MCAT General Chemistry Review
Part I Review
Chapter 11: Redox Reactions and Electrochemistry
The mitochondria are the power company of your body. No, this doesn’t mean that they send vaguely hostile letters threatening to cut off your electricity if you don’t pay your overdue bill (although cells certainly do send loud and clear messages when there is an oxygen debt). Rather, like actual power companies, the primary purpose of the mitochondria is to manufacture a deliverable and usable form of energy. Certainly by now you are generally well aware of the complex processes by which the potential energy of chemical bonds, which is really just electric potential energy, in food molecules (carbohydrates, amino acids, and lipids) is converted into the potential energy of the phosphate bond in adenosine triphosphate (ATP). ATP is then delivered to different regions of the cell, where it is used to energize all of the processes essential to the maintenance of life.
The mitochondria generate tremendous amounts of ATP. (In humans, the average daily turnover of ATP is more than 50 kilograms!) Without a continuous supply and replenishment of ATP, we couldn’t live for even a second in its absence: It powers the contraction of our heart muscle and maintains the membrane potential essential for neurological function ( just to name a couple of life-essential roles of ATP). How do the mitochondria manufacture these packets of life-sustaining energy? You have read of the double-membrane structure of the mitochondria and their electron transport chain and F1Fo ATP synthase. You have learned about oxidative phosphorylation and the role of O2. But has it ever occurred to you that the mitochondria in more or less literal ways act as the batteries of the cell? Have you ever wondered at the similarity of the phrase proton motive forceto another term that you have learned and used in the context of electrochemistry and circuits? Are proton motive force (pmf) and electromotive force (emf) the same thing or, at the very least, similar in nature?
In fact, the mitochondria do function in ways similar to batteries: There is separation and buildup of a charge gradient; there is potential difference (voltage) between separated compartments; there is movement of charge and dissipation of energy. We could say that mitochondria function in ways most similar to a particular type of electrochemical cell called the concentration cell. In both concentration cells and mitochondria, a concentration gradient of ions between two separated compartments connected to each other by some means of charge conduction establishes an electrical potential difference (a voltage). This voltage, called electromotive force in a concentration cell and proton motive force in the mitochondria, provides the “pressure to move” charge (that is, creates current) from one compartment to the other. In the concentration cell, a redox reaction takes place, and electrons move in the direction that causes the concentration gradient to be dissipated. In the mitochondria, the charge buildup is in the form of a hydrogen ion (proton) gradient between the intermembrane space and the matrix. Embedded in the inner membrane is the F1Fo ATP synthase protein, which serves the dual role of proton channel (the conductive pathway) and catalyst (the electric motor) for the formation of the high-energy phosphate bond of ATP. As the hydrogen ions flow down their chemical-electrical gradient, energy is dissipated (remember, the positively charged ions are moving from high potential to low potential), and this energy is harnessed by the ATP synthase for the formation of ATP.
In this, the final chapter of our review of general chemistry for the MCAT, we will focus our attention on the study of the movement of electrons in chemical reactions. Such reactions are called oxidations and reductions, and because they always occur in pairs, they are usually referred to, in shorthand, as redox reactions. Electrochemistry is the study of the relationships between chemical reactions and electrical energy. We will learn of the ways in which the principles of electrochemistry can be applied to create different types of electrochemical cells, including galvanic (voltaic), electrolytic, and concentration cells. Regarding the thermodynamics of electrochemistry, we will focus on the significance of reduction potentials and examine the relationship among electromotive force, the equilibrium constant, and Gibbs function.
Reactions that involve the transfer of electrons from one chemical species to another can be classified as oxidation-reduction reactions. Now, that’s quite a mouthful to say, so these reactions are commonly called “redox” reactions as a shorthand way of noting that oxidation and reduction are always coupled.
OXIDATION AND REDUCTION
The law of conservation of charge states that an electrical charge can be neither created nor destroyed. Thus, an isolated loss or gain of electrons cannot occur; oxidation (loss of electrons) and reduction (gain of electrons) must occur simultaneously, resulting in an electron transfer called a redox reaction. An oxidizing agent causes another atom in a redox reaction to undergo oxidation and is itself reduced. A reducing agent causes the other atom to be reduced and is itself oxidized. There are various memory devices designed to help you remember these terms. One that is especially well known is OIL RIG, which stands for “Oxidation Is Loss; Reduction Is Gain.”
OIL RIG stands for “Oxidation Is Loss, Reduction Is Gain,” because, as we do often, we are talking about those all-important electrons. Alternatively, reduction is just what it sounds like: reduction of charge.
ASSIGNING OXIDATION NUMBERS
It is important, of course, to know which atom is oxidized and which is reduced. Oxidation numbers are assigned to atoms in order to keep track of the redistribution of electrons during chemical reactions. From the oxidation numbers of the reactants and products, it is possible to determine how many electrons are gained or lost by each atom. The oxidation number of an atom in a compound is assigned according to the following rules:
1. The oxidation number of free elements is zero. For example, the atoms in N2, P4, S8, and He all have oxidation numbers of zero.
2. The oxidation number for a monatomic ion is equal to the charge of the ion. For example, the oxidation numbers for Na+, Cu2+, Fe3+, Cl–, and N3– are +1, +2, +3, -1, and -3, respectively.
Don’t forget that you have the periodic table available to you on Test Day, so use it to organize your thoughts with regard to these rules rather than memorizing these on their own. Beware of transition metals but realize we can often figure their oxidation by default.
3. The oxidation number of each Group IA element in a compound is +1. The oxidation number of each Group IIA element in a compound is +2.
4. The oxidation number of each Group VIIA element in a compound is -1, except when combined with an element of higher electronegativity. For example, in HCl, the oxidation number of Cl is -1; in HOCl, however, the oxidation number of Cl is +1.
5. The oxidation number of hydrogen is -1 in compounds with less electronegative elements than hydrogen (Groups IA and IIA). Examples include NaH and CaH2. The more common oxidation number of hydrogen is +1.
