The chemistry of electron transfer – oxidation & reduction
Chemical reactions in which electrons are transferred from one atom/molecule to another are called reduction-oxidation reactions, or redox reactions, for short. Redox reactions are a very important class of chemical reactions in industry, biology and elsewhere.
Oxidation, a loss of electrons by a substace, and reduction, a gain of electrons, always occur together. There aren't any free electrons in a chemical system, so the loss of an electron somewhere has to create a gain somewhere else.
If you need to, you should review the section on oxidation numbers before going on with this section.
In this section we'll learn how to balance redox reactions, a process a little more complicated than our usual mass balancing. In a future section, we'll learn about electrochemical cells – batteries and other useful devices.
There are a few techniques for balancing redox reactions. The one I use is the one I learned from a great teacher of chemistry, Prof. Stephen Thompson at Colorado State University.
Oxidation & reduction
Oxidation is a loss of electron(s) and Reduction is a gain of electron(s).
Oxidation and reduction always occur together in any redox reaction.
Balancing redox reactions
Outline of the procedure
Our approach to balancing redox reactions will look like this (There might be some terminology you're not familiar with yet here, but we'll cover it below):
- Assign oxidation numbers to each atom.
- Separate the reaction into oxidation and reduction "half reactions."
- Balance each half reaction with respect to the atoms undergoing oxidation and reduction.
- Balance any unbalanced oxygen or hydrogen atoms using water; compensate for added hydrogens by adding H+ to the other side.
- Balance the charges by adding electrons to the appropriate side of each half reaction.
- If you're working in basic solution, add enough OH- to half-reaction sides containing any H+ (that just wouldn't be hanging around in a basic solution). You'll have to add the same number of OH- to the other side, too. Then H+ + OH- → H2O.
- Multiply the half reactions by coefficients that make the number of transferred electrons the same in each reaction.
- Add the half reactions. The electrons should cancel now.
- Cancel any atoms or molecules common to both sides, as you would for an algebraic equation.
- Reduce to the simplest coefficients.
This can seem like a lot of steps, but once you get the hang of it, it flows pretty naturally, so don't give up. Work through a few examples and you'll see that it gets easier. Also, this isn't the only approach to balancing redox reactions.
One more thing: we usually begin balancing a redox reaction using a net ionic equation that need not be entirely complete. As long as the atoms that are oxidized and reduced, the balancing procedure will pull out the necessary details.
The best way to go about learning to balance redox reactions is by example, so let's dive into a couple of them. Here's the first.
Example 1
Balance the reaction:
$HCl + KMnO_4 \longrightarrow Cl_2 + MnCl_2 + H_2O$ in acidic solution
Now, you'll notice that this reaction isn't even balanced by atoms ("mass balanced"). There is a potassium (K) atom on the left but none on the right. We'll get to that in time.
To better prepare, it's often best to write the net ionic reaction, by assuming that everything in solution, except for the solvent, water, and covalently-bonded species, like Cl2, will ionize. It looks like this:
$Cl^- + H^+ + MbO_4^- \longrightarrow Cl_2 + Mn^{2+} + H_2O$
1. Assign oxidation numbers.
Now it's easy to see that chlorine (Cl-) is oxidized (loses electrons) and manganese (Mn) is reduced (gains electrons):
2. Separate into half-reactions
Now we focus only on the chemical species being oxidized and reduced, and write oxidation and reduction half-reactions, respecitvely:
$$ \begin{align} Cl^- &\longrightarrow Cl_2 \phantom{000} \color{steelblue}{\text{oxidation}} \\[5pt] MnO_4^- &\longrightarrow Mn^{2+} \phantom{000} \color{#C5D82F}{\text{reduction}} \end{align}$$
3. Balance the oxidized and reduced atoms with coefficients.
Only the chlorine ion needs a coefficient here because there is only one Cl on the left and two on the right.
$$ \begin{align} 2 \, Cl^- &\longrightarrow Cl_2 \\[5pt] MnO_4^- &\longrightarrow Mn^{2+} \end{align}$$
4. Balance any unbalanced oxygen or hydrogen atoms using water; compensate for added hydrogens by adding H+ to the other side.
