MCAT Electrochemistry Practice Questions with Answers

MCAT Electrochemistry Practice Questions with Answers

Mastering MCAT Electrochemistry Practice Questions with Answers is essential for any pre-medical student aiming for a top-tier score on the Chemical and Physical Foundations of Biological Systems section. Electrochemistry bridges the gap between chemical reactions and electrical energy, governing everything from the mitochondrial electron transport chain to the batteries in your smartphone. By understanding how electrons move between species, you can predict reaction spontaneity and calculate the work potential of chemical systems.

Concept Explanation

Electrochemistry is the study of chemical reactions that cause electrons to move, creating a flow of electricity through the transfer of charge in oxidation-reduction (redox) reactions. At its core, this field focuses on two main types of electrochemical cells: galvanic (voltaic) cells and electrolytic cells. Galvanic cells harness spontaneous chemical reactions to generate electrical energy, while electrolytic cells use electrical energy to drive non-spontaneous chemical reactions.

Key concepts you must master for the MCAT include:

• Oxidation and Reduction: Oxidation is the loss of electrons (increase in oxidation state), and reduction is the gain of electrons (decrease in oxidation state). Remember the mnemonic "OIL RIG" (Oxidation Is Loss, Reduction Is Gain).
• Cell Potential ($E^{\circ}_{cell}$): This measures the "push" or electromotive force (emf) of the cell. It is calculated as $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$. A positive $E^{\circ}_{cell}$ indicates a spontaneous reaction.
• Gibbs Free Energy ($\Delta G$): The relationship between spontaneity and electrical work is defined by the equation $\Delta G^{\circ} = -nFE^{\circ}_{cell}$, where $n$ is the moles of electrons and $F$ is Faraday’s constant ($\approx 96,485 \text{ C/mol } e^{-}$).
• The Nernst Equation: This allows you to calculate the cell potential under non-standard conditions: $E = E^{\circ} - \frac{0.0592}{n} \log Q$ at $25^{\circ} \text{C}$.
• Anodes and Cathodes: In all cells, oxidation occurs at the anode and reduction occurs at the cathode ("An Ox, Red Cat"). However, the signs change: in galvanic cells, the anode is negative, while in electrolytic cells, the anode is positive.

To effectively prepare, many students find that using retrieval practice for medical students helps solidify these complex formulas and sign conventions into long-term memory. Understanding these fundamentals is the first step toward solving complex passage-based problems on exam day.

Solved Examples

Example 1: Calculating Standard Cell Potential
Consider a galvanic cell with the following half-reactions:
$Ag^{+} + e^{-} \rightarrow Ag(s) \quad E^{\circ} = +0.80 \text{ V}$
$Cu^{2+} + 2e^{-} \rightarrow Cu(s) \quad E^{\circ} = +0.34 \text{ V}$

1. Identify the cathode and anode. In a galvanic cell, the reaction with the higher reduction potential occurs at the cathode. Therefore, Silver (Ag) is the cathode and Copper (Cu) is the anode.
2. Reverse the anode reaction to represent oxidation: $Cu(s) \rightarrow Cu^{2+} + 2e^{-}$.
3. Use the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
4. Substitute the values: $E^{\circ}_{cell} = 0.80 \text{ V} - 0.34 \text{ V} = +0.46 \text{ V}$.

Example 2: Relating Cell Potential to Gibbs Free Energy
Calculate the standard Gibbs free energy change ($\Delta G^{\circ}$) for a reaction where $n = 2$ and $E^{\circ}_{cell} = +1.10 \text{ V}$. Use $F \approx 10^{5} \text{ C/mol}$ for MCAT estimation.

1. State the formula: $\Delta G^{\circ} = -nFE^{\circ}$.
2. Plug in the knowns: $\Delta G^{\circ} = -(2)(10^{5})(1.10)$.
3. Perform the calculation: $\Delta G^{\circ} = -2.2 \times 10^{5} \text{ J}$.
4. Convert to kJ: $-220 \text{ kJ}$. Because $\Delta G$ is negative, the reaction is spontaneous.

