MCAT Redox Practice Questions with Answers

Mastering MCAT Redox is essential for success on the Chemical and Physical Foundations of Biological Systems section, as reduction-oxidation reactions underpin everything from cellular respiration to electrochemical cells. Oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species, where one species loses electrons (oxidation) and another gains them (reduction). Understanding these concepts allows students to predict the spontaneity of biological reactions and calculate the voltage of batteries. To excel, students often use retrieval practice for medical students to cement these complex chemical rules into long-term memory.

Concept Explanation

MCAT Redox refers to the study of chemical reactions involving the transfer of electrons, characterized by changes in the oxidation states of the participating atoms. The most fundamental mnemonic for this topic is "OIL RIG": Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Alternatively, "LEO the lion says GER": Lose Electrons Oxidation, Gain Electrons Reduction.

To identify redox reactions, one must assign oxidation numbers based on specific rules. For example, an atom in its elemental state (like $\text{O}_2$ or $\text{Fe}$) has an oxidation state of 0, while oxygen in a compound is typically -2 and hydrogen is +1. In any neutral compound, the sum of oxidation states must equal zero; in a polyatomic ion, the sum must equal the ion's charge. These principles are vital when analyzing the electron transport chain, where complexes are sequentially reduced and oxidized.

Key terms to distinguish include:

• Oxidizing Agent: The species that gets reduced (it "takes" electrons from another, causing that other species to be oxidized).
• Reducing Agent: The species that gets oxidized (it "gives" electrons to another, causing that other species to be reduced).
• Disproportionation: A specific type of redox reaction where the same element is both oxidized and reduced.

Electrochemical cells further apply these concepts. In a galvanic (voltaic) cell, spontaneous reactions ($\Delta G < 0$) generate electricity, with oxidation occurring at the anode and reduction at the cathode. In electrolytic cells, non-spontaneous reactions ($\Delta G > 0$) are driven by an external power source. Calculating the standard electromotive force ($E^\circ_{ \text{cell}}$) involves subtracting the reduction potential of the anode from the reduction potential of the cathode: $$E^\circ_{ \text{cell}} = E^\circ_{ \text{red, cathode}} - E^\circ_{ \text{red, anode}}$$

Solved Examples

Example 1: Assigning Oxidation Numbers
Determine the oxidation state of sulfur in the sulfate ion $\text{SO}_4^{2-}$.

1. Identify the known oxidation states: Oxygen is almost always -2.
2. Set up the equation based on the total charge: Let $x$ be the oxidation state of sulfur. There are 4 oxygen atoms.
3. Write the expression: $x + 4(-2) = -2$.
4. Solve for $x$: $x - 8 = -2 \rightarrow x = +6$.
5. Conclusion: Sulfur has an oxidation state of +6 in sulfate.

Example 2: Identifying Agents
In the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$, identify the oxidizing and reducing agents.

1. Look at the change for Zinc: $\text{Zn}$ goes from 0 to +2. This is a loss of electrons (oxidation).
2. Look at the change for Copper: $\text{Cu}^{2+}$ goes from +2 to 0. This is a gain of electrons (reduction).
3. Identify the agents: Since Zinc is oxidized, it is the reducing agent. Since Copper is reduced, it is the oxidizing agent.

Example 3: Calculating Cell Potential
Given the standard reduction potentials $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ ($E^\circ = +0.80 \text{V}$) and $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ ($E^\circ = +0.34 \text{V}$), calculate the $E^\circ_{ \text{cell}}$ for a spontaneous galvanic cell.

1. Determine the cathode and anode: In a galvanic cell, the reaction with the higher reduction potential occurs at the cathode. Therefore, Silver is the cathode ($+0.80 \text{V}$) and Copper is the anode ($+0.34 \text{V}$).
2. Apply the formula: $$E^\circ_{ \text{cell}} = E^\circ_{ \text{cathode}} - E^\circ_{ \text{anode}}$$
3. Calculate: $E^\circ_{ \text{cell}} = 0.80 \text{V} - 0.34 \text{V} = +0.46 \text{V}$.
4. Conclusion: The cell potential is +0.46V.

Practice Questions

1. What is the oxidation state of Manganese in the permanganate ion $\text{MnO}_4^-$?

2. In the following reaction, which species is being reduced?
$$\text{Fe}_2 \text{O}_3 + 3 \text{CO} \rightarrow 2 \text{Fe} + 3 \text{CO}_2$$

3. Identify the reducing agent in the reaction: $2 \text{Na} + \text{Cl}_2 \rightarrow 2 \text{NaCl}$.

4. Balancing in acidic solution: Balance the following half-reaction: $\text{Cr}_2 \text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$.

