}\)
\(+\)\(\text{Cl}_{2} + 2e^{-}\)\(\rightarrow\)\(2\text{Cl}^{−}\)
 \(2\text{Fe}^{2+} + \text{Cl}_{2} + 2e^{-}\)\(\rightarrow\)\(2\text{Fe}^{3+} + 2\text{Cl}^{−} + 2e^{-}\)

Step 6: Write the final answer

Cancelling out the electrons gives: \[2\text{Fe}^{2+}\text{(aq)} + \text{Cl}_{2}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{Cl}^{−}\text{(aq)}\]

Note that we leave the co-efficients in front of the iron ions since the charge on the left hand side has to be the same as the charge on the right hand side in a redox reaction.

In the example above we did not need to know if the reaction was taking place in an acidic or basic medium (solution). However if there is hydrogen or oxygen in the reactants and not in the products (or if there is hydrogen or oxygen in the products but not in the reactants) then we need to know what medium the reaction is taking place in. This will help us to balance the redox reaction.

If a redox reaction takes place in an acidic medium then we can add water molecules to either side of the reaction equation to balance the number of oxygen atoms. We can also add hydrogen ions to balance the number of hydrogen atoms. We do this because we are writing the net ionic equation (showing only the ions involved and often only the ions containing the elements that change oxidation number) for redox reactions and not the net reaction equation (showing all the compounds that are involved in the reaction). If a Bronsted acid is dissolved in water then there will be free hydrogen ions.

If a redox reaction takes place in an basic medium then we can add water molecules to either side of the reaction equation to balance the number of oxygen atoms. We can also add hydroxide ions (\(\text{OH}^{-}\)) to balance the number of hydrogen atoms. We do this because we are writing the net ionic equation (showing only the ions involved and often only the ions containing the elements that change oxidation number) for redox reactions and not the net reaction equation (showing all the compounds that are involved in the reaction). If a Bronsted base is dissolved in water then there will be free hydroxide ions.

Example 2: Balancing Redox Reactions

Question

Balance the following redox reaction: \[\text{Cr}_{2}\text{O}_{7}^{2-}\text{(aq)} + \text{H}_{2}\text{S (aq)} \rightarrow \text{Cr}^{3+}\text{(aq)} + \text{S (s)}\]

The reaction takes place in an acidic medium.

Step 1: Write a reaction for each compound

\[\text{Cr}_{2}\text{O}_{7}^{2-} \rightarrow \text{Cr}^{3+}\] \[\text{H}_{2}\text{S} \rightarrow \text{S}\]

Step 2: Balance the atoms on either side of the arrow

We check that the atoms on both sides of the arrow are balanced.

In the first reaction we have \(\text{2}\) chromium atoms and \(\text{7}\) oxygen atoms on the left hand side. On the right hand side we have \(\text{1}\) chromium atom and no oxygen atoms. Since we are in an acidic medium we can add water to the right hand side to balance the number of oxygen atoms. We also multiply the chromium by \(\text{2}\) on the right hand side to make the number of chromium atoms balance. \[\text{Cr}_{2}\text{O}_{7}^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}\]

Now we have hydrogen atoms on the right hand side, but not on the left hand side so we must add \(\text{14}\) hydrogen ions to the left hand side (we can do this because the reaction is in an acidic medium): \[\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}\] We do not use water to balance the hydrogens as this will make the number of oxygen atoms unbalanced.

For the second part of the reaction we need to add \(\text{2}\) hydrogen ions to the right hand side to balance the number of hydrogens: \[\text{H}_{2}\text{S} \rightarrow \text{S} + 2\text{H}^{+}\]

Step 3: Add electrons to balance the charges

We now add electrons to each reaction so that the charges balance.

We add the electrons to the side with the greater positive charge.

\[\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}\] \[\text{H}_{2}\text{S} \rightarrow \text{S} + 2\text{H}^{+} + 2\text{e}^{-}\]

Step 4: Balance the number of electrons

We now make sure that the number of electrons in both reactions is the same.

The reaction for chromium has \(\text{6}\) electrons, while the reaction for sulfur has \(\text{2}\) electrons. So we must multiply the reaction for sulfur by \(\text{3}\) to ensure that the charges balances.

\[\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}\] \[3\text{H}_{2}\text{S} \rightarrow 3\text{S} + 6\text{H}^{+} + 6\text{e}^{-}\]

We now have the two half-reactions for this redox reaction.

The reaction involving sulfur is the oxidation half-reaction as sulfur became more positive (lost electrons). The reaction for chromium is the reduction half-reaction as chromium has become more negative (gained electrons).

Step 5: Combine the two half-reactions

We combine the two half-reactions:

 \(\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} + 6\text{e}^{-}\)\(\rightarrow\)\(2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}\)
\(+\)\(3\text{H}_{2}\text{S}\)\(\rightarrow\)\(3\text{S} + 6\text{H}^{+} + 6\text{e}^{-}\)
 \(\text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^{+} + 6\text{e}^{-} + 3\text{H}_{2}\text{S}\)\(\rightarrow\)\(2\text{Cr}^{3+} + 7\text{H}_{2}\text{O} + 3\text{S} + 6\text{H}^{+} + 6\text{e}^{-}\)

Step 6: Write the final answer

Crossing off the electrons and hydrogen ions gives: \[\text{Cr}_{2}\text{O}_{7}^{2-}\text{(aq)} + 3\text{H}_{2}\text{S (aq)} + 8\text{H}^{+}\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{S (s)} + 7\text{H}_{2}\text{O (l)}\]

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

[Attributions and Licenses]


This is a lesson from the tutorial, Acid-Base and Redox Reactions and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts

  • This is the first time am coming in contact with a teaching that makes me understand this redox reaction ...may God bless you guys for this teaching.... Excellent job guys 🙌🙌