Electrochemistry for the MCAT: Everything You Need to Know

A complete guide to MCAT electrochemistry, covering key formulas, conceptual strategies, and practice questions to boost your chemistry score.

(Note: This guide is part of our MCAT General Chemistry series.)

Table of Contents

Part 1: Introduction to electrochemistry

Part 2: Cell potentials

a) Oxidation and reduction potentials

b) Electromotive force

Part 3: Concentration Cells

a) Calculation of electric potential

b) Nernst equation

Part 4: Electrochemical Cells

a) Voltaic cells

b) Electrolytic cells

c) Batteries

Part 5: High-Yield Terms

Part 6: Passage-Based Questions and Answers

Part 7: Standalone Questions and Answers

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Part 1: Introduction to electrochemistry

Electrochemistry drives the batteries in your car, runs your cell phone, and even powers cells within your body! Thus, electrochemistry has high biological and medical importance.

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Part 2: Cell potentials

SAC343 code block

a) Oxidation and reduction potentials

Recall that reduction and oxidation reactions are types of reactions in which atoms exchange electrons. Oxidation refers to the loss of electrons from an atom, whereas reduction refers to the gain of electrons by an atom. The mnemonic “OIL RIG,” Oxidation Is Loss and Reduction Is Gain, may be helpful in remembering this distinction. (For more information on these topics, be sure to refer to our guide on oxidation and reduction reactions.)

The standard oxidation potential tells us how likely a species is to be oxidized, or lose electrons, under standard conditions (1M concentrations, 1 atm pressure, 298K). Similarly, the standard reduction potential tells us how likely it is for a species to be reduced, or gain electrons, under standard conditions. Both of these are measured in volts (V). 

Reduction and oxidation potentials are measured relative to a reference electrode, such as the standard hydrogen electrode (SHE) or the saturated calomel electrode (SCE). These reference electrodes allow us to calculate the emf of half-reactions in relation to a benchmark value. Thus, a reference electrode is assumed to have zero potential (0 V) in relation to itself. 

Oxidation and reduction potentials are critical considerations in analyzing electrochemical cells. Electrochemical cells contain two sets of electrodes: an anode and a cathode. Electrons flow from the anode and toward the cathode. Thus, oxidation occurs at the anode and reduction occurs at the cathode. (Another useful mnemonic is “AN OX, RED CAT”: the first few syllables of the words anode—oxidation, and reduction—cathode.)

b) Electromotive force

The standard electromotive force (E° _cell), or emf, is the difference between the reduction potentials of the cathode and anode of an oxidation-reduction reaction. A spontaneous reaction will have a positive E° _cell, while a nonspontaneous reaction will have a negative E° _cell. Therefore, the free energy change (ΔG) of a reaction and the standard electromotive force of a reaction always has opposite signs.

E° _cell = E° _{reduction of cathode} – E° _{reduction of anode}

A half-reaction shows the chemical oxidation or reduction that takes place at a single electrode. Thus, the net overall equation for an electrochemical cell requires two half-reactions. 

When presented with a problem where you have to calculate the standard electromotive force, you will most likely be given a set of half-reactions and their standard reduction potentials.  For example, you might see something like this:

A⁺ + e^- →A (E°_reduction = 1.2 V)
B⁺ + e^- →B (E°_reduction = 0.6 V)

How can these half-reactions be put together to represent the overall electrochemical cell? 

These are both reduction reactions. However, when two half-reactions are found in a single electrochemical cell, the one with a higher E°_reduction is the one that can be spontaneously reduced in the overall reaction. The one with a lower E°_reduction is the one that can be spontaneously oxidized in the overall reaction. In our example reactions above, A will be spontaneously reduced by B, and B will be spontaneously oxidized by A.