6. In most compounds, the oxidation number of oxygen is -2. This is not the case, however, in molecules such OF2. Here, because F is more electronegative than O, the oxidation number of oxygen is +2. Also, in peroxides, such as BaO2, the oxidation number of O is -1 instead of -2 because of the structure of the peroxide ion, [O-O]2–. (Note that Ba, a Group IIA element, cannot be a +4 cation.)
7. The sum of the oxidation numbers of all the atoms present in a neutral compound is zero. The sum of the oxidation numbers of the atoms present in a polyatomic ion is equal to the charge of the ion. Thus, for SO42-, the sum of the oxidation numbers must be -2.
The conventions of formula writing put cation first and anion second. Thus NaH implies H– while HCl implies H+. So use the way the compound is written on the MCAT along with the periodic table to help you determine oxidation states.
Example: Assign oxidation numbers to the atoms in the following reaction in order to determine the oxidized and reduced species and the oxidizing and reducing agents.
SnCl2 + PbCl4 SnCl4 + PbCl2
Solution: All these species are neutral, so the oxidation numbers of each compound must add up to zero. In SnCl2, because there are two chlorines present and chlorine has an oxidation number of -1, Sn must have an oxidation number of +2. Similarly, the oxidation number of Sn in SnCl4 is +4; the oxidation number of Pb is +4 in PbCl4 and +2 in PbCl2. Notice that the oxidation number of Sn goes from +2 to +4; it loses electrons and thus is oxidized, making it the reducing agent. Because the oxidation number of Pb has decreased from +4 to +2, it has gained electrons and been reduced. Pb is the oxidizing agent. The sum of the charges on both sides of the reaction is equal to zero, so charge has been conserved.
BALANCING REDOX REACTIONS
Okay, okay. We know you don’t like balancing redox reactions. We know that there are many steps involved and that it can sometimes be difficult to remember how to balance not only mass but charge as well. And we recognize the unfortunate fact that we’ve just presented you with seven rules of assigning oxidation numbers to understand and remember. And now we’re giving you five steps to remember. And—hey, don’t think we didn’t just catch you rolling your eyes! Sigh all you want, but the truth of the matter is, the process of balancing redox reactions is tested on the MCAT. We might suggest that balancing redox reactions is sort of like balancing a checkbook, but we are hesitant to do so for at least two reasons: (1) balancing a checkbook isn’t exactly “fun,” and (2) aside from accountants (for whom it is simply part of their nature), we don’t know anyone who actually balances his checkbook.
By assigning oxidation numbers to the reactants and products, you can determine how many moles of each species are required for conservation of charge and mass, which is necessary to balance the equation. To balance a redox reaction, both the net charge and the number of atoms must be equal on both sides of the equation. The most common method for balancing redox equations is the half-reaction method, also known as the ion-electron method, in which the equation is separated into two half-reactions—the oxidation part and the reduction part. Each half-reaction is balanced separately, and they are then added to give a balanced overall reaction. As we review the steps involved, let’s consider a redox reaction between KMnO4 and HI in an acidic solution.
Methodical, step-by-step approaches like this one are great for the MCAT. Most often, you will not have to get through all of these steps before you can narrow down your answer choices. Often you will be able to find the correct answer halfway through the steps.
MnO4- + I- I2 + Mn2+
Step 1: Separate the two half-reactions.
Step 2: Balance the atoms of each half-reaction. First, balance all atoms except H and O. Next, in an acidic solution, add H2O to balance the O atoms and then add H+ to balance the H atoms. (In a basic solution, use OH- and H2O to balance the O’s and H’s.)
To balance the iodine atoms, place a coefficient of 2 before the I- ion.
2 I- I2
For the permanganate half-reaction, Mn is already balanced. Next, balance the oxygens by adding 4 H2O to the right side.
MnO4- Mn2+ + 4 H2O
Finally, add H+ to the left side to balance the 4 H2Os. These two half-reactions are now balanced.
MnO4- + 8 H+ Mn2+ + 4 H2O
Step 3: Balance the charges of each half-reaction. The reduction half-reaction must consume the same number of electrons as are supplied by the oxidation half. For the oxidation reaction, add 2 electrons to the right side of the reaction:
2 I- I2 + 2 e-
For the reduction reaction, a charge of +2 must exist on both sides. Add 5 electrons to the left side of the reaction to accomplish this:
5 e- + 8 H+ + MnO4- Mn2+ + 4 H2O
Next, both half-reactions must have the same number of electrons so that they will cancel. Multiply the oxidation half by 5 and the reduction half by 2.
5(2I- I2 + 2e-)
2(5e- + 8H+ + MnO4- Mn2+ + 4 H2O)
Step 4: Add the half-reactions:
10 I- 5 I2 + 10 e-
16 H+ + 2 MnO4- + 10 e- 2 Mn2+ + 8 H2O
The final equation is this:
10 I- + 10 e- + 16 H+ + 2 MnO4- 5 I2 + 2 Mn2+ + 10 e- + 8 H2O
To get the overall equation, cancel out the electrons and any H2Os, H+s, or OH-s that appear on both sides of the equation.
10 I- + 16 H+ + 2 MnO4- 5 I2 + 2 Mn2+ + 8 H2O
Step 5: Finally, confirm that mass and charge are balanced. There is a +4 net charge on each side of the reaction equation, and the atoms are stoichiometrically balanced.
Now we get to the fun part of electrochemistry! Here, we will review the way in which redox reactions can be used to supply energy that can be used to do work. We’ll also cover the way in which energy can be used to drive certain useful redox reactions. You don’t have to look too hard to find a battery in your immediate surroundings: a flashlight, your watch, the smoke detector, a cell phone, your cordless toothbrush—all these devices contain and run on batteries. You’ve even got lots of little chemiosmotic “batteries,” as we discussed in the introduction, inside your cells.
