Oxidation and Reduction Reactions for the MCAT: Everything You Need to Know
Learn key MCAT concepts about oxidation and reduction reactions, plus practice questions and answers
Table of Contents
Part 1: Introduction to Oxidation and Reduction Reactions
Part 2: Overview of Atomic Structure
Part 3: Oxidation and Reduction
a) Definitions
b) Assigning oxidation states
c) Hydrides
Part 4: Applications of Redox Reactions
a) Writing and balancing redox reactions
b) Electrochemical cells
c) Electron transport chain
Part 5: High-Yield Terms
Part 6: Passage-Based Questions and Answers
Part 7: Standalone practice questions
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Part 1: Introduction to Oxidation and Reduction Reactions
Redox reactions constitute fundamental processes in our daily lives. Oxidation and reduction reactions are key in creating energy from the food we eat. Additionally, these processes are key to the operations behind batteries, in which reduction and oxidation reactions generate the power we need to drive our cars.
Redox reactions refer to chemical reactions in which the exchange of electrons results in the oxidation of some atoms and the reduction of others. Success on this topic requires a detailed understanding of both oxidation and reduction and how oxidation states change over the course of a given reaction.
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Part 2: Overview of Atomic Structure
How are electrons spatially distributed in an atom? The electron cloud model provides us with an answer and proves to be useful in visualizing how electrons are lost (oxidation) and gained (reduction) in redox reactions.
The electron cloud model, developed by Erwin Schrodinger in the 1920s, describes electrons as being distributed within certain regions of probability around the central nucleus. The electron cloud model-which is the preferred model of modern scientists-is distinct from Bohr's atomic model in which electrons were imagined to move in discrete concentric orbits around the nucleus-much like satellites around the Earth or the cabins of a Ferris wheel.
The electron cloud model emphasizes the indistinct nature of electron distribution. The electron cloud model theorized that electrons do not move in static orbits around the central nucleus, such that electrons are always a specific and discrete distance away from the center of the atom. Instead, we can only guess where an electron might be-that guess is mathematically computed and described as a region of probability called the electron cloud. These regions of probability can be visualized and illustrated as certain shapes, which gives rise to the atomic theory of subshells and electron orbitals.
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Part 3: Oxidation and Reduction
a) Definitions
You may already know that oxidation refers to a “loss of electrons,” while reduction refers to a “gain of electrons.” A useful mnemonic to use is: “LEO says GER”—which stands for “Losing Electrons is Oxidation and Gaining Electrons is Reduction.” The mnemonic “OIL RIG” can be used to a similar effect: “Oxidation Is Loss, Reduction Is Gain.” The word redox encompasses both of these processes: it is a portmanteau of the words “reduction” and “oxidation.”
It’s important to note the distinction between an oxidizing agent and a compound that is oxidized. The adjectives “oxidizing” and “reducing” describe the change that the agents or compounds are provoking in a different compound. They do not mean that the compound itself is becoming oxidized or reduced, respectively. Consequently, an oxidizing agent (or oxidant) is a compound that is reduced while enabling the oxidation of a different compound). A reducing agent (or reductant) is a compound that is oxidized, as it enables the reduction of a different compound (by donating electrons).
Due to the nature of exchanging electrons, oxidation and reduction reactions are always coupled together. This implies that as one element is oxidized (or loses electrons), another element must be reduced (receives those electrons).
The exception is in disproportion reactions, in which one element can undergo both oxidation and reduction. In this case, atoms from the same element simultaneously act as the oxidizing agent and the reducing agent.
b) Assigning oxidation states
An atom’s oxidation state is a number that describes its degree of oxidation, or amount of electron loss. Thus, an additional definition of oxidation and reduction may utilize oxidation states. When an atom is oxidized, its oxidation state increases. When an atom is reduced, its oxidation state decreases.
There are several rules to follow when assigning oxidation states. (There are several exceptions, but it is rare you will encounter them in your studying.) This may seem like a long list, but fortunately, many of the rules follow trends that are easily learned in relation to the periodic table.
- The oxidation state of uncharged, free elements—or elements by themselves—is always 0. This includes all of the naturally occurring diatomic elements (H2, N2, O2, F2, Cl2, Br2, I2)
- Alkali metals in group 1 always have an oxidation state of +1.
- Alkaline earth metals in group 2 always have an oxidation state of +2.
- Halogens in group 17 almost always have an oxidation state of -1.
- Hydrogen always has an oxidation state of +1.
- Oxygen usually has an oxidation state of -2, except in peroxides such as H2O2. When found in peroxides, oxygen has an oxidation state of -1.