Thus, the half-reaction involving B would actually occur in the reverse direction. As a result, a better representation of the oxidation of species B is as follows:

B →B⁺ + e^- (E°_oxidation = -0.6 V)
Note that the E°_oxidation of a half-reaction is equal to -1 x E°_reduction, and vice-versa.
Thus, the net reaction for our example electrochemical cell is as follows:

A⁺ + B →A + B⁺

The table below provides examples of reduction potentials for several different half-reactions. You do not need to memorize these for test day, as any required values will be provided when needed.

Table 1   Examples of common reactions and their reduction potentials (don't memorize)

Reaction	E° (V)
Al3+ + 3e- →Al	-1.66
Zn2+ + 2e- →Zn	-0.76
Cu2+ + 2e- →Cu	0.34
Fe3+ + e- →Fe2+	0.77
Ag+ + e- →Ag	0.80

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Part 3: Concentration cells

a) Calculation of electric potential

Concentration cells are a type of electrochemical cell that can spontaneously generate current via a concentration gradient. These types of cells usually contain two or more compartments, which are separated by a semipermeable membrane. A gradient of particles with positive or negative charge is created across the semipermeable membrane or film. 

This concentration gradient drives the flow of electrons across the semipermeable membrane. Since electrons and net charge are moving in the same direction down a concentration gradient, this creates an electromotive force. It’s important to note that the flow of electrons and the direction of current are opposite of one another. (For more information on current and circuits, be sure to refer to our guide on electrostatics, magnetism, and circuits.)  Finally, once the gradient is lost and the concentrations become equalized, the electromotive force returns to zero. 

The neurons within our body are an excellent example of a concentration cell. Recall that electrochemical gradients are crucial to the function of a neuron. For example, the neuron requires the flow of cations and anions to maintain its resting potential. (For more information on this, be sure to refer to our guide on the nervous system.) This has a value of -70 mV, meaning that the inside of the cell is more negative relative to the outside of the cell. However, when this is disrupted to a sufficient degree, the neuron may hit its threshold value of -55 mV. This will trigger an action potential to occur. 

Figure 1    An illustration of how a cell acts as a concentration cell via the exchange of cations and anions. The purple molecules represent a variety of anions in the cell that the potassium cation can bind with.

b) Nernst equation

Recall that the electromotive force values listed in tables are only provided at standard conditions. What if the electrochemical cell is placed in non-standard conditions: at a different temperature than 298K, or different pressure than 1 atm?

The Nernst equation allows us to calculate the electromotive force generated by a concentration cell under non-standard conditions.

E_cell = E° _cell – (0.0592/n)log(Q)
E_cell: electromotive force under non-standard conditions
E° _cell: electromotive force under standard conditions
n: number of moles of electrons
Q: reaction quotient
The reaction quotient is found using the concentration of products and reactants for a reaction. Remember, only the aqueous species are included. If a reaction had the formula aA_(s) + bB_(aq) → cC_(s) + dD_(aq), then the reaction quotient would be:

Q = [D]^d/ [B]^b

For test day, the Nernst equation will most likely be given to you. So, it will not be necessary to memorize the equation. Focus on understanding each of the terms. To put it simply, the Nernst equation illustrates that the electromotive force is a function of the concentration of charged particles within the cell.

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Part 4: Electrochemical cells

a) Voltaic cells

Voltaic cells, or galvanic cells, perform spontaneous reactions and have a positive electromotive force. Thus, the anode within a voltaic cell is negative, while the cathode is positive. In a voltaic cell, the cathode has a higher reduction potential than the anode.

Let’s take what we’ve learned so far and apply it to an example. 

The half-reactions below are carried out using a galvanic cell. Find the electromotive force of the overall reaction.

A⁺ + e^- →A_(s) E°_reduction = 0.28 V
B⁺ + e^- →B_(s) E°_reduction = -0.68 V
Since we have a galvanic cell, there are a few things we know right off the bat.

1. The overall electromotive force of the cell must be positive.
2. The half-reaction involving species A must occur at the cathode because it has a higher reduction potential. Therefore, A⁺ is spontaneously reduced to A_(s).
3. At the anode, oxidation must occur. Thus, the reverse of the presented half-reaction will occur because it has a lower reduction potential. Therefore, B_(s) is spontaneously oxidized to B⁺.