A way to keep the electrodes straight: AN OX and a RED CAT. Another easy way to remember this is by the spelling of the words: oxidAtion and reduCtion.
Electrochemical cells are contained systems in which redox reactions occur. There are three types of electrochemical cells, galvanic cells (also known as voltaic cells), electrolytic cells, and concentration cells. Spontaneous reactions occur in galvanic cells and concentration cells, and nonspontaneous reactions in electrolytic cells. All three types contain electrodes at which oxidation and reduction occur. For all electrochemical cells, the electrode at which oxidation occurs is called the anode, and the electrode where reduction occurs is called the cathode. Furthermore, we can also generally state that for all electrochemical cells, the movement of electrons is from anode to cathode and current i runs from cathode to anode.
GALVANIC (VOLTAIC) CELLS
All of the nonrechargeable batteries that you have lying around your house or apartment, in battery-operated devices or stored away in their packaging along with the butter and cream cheese in your refrigerator (yes, many people do keep their batteries in the refrigerator, and there is some thermodynamic justification for the practice—although we can’t prove that there is any particular benefit to keeping them with the butter and cream cheese—but we digress) are galvanic cells, also called voltaic cells. Realizing this, you will have no problem remembering the key principles of the operation of galvanic cells. Don’t remember whether galvanic cells house a spontaneous or nonspontaneous redox reaction? Just think about the fact that you use household batteries to supply energy to do something useful, like power a flashlight. If energy is being supplied by the battery, then it must be the case that the redox reaction taking place is giving off energy, which means the reaction’s free energy must be decreasing (–G) and the reaction must, therefore, be spontaneous. Don’t remember whether galvanic cells have positive or negative electromotive forces? Again, just think about the battery in the flashlight. You know (from the thought sequence just described) that the change in Gibbs function is negative; emf always has the sign opposite to the change in free energy.
Galvanic cells are commonly used as batteries; to be economically viable, batteries must be spontaneous!
Let’s examine the “inner workings” of a galvanic (voltaic) cell. Two electrodes of distinct chemical identity are placed in separate compartments, which we call half-cells. The two electrodes are connected to each other by a conductive material, such as a copper wire. Along the wire, you may find various other components of a circuit, such as resistors or capacitors, but for now, let’s keep our focus on the battery itself. Surrounding each of the electrodes is an electrolyte solution (aqueous), composed of cations and anions. As in Figure 11.1, the illustration of a Daniell cell, the cations in each of the two half-cell solutions may be of the same element as the respective metal electrode. Connecting the two solutions is a structure called the salt bridge, which consists of an inert salt.
When the electrodes are connected to each other by a conductive material, charge will begin to flow as the result of a redox reaction that is taking place between the two half-cells. The redox reaction in a galvanic cell is spontaneous, and therefore the change in Gibbs function for the reaction is negative (–G). As the spontaneous redox reaction proceeds (toward equilibrium, which we’ll discuss next), the movement of charge (electrons) results in a conversion of electric potential energy into kinetic energy (the electrons are moving, after all, and small as they are, electrons do have mass). By separating the reduction and oxidation half-reactions into two compartments, we are able to harness this energy and use it to do work by connecting various electrical machines or devices into the circuit between the two electrodes. If you’ve ever stuck a 9-volt battery (the rectangular type with the + and - ends on the same side) to your tongue, you know full well the reality of energy transfer by redox reaction. If you’ve never stuck a 9-volt battery to your tongue, we recommend that you consider this an experience you can live without. Seriously, don’t try it.
In the Daniell cell, a zinc bar is placed in an aqueous ZnSO4 solution, and a copper bar is placed in an aqueous CuSO4 solution. The anode of this cell is the zinc bar where Zn (s) is oxidized to Zn2+ (aq). The cathode is the copper bar, and it is the site of the reduction of Cu2+ (aq) to Cu (s). The half-cell reactions are written as follows:
Zn (s) Zn2+ (aq) + 2 e- (anode)
Cu2+ (aq) + 2 e- Cu (s) (cathode)
If the two half-cells were not separated, the Cu2+ ions would react directly with the zinc bar, and no useful electrical work would be obtained. Since the solutions (and electrodes) are physically separated, they must be connected to complete the circuit. Without connection, the electrons from the zinc oxidation half-reaction would not be able to get to the copper ions; thus, a wire (or other conductor) is necessary. However, if only a wire were provided for this electron flow, the reaction would soon cease, because an excess negative charge would build up in the solution surrounding the cathode and an excess positive charge would build up in the solution surrounding the anode. Eventually, the excessive charge accumulation would provide a counter voltage large enough to prevent the redox reaction from taking place, and the current would cease. This charge gradient is dissipated by the presence of a salt bridge, which permits the exchange of cations and anions. The salt bridge contains an inert electrolyte, usually KCl or NH4NO3, whose ions will not react with the electrodes or with the ions in solution. At the same time that the anions from the salt bridge (e.g., Cl-) diffuse into the ZnSO4 solution to balance out the charge of the newly created Zn2+ ions, the cations of the salt bridge (e.g., K+) flow into the CuSO4 solution to balance out the charge of the SO42- ions left in solution when the Cu2+ ions are reduced to Cu and precipitate out of solution (“plate out”) onto the copper cathode.
The purpose of the salt bridge is to exchange anions and cations to balance, or dissipate, newly generated charges.
During the course of the reaction, electrons flow from the zinc bar (anode) through the wire and the voltmeter (if one is connected) toward the copper bar (cathode). The anions (Cl-) flow externally (via the salt bridge) into the ZnSO4, and the cations (K+) flow into the CuSO4. This flow depletes the salt bridge and, along with the finite quantity of Cu2+ in the solution, accounts for the relatively short lifetime of the cell.
A cell diagram is a shorthand notation representing the reactions in an electrochemical cell. A cell diagram for the Daniell cell is as follows:
Zn (s) | Zn2+ (xM SO42- ) || Cu2+ (yM SO42-) | Cu (s)
Know the shorthand notation for cells because it is possible that on Test Day they will not spell it all out for us.
