Medium MCAT Electrochemistry Practice Questions

Mastering electrochemistry is essential for any pre-medical student aiming for a high score on the Chemical and Physical Foundations of Biological Systems section of the MCAT. This guide provides Medium MCAT Electrochemistry Practice Questions designed to bridge the gap between basic definitions and the complex, multi-step problems found on the actual exam. By engaging with these problems, you can refine your understanding of redox reactions, cell potentials, and the Nernst equation.

Concept Explanation

Electrochemistry is the study of the relationship between chemical reactions and electricity, specifically focusing on the movement of electrons through oxidation-reduction (redox) processes. In these systems, oxidation occurs at the anode (loss of electrons) and reduction occurs at the cathode (gain of electrons), a mnemonic often remembered as "An Ox" and "Red Cat." These reactions take place in two primary types of cells: galvanic (voltaic) cells, which use spontaneous reactions to generate electrical energy, and electrolytic cells, which require an external power source to drive non-spontaneous reactions. For students looking to optimize their preparation, utilizing retrieval practice for medical students can significantly enhance the retention of these complex electrochemical conventions.

The driving force behind electron flow is the electromotive force (EMF) or cell potential $E_{cell}$. This is calculated using standard reduction potentials $E^{\circ}$, which are measured under standard conditions (1 M concentration, 1 atm pressure, and 298 K). The relationship between the Gibbs free energy change $\Delta G$ and the cell potential is given by the equation:

$$\Delta G = -nFE_{cell}$$

Where $n$ is the number of moles of electrons transferred and $F$ is Faraday’s constant (approximately $96,485 \text{ C/mol } e^-$). According to LibreTexts Chemistry, a positive cell potential indicates a spontaneous reaction ($\Delta G < 0$), while a negative cell potential indicates a non-spontaneous reaction. Understanding these thermodynamic relationships is a key component of mastering STEM subjects like biochemistry and physics on the MCAT.

Solved Examples

Example 1: Calculating Standard Cell Potential
Given the following half-reactions, calculate the standard cell potential for a galvanic cell using Zinc and Copper:
$$Zn^{2+} + 2e^- \rightarrow Zn \quad E^{\circ} = -0.76 \text{ V}$$
$$Cu^{2+} + 2e^- \rightarrow Cu \quad E^{\circ} = +0.34 \text{ V}$$

1. Identify the cathode and anode. In a galvanic cell, the half-reaction with the more positive reduction potential occurs at the cathode. Therefore, Copper is the cathode ($+0.34 \text{ V}$) and Zinc is the anode.
2. Reverse the anode reaction to find the oxidation potential: $Zn \rightarrow Zn^{2+} + 2e^- \quad E_{ox}^{\circ} = +0.76 \text{ V}$.
3. Sum the potentials: $E_{cell}^{\circ} = E_{red,cathode}^{\circ} - E_{red,anode}^{\circ}$.
4. Calculate: $E_{cell}^{\circ} = 0.34 \text{ V} - (-0.76 \text{ V}) = +1.10 \text{ V}$.

Example 2: Determining Gibbs Free Energy
Determine the standard Gibbs free energy change for a reaction where $n = 2$ and $E_{cell}^{\circ} = +0.50 \text{ V}$.

1. Use the formula $\Delta G^{\circ} = -nFE_{cell}^{\circ}$.
2. Substitute the values: $\Delta G^{\circ} = -(2 \text{ mol } e^-)(96,485 \text{ C/mol } e^-)(0.50 \text{ J/C})$.
3. Simplify: $\Delta G^{\circ} = -96,485 \text{ J}$ or approximately $-96.5 \text{ kJ}$.
4. Since $\Delta G^{\circ}$ is negative, the reaction is spontaneous.

Example 3: Applying the Nernst Equation
Calculate the cell potential at 298 K for the following reaction when $[Zn^{2+}] = 0.1 \text{ M}$ and $[Cu^{2+}] = 1.0 \text{ M}$. The standard potential is $+1.10 \text{ V}$.

1. Write the Nernst Equation: $E = E^{\circ} - \frac{0.0592}{n} \log Q$.
2. Determine $Q$: $Q = \frac{[Zn^{2+}]}{[Cu^{2+}] \text{ (products over reactants)}} = \frac{0.1}{1.0} = 0.1$.
3. Substitute values: $E = 1.10 - \frac{0.0592}{2} \log(0.1)$.
4. Solve: Since $\log(0.1) = -1$, the term becomes $+0.0296$.
5. Final result: $E = 1.10 + 0.0296 = 1.1296 \text{ V}$.