To calculate the electromotive force of the cell, simply plug in the reduction potentials into the formula discussed earlier:

E° _cell = E° _{reduction of cathode} – E° _{reduction of anode}
E° _cell = 0.28 V – (-0.68 V)
E° _cell = 0.96 V

A voltaic cell also requires the presence of a salt bridge. This is a crucial structure that contains an inert salt. 

As the cathode in a voltaic cell acquires more electrons, it accumulates a negative charge. Similarly, as the anode loses more electrons, it accumulates a positive charge. As the two electrodes become more and more charged, the reaction becomes less efficient. (In other words, it becomes harder to add negatively charged electrons to an already negatively charged region.) To address this, the salt bridge sends anions to the anode and cations to the cathode to neutralize the charges in the aqueous solutions. These anions and cations may be any charged chemical species that are not readily oxidized or reduced.

Figure 2    A voltaic/galvanic cell.

While it may be difficult to identify which end is the anode and which end is the cathode, remember that the overall emf of a voltaic cell must be positive. This allows the current to flow spontaneously. 

b) Electrolytic cells

Electrolytic cells carry out nonspontaneous reactions and have a negative electromotive force. In other words, a source of electrical energy is required to perform the chemical half-reactions. This process is referred to as electrolysis, where the chemical species that undergo decomposition are known as electrolytes.

Electrolytic cells are in direct contrast to voltaic cells. The anode of an electrolytic cell is positive, whereas the cathode is negative. In an electrolytic cell, an external voltage source makes the anode have a higher reduction potential than the cathode. 

Note that although the charges are different for anodes and cathodes when comparing electrolytic and galvanic cells, the sites of oxidation and reduction remain the same. This is important: for all electrochemical cells, the anode is the site of oxidation and the cathode is the site of reduction. (The mnemonic AN OX, RED CAT applies to all electrochemical cells.)

Let’s take what we’ve learned so far and apply it to an example. 

The half-reactions below are carried out using an electrolytic cell. Find the electromotive force of the overall reaction.

A⁺ + e^- →A_(s) E°_reduction = 0.28 V
B⁺ + e^- →B_(s) E°_reduction = -0.68 V
There are some key things we are able to determine immediately due to its identity as an electrolytic cell.

1. The electromotive force of the cell will be negative.
2. The half-reaction involving species A is the one occurring in our anode because it has a higher reduction potential. Since oxidation occurs at the anode, we will have the reverse of the reaction occurring. Thus, A_(s) will be oxidized to A⁺.
3. The half-reaction involving species B will occur in our cathode because it has a lower reduction potential. Since reduction always occurs at the cathode, B⁺ will be reduced to B_(s).

Like before, enter the reduction potentials into the formula to find the electromotive force of the cell.

E° _cell = E° _{reduction of cathode} – E° _{reduction of anode}
E° _cell = -0.68 V – (0.28 V)
E° _cell = -0.96 V

Figure 3    An electrolytic cell.

In summary, voltaic and electrolytic cells are different chemical systems that follow the same chemical principles. Both possess anodes and cathodes, and both harness the reduction-oxidation reactions to generate the flow of electrons.

Table 2    Summary table of Voltaic and electrolytic cells

	Anode	Cathode	Electron Flow	Spontaneity
Voltaic Cell	-	+	Anode → Cathode	Spontaneous
Electrolytic Cell	+	-	Anode → Cathode	Nonspontaneous

c) Batteries

Batteries are rechargeable electrochemical cells. When batteries charge, they function as electrolytic cells. When batteries discharge, they function as galvanic cells. 

For the purposes of the MCAT, you should be familiar with two important batteries. 

The first is the lead storage battery. When discharging, its two cells are a lead anode and a lead (IV) oxide cathode that are connected by sulfuric acid. When it is completely discharged, the battery has two identical lead sulfate (PbSO₄) electrodes. To bring the battery back to its charged state, these electrodes dissociate to bring back the original anode (Pb) and cathode (PbO₂).

