4.2: Balancing redox reactions

Learning how to write balanced redox reactions by hand is an effective way to reinforce redox concepts. To start with, consider the following simple redox reaction:

\[\text{Zn} \ (s) + \text{Ag}^{+} \longleftrightarrow \text{Zn}^{2+} + \text{Ag} \ (s)\]

Is this reaction balanced? In terms of elements, yes, but there is more to it than that. Redox reactions must also be balanced in terms of charge. The amount of oxidation must equal the amount of reduction. In other words, electron donation must equal electron acceptance. Electrons do not just swim about freely in aqueous solutions.

Redox reactions can be quite difficult to balance, but there is an easy procedure that you can follow to write and balance simple redox reactions. For the purposes of this book, we will use the following steps:

1. Write half reactions in term of oxidation and reduction.
2. Balance electrons and the species being oxidized and reduced.
3. Combine half reactions.
4. Balance oxygens by adding water molecules and then balance hydrogens by adding hydrogen ions \((\text{H}^{+})\)
5. Confirm charge and elemental balance.

For our first example, let’s apply these rules to the reaction above \((\PageIndex{1})\).

1. Write half reactions in term of oxidation and reduction. For this step, you need to identify oxidation states for each element in the reaction. In doing so, we can see that the oxidation state of zinc increases from \(0\) to \(+2\), indicating that it loses two electrons as the reaction goes forward. Similarly, we can see that the oxidation state of silver decreases from \(+1\) to \(0\) as the reaction goes forward, indicating that it gains one electron. Based on this information, we can write the following half reactions: \[\begin{align} \text{Oxidation half} \quad\quad\quad & \text{Zn} \longleftrightarrow \text{Zn}^{2+} + 2 \ e^{-} \\ \text{Reduction half} \quad\quad\quad & \text{Ag}^{+} + e^{-} \longleftrightarrow \text{Ag} \end{align}\]
2. Balance electrons and the species being oxidized and reduced. We notice that the oxidation half reaction includes two electrons as written and the reduction half includes only one. Therefore, we need to multiply the reduction half reaction by two: \[\begin{align} \text{Oxidation half} \quad\quad\quad & \text{Zn} \longleftrightarrow \text{Zn}^{2+} + 2 \ e^{-} \\ \text{Reduction half} \quad\quad\quad & 2 \ \text{Ag}^{+} + 2 \ e^{-} \longleftrightarrow 2 \ \text{Ag} \end{align}\]
3. Combine half reactions by adding them together. The two electrons on the product side cancel out with the two electrons on the reactant side. \[\text{Zn}^{2+} + 2 \ \text{Ag}^{+} \longleftrightarrow \text{Zn}^{2+} + 2 \ \text{Ag}\]
4. Balance oxygens and hydrogens. None are present in this reaction so we can skip this step.
5. Confirm charge balance. The net charge on both the reactant and product sides of the equation is \(+2\). Balance in charge and quantities of elements helps us confirm that we have balanced the reaction correctly.

For a second example, we consider a more complicated reaction in which the carbon in acetate \(\left(\text{CH}_{3} \text{COO}^{-} \right)\) is oxidized to bicarbonate \(\left(\text{HCO3}^{-}\right)\) and iron in the mineral goethite \((\text{FeOOH})\) is reduced to ferrous iron \(\left(\text{Fe}^{2+}\right)\). This reaction is used by some metal-reducing microorganisms as a source of energy.

1. Define the half reactions. In the oxidation half reaction, acetate includes two carbons and bicarbonate includes one. As such, two bicarbonates are needed in the products for every acetate in the reactants. The carbons have an average oxidation state of \(0\) in acetate and \(+4\) in bicarbonate. Therefore, carbon releases eight electrons overall in the oxidation half reaction. In the reduction half, the oxidation state of iron in goethite is \(+3\). Therefore, it must gain one electron to become \(\text{Fe}^{2+}\) \[\begin{align} \text{Oxidation half} \quad\quad\quad & \text{CH}_{3} \text{COO}^{-} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + 8 \ e^{-} \\ \text{Reduction half} \quad\quad\quad & \text{FeOOH} + e^{-} \longleftrightarrow \text{Fe}^{2+} \end{align}\]
2. Balance the electrons by multiplying the reduction half reaction by eight. \[\begin{align} \text{Oxidation half} \quad\quad\quad & \text{CH}_{3} \text{COO}^{-} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + 8 \ e^{-} \\ \text{Reduction half} \quad\quad\quad & 8 \ \text{FeOOH} + 8 \ e^{-} \longleftrightarrow 8 \ \text{Fe}^{2+} \end{align}\]
3. Combine the half reactions. \[\text{CH}_{3} \text{COO}^{-} + 8 \ \text{FeOOH} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + 8 \ \text{Fe}^{2+}\]
4. Balance the oxygens first by adding water and then balance the hydrogens. In reaction \((\PageIndex{11})\), there are 18 oxygens on the reactant side and only six on the product side. Therefore, we can balance oxygens by adding 12 waters to the product side. \[\text{CH}_{3} \text{COO}^{-} + 8 \ \text{FeOOH} + 15 \ \text{H}^{+} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + 8 \ \text{Fe}^{2+} + 12 \text{H}_{2} \text{O}\]
5. Do charges and elements balance? Yes, the net charge is \(+14\) on each side of the reaction and the elements are balanced.