5. Calculate the oxidation state of Carbon in glucose ($\text{C}_6 \text{H}_{12} \text{O}_6$).

6. A cell is constructed with a Lead electrode ($E^\circ_{ \text{red}} = -0.13 \text{V}$) and a Zinc electrode ($E^\circ_{ \text{red}} = -0.76 \text{V}$). If this is a galvanic cell, which metal acts as the anode?

7. What is the standard free energy change ($\Delta G^\circ$) for a reaction where $E^\circ_{ \text{cell}} = +1.20 \text{V}$ and 2 moles of electrons are transferred? (Use $F \approx 96,500 \text{ C/mol}$).

8. In the reaction $2 \text{H}_2 \text{O}_2 \rightarrow 2 \text{H}_2 \text{O} + \text{O}_2$, what is the oxidation state of Oxygen in the reactant, and what type of reaction is this?

9. Which of the following is the strongest oxidizing agent based on reduction potentials: $\text{F}_2$ ($+2.87 \text{V}$), $\text{Li}^+$ ($-3.05 \text{V}$), $\text{Au}^{3+}$ ($+1.50 \text{V}$)?

10. During the electrolysis of molten $\text{NaCl}$, what product is formed at the cathode?

Answers & Explanations

1. Answer: +7. Explanation: Oxygen is -2. $x + 4(-2) = -1 \rightarrow x - 8 = -1 \rightarrow x = +7$.
2. Answer: Iron (in $\text{Fe}_2 \text{O}_3$). Explanation: Iron starts at +3 in $\text{Fe}_2 \text{O}_3$ and goes to 0 in $\text{Fe}$. Gaining electrons is reduction.
3. Answer: Na (Sodium). Explanation: Sodium goes from 0 to +1. Since it is oxidized, it is the reducing agent.
4. Answer: $\text{Cr}_2 \text{O}_7^{2-} + 14 \text{H}^+ + 6e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2 \text{O}$. Explanation: First balance Cr (2 on each side), then O by adding 7 $\text{H}_2 \text{O}$, then H by adding 14 $\text{H}^+$, and finally charge by adding 6 electrons to the left.
5. Answer: 0. Explanation: $6(x) + 12(+1) + 6(-2) = 0 \rightarrow 6x + 12 - 12 = 0 \rightarrow 6x = 0 \rightarrow x = 0$.
6. Answer: Zinc. Explanation: In a galvanic cell, the half-reaction with the lower (more negative) reduction potential is the anode. $-0.76 \text{V} < -0.13 \text{V}$.
7. Answer: $\approx -231.6 \text{ kJ}$. Explanation: Use $\Delta G^\circ = -nFE^\circ$. $\Delta G^\circ = -(2)(96,500)(1.20) = -231,600 \text{ J} = -231.6 \text{ kJ}$.
8. Answer: -1; Disproportionation. Explanation: In peroxides, Oxygen is -1. It goes to -2 in water (reduction) and 0 in $\text{O}_2$ (oxidation). Since the same element is both oxidized and reduced, it is disproportionation.
9. Answer: $\text{F}_2$. Explanation: The most positive reduction potential indicates the strongest tendency to be reduced, making it the strongest oxidizing agent.
10. Answer: Sodium metal ($\text{Na}$). Explanation: Reduction always occurs at the cathode. $\text{Na}^+ + e^- \rightarrow \text{Na}(l)$.

1. Which of the following describes the function of a salt bridge in a galvanic cell?

• ATo provide electrons to the cathode
• BTo maintain electrical neutrality by allowing ion flow
• CTo increase the rate of oxidation at the anode
• DTo prevent the two solutions from ever coming into contact

Frequently Asked Questions

What is the difference between an oxidizing agent and a reducing agent?

An oxidizing agent facilitates the oxidation of another substance by gaining electrons itself, whereas a reducing agent facilitates the reduction of another substance by losing its own electrons.

How do you identify a disproportionation reaction?

A disproportionation reaction occurs when a single reactant element is simultaneously oxidized and reduced, resulting in two different products with different oxidation states for that same element.

Why is the anode negative in a galvanic cell but positive in an electrolytic cell?

In a galvanic cell, the anode is the source of electrons released by spontaneous oxidation, while in an electrolytic cell, the anode is attached to the positive terminal of an external battery to pull electrons away from the species being oxidized.

Does the MCAT require memorizing specific reduction potentials?

No, the MCAT will provide the necessary reduction potential values in a table or passage, but you must know how to use them to calculate total cell potential and determine spontaneity.

What is the relationship between E°cell and the equilibrium constant Keq?

A positive standard cell potential corresponds to a spontaneous reaction where the equilibrium constant Keq is greater than 1, calculated using the equation E^\circ = \frac{RT}{nF} \ln K_{ \text{ \neq}}.

How is retrieval practice used to master redox?

Students can use retrieval practice by self-testing on oxidation state rules and drawing electrochemical cells from memory to strengthen neural pathways related to these concepts.