Practice Questions

1. A galvanic cell is constructed using a silver electrode ($Ag^+/Ag, E^{\circ} = +0.80 \text{ V}$) and a magnesium electrode ($Mg^{2+}/Mg, E^{\circ} = -2.37 \text{ V}$). What is the standard cell potential?

2. In an electrolytic cell, a current of 5.0 Amperes is passed through a solution of $Cu(NO_3)_2$ for 1,930 seconds. How many moles of copper metal are deposited at the cathode? (Faraday's constant $\approx 96,500 \text{ C/mol}$)

3. Which of the following changes would increase the cell potential of a concentration cell where both compartments contain $Ni/Ni^{2+}$?
I. Increasing the concentration of $Ni^{2+}$ in the cathode compartment.
II. Increasing the concentration of $Ni^{2+}$ in the anode compartment.
III. Increasing the size of the Ni electrodes.

4. A redox reaction has an equilibrium constant $K_{ \neq} = 1.0 \times 10^{-5}$. What can be inferred about the standard cell potential $E_{cell}^{\circ}$ and the spontaneity of the reaction under standard conditions?

5. Lead-acid batteries are commonly used in vehicles. During discharge, the following reaction occurs:
$$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$$
What is the oxidation state of lead in $PbO_2$ and $PbSO_4$, respectively?

6. If the reduction of one mole of $Al^{3+}$ to $Al(s)$ requires 3 moles of electrons, how many Coulombs of charge are required to produce 9.0 grams of Aluminum? (Atomic mass of $Al = 27 \text{ g/mol}$)

7. Consider the following reduction potentials:
$$Fe^{3+} + e^- \rightarrow Fe^{2+} \quad E^{\circ} = +0.77 \text{ V}$$
$$Sn^{4+} + 2e^- \rightarrow Sn^{2+} \quad E^{\circ} = +0.15 \text{ V}$$
If these are combined into a spontaneous cell, which species acts as the reducing agent?

8. How does the addition of a salt bridge affect a galvanic cell, and what would happen if it were removed while the circuit was closed?

9. A researcher finds that a specific electrochemical reaction has a $\Delta G^{\circ} > 0$. If the concentration of reactants is significantly increased such that $Q < K_{ \neq}$, can the reaction become spontaneous?

10. What is the standard potential for the disproportionation of $Cu^+$ into $Cu(s)$ and $Cu^{2+}$ given:
$$Cu^+ + e^- \rightarrow Cu(s) \quad E^{\circ} = +0.52 \text{ V}$$
$$Cu^{2+} + e^- \rightarrow Cu^+ \quad E^{\circ} = +0.15 \text{ V}$$

Answers & Explanations

1. Answer: +3.17 V.
To find the cell potential, use $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$. The cathode is the electrode with the higher reduction potential (Silver, +0.80 V). The anode is Magnesium (-2.37 V). Calculation: $0.80 - (-2.37) = 3.17 \text{ V}$.

2. Answer: 0.05 moles.
First, calculate total charge: $Q = I \times t = 5.0 \text{ A} \times 1,930 \text{ s} = 9,650 \text{ C}$. Next, convert charge to moles of electrons: $9,650 \text{ C} / 96,500 \text{ C/mol} = 0.1 \text{ moles of electrons}$. The reduction of $Cu^{2+}$ is $Cu^{2+} + 2e^- \rightarrow Cu$, meaning 2 moles of electrons produce 1 mole of Cu. Therefore, $0.1 / 2 = 0.05 \text{ moles of Cu}$.

3. Answer: I only.
A concentration cell functions based on the concentration gradient. The Nernst equation shows that $E_{cell} = E^{\circ} - \frac{0.059}{n} \log \frac{[Anode]}{[Cathode]}$. Since $E^{\circ}$ for a concentration cell is 0, increasing the cathode concentration decreases the ratio $Q$, making $\log Q$ more negative and thus $E_{cell}$ more positive. Electrode size does not affect cell potential.

4. Answer: $E_{cell}^{\circ}$ is negative; the reaction is non-spontaneous.
The relationship between $E_{cell}^{\circ}$ and $K_{ \neq}$ is $E_{cell}^{\circ} = \frac{RT}{nF} \ln K_{ \neq}$. If $K_{ \neq} < 1$, then $\ln K_{ \neq}$ is negative, resulting in a negative $E_{cell}^{\circ}$. A negative standard potential corresponds to a positive $\Delta G^{\circ}$, indicating a non-spontaneous reaction under standard conditions.