The following rules are used in constructing a cell diagram:
1. The reactants and products are always listed from left to right in this form:
anode | anode solution || cathode solution | cathode
2. A single vertical line indicates a phase boundary.
3. A double vertical line indicates the presence of a salt bridge or some other type of barrier.
Electrolytic cells, in almost all of their characteristics and behavior, are the opposite of galvanic cells. Whereas galvanic cells house spontaneous redox reactions, which when allowed to proceed, generate current and deliver electrical energy, electrolytic cells house nonspontaneous reactions, which require the input of energy to proceed. Galvanic cells supply energy; electrolytic cells consume it. The change in Gibbs function for the redox reaction of an electrolytic cell is positive. The particular type of redox reaction that is driven by the external voltage source is called electrolysis, in which chemical compounds are decomposed. For example, electrolytic cells can be used to drive the nonspontaneous decomposition of water into oxygen and hydrogen gas. Another example, the electrolysis of molten NaCl, is illustrated in Figure 11.2.
Anions are negatively charged and travel to the anode. Cations are positively charged and travel to the cathode. The same principles apply to electrophoresis, a laboratory technique used to purify, separate, or identify compounds.
In this electrolytic cell, molten NaCl is decomposed into Cl2 (g) and Na (l). The external voltage source (“battery” in Figure 11.2) supplies energy sufficient to drive a redox reaction in the direction that is thermodynamically unfavorable (i.e., nonspontaneous). In this example, Na+ ions migrate toward the cathode, where they are reduced to Na (l). At the same time, Cl- ions migrate toward the anode, where they are oxidized to Cl2 (g). You’ll notice that the half-reactions do not need to be separated into different compartments because the desired reaction is nonspontaneous. Furthermore, irrespective of the fact that a nonspontaneous reaction is being driven by an external voltage, it is still the case—as it always is for every type of electrochemical cell—that oxidation occurs at the anode and reduction occurs at the cathode. (Note that sodium is a liquid at the temperature of molten NaCl; it is also less dense than the molten salt and, thus, is easily removed as it floats to the top of the reaction vessel.) This cell is used in industry as the major means of sodium and chlorine production. You may be wondering why industry goes to all this trouble to produce these compounds and why these compounds just can’t be dug up or pumped up from somewhere, rather than manufactured. Well, think about the thermodynamics of electrolysis. Energy is supplied to drive a nonspontaneous process. This means that the products of the reaction are not naturally favored. In fact—and we’ve mentioned this before (see Chapter 2, The Periodic Table)—sodium and chlorine are never found naturally in their elemental form because they are so reactive. So before we can use elemental sodium or chlorine gas, we have to make it ourselves first.
Michael Faraday was the first to define certain quantitative principles governing the behavior of electrolytic cells. He theorized that the amount of chemical change induced in an electrolytic cell is directly proportional to the number of moles of electrons that are exchanged during a redox reaction. The number of moles exchanged can be determined from the balanced half-reaction. In general, for a reaction that involves the transfer of n electrons per atom, M,
Mn+ + ne- M (s)
According to this equation, one mole of M(s) will be produced if n moles of electrons are supplied. Additionally, the number of moles of electrons needed to produce a certain amount of M(s) can now be related to the measurable electrical property of charge. One electron carries a charge of 1.6 × 10-19 coulombs (C). The charge carried by one mole of electrons can be calculated by multiplying this number by Avogadro’s number, as follows:
(1.6 × 10-19 C/e-)(6.022 × 1023 e-/mol e-) = 96,487 C/mol e-
(1.6 × 10–19)(6.022 × 1023) = 96,487 C/mol e– (~105C/mol e–)
This number is called Faraday’s constant, and one faraday (F) is equivalent to the amount of charge contained in one mole of electrons (1 F = 96,487 coulombs, or J/V).
This number is called Faraday’s constant, and one faraday (F) is equivalent to the amount of charge contained in one mole of electrons (1 F = 96,487 coulombs, or J/V) or one equivalent. On the MCAT, you should round up this number to 100,000 C/mol e- to make calculations more manageable.
A concentration cell is a special type of voltaic cell: two half-cells connected by a conductive material, allowing a spontaneous redox reaction to proceed, generating a current and delivering energy. Just like a galvanic cell, the concentration cell houses a redox reaction that has a negative G. The distinguishing characteristic of a concentration cell is in its design: The electrodes are chemically identical. For example, if both electrodes are copper metal (they are chemically identical), they have the same reduction potential so current is generated as a function of a concentration gradient established between the two solutions surrounding the electrodes. The concentration gradient results in a potential difference between the two compartments and drives the movement of electrons in the direction that results in equilibration of the ion gradient. Current will cease when the concentrations of ion species in the half-cells are equal. This implies that the voltage (V) or emf of a concentration cell is zero when the concentrations are equal; the voltage as a function of concentrations can be calculated using the Nernst equation.
ELECTRODE CHARGE DESIGNATION
In a galvanic cell, current is spontaneously generated as electrons are released by the oxidized species at the anode and travel through the conductive material to the cathode, where the reduction occurs. Because the anode of a galvanic cell is the source of electrons, it is considered the negative electrode; the cathode is considered the positive electrode. Electrons, therefore, move from negative (low electric potential) to positive (high electric potential), while the current (by convention, the movement of positive charge) is from positive (high electric potential) to negative (low electric potential).
Conversely, the anode of an electrolytic cell is considered positive, because it is attached to the positive pole (the cathode) of the external voltage source (external battery) and so attracts anions from the solution. The cathode of an electrolytic cell is considered negative, because it is attached to the negative pole (the anode) of the external voltage source and so attracts cations from the solution.
In an electrolytic cell, the anode is positive and the cathode is negative. In a galvanic cell, the anode is negative and the cathode is positive. However, in both types of cells, reduction occurs at the cathode, and oxidation occurs at the anode.