For a third and final example, we consider oxidation of elemental hydrogen \(\left(\text{H}_{2}\right)\) coupled with reduction of oxygen \(\left(\text{O}_{2}\right)\), with water as the product in both half reactions.

1. Define the half reactions. In the oxidation half reaction, the oxidation state of each hydrogen goes from \(0\) to \(+1\) and there are two hydrogens on each side. Therefore, two electrons are released in the oxidation half reaction. In the reduction half, the oxidation state of oxygen goes from \(0\) to \(-2\) and there are two oxygens on each side. Therefore, four electrons are gained in the reduction half reaction. \[\begin{align} \text{Oxidation half} \quad\quad\quad & \text{H}_{2} \longleftrightarrow \text{H}_{2} \text{O} + 2 \ e^{-} \\ \text{Reduction half} \quad\quad\quad & \text{O}_{2} + 4 \ e^{-} \longleftrightarrow 2 \ \text{H}_{2} \text{O} \end{align}\]
2. Balance the electrons by multiplying the reduction half reaction by two. \[\begin{align} \text{Oxidation half} \quad\quad\quad & 2 \ \text{H}_{2} \longleftrightarrow 2 \ \text{H}_{2} \text{O} + 4 \ e^{-} \\ \text{Reduction half} \quad\quad\quad & \text{O}_{2} + 4 \ e^{-} \longleftrightarrow 2 \ \text{H}_{2} \text{O} \end{align}\]
3. Combine the half reactions. \[2 \ \text{H}_{2} + \text{O}_{2} \longleftrightarrow 2 \ \text{H}_{2} \text{O}\]
4. Balance the oxygens and then hydrogens. In reaction \((\PageIndex{17})\), there are two oxygens on the reactant side and four on the product side. Therefore, two waters must be added to the reactant side. \[2 \ \text{H}_{2} + \text{O}_{2} + 2 \text{H}_{2} \text{O} \longleftrightarrow 4 \ \text{H}_{2} \text{O}\]In reaction \((\PageIndex{18})\), there are eight hydrogens on the reactant side and eight on the product side, so hydrogen is already balanced.
5. Do charges and elements balance? Yes, the net charge is \(0\) on each side of the reaction and the elements are balanced. However, before we finalize this reaction, we can simplify it by removing some of the waters. Water molecules are present in both the product and reactant side of the reaction. We can delete two on each side to eliminate the redundancy. Doing so, gives us the following reaction, which is balanced: \[2 \ \text{H}_{2} + \text{O}_{2} \longleftrightarrow 2 \ \text{H}_{2} \text{O}\]

Practice \(\PageIndex{1}\)

Write redox reactions defined by the following electron transfers:

1. Reduction of \(\text{Fe}\) in \(\text{Fe(OH)}_{3}\) to \(\text{Fe}^{2+}\) coupled with oxidation of \(\text{C}\) in \(\text{CH}_{4}\) to \(\text{HCO}_{3}^{-}\)
2. Reduction of \(\text{S}\) in \(\text{SO}_{4}^{2-}\) to \(\text{H}_{2} \text{S}\) coupled with oxidation of \(\text{C}\) in \(\text{CH}_{4}\) to \(\text{HCO}_{3}^{-}\)
3. Reduction of \(\text{N}\) in \(\text{NO}_{2}^{-}\) to \(\text{N}_{2}\) coupled with oxidation of \(\text{H}\) in \(\text{H}_{2}\) to \(\text{H}_{2} \text{O}\)
4. Reduction of \(\text{O}\) in \(\text{O}_{2}\) to \(\text{H}_{2} \text{O}\) coupled with oxidation of \(\text{N}\) in \(\text{NH}_{4}^{+}\) to \(\text{NO}_{3}^{-}\)