The second is the nickel-cadmium battery. When this battery discharges, the anode is composed of cadmium, and the cathode is composed of nickel (III) oxide-hydroxide. Both electrodes are in contact with a potassium hydroxide (KOH) solution. When fully discharged, the anode consists of cadmium hydroxide, and the cathode consists of nickel (II) hydroxide. To charge the nickel-cadmium battery, the electrodes are dissociated to return to the original compositions of the anode (Cd) and cathode (NiO(OH)).

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Part 5: High-yield terms

Standard oxidation potential: voltage resulting from the oxidation of a species
Standard reduction potential: voltage resulting from the reduction of a species
Standard Conditions: 1M concentrations, 1 atmosphere pressure, temperature of 298K
Electromotive Force (emf): calculated by E° _cell = E° _{reduction of cathode} – E° _{reduction of anode}
Concentration cell: a spontaneous electrochemical cell that generates current through concentration gradient
Nernst equation: E _cell = E° _cell – (0.0592/n)log(Q)
Voltaic (galvanic) cell: Electrochemical cell that carries out spontaneous redox reactions
Salt bridge: maintains the efficiency of a voltaic cell using an inert salt to neutralize the charges in the aqueous solutions
Electrolytic cell: electrochemical cell that carries out nonspontaneous reactions
Lead storage battery: rechargeable battery consisting of a Pb anode and PbO₂ cathode when discharging; when charging, both electrodes consist of PbSO₄
Nickel-cadmium battery: a rechargeable battery consisting of a Cd anode and NiO(OH) cathode; when charging, the anode consists of Cd(OH)₂, and the cathode consists of Ni(OH)₂

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Part 6: Passage-based questions and answers

The Daniell cell is a kind of zinc-copper battery previously widely used in the European telegraph industry. A typical Daniell cell contains a salt bridge connecting the anode electrode of a zinc sulfate solution and an immersed zinc plate, as well as the cathode electrode of a copper sulfate solution and an immersed copper plate. A porous barrier exists between the metals. When discharged, the zinc anode is oxidized according to equation 1.

Zn_(s) →Zn²⁺_(aq) + 2e^-
Equation 1
Simultaneously, the copper cathode reduced according to equation 2. Cu²⁺_(aq) + 2e^- →Cu_(s)
Equation 2
Owing to the serious self-discharge and inability to be recharged, the Daniell cell was replaced by modern rechargeable battery technologies. Modern solutions include the lead-storage battery developed in 1859 and the nickel-cadmium battery developed in 1909. The half-reactions of a discharging lead-storage battery are expressed by equations 3 and 4.

Pb_(s) + HSO4-_(aq) →PbSO4_(s) + H⁺ + 2e^- (E°_reduction = -0.36 V)
Equation 3

PbO2_(s) + SO42-_(aq) + 4H⁺ + 2e^- →PbSO4_(s) + 2H2O (E°_reduction = 1.69 V)
Equation 4 _{DONG, X., WANG, Y. & XIA, Y. RE-BUILDING DANIELL CELL WITH A LI-ION EXCHANGE FILM. SCI REP 4, 6916 (2014). THE ARTICLE’S FULL TEXT IS AVAILABLE HERE: HTTPS://WWW.NATURE.COM/ARTICLES/SREP06916. THE ARTICLE IS NOT COPYRIGHTED BY SHEMMASSIAN ACADEMIC CONSULTING. DISCLAIMER: SHEMMASSIAN ACADEMIC CONSULTING DOES NOT OWN THE PASSAGE PRESENTED HERE. CREATIVE COMMON LICENSE: HTTP://CREATIVECOMMONS.ORG/LICENSES/BY/4.0/. CHANGES WERE MADE TO ORIGINAL ARTICLE TO CREATE AN MCAT-STYLE PASSAGE.}

Question 1: What is the net reaction occurring in the Daniell cell as it discharges?