In spite of this difference in designating charge (sign), oxidation takes place at the anode and reduction takes place at the cathode in both types of cells, and electrons always flow through the wire from the anode to the cathode. A simple mnemonic is that the CAThode attracts the CATions and the ANode attracts the ANions. In the Daniell cell, for example, the electrons created at the anode by the oxidation of the elemental zinc travel through the wire to the copper half-cell. There they attract copper(II) cations to the cathode, resulting in the reduction of the copper ions to elemental copper, and cations out of the salt bridge into the compartment. The anode, having lost electrons, attracts anions from the salt bridge into the compartment at the same time the Zn2+ ions formed by the oxidation process move away from the anode and toward the cathode.
Realize that this is an important distinction to understand not just for electrochemistry in the Physical Sciences section of Test Day but also for applications of electrochemistry in the Biological Sciences section. This distinction arises, for example, in a variant of electrophoresis, called isoelectric focusing, a technique often used to separate amino acids based on the unique isoelectric point (pI) of each amino acid. The positively charged amino acids (those that are protonated at the pH of the solution) will migrate toward the cathode; negatively charged amino acids (those that are deprotonated at the solution pH) will migrate instead toward the anode.
Reduction Potentials and the Electromotive Force
For galvanic cells, the direction of spontaneous movement of charge is from anode, the site of oxidation, to cathode, the site of reduction. This is simple enough to remember, but it begs the question; How do we determine which electrode species will be oxidized and which will be reduced? The relative tendencies of different chemical species to be reduced have been determined experimentally, using the tendency of the hydrogen ion (H+) to be reduced as an arbitrary zero reference point.
A reduction potential is measured in volts (V) and defined relative to the standard hydrogen electrode (SHE), which is arbitrarily given a potential of 0.00 volts. The species in a reaction that will be oxidized or reduced can be determined from the reduction potential of each species, defined as the tendency of a species to acquire electrons and be reduced. Each species has its own intrinsic reduction potential; the more positive the potential, the greater the species’ tendency to be reduced. Standard reduction potential, (Eredo), is measured under standard conditions: 25°C, 1 M concentration for each ion participating in the reaction, a partial pressure of 1 atm for each gas that is part of the reaction, and metals in their pure state. The relative reactivities of different half-cells can be compared to predict the direction of electron flow. A higher Eredomeans a greater relative tendency for reduction to occur, while a lower Eredomeans a greater relative tendency for oxidation to occur.
A reduction potential is exactly what it sounds like. It tells us how likely a compound is to be reduced. The higher the value, the more likely it is to be reduced.
In galvanic cells, the electrode species with the higher reduction potential is the cathode, and the electrode species with the lower reduction potential is the anode. Since the species that has a stronger tendency to gain electrons is actually gaining electrons, the redox reaction is spontaneous, and the G is negative, as we’ve seen. In electrolytic cells, the electrode species with the higher reduction potential is “forced” (by the external voltage source) to be oxidized and is, therefore, the anode. The electrode species with the lower reduction potential is “forced” to be reduced and is, therefore, the cathode. Since the movement of electrons is in the direction against the “tendency” of the respective electrochemical species, the redox reaction is nonspontaneous, and the G is positive.
Example: Given the following half-reactions and E° values, determine which species would be oxidized and which would be reduced.
Ag+ + e- Ag (s) E° = + 0.8 V
Tl+ + e- Tl (s) E° = -0.34 V
Solution: Ag+ would be reduced to Ag(s) and Tl(s) would be oxidized to Tl+, because Ag+ has the higher E°. Therefore, the reaction equation would be
Ag+ + Tl (s) Tl+ + Ag (s)
which is the sum of the two spontaneous half-reactions.
It should be noted that reduction and oxidation are opposite processes. Therefore, to obtain the oxidation potential of a given half-reaction, the reduction half-reaction and the sign of the reduction potential are both reversed. For instance, from the example above, the oxidation half-reaction and oxidation potential of Tl(s) are
Tl (s) Tl+ + e- E° = + 0.34 V
THE ELECTROMOTIVE FORCE
Standard reduction potentials are also used to calculate the standard electromotive force (emf or E°cell) of a reaction, the difference in potential between two half-cells at standard conditions. The emf of a reaction is determined by adding the standard reduction potential of the reduced species and the standard oxidation potential of the oxidized species. When adding standard potentials, do not multiply them by the number of moles oxidized or reduced.
emf ° = E°cath - E°anode = E°red + E°ox
where E°ox is the oxidation potential of the anode, which is the negative of the reduction potential. The standard emf of a galvanic cell is positive, while the standard emf of an electrolytic cell is negative.
Example: Given that the standard reduction potentials for Sm3+ and [RhCl6]3- are -2.41 V and +0.44 V, respectively, calculate the emf of the following reaction:
Sm3+ + Rh + 6 Cl- [RhCl6]3- + Sm
Solution: First, determine the oxidation and reduction half-reactions. As written, the Rh is oxidized, and the Sm3+ is reduced. Thus, the Sm3+ reduction potential is used as is, while the reverse reaction for Rh, [RhCl6]3- Rh + 6 Cl-, applies and the oxidation potential of [RhCl6]3- must be used. Then, using the equation given, the emf can be calculated to be (-2.41 V) + (-0.44 V) = -2.85 V. The cell is thus electrolytic as written. From this result, it is evident that the reaction would proceed spontaneously to the left, in which case the Sm would be oxidized while [RhCl6]3- would be reduced.
Note that there are two ways to express emf. The first allows us to use only reduction potentials. The second asks us to change the sign of the “Ered” value in order to use an oxidation potential, which is the exact opposite of a reduction potential. Be careful on the MCAT to pay close attention to the value and sign of the numbers you are using ... and DON’T multiply by the coefficients!