Now the reduction equation has oxygens on the left but not on the right. To balance them, add four waters to the right, and compensate for the added hydrogens by adding 8 protons on the left. We're working in acid solution, so this is appropriate.
$$ \begin{align} 2 \, Cl^- &\longrightarrow Cl_2 \\[5pt] 8 \, H^+ + MnO_4^- &\longrightarrow Mn^{2+} + 4 \, H_2O \end{align}$$
5. Balance the charge on right and left with electrons
In the oxidation reaction, there are two negative charges on the left, and none on the right, so we add two electrons on the right. In the reduction reaction, there are 7 positive charges on the left and 2 positives on the right, so we need to add 5 electrons on the left.
$$ \begin{align} 2 \, Cl^- &\longrightarrow Cl_2 + 2 \, e^- \\[5pt] 5 \, e^- + 8 \, H^+ + MnO_4^- &\longrightarrow Mn^{2+} + 4 \, H_2O \end{align}$$
Notice that where we put the electrons makes sense in terms of oxidation (loss of electrons) and reduction (gain of electrons).
6. If necessary, multiply the equations through by coefficients that make the number of electrons the same in both half reactions.
The number of electrons in the respective equations are 2 and 5, so we'll multiply the oxidation reaction through by 5 and the reduction reaction through by 2:
$$ \begin{align} 10 \, Cl^- &\longrightarrow 5 \, Cl_2 + 10 \, e^- \\[5pt] 10 \, e^- + 16 \, H^+ + 2 \, MnO_4^- &\longrightarrow 2\, Mn^{2+} + 8 \, H_2O \end{align}$$
7. Finally, add the equations, cancelling anything that appears on both sides:
$$ \require{cancel} \begin{align} 10 \, Cl^- &\longrightarrow 5 \, Cl_2 + \cancel{10 \, e^-} \\[5pt] \cancel{10 \, e^-} + 16 \, H^+ + 2 \, MnO_4^- &\longrightarrow 2\, Mn^{2+} + 8 \, H_2O \\ \hline \\ 16 \, H^+ + 2 \, MNO_4^- + 10 \, Cl^- &\longrightarrow 2 \, Mn^{2+} + 5 \, Cl_2 + 8 \, H_2O \end{align}$$
Take a look at the result. All of the atoms are mass balanced, and the total charge is the same on the left and right. Finally, all of the coefficients are reduced to their smallest values — there are no common factors other than 1.
Example 2
$Cn^- + MnO_4^- \longrightarrow CNO + MNO_2 + H_2O$ in basic solution
First assign oxidation numbers to this net ionic equation. In this case, It's difficult to determine oxidation numbers for the carbon and nitrogen in CN- and CNO, but if we group them as below, it's easy to see that the CN moeity is oxidized (loses electrons).
Now our half reactions are
$$ \begin{align} CN^- &\longrightarrow CNO \\[5pt] MNO_4^- &\longrightarrow MNO_2 \end{align}$$
Both will require addition of water and H+ ions to balance the masses. We're adding water and H+ for now, and we'll reconcile this with the fact that we mean for this reaction to be in basic solution later,
$$ \begin{align} H_2O + CN^- &\longrightarrow CNO + 2 \, H^+ \\[5pt] 4 \, H^+ + MNO_4^- &\longrightarrow MNO_2 + 2 \, H_2O \end{align}$$
Adding electrons to balance the charge gives
$$ \begin{align} H_2O + CN^- &\longrightarrow CNO + 2 \, H^+ + 3 \, e^-\\[5pt] 3 \, e^- + 4 \, H^+ + MNO_4^- &\longrightarrow MNO_2 + 2 \, H_2O \end{align}$$
We already have the same number of electrons on both sides, so we won't need to multiply through by anything to balance those. The next step is to understand that in basic solution there won't be any appreciable amount of H+ around, so we add enough OH- to each side of the equation to neutralize any of that (H+ + OH- → H2O)
$$ \begin{align} 2 \, OH^- + H_2O + CN^- &\longrightarrow CNO + 2 \, H^+ + 3 \, e^- + 2 \, OH^-\\[5pt] 4 \, OH^- 3 \, e^- + 4 \, H^+ + MNO_4^- &\longrightarrow MNO_2 + 2 \, H_2O + 4 \, OH^- \end{align}$$
Now we can cancel anything common on the collective left and right sides of these equations, preparing to add them:
$$ \begin{align} \cancel{2 \, OH^-} + H_2O + CN^- &\longrightarrow CNO + \cancel{2 \, H_2O} + \cancel{3 \, e^-} \\[5pt] \cancel{4 \, H_2O} + \cancel{3 \, e^-} + MNO_4^- &\longrightarrow MnO_2 + \cancel{2 \, H_2O} + \cancel{4} 2 \, OH^- \\ \hline \\ 4 \, H_2O + CN^- + MNO_4^- &\longrightarrow CNO + MnO_2 + 2 \, OH^- \end{align}$$
That's the balanced redox equation. It's both mass (number & kind of atoms) and charge balanced.