Example 3: Faraday’s Law of Electrolysis
How many moles of electrons are required to plate 112 grams of Iron from a solution of $FeCl_{2}$? (Atomic weight of Fe = 56 g/mol).

1. Calculate moles of Iron: $\frac{112 \text{ g}}{56 \text{ g/mol}} = 2 \text{ moles of Fe}$.
2. Identify the oxidation state: In $FeCl_{2}$, Iron is $Fe^{2+}$.
3. Determine the electron ratio: The reduction is $Fe^{2+} + 2e^{-} \rightarrow Fe(s)$. This means 2 moles of $e^{-}$ are needed for every 1 mole of Fe.
4. Calculate total electrons: $2 \text{ moles Fe} \times 2 \text{ moles } e^{-}/ \text{mol Fe} = 4 \text{ moles of electrons}$.

Practice Questions

1. A concentration cell is constructed using two copper electrodes in solutions of $0.1 \text{ M } Cu^{2+}$ and $0.001 \text{ M } Cu^{2+}$. In which direction will electrons flow?
2. Which of the following changes will increase the cell potential of a spontaneous reaction where the reaction quotient $Q$ is less than the equilibrium constant $K$?
3. An electrolytic cell contains molten $NaCl$. If a current of 5 Amperes is applied for 1,930 seconds, how many moles of $Cl_{2}$ gas are produced?

4. In a galvanic cell, what is the purpose of the salt bridge, and what would happen if it were removed while the circuit was closed?
5. Compare the reduction potentials: $Zn^{2+} + 2e^{-} \rightarrow Zn(s)$ ($E^{\circ} = -0.76 \text{ V}$) and $Ni^{2+} + 2e^{-} \rightarrow Ni(s)$ ($E^{\circ} = -0.25 \text{ V}$). Which species is the strongest oxidizing agent?
6. If a reaction has a $K_{ \neq}$ of $10^{10}$, what can be inferred about the sign of $E^{\circ}_{cell}$ and the magnitude of $\Delta G^{\circ}$?
7. According to the Nernst equation, what happens to the cell potential as the reaction approaches equilibrium?
8. A lead-acid battery is being recharged. Does it act as a galvanic or electrolytic cell during this process, and what reaction occurs at the positive terminal?
9. Calculate the standard cell potential for the reaction: $3Mg(s) + 2Al^{3+}(aq) \rightarrow 3Mg^{2+}(aq) + 2Al(s)$ given $Mg^{2+} + 2e^{-} \rightarrow Mg$ ($E^{\circ} = -2.37 \text{ V}$) and $Al^{3+} + 3e^{-} \rightarrow Al$ ($E^{\circ} = -1.66 \text{ V}$).
10. How does increasing the surface area of the electrodes affect the standard cell potential ($E^{\circ}$) and the current of the cell?