The rules listed above are a good starting place for determining the oxidation states of individual atoms. Thus, when presented with a molecular compound, it is easiest to assign oxidation states to any hydrogen (+1) or oxygen (-2 or -1) atoms first. Then, proceed to assign oxidation states for atoms in the remainder of the molecule.
When calculating the oxidation states of atoms within compounds, stoichiometric coefficients must be accounted for. Multiply each atom’s oxidation state by its associated stoichiometric coefficient when computing calculations for the total ionic charge. The total charge yielded by this calculation should equal the net charge of the molecule.
In a neutral compound, the oxidation states of each of its constituent atoms will add up to zero, the overall charge of the compound. Here is an example for assigning oxidation states to potassium permanganate, KMnO4. Note that it is a neutrally charged compound.
- K is an alkali metal—its oxidation state is always +1.
- Mn is a transition metal—let’s represent its oxidation state as an unknown variable “x”.
- Oxygen usually has an oxidation state of -2
- Using the molecular formula, we can set up an algebraic equation to solve for the value of x: (1)(+1) + (1)(x) + (4)(-2) = 0
Figure 1 Labeled oxidation states of potassium permanganate
- Oxygen usually has an oxidation state of -2
- The oxidation state of phosphorus is unknown—let’s represent its oxidation state as an unknown variable “x”
- Using the molecular formula, we can set up an algebraic equation to solve for the value of x: x + (4)(-2) = -3
Figure 2 Labelled oxidation states of the phosphate ion
c) Hydrides
Although exceptions to the oxidation state rules do not often arise, they may arisedue to differences in electronegativity. For instance, hydrogen acts as a hydride when bound to the less electronegative metals (metal hydrides) and thus possesses an oxidation state of -1 rather than its customary +1.
Hydrides are negatively charged ions of the hydrogen atom (H-). They can be considered protons with a pair of free electrons. Thus, the gain of a hydride ion always results in a reduction, while the loss of a hydride ion always results in oxidation. (In biochemical contexts, it is often easier to think of reduction as the gain of a bonded hydrogen atom, and oxidation as the loss of a bonded hydrogen atom.)
Hydrides play a significant role in cellular processes, particularly in the context of electron-carrying molecules such as NADH and FADH2. Recall that during carbohydrate metabolism, NAD+ is reduced to NADH. This occurs through the addition of a hydride ion. This reduction reaction is always coupled to the oxidation of another molecule, such as a hydrocarbon substrate.
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Part 4: Applications of redox reactions
Let’s take a closer look at redox reactions in different applications. There are several important items to keep in mind, namely, to track changes in oxidation states and to balance any reactions that occur.
a) Writing and balancing redox reactions
Recall that the reduction of one compound is always tied to the oxidation of another. How do we determine how much of a certain quantity is reduced or oxidized?
There are several steps to balancing a redox reaction. There are more steps involved than in balancing an ordinary chemical reaction, but the principle of conservation of mass still applies.
To balance redox reactions, first assign oxidation numbers to each of the atoms involved in the reaction. Then, rewrite the reactants and products involved into two half-reactions, in which one is an oxidation reaction, and the other is a reduction reaction.
In summary:
Find oxidation numbers
Break reaction into half-reactions via identifying oxidation vs. reduction components
Balance masses via coefficients and/or adding water and H+
Balance charges via adding electrons
Balance electrons via multiplying each reaction so that net electrons equal zero
Add the half-reactions
Let’s work on an example.
First, determine oxidation numbers for each element. The oxidation state for copper changes from 0 to 2+, while the oxidation state for silver changes from 1+ to 0. Copper has lost electrons, thus resulting in a more positive oxidation number. Silver has gained electrons, thus resulting in a less negative oxidation number. Thus, copper has undergone an oxidation reaction, while silver has undergone a reduction reaction. The two half-reactions are as follows:
Reduction: Ag+ (aq) → Ag (s)
These two half-reactions both already follow the conservation of mass: there is one copper and silver on both sides of the equation. However, the charges are not yet balanced.
In the oxidation reaction, copper goes from 0 charge to 2+. That means that two electrons must be added on the right-hand side of the equation to achieve an overall net charge of zero. Similarly, the reduction reaction has a 1+ charge on the reactants side of the equation and 0 on the product side. Thus, one electron must be added to the reactants.
Reduction: 1e- + Ag+ (aq) → Ag (s)
Now that electrons have been added to both half-reactions, the number of electrons used in each half-reaction must equal each other. There must be an equal amount of electrons participating in both the oxidation and reduction reactions.
In this case, the reduction half-reaction can be modified by a whole-number coefficient. By multiplying each coefficient in the reaction by 2, we obtain:
Thus, every two electrons produced in the reduction half-reaction are balanced by two electrons produced in the oxidation half-reaction.