Thermodynamics of Redox Reactions
Throughout our discussion of electrochemistry and the different types of electrochemical cells, we have been making references to the spontaneity or nonspontaneity of the redox reactions housed in each of the different cell types. Let’s now look more formally at this topic by relating the state function of free energy to emf and the concentrations of the redox reactants and products to the voltage of a cell at a given point in time.
Recall that if G is positive, the reaction is not spontaneous; if G is negative, the reaction is spontaneous. Go back to Chapter 6 if you need a review.
emf AND GIBBS FREE ENERGY
As you know well by now, the thermodynamic criterion for determining the spontaneity of a reaction is the change in Gibbs free energy, G, which is the change in the chemical potential of a reaction or the change in the amount of energy of a chemical system available to do work. In an electrochemical cell, the work done is dependent on the number of coulombs and the energy available. Thus, G and emf are related as follows:
G = -nFEcell
where n is the number of moles of electrons exchanged, F is Faraday’s constant, and Ecell is the emf of the cell. Keep in mind that if Faraday’s constant is expressed in coulombs (J/V), then G must be expressed in J, not kJ. (The astute student will notice the similarity of this relationship to that expressed by the physics formula W = q V for the amount of work available or needed in the transport of a charge q across a potential difference V! And if you didn’t notice it before, you do now, so that means you are now astute!)
If the reaction takes place under standard conditions (298 K, 1 atm pressure, and all solutions at 1 M concentration), then the G is the standard Gibbs free energy, and Ecell is the standard cell potential. The above equation then becomes
G° = -nFE°cell
You should notice the significance of the negative sign on the right side of the equation. Being mindful of it will help you eliminate wrong answer choices that have the wrong sign for either the G° or the E°cell. For example, if you are asked to calculate the change in Gibbs function for a galvanic cell, you will immediately be able to eliminate the answer choices that have positive values, because you know that voltaic cells have positive emfs and the equation tells you the change in Gibbs will have the opposite sign.
THE EFFECT OF CONCENTRATION ON emf
So far, we have considered the calculation of cell emf only under standard conditions (all the ionic species are 1 M, and all gases are at a pressure of 1 atm). However, electrochemical cells may have ionic concentrations that are greater or lesser than 1 M. In fact, for the concentration cell, the concentrations of the ion in the two compartments must be different, even if one of them is 1 M, for there to be a voltage and current. Concentration does have an effect on the emf of a cell: emf varies with the changing concentrations of the species involved. It can be determined by the use of theNernst equation:
where Q is the reaction quotient for the reaction at a given point in time. For example, for the following reaction,
aA + bB cC + dD
the reaction quotient can be calculated as follows:
Although the expression for the reaction quotient Q has two terms for the concentrations of reactants and two terms for the concentrations of products, you need to remember that only the species in solution are included. When considering the case of the Daniell cell, for example, we need to think about which species of the redox reaction are in solution. Upon oxidation, the resulting cation will enter into solution, so the product concentration is the concentration of the oxidized species. Because the electrons are captured by the cations that surround the cathode in the reduction half-reaction, these cations are the reactants of the redox reaction, so the reactant concentration is the concentration of the species that gets reduced.
The emf of a cell can be measured by a voltmeter. A potentiometer is a kind of voltmeter that draws no current, and it gives a more accurate reading of the difference in potential between two electrodes.
emf AND THE EQUILIBRIUM CONSTANT (Keq)
For reactions in solution, G° can determined in another manner, as follows:
G° = -RT ln Keq
where R is the gas constant 8.314 J/(K•mol), T is the absolute temperature in K, and Keq is the equilibrium constant for the reaction.
Combining the two equations for standard free energy change (above), we see that
G° = -nFE°cell = -RT ln Keq
nFE°cell = RT ln Keq
If E°cell is positive, ln K is positive. This means that K must be greater than one and that the equilibrium lies toward the right (i.e., products are favored).
If the values for n, T, and Keq are known, then the E° cell for the redox reaction can be readily calculated. On the MCAT, you will not be expected to calculate (or even approximate) natural log values in your head. That being said, these equations will still be tested but in a more conceptual way. Analysis of the equations shows us that for redox reactions that have equilibrium constants less than 1 (which tells us that the equilibrium state favors the reactants), the E°cell will be negative, because the natural log of any number less than 1 is negative. These properties are characteristic of electrolytic cells, which house nonspontaneous redox reactions. If the Keq for the redox reaction is greater than 1 (which tells us that the equilibrium state favors the products), the E°cell will be positive, because the natural log of any number greater than 1 is positive. These properties are characteristic of galvanic cells, which house spontaneous redox reactions. If the Keq is equal to 1 (which tells us that the concentrations of the reactants and products are equal at equilibrium), the E°cell will be equal to zero, because by definition of standard conditions (all ionic species at the same concentration, 1 M), the reaction is already at equilibrium. An easy way to remember this is E°cell = 0 V for any concentration cell, because, by definition, the equilibrium state of a concentration cell is when the concentrations of the ion in the two half-cells are equal (Keq = 1 for concentration cell).
In this chapter, we covered the essential MCAT topics of redox reactions and electrochemistry. We reviewed the rules for assigning oxidation numbers to help us keep track of the movement of electrons from the species that are oxidized (reducing agents) to the species that are reduced (oxidizing agents). We also covered the sequence of steps involved in balancing redox reactions through the half-reaction method and the properties and behavior of the different types of electrochemical cells. Galvanic cells rely on spontaneous redox reactions to produce current and supply energy. The concentration cell is a special type of galvanic cell for which the current is dependent upon an ion concentration gradient rather than a difference in reduction potential between two chemically distinct electrodes. Electrolytic cells rely on external voltage sources to drive a nonspontaneous redox reaction called electrolysis. Finally, we considered the thermodynamics of the different cell types. Galvanic and concentration cells have positive emf and negative free energy changes. Electrolytic cells have negative emf and positive free changes.