Answers & Explanations

1. Answer: From the anode (lower concentration) to the cathode (higher concentration). In a concentration cell, the electrodes are the same, but the ion concentrations differ. Electrons flow to the side where they can reduce ions to decrease the concentration gradient. The compartment with $0.001 \text{ M }$ is the anode (oxidation), and the compartment with $0.1 \text{ M }$ is the cathode (reduction).
2. Answer: Decreasing the concentration of products or increasing the concentration of reactants. According to the Nernst equation, $E = E^{\circ} - (0.0592/n) \log Q$. To increase $E$, we must decrease $Q$. Since $Q = [ \text{products}]/[ \text{reactants}]$, reducing products or increasing reactants achieves this.
3. Answer: 0.05 moles. Use the formula $It = nn_{e}F$. $(5 \text{ A})(1930 \text{ s}) = 9650 \text{ Coulombs}$. Moles of electrons $= 9650 / 96500 = 0.1 \text{ moles } e^{-}$. The reaction is $2Cl^{-} \rightarrow Cl_{2} + 2e^{-}$. Since 2 moles of electrons produce 1 mole of $Cl_{2}$, then $0.1 / 2 = 0.05 \text{ moles}$.
4. Answer: It maintains electrical neutrality; removing it stops the current. The salt bridge allows anions and cations to migrate between half-cells to prevent charge buildup. Without it, the solution would quickly become polarized, creating a counter-potential that stops electron flow immediately.
5. Answer: $Ni^{2+}$. The strongest oxidizing agent is the species most easily reduced, which corresponds to the most positive (or least negative) reduction potential. $-0.25 \text{ V}$ is greater than $-0.76 \text{ V}$, so $Ni^{2+}$ is the stronger oxidizing agent.
6. Answer: $E^{\circ}_{cell} > 0$ and $\Delta G^{\circ} < 0$. A large $K_{ \neq}$ ($> 1$) implies the reaction is spontaneous at standard conditions. Spontaneity requires a negative Gibbs free energy and a positive cell potential according to $\Delta G^{\circ} = -RT \ln K$.
7. Answer: The cell potential ($E$) approaches zero. As a reaction reaches equilibrium, $Q = K$. At this point, the chemical driving force vanishes, the battery is "dead," and $E = 0$. Note that $E^{\circ}$ remains constant as it is a standard state property.
8. Answer: Electrolytic cell; oxidation of $PbSO_{4}$ to $PbO_{2}$. Recharging is a non-spontaneous process requiring an external power source, making it electrolytic. In a lead-acid battery, the positive terminal is the cathode during discharge but becomes the anode (oxidation) during recharging.
9. Answer: +0.71 V. Magnesium is being oxidized (anode) and Aluminum is being reduced (cathode). $E^{\circ}_{cell} = E^{\circ}_{red, cathode} - E^{\circ}_{red, anode} = (-1.66) - (-2.37) = +0.71 \text{ V}$. The coefficients in the balanced equation do not change the voltage.
10. Answer: $E^{\circ}$ remains unchanged, but current increases. Cell potential is an intrinsic property (like density) and does not depend on the amount of material. However, increasing surface area allows for more simultaneous reactions, which increases the rate of electron flow (current).

Frequently Asked Questions

What is the difference between a galvanic and an electrolytic cell?

A galvanic cell converts chemical energy into electrical energy through spontaneous reactions ($\Delta G < 0$), while an electrolytic cell uses electrical energy to drive non-spontaneous reactions ($\Delta G > 0$). In galvanic cells, the anode is negative, whereas in electrolytic cells, the anode is positive.

Does the stoichiometric coefficient affect the cell potential?

No, the standard cell potential ($E^{\circ}$) is an intensive property and does not change regardless of how many times you multiply the half-reactions to balance the equation. This is a common trap on the MCAT, as students often try to multiply the voltage like they do for enthalpy changes.

How do I remember where oxidation and reduction occur?

The most reliable way is the mnemonic "An Ox, Red Cat," which stands for Anode = Oxidation and Reduction = Cathode. This rule applies to every single electrochemical cell, including galvanic, electrolytic, and concentration cells, providing a consistent reference point for students.

What is a concentration cell?

A concentration cell is a specialized type of galvanic cell where both electrodes are made of the same material, but they are immersed in solutions of differing concentrations. The potential is driven by the entropy increase associated with equalizing the concentrations in the two compartments.

Why is the salt bridge necessary in a galvanic cell?

The salt bridge completes the circuit by allowing the flow of ions to maintain electrical neutrality in the half-cells. Without it, the anode would become too positive and the cathode too negative, causing the electron flow to cease almost instantly due to electrostatic repulsion.

How does the MCAT test the Nernst Equation?

The MCAT usually tests the Nernst Equation conceptually by asking how changes in concentration ($Q$) affect the cell potential ($E$). You should know that as reactants are consumed and products are formed, $Q$ increases, which mathematically causes the cell potential to decrease until it reaches zero at equilibrium.