A) Zn_(s) + Cu²⁺_(aq) →Zn²⁺_(aq) + Cu_(s)

B) Zn²⁺_(aq) + Cu_(s) →Zn_(s) + Cu²⁺_(aq)

C) Zn_(aq) + Cu²⁺_(aq) →Zn²⁺_(s) + Cu_(s)

D) Zn²⁺_(s) + Cu_(s) →Zn_(s) + Cu²⁺_(s)

Question 2: What is the reaction quotient for the overall reaction when the Daniell cell is discharging?

A) Q = [Zn²⁺][Cu] / [Zn][Cu²⁺]

B) Q = [Zn²⁺]/[Cu²⁺]

C) Q = [Zn][Cu²⁺] / [Zn²⁺][Cu]

D) Q = [Zn²⁺]/[Zn]

Question 3: Which of the following accurately describes the processes occurring in a lead-storage battery?

A) When discharging, both electrodes are identical

B) When discharging, the anode consists of elemental lead and the cathode consists of lead (IV) oxide

C) When discharging, the anode consists of lead (IV) oxide and the cathode consists of elemental lead

D) The electrodes are placed in a solution of potassium hydroxide

Question 4: Which of the following accurately characterizes a nickel-cadmium battery?

A) When completely discharged, the two electrodes are identical

B) When discharging, the anode consists of elemental cadmium and the cathode consists of elemental nickel

C) To recharge the battery, it must function as an electrolytic cell

D) The electrodes are in a solution of sulfuric acid

Question 5: What is the electromotive force of a discharging lead-storage battery?

A) -1.33 V

B) 1.33 V

C) -2.05 V

D) 2.05 V

Answer key for passage-based questions

1. Answer choice A is correct. According to the information presented in the passage, the zinc electrode is oxidized, and the copper electrode is reduced as the cell discharges. Therefore, the resulting products should include oxidized zinc (Zn²⁺) and reduced copper (Cu) (choices B and D are incorrect. Additionally, the oxidized zinc ion would remain in an aqueous state, whereas the reduced copper would be a solid (choice C is incorrect).

2. Answer choice B is correct. The expression for the reaction quotient involves dividing the product of the concentrations of products by the product of the concentrations of reactants. (For more information on this, be sure to refer to our guide on chemical equilibrium and kinetics.) Additionally, the concentration of each species is raised to a power equal to its coefficient in the balanced equation. The overall reaction for the Daniell cell is Zn_(s) + Cu²⁺_(aq) →Zn²⁺_(aq) + Cu_(s). The reaction quotient only includes the concentrations of aqueous species, the presence of solids or solvent is ignored (choices A, C, and D are incorrect).

3. Answer choice B is correct. A discharging lead-storage battery consists of a lead anode and a lead (IV) oxide cathode (choices A and C are incorrect). Elemental lead refers to Pb that has zero charge and an oxidation state of zero. Sulfuric acid acts as the bridge connecting the two electrodes. In a nickel-cadmium battery, both electrodes are immersed in a solution of potassium hydroxide (choice D is incorrect). 

4. Answer choice C is correct. For all batteries, they must function as electrolytic cells when charging. However, when discharging, they function as galvanic cells. When completely discharged, the anode of a nickel-cadmium battery consists of cadmium hydroxide, and the cathode consists of nickel (II) hydroxide (choice A is incorrect). When discharging, the anode consists of cadmium, and the cathode consists of nickel (III) oxide-hydroxide (choice B is incorrect). Both electrodes are in a solution of potassium hydroxide (choice D is incorrect). 

5. Answer choice D is correct. The relevant oxidation and reduction half-reactions are provided in the passage. Since oxidation always occurs at the anode, and reduction always occurs at the cathode, these values can be entered into the formula for standard electromotive force:

E° _cell = E° _{reduction of cathode} – E° _{reduction of anode}
E° _cell = 1.69 V – (-0.36 V)
E° _cell = 2.05 V

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Part 7: Standalone questions and answers

Question 1: Which of the following accurately characterizes standard conditions?