Lastly, the two balanced oxidation and reduction reactions can be added together.
Since two electrons are present on both sides of the reaction, they can be eliminated. The resulting reaction represents the balanced, total reaction between the oxidation of copper and the reduction of silver.
Figure 3 Changes in oxidation state
In an experimental setting, redox titrations are performed to determine the concentration of an analyte. Similar to acid-base titrations, a reducing or oxidizing solution is added to the analyte until an endpoint is reached. When the endpoint—or equivalence point—is reached, the known concentrations and volumes can be used to determine the concentration of the original analyte.
For a review of titrations, be sure to refer to our guide on acids and bases.
b) Electrochemical cells
Recall that galvanic cells (also known as voltaic cells) and electrolytic cells are two major types of electrochemical cells. In galvanic cells, spontaneous redox reactions take place to generate electric energy (ΔG < 0). In electrolytic cells, an external voltage source is used to drive nonspontaneous redox reactions (ΔG > 0), which generate the desired chemical product. (For more information on this, be sure to refer to our guide on electrochemistry.)
Figure 4 Setup of a galvanic cell
Without being given the two half-reactions of the cell, we can still determine what the half-reactions are. Recall that oxidation takes place at the anode, while reduction occurs at the cathode. (The mnemonic “AN OX; RED CAT” may be helpful in remembering this.) Further, since the oxidation state of Cu changes to +2, the element must be oxidized. Thus, the half-reaction at the anode is:
The oxidation state of silver changes from +1 to 0. Thus, the element must be reduced. The half-reaction at the cathode is the reduction of 2Ag+.
If the number of moles of electrons needed to drive an electrolytic reaction is known, the quantity can be converted to charge using Faraday’s constant (96,485.3365 Coulombs per mole).
c) Electron transport chain
NAD+ and FAD are coenzymes—small organic molecules that are essential to the functioning of enzymes. As “electron carriers,” NAD+ and FAD are both capable of accepting two electrons each in the form of hydrides. After gaining two electrons, NAD+ and FAD are in their reduced forms: NADH, and FADH2. NADH and FADH2 can also lose two electrons to convert to their oxidized forms: NAD+ and FAD.
Figure 5 Oxidized and reduced forms of key electron carriers
This electron transfer is coupled to the movement of protons into the intermembrane space, generating the proton gradient that powers ATP synthase and its oxidative phosphorylation of ADP. Electrons are transferred between complexes embedded in the inner mitochondrial membrane until they are finally deposited onto the final electron acceptor: oxygen. This final oxidation-reduction reaction results in the production of water (H2O), a waste product of aerobic respiration.
A big part of understanding the big picture of metabolism is appreciating when and where ATP, NADH, and FADH2 are made. As you’ve learned before, oxidative phosphorylation—not substrate-level phosphorylation—is the key source of ATP. In other words, the electron-rich NADH and FADH2 coenzymes that are oxidized at the electron transport chain are essential to generating the chemiosmotic conditions necessary for oxidative phosphorylation, and ultimately for ATP generation.
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Part 5: High-yield terms
Combustion: a type of redox reaction in which a hydrocarbon fuel and oxygen gas react to produce CO2, H2O, and heat
Coenzymes: small organic molecules that are essential to the functioning of enzymes; examples include various vitamins, NAD+, and FAD
Electron cloud model: a theory developed by Erwin Schrödinger in 1926, which describes electrons as being distributed within “clouds” or regions of probability around the central nucleus; as the preferred model of modern science, it is often compared to Bohr’s (outdated) atomic model
Galvanic cells: electrochemical cells in which spontaneous redox reactions take place to generate electric energy (ΔG < 0)
Electrolytic cells: electrochemical cells in which an external voltage source is used to drive nonspontaneous redox reactions (ΔG > 0), which generate a desired chemical product
Free elements: pure neutral elements that are found by themselves, uncombined with other elements; their oxidation states are always 0
Oxidation: loss of electrons; oxidation is indicated by an increase in oxidation state
Oxidation state: a number that describes its degree of oxidation, or amount of electron loss
Oxidizing agent (or oxidant): a compound that is reduced while enabling the oxidation of a different compound
Reduction: gain of electrons; reduction is indicated by a decrease in oxidation state
Reducing agent (or reductant): a compound that is oxidized as it enables the reduction of a different compound
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Part 6: Passage-based questions
Rechargeable batteries use specialized chemical electrodes to enhance the capacity of its electrochemical charge and improve longevity of the battery. Among these batteries, lithium-ion batteries (LIBs) are widely used.