5. Answer: +4 and +2.
In $PbO_2$, oxygen is typically -2. With two oxygens ($-4$), lead must be +4 to balance the neutral molecule. In $PbSO_4$, the sulfate ion ($SO_4^{2-}$) has a charge of -2, so lead must be +2.

6. Answer: 96,500 C.
First, find the moles of Aluminum: $9.0 \text{ g} / 27 \text{ g/mol} = 1/3 \text{ mole of Al}$. Since 1 mole of Al requires 3 moles of electrons, $1/3$ mole of Al requires 1 mole of electrons. One mole of electrons carries a charge of approximately 96,500 Coulombs (Faraday's constant).

7. Answer: $Sn^{2+}$.
The spontaneous reaction will involve the reduction of the species with the higher potential ($Fe^{3+}$) and the oxidation of the species with the lower potential ($Sn^{2+}$). The species that is oxidized is the reducing agent. Therefore, $Sn^{2+}$ reduces $Fe^{3+}$.

8. Answer: The salt bridge maintains electrical neutrality; removing it stops the reaction.
As the reaction proceeds, charge builds up (positive at the anode, negative at the cathode). The salt bridge allows ions to flow and neutralize this buildup. Without it, the charge imbalance would immediately halt the flow of electrons, stopping the current.

9. Answer: Yes.
Spontaneity is determined by $\Delta G$, not just $\Delta G^{\circ}$. The equation $\Delta G = \Delta G^{\circ} + RT \ln Q$ shows that if $Q$ is made very small (by increasing reactants or decreasing products), the $RT \ln Q$ term can become sufficiently negative to make $\Delta G$ negative, even if $\Delta G^{\circ}$ is positive.

10. Answer: +0.37 V.
The disproportionation reaction is $2Cu^+ \rightarrow Cu(s) + Cu^{2+}$. This consists of reduction: $Cu^+ + e^- \rightarrow Cu(s) \quad (E^{\circ} = +0.52 \text{ V})$ and oxidation: $Cu^+ \rightarrow Cu^{2+} + e^- \quad (E^{\circ} = -0.15 \text{ V})$. Summing these gives $0.52 - 0.15 = 0.37 \text{ V}$. Since it is positive, the disproportionation is spontaneous.

1. Which of the following describes the flow of electrons in a galvanic cell?

• AFrom the cathode to the anode through the salt bridge
• BFrom the anode to the cathode through the external wire
• CFrom the cathode to the anode through the external wire
• DFrom the anode to the cathode through the salt bridge

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 redox reactions, while an electrolytic cell uses electrical energy to drive non-spontaneous chemical reactions. In galvanic cells, the anode is negative, whereas in electrolytic cells, the anode is connected to the positive terminal of the power source.

How do you identify which species will be reduced on the MCAT?

You should look at the standard reduction potentials ($E^{\circ}$) provided in the passage or question. The species with the more positive (higher) reduction potential has a greater tendency to gain electrons and will be reduced at the cathode in a spontaneous cell.

Why is a salt bridge necessary in a voltaic cell?

A salt bridge is necessary to maintain electrical neutrality within the half-cells by allowing the migration of ions. Without it, a charge imbalance would quickly build up, creating a counter-potential that stops the flow of electrons and halts the reaction.

What does a negative cell potential indicate about spontaneity?

A negative cell potential ($E_{cell} < 0$) indicates that the reaction is non-spontaneous in the forward direction under the given conditions. This corresponds to a positive Gibbs free energy change ($\Delta G > 0$), meaning work must be done on the system to make the reaction occur.

How is Faraday's constant derived for electrochemical calculations?

Faraday's constant represents the total electrical charge carried by one mole of electrons. It is calculated by multiplying the charge of a single electron ($1.602 \times 10^{-19} \text{ C}$) by Avogadro's number ($6.022 \times 10^{23} \text{ mol}^{-1}$), resulting in approximately $96,485 \text{ C/mol}$.

Can concentration affect the cell potential of a redox reaction?

Yes, the cell potential is dependent on the concentrations of the reactants and products as described by the Nernst equation. If the concentration of reactants is increased or products decreased, the reaction becomes more favorable, increasing the cell potential.