Without further delay, we want to offer you our heartiest congratulations for completing this review of general chemistry for the MCAT. The hard work, time, and energy you have invested in a careful, thorough, and thoughtful review of the topics covered within the pages of this book will pay off in points on Test Day—you can be sure of that. We hope that we have been successful in meeting our goals in writing these review notes: to assess the general concepts and principles essential to answering correctly the general chemistry questions on the MCAT; to guide you in the development of critical thinking skills necessary for analyzing passages, question stems, and answer choices; and to provide holistic preparation for your Test Day experience. In addition to all of these, we had in mind the particular goals of helping you relate the science to your everyday life experiences and future experiences as a physician, demystify it, and have some fun in the process. We are grateful for the opportunity to partner with you in your journey to success on the MCAT and, beyond that, success in your medical education and future practice as a great physician!
CONCEPTS TO REMEMBER
Oxidation is loss of electrons; reduction is gain of electrons (OIL RIG). Oxidation and reduction always occur as paired of reactions, in accordance with the law of conservation of charge.
The oxidizing agent is the chemical species that causes another species to be oxidized and is itself reduced. The reducing agent is the chemical species that causes another species to be reduced and is itself oxidized. Assign oxidation numbers to identify the oxidizing and reducing agents.
Redox reactions can be balanced by the half-reaction method:
—Separate the half-reactions.
—Balance the atoms in each half-reaction.
—Balance the charges in each half-reaction.
—Add the half-reactions.
—Confirm balance of mass and charge in redox reaction.
There are two basic types of electrochemical cells: galvanic (voltaic) cells and electrolytic cells. Concentration cells are a special type of galvanic cell.
Galvanic cells have spontaneous redox reactions, generate current, and supply energy. The G is negative, and the Ecell is positive.
Concentration cells have spontaneous redox reactions, generate current, and supply energy. Current is dependent upon ion concentration gradient, not the difference in reduction potential between two different electrodes. The G° and the E°cell are both 0, because the current ceases when the concentrations of the ion are equal in both compartments.
Electrolytic cells have nonspontaneous redox reactions, require external voltage to generate current, and consume energy. The G is positive, and the Ecell is negative.
Reduction potential, measured in volts (V), is a measure of a chemical species’ tendency to be reduced (gain electrons) relative to the standard hydrogen electrode (SHE). The higher the reduction potential, the greater the tendency to gain electrons and be reduced.
The standard emf of a cell is the difference between the standard reduction potential of the cathode minus the standard reduction potential of the anode.
The Nernst equation is useful for calculating Ecell for all ion concentrations.
EQUATIONS TO REMEMBER
1. An electrolytic cell is filled with water. Which of the following will move toward the cathode of such a cell?
I. H+ ions
II. O2- ions
A. I only
B. II only
C. I and III
D. II and III
2. Consider the following data:
Hg2+ + 2 e– Hg
Cu+ + e– Cu
Zn2+ + 2 e– Zn
Al3+ + 3 e– Al
The anode of a certain galvanic cell is composed of copper. Which of the metals from the data table can be used at the cathode?
D. None of the above
3. Consider the following equation:
3 Na (s) + H3N (aq) Na3N (s) + H2 (g)
Which species acts as an oxidizing agent?
4. How many electrons are involved in the following unbalanced reaction?
Cr2O72- + H+ + e- Cr2+ + H2O
Questions 5 and 6 refer to the following two half-reactions:
O2 + 4 H+ + 4 e- 2 H2O +1.23 V (Reaction 1)
PbO2 + 4 H+ + 2 e- Pb2+ + 2 H2O +1.46 V (Reaction 2)
5. If the two half-reactions combine to form a spontaneous system, what is the net balanced equation of the full reaction?
A. 2 PbO2 + 4 H+ 2 Pb2+ + O2 + 2 H2O
B. PbO2 Pb2+ + O2 + 2 e-
C. Pb2+ + O2 + 2 H2O 2 PbO2 + 4 H+
D. Pb2+ + O2 + 2 e- PbO2
6. Find the standard potential of the following reaction:
Pb2+ + O2 + 2 H2O 2 PbO2 + 4 H+
A. +0.23 V
B. -0.23 V
C. -1.69 V
D. +2.69 V
7. Rusting occurs due to the oxidation-reduction reaction of iron with environmental oxygen:
4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)
Some metals, such as copper, are unlikely to react with oxygen. Which of the following best explains this observation?
A. Iron has a more positive reduction potential, making it more likely to donate electrons to oxygen.
B. Iron has a more positive reduction potential, making it more likely to accept electrons from oxygen.
C. Iron has a less positive reduction potential, making it more likely to donate electrons to oxygen.
D. Iron has a less positive reduction potential, making it more likely to accept electrons from oxygen.
8. Lithium aluminum hydride (LiAlH4) is often used in laboratories because of its tendency to donate a hydride ion. Which of the following properties does LiAlH4 exhibit?
A. Strong reducing agent
B. Strong oxidizing agent
C. Strong acid
D. Strong base
9. If the value of E°cell is known, what other data is needed to calculate G?
A. Equilibrium constant
B. Reaction quotient
C. Temperature of the system
D. Number of moles of electrons transferred
10. Which of the following compounds is least likely to be found in the salt bridge of a galvanic cell?
11. What is the oxidation number of chlorine in NaClO?
12. Which of the following is most likely to increase the rate of an electrolytic reaction?
A. Increasing the resistance in the circuit
B. Increasing the volume of electrolyte
C. Increasing the current
D. Increasing the pH
13. The following electronic configurations represent elements in their neutral form. Which element is the strongest oxidizing agent?
Small Group Questions
1. How can a halogen have a positive oxidation number?
2. When adding standard potentials, why are we instructed not to multiply them by molar coefficients?
Explanations to Practice Questions
In an electrolytic cell, ionic compounds are broken up into their constituents; the cations (positively charged ions) migrate toward the cathode, and the anions (negatively charged ions) migrate toward the anode. In this case, the cations are H+ ions (protons), so option I is correct. To balance charge, electrons are transported from the anode to the cathode, meaning that option III is also correct. Option II is incorrect for two reasons. First, it’s unlikely that the anions would be O2- rather than OH-. Second, these anions would flow to the anode, not the cathode.