There are other methods of balancing redox reactions in basic solutions. You don't always have to use this backtracking method — add acid, then base. You can sometimes just add base (as in the next example), but the acid-then-base method will always work. If you work through enough examples, you'll develop an intuition for it and your own style.
Example 3
$Cl_{2 \; (g)} \longrightarrow Cl^-_{(aq)} + ClO^-_{(aq)}$ in basic solution
Assigning oxidation numbers tells us that in this reaction, chlorine gas (Cl2) is both oxidized and reduced:
The resulting half reactions are then
$$ \begin{align} Cl_2 &\longrightarrow Cl^- \\[5pt] Cl_2 &\longrightarrow ClO^- \end{align}$$
Balancing the atoms is easy with the reduction half reaction. The oxidation half reaction requires addition of water and H+ ions to balance the oxygens. Remember, we're proceeding as though this reaction were being run in acidic solution, and we'll adjust to basic later.
$$ \begin{align} Cl_2 &\longrightarrow 2 \, Cl^- \\[5pt] 2 \, H_2O + Cl_2 &\longrightarrow 2 \, ClO^- + 4 \, H^+ \end{align}$$
Adding electrons to balance the charges gives
$$ \begin{align} 2 \, e^- + Cl_2 &\longrightarrow 2 \, Cl^- \\[5pt] 2 \, H_2O + Cl_2 &\longrightarrow 2 \, ClO^- + 4 \, H^+ + 2 \, e^- \end{align}$$
Now we add four OH- ions to both sides of the oxidation reaction in order to neutralize the 4H+ acidic protons. Adding the hydroxides to both sides maintains our hard-won mass and charge balance.
$$ \begin{align} 2 \, e^- + Cl_2 &\longrightarrow 2 \, Cl^- \\[5pt] 4 \, OH^- + 2 \, H_2O + Cl_2 &\longrightarrow 2 \, ClO^- + 4 \, H^+ + 2 \, e^- + 4 \, OH^- \end{align}$$
Converting to water on the right side of the oxidation reaction and cancelling things in common across the arrows of both reactions gives us the sum of the half reactions.