A) 1M concentrations, 760mmHg pressure, 298K

B) 1M concentrations, 1 atm, 20˚C

C) 1M concentrations, 1 torr, 298K

D) 1M concentrations, 1 atm, 273K

Question 2: Which of the following statements most accurately describes concentration cells?

A) Nonspontaneous electrochemical cells that generate a current using a concentration gradient

B) Spontaneous electrolytic cells that generate a current via a concentration gradient

C) Galvanic cells that generate a current via a concentration gradient

D) Batteries that store energy by creating concentration gradients

Question 3: Which of the following statements regarding the anode of a voltaic cell is false?

A) The anode is the site of oxidation

B) Current flows away from the anode

C) The anode has a lower reduction potential than the cathode

D) Electrons flow away from the anode

Question 4: The half-reactions below occur in a voltaic cell under standard conditions. What is the electromotive force generated by the electrochemical cell?

Ag⁺ + e^- →Ag (E°_reduction = 0.34 V)
Zn²⁺ + 2e^- →Zn (E°_reduction = -0.76 V)

A) -0.42 V

B) 1.1 V

C) 0.42 V

D) -1.1 V

Question 5: Which of the following accurately describes the differences between a voltaic cell and an electrolytic cell?

A) A voltaic cell is nonspontaneous, while an electrolytic cell is spontaneous

B) In a voltaic cell, the anode is the site of oxidation, whereas in an electrolytic cell, it is the site of reduction

C) In a voltaic cell, the cathode is negative, whereas in an electrolytic cell, it is positive

D) In a voltaic cell, the anode is negative, whereas in an electrolytic cell, it is positive

Answer key for standalone practice questions

1. Answer choice A is correct. Standard conditions are defined as 1M concentrations, 1 atm of pressure, and a temperature of 298K (choice D is incorrect). 1 atm of pressure is equivalent to 760mmHg or 760 torr (choice C is incorrect). A temperature of 298 K is equal to 25˚C (choice B is incorrect). 

2. Answer choice C is correct. Concentration cells are spontaneous electrochemical cells that generate current using concentration gradients. Concentration gradients of charged particles are created and separated using a film of semipermeable membrane. The movement of these charged particles down a concentration gradient generates a current (choice A is incorrect). Concentration cells are spontaneous, as they rely on the spontaneous movement of particles down a concentration gradient (choice D is incorrect). Electrolytic cells are strictly nonspontaneous (choice B is incorrect). 

3. Answer choice B is correct. In all electrochemical cells, the anode is the site of oxidation (choice A is true). Additionally, electrons always flow from anode to cathode (choice D is true). Since current flows in the opposite direction of the flow of electrons, it flows from cathode to anode (choice B is false). In a voltaic cell, or galvanic cell, the cathode has a higher reduction potential than the anode (choice C is true). 
4. Answer choice B is correct. Since this reaction occurs in a voltaic cell, the half-reaction involving silver occurs at the cathode as it has a higher reduction potential. Thus, the half-reaction involving zinc will occur at the anode. These reduction potentials can be entered into the following formula to find the standard cell potential:

E° _cell = E° _{reduction of cathode} – E° _{reduction of anode}
E° _cell = 0.34 V – (-0.76 V)
E° _cell = 1.1 V

Recall that voltaic/galvanic cells are spontaneous electrochemical cells. Thus, the emf generated by these cells is always greater than zero (choices A and D are incorrect). Note that if this reaction occurs under non-standard equations (e.g., concentrations differing from 1M), the Nernst equation must be applied. 

5. Answer choice D is correct. In a voltaic cell, the anode is negatively charged, and the cathode is positively charged. In an electrolytic cell, the anode is positively charged, and the cathode is negatively charged (choice C is incorrect). A voltaic cell is spontaneous, while an electrolytic cell is nonspontaneous (choice A is incorrect). Regardless of the type of electrochemical cell, the anode is the site of oxidation (choice B is incorrect).

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