Oxidation occurs at the anode, and reduction occurs at the cathode. Since Cu is at the anode, it must be oxidized. Its standard reduction potential is +0.52 V, so its standard oxidation potential is -0.52 V. The half-reaction potentials in a feasible galvanic cell must add up to a value greater than 0, so we can answer this question by adding the oxidation potential of copper to the reduction potentials of the other metals and finding a reasonable match.
Mercury has a reduction potential of 0.85 V, which is enough to outweigh the potential contributed by copper (-0.52 V). The cell’s emf would be +0.33 V. Zinc and aluminum both have negative reduction potentials, so the overall potential of the cell will be even lower than that of copper (-1.28 V and -2.18 V, respectively). Zinc and aluminum would only be viable options if we were dealing with an electrolytic cell.
The oxidizing agent is the species that is reduced in any given equation. In this problem, two H+ ions from H3N are reduced to one neutral H2 atom. H3N is not the reducing agent because the H+ ions and the N3- ions are independent of one another in solution.
First, let’s balance the equation, first making sure all the atoms are present in equal quantities on both sides.
Cr2O72- + 14 H+ + e-2 Cr 2+ + 7 H2O
Now, let’s adjust the number of electrons to balance the charge. Currently, the left side has a charge of +12 (-2 from dichromate and +14 from protons). The right side has a charge of +4 (+2 from each chromium cation). To decrease the charge on the left side from +12 to +4, we should add 8 electrons.
Cr2O72- + 14 H+ + 8 e- 2 Cr2+ + 7 H2O
Both half-reactions are written as reductions, so one of the two must be reversed to perform oxidation. If a reaction represents a spontaneous process, then the overall potential must be positive. The only way to create a spontaneous oxidation-reduction system is by reversing Reaction 1 to produce a positive net potential.
2 H2O O2 + 4 H+ + 4 e- -1.23 V (oxidation)
The half-reactions are each already balanced for mass and charge. Before we add them together, we must make sure each of them transfers the same number of electrons. Multiply all of Reaction 2’s coefficients by 2 so that both reactions transfer 4 moles of electrons.
2 H2O O2 + 4 H+ + 4 e- -1.23 V (oxidation)
2 PbO2 + 8 H+ + 4 e- 2 Pb2+ + 4 H2O +1.46 V (reduction)
Now, add the two reactions together.
2 H2O + 2 PbO2 + 8 H+ + 4 e- O2 + 4 H+ + 4 e- + 2 Pb2+ + 4 H2O
Finally, let’s erase duplicate instances of H+, e-, and H2O.
2 PbO2 + 4 H+ O2 + 2 Pb2+ + 2 H2O
This yields the reaction in (A).
This reaction is simply the reverse of the net balanced equation we found in question 5. Since that equation referred to a spontaneous system, this equation must refer to a nonspontaneous system. Nonspontaneous systems have negative Ecell values, so we can eliminate (A) and (D). Here, Reaction 1 represents reduction, so its given reduction potential should remain the same (+1.23 V). Reaction 2 represents oxidation, so its given reduction potential needs to be multiplied by -1 (to yield -1.46 V). The sum of those two values is -0.23 V.
In the oxidation-reduction reaction of a metal with oxygen, the metal will be oxidized (donate electrons), and oxygen will be reduced (accept electrons). This fact allows us to immediately eliminate (B) and (D). You should also know that a species with a higher reduction potential is more likely to be reduced and a species with a lower reduction potential is more likely to be oxidized. Based on the information in the question, iron is oxidized more readily than copper; this means that iron has a lower reduction potential.
To answer this question, you must know that a hydride ion is composed of a hydrogen nucleus with two electrons, thereby giving it a negative charge and a considerable tendency to donate its extra electron. This means that LiAlH4 is a strong reducing agent.
This answer comes directly from the equation relating Gibbs free energy and E°cell. G = -nFE°cell, where n is the number of moles of electrons transferred and F is the Faraday constant, 96,487 J•V-1mol-1.
Salt bridges contain inert electrolytes. Ionic compounds, such as (A), (C), and (D), are known to be strong electrolytes because they completely dissociate in solution. (B) cannot be considered an electrolyte because its atoms are covalently bound and will not dissociate in aqueous solution. (B) and (C) may appear similar, but there is an important distinction to be made. (C) implies that Mg2+ and SO32- are the final, dissociated ionic constituents, while (B) implies that SO32- might want to dissociate into smaller elements.
In NaClO (sodium hypochlorite), sodium carries its typical +1 charge, and oxygen carries its typical -2 charge. This means that the chlorine atom must carry a +1 charge in order to balance the overall -1 charge. Although this may seem atypical, it is not uncommon (NaClO, for instance, is the active ingredient in household bleach).
Current is defined in amperes, a unit that breaks down to coulombs per second. An increased current will mean that more electrons (coulombs) will be transported per second. (A) is incorrect because an increase in resistance will decrease the current, thereby producing the opposite of the desired effect. (B) is incorrect because the amount of electrolyte will only affect the amount of final product produced; it does not limit the rate. (D) is not always relevant because the pH only affects electrolytic reactions that involve acids and bases.
A strong oxidizing agent will be easily reduced, meaning that it will have a tendency to gain electrons. Atoms usually gain electrons if they are one or two electrons away from filling up their valence shell. (A) has a full 4s-orbital, meaning that it can only gain an electron if it gains an entire psubshell. (B) has a stable, half-full d-orbital, so it is unlikely to pick up electrons unless it can gain five. (C) has only a single electron in the outer shell, which can easily be lost upon ionization. (D) would fill up its p-orbital by gaining one electron, but the five electrons currently present would not be easily lost through ionization. (D) is the correct answer.