$$ \begin{align} \cancel{2 \, e^-} + Cl_2 &\longrightarrow 2 \, Cl^- \\[5pt] 4 \, OH^- + \cancel{2 \, H_2O} + Cl_2 &\longrightarrow 2 \, ClO^- + \cancel{4}\color{red}{2} \, H^+ + \cancel{2 \, e^-} +4 \, OH^- \\ \hline \\ 4 \, OH^- + 2 \, Cl_2 &\longrightarrow 2 \, Cl^- + 2 \, ClO^- + 2 \, H_2O \end{align}$$
That reaction has coefficients all divisible by two, so we should reduce them to get the final balanced equation:
$$2 \, OH^- + Cl_2 \longrightarrow Cl^- + ClO^- + H_2O$$
Practice problems
Balance these redox reactions:
-
$KMnO_4 + C_2H_5OH \rightleftharpoons Mn^{2+} + C_2H_4O + K^+$ in acidic solution
Solution
The half reactions:
$$ \begin{align} C_2H_5OH &\longrightarrow C_2H_4O \phantom{00} \color{steelblue}{\text{oxidation}} \\[5pt] KMnO_4 &\longrightarrow Mn^{2+} \phantom{00} \color{#C5D82F}{\text{reduction}} \end{align}$$
Balance with protons:
$$ \begin{align} C_2H_5OH &\longrightarrow C_2H_4O + 2 \, H^+ \\[5pt] 8 \, H^+ + KMnO_4 &\longrightarrow Mn^{2+} + 4\, H_2O \end{align}$$
Balance the charge with electrons:
$$ \begin{align} C_2H_5OH &\longrightarrow C_2H_4O + 2 \, H^+ + 4 \, e^- \\[5pt] 6 \, e^- + 8 \, H^+ + KMnO_4 &\longrightarrow Mn^{2+} + 4\, H_2O \end{align}$$
Balance and add:
$$ \begin{align} 6(C_2H_5OH &\longrightarrow C_2H_4O + 2 \, H^+ + 4 \, e^-) \\[5pt] 4(6 \, e^- + 8 \, H^+ + KMnO_4 &\longrightarrow Mn^{2+} + 4 \, H_2O) \\ \hline 6 \, C_2H_5OH + \cancel{24 \, e^-} &+ \cancel{32}\color{red}{20} \, H^+ + 4 \, KMnO_4 \longrightarrow \\[5pt] 6 \, C_2H_4O + \cancel{12 \, H^+} &+ \cancel{24 \, e^-} + 4 \, Mn^{2+} + 16 \, H_2O \\ \hline 6 \, C_2H_5OH + 4 \, KMnO_4 + 20 \, H^+ &\longrightarrow 6 \, C_2H_4O + 4 \, Mn^{2+} + 16 \, H_2O \end{align}$$
Dividing all coefficients by 2 gives our final, balanced equation:
$$ \begin{align} 3 \, C_2H_4OH &+ 2 \, KMnO_4 + 10 \, H^+ \longrightarrow \\[5pt] 3 \, C_2H_4O &+ 2 \, Mn^{2+} + 8 \, H_2O \end{align}$$
-
$KMnO_4 + C_2H_5OH \rightleftharpoons Mn^{2+} + C_2H_4O + K^+$ in basic solution
Solution
The half reactions:
$$ \begin{align} C_2H_5OH &\longrightarrow C_2H_4O \phantom{00} \color{steelblue}{\text{oxidation}} \\[5pt] KMnO_4 &\longrightarrow Mn^{2+} + K^+ \phantom{00} \color{#C5D82F}{\text{reduction}} \end{align}$$
Balance with protons:
$$ \begin{align} C_2H_5OH &\longrightarrow C_2H_4O + 2 \, H^+ \\[5pt] 8 \, H^+ + KMnO_4 &\longrightarrow Mn^{2+} + K^+ + 4\, H_2O \end{align}$$
Balance the charge with electrons:
$$ \begin{align} C_2H_5OH &\longrightarrow C_2H_4O + 2 \, H^+ + 2 e^-\\[5pt] 5 \, e^- + 8 \, H^+ + KMnO_4 &\longrightarrow Mn^{2+} + K^+ + 4\, H_2O \end{align}$$
Balance and add:
$$ \begin{align} 5(C_2H_5OH &\longrightarrow C_2H_4O + 2 \, H^+ + 2 \, e^-) \\[5pt] 2(5 \, e^- + 8 \, H^+ + KMnO_4 &\longrightarrow Mn^{2+} + K^+ + 4 \, H_2O) \\ \hline \\ \cancel{10 \, e^-} _ 16 \, H^+ + 2 \, KMnO4 &\longrightarrow 2 \, Mn^{2+} + 2 \, K^+ + 8 \, H_2O \\[5pt] 5 \, C_2H_5OH &\longrightarrow 5 \, C_2H_4O + 10 \, H^+ + \cancel{10 \, e^-} \end{align}$$
Now add 16 OH- to the left side to create 16 waters with the 16 H^s, and the same number to the right side (everything has to balance. That will leave 6 OH- and 10 H2O on the right:
$$ \begin{align} 16 \, H_2O &+ 2 \, KMnO_4 + 5 \, C_2H_5OH \longrightarrow \\[5pt] & 2 \, Mn^{2+} + 2 \, K^+ + 10 \, H_2O + 6 \, OH^- + 5 \, C_2H_4O \\[5pt] \cancel{16}\color{red}{6} \, H_2O &+ 2 \, KMnO_4 + 5 \, C_2H_5OH \longrightarrow \\[5pt] & 2 \, Mn^{2+} + 2 \, K^+ + \cancel{10 \, H_2O} + 6 \, OH^- + 5 \, C_2H_4O \\[5pt] \end{align}$$
$$ \begin{align} 6 \, H_2O &+ 2 \, KMnO_4 + 5 \, C_2H_5OH \longrightarrow \\[5pt] & 2 \, Mn^{2+} + 2 \, K^+ + 5 \, C_2H_4O + 6 \, OH^- \end{align}$$
-
$Cr_2O_7^{2-} + Cl^- \rightleftharpoons Cr^{3+} + Cl_2$ in acidic solution
Solution
The half reactions:
$$ \begin{align} Cl^- &\longrightarrow Cl_2 \phantom{00} \color{steelblue}{\text{oxidation}} \\[5pt] Cr_2O_7^{2-} &\longrightarrow Cr^{3+} \phantom{00} \color{#C5D82F}{\text{reduction}} \end{align}$$
Balance with protons:
$$ \begin{align} 14 \, H^+ + Cr_2O_7^{2-} &\longrightarrow Cr^{3+} + 7 \, H_2O \\[5pt] 2 \, Cl^- &\longrightarrow Cl_2 \end{align}$$
Balance the charge with electrons:
$$ \begin{align} 9 \, e^- + 14 \, H^+ + Cr_2O_7^{2+} &\longrightarrow Cr^{3+} + 7 \, H_2O\\[5pt] 2 \, Cl^- &\longrightarrow Cl_2 + 2 \, e^- \end{align}$$
Balance and add:
$$ \begin{align} 2(9 \, e^- + 14 \, H^+ + Cr_2O_7^{2+} &\longrightarrow Cr^{3+} + 7 \, H_2O)\\[5pt] 9(2 \, Cl^- &\longrightarrow Cl_2 + 2 \, e^-) \\ \hline \\ \cancel{18 \, e^-} + 28 \, H^+ + 2 \, Cr_2O_7^{2-} &\longrightarrow 2 \, Cr^{3+} + 14 \, \, H_2O \\[5pt] 18\, Cl^- &\longrightarrow 9 \, Cl_2 + \cancel{18 \, e^-} \end{align}$$
Adding the last two equations gives us the final equation:
$$ \begin{align} 28 \, H^+ &+ 2 \, Cr_2O_7^{2-} + 18 \, Cl^- \longrightarrow \\[5pt] & 2 \, Cr^{3+} + 9 \, Cl_2 + 14 \, H_2O \end{align}$$
-
$Au_{(aq)}^{3+} + I_{(aq)}^- \rightleftharpoons Au_{(s)} + I_{2 \; (s)}$ in acidic solution
Solution
The half reactions:
$$ \begin{align} 2 \, I^- &\longrightarrow I_2 \phantom{00} \color{steelblue}{\text{oxidation}} \\[5pt] Au^{3+} &\longrightarrow Au \phantom{00} \color{#C5D82F}{\text{reduction}} \end{align}$$
Balance with electrons:
$$ \begin{align} 3 \, e^- + Au^{3+} &\longrightarrow Au \\[5pt] 2 \, I^- &\longrightarrow I_2 + 2 \, e^- \end{align}$$