It has been shown that using organic compounds in place of metal-based electrode materials may increase electrochemical capacity and maximize environmental sustainability. To investigate this, a team of researchers seeks to characterize the behavior of a synthetic dimer compound derived from a naphthazarin (5,8-dihydroxy-1,4-naphthoquinone) monomer skeleton.
When this newly synthesized naphthazarin dimer was tested in a battery electrode, it showed a lengthened cycle-life. This increased performance did not appear to sacrifice the initial high capacity and higher energy density compared to the naphthazarin monomer or to a conventional lithium electrode. The behavior of the naphthazarin monomer was characterized and is shown in Figure 1.
Figure 1 Battery performance of the electrode using the naphthazarin monomer 1.
The team also noted the bipolar characteristic of the dimer, which suggests that its lithium salt may function as a material in either the positive or the negative electrode. This is a property that has been demonstrated in only one other organic material, which suggests that using the lithium salt of the dimer may provide a low-cost and efficient alternative. The resulting half-cell reactions are shown in Figure 2.
Figure 2 Charge/discharge behavior of the symmetric full cell using the naphthazarin dimer in both the positive and negative electrodes.
Question 1: Based on Figure 1, which of the following pair of structures best represents the forms of the naphthazarin-derivative most likely to dominate under reducing conditions and oxidizing conditions, respectively?
A)
B)
C)
D)
A) +5
B) +3
C) -3
D) -5
A) Eight
B) Four
C) Two
D) One
Question 4: What is the oxidation state of the sulfur compound described in paragraph 4?
A) -2
B) -1
C) 0
D) 2
Answer key for passage-based questions
1.Answer choice C is correct. Under reducing conditions, the compound will likely be in its most reduced form. Reduction results in a gain of electrons, which the compound has acquired in the form of covalent electrons donated by lithium (choices A and D are incorrect). Under oxidizing conditions, the compound will likely be in its most oxidized form. Oxidation results in a loss of electrons. Note that these structures are also presented in Figure 1a. The diquinone form of the molecule possesses the most bonds to oxygen, indicating a high oxidation state. As the reaction progresses leftward, the compound is successively reduced—two bonds to oxygen and two electrons are successively lost (choices A, B, and D are incorrect).
3. Answer choice C is correct. Each FADH2 molecule is capable of losing 2 electrons to become FAD+. Figure 2 shows that the negative electrode half-reaction involves the exchange of four electrons (choices A, B, and D are incorrect).
4. Answer choice C is correct. The paragraph specifies the sulfur as “elemental sulfur,” which means sulfur in its naturally occurring, uncombined form. All elemental compounds have an oxidation state of 0 (choice C is correct). Sulfur would take on other oxidation states (usually -2) if it were in solution or as a constituent atom within a larger compound (choice A is incorrect).
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Part 7: Standalone practice questions
Question 1: The explosive reaction that occurs when sodium contacts water is described by the following chemical reaction:
What are the observed oxidation states of hydrogen?
A) +1, +1
B) -1, 0
C) +1, 0
D) +2, +1
Question 2: Which of the following is NOT a redox reaction?
A) Fe2O3 + 3CO ⇋ 2Fe + 3CO2
B) 2H2 + O2 ⇋ 2H2O
C) CH4 + 2O2 ⇋ CO2 + 2H2O
D) CuO +H2SO4 ⇋ CuSO4 + H2O
A) -2
B) -1
C) 0
D) +1
Question 4: What are the stoichiometric coefficients for the balanced redox equation?
A) 1, 2, 3, 4
B) 1, 2, 2, 1
C) 1, 3, 2, 3
D) 1, 2, 3, 2
Question 5: What is the oxidizing agent in the combustion of methane (shown below)?
A) CH4
B) O2
C) CO2
D) H2O
Answer key for standalone questions
2. Answer choice D is correct. Notice that there is no change in oxidation state for any of the elements present in this equation. Instead, this is a double-displacement reaction. (For more information on this, be sure to refer to our guide on molecules and stoichiometry.) In this reaction, the oxidation state of Cu remains at +2; the oxidation state of O remains at -2; the oxidation state of H remains at +1; and the oxidation state of S remains at +6 (choices A, B, and C are incorrect).
3. Answer choice B is correct. Hydrogen always has a +1 oxidation state (unless it is bound to a metal). Oxygen has an oxidation state of -2, except in cases where oxygen is present in peroxides. Thus, oxygen has a -1 oxidation state in this molecule, which is hydrogen peroxide (choices A, C, and D are incorrect).
4. Answer choice C is correct. This question can be solved through simple mass balance. To ensure that the conservation of mass is fulfilled, ensure that the same number of atoms of each element is present on both the right and the left-hand sides of the equation. This question can also be solved by solving the half-reactions for the oxidation and reduction of iron and carbon.