Balance and add:
$$ \begin{align} 2( 3 \, e^- + Au^{3+} &\longrightarrow Au)\\[5pt] 3(2 \, I^- &\longrightarrow I_2 + 2 \, e^-) \\ \hline \\ \cancel{6 \, e^-} + 2 \, Au^{3+} + 6 \, I^- &\longrightarrow 2 \, Au + 3 \, I_2 + \cancel{6 \, e^-} \end{align}$$
The balanced equation is:
$$2 \, Au^{3+} + 6 \, I^- \longrightarrow 3 \, I_2 + 2 \, Au$$
-
$SO_{3 \; (aq)}^{2-} + MnO_{4 \; (aq)}^- \rightleftharpoons SO_{4 \; (aq)}^{2-} + Mn_{(aq)}^{2+}$ in acidic solution
Solution
The half reactions:
$$ \begin{align} SO_3^{2-} &\longrightarrow SO_4^{2-} \phantom{00} \color{steelblue}{\text{oxidation}} \\[5pt] MnO_4^- &\longrightarrow Mn^{2+} \phantom{00} \color{#C5D82F}{\text{reduction}} \end{align}$$
Balance with protons:
$$ \begin{align} H_2O + SO_3^{2-} &\longrightarrow SO_4^{2-} + 2 \, H^+ \\[5pt] 8 \, H^+ + MnO_4^- &\longrightarrow Mn^{2+} + 4 \, H_2O \end{align}$$
Balance with electrons, mass-balance and add:
$$ \begin{align} 5(H_2O + SO_3^{2-} &\longrightarrow SO_4^{2-} + 2 \, H^+ + 2 \, e^-) \\[5pt] 2(5 e^- + 8 \, H^+ + MnO_4^- &\longrightarrow Mn^{2+} + 4 \, H_2O) \\ \hline \cancel{5 \, H_2O} + 5 \, SO_3^{2-} &+ \cancel{10 \, e^-} + \cancel{16}\color{red}{6} \, H^+ + 2 \, MnO_4^- \longrightarrow \\[5pt] 5 SO_4^{2-} + \cancel{10 \, H^+} &+ \cancel{10 \, e^-} + 2 \, Mn^{2+} + \cancel{8}\color{red}{3} \, H_2O \end{align}$$
The balanced equation is:
$$ \begin{align} 5 \, SO_3^{2-} &+ 6 \, H^+ + 2 \, MnO_4^- \longrightarrow \\[5pt] 5 \, SO_4^{2-} &+ 3 \, H_2O + 2 \, Mn^{2+} \end{align}$$
-
$ClO^- + Fe(OH)_3 \rightleftharpoons Cl^- + FeO_4^{2-}$ in basic solution
Solution
The half reactions:
$$ \begin{align} Fe(OH)_3 &\longrightarrow FeO_4^{2-} \phantom{00} \color{steelblue}{\text{oxidation}} \\[5pt] ClO^- &\longrightarrow Cl^- \phantom{00} \color{#C5D82F}{\text{reduction}} \end{align}$$
Balance with protons and water:
$$ \begin{align} 2 \, H^+ + ClO^- &\longrightarrow Cl^- + H_2O \\[5pt] H_2O + Fe(OH)_3 &\longrightarrow FeO_4^{2-} + 5 \, H^+ \end{align}$$
Balance with electrons, mass-balance and add:
$$ \begin{align} 3(2 \, e^- + 2 \, H^+ + ClO^- &\longrightarrow Cl^- + H_2O) \\[5pt] 2(H_2O + Fe(OH)_3 &\longrightarrow FeO_4^{2-} + 5 \, H^+ + 3 \, e^-\\ \hline \cancel{6 \, e^-} + \cancel{6 \, H^+} + 3 \, ClO^- &+ \cancel{2 \, H_2O} + 2 \, Fe(OH)_3 \longrightarrow \\[5pt] 3 \, Cl^- + \cancel{3}\color{red}{1} \, H_2O &+ 2 \, FeO_4^{2-} + \cancel{10}\color{red}{4} \, H^+ + \cancel{6 \, e^-} \end{align}$$
The balanced equation is:
$$ \begin{align} 4 \, OH^- &+ 3 \, ClO^- + 2 \, Fe(OH)_3 \longrightarrow \\[5pt] 3 \, Cl^- &+ 5 \, H_2O + 2 \, FeO_4^{2-} \end{align}$$
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