AP Chemistry Quiz: Cell Potential And Free Energy

What this quiz covers

This quiz focuses on Cell Potential And Free Energy, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
2. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
5. The reaction is spontaneous because Al is a more active metal than Ag.

Explanation: This question tests understanding of cell potential and free energy. When the anode and cathode are already labeled, we know that oxidation occurs at the anode (Al → Al³⁺) and reduction occurs at the cathode (Ag⁺ → Ag). Since aluminum has a very negative reduction potential (around -1.66 V) and silver has a positive reduction potential (around +0.80 V), the cell potential E°cell = E°cathode - E°anode will be positive. A positive E°cell indicates that the reaction is spontaneous as written. A common error is assuming that just because aluminum is more active, the reaction direction could be reversed, but the electrode labels define the reaction direction. The key principle is that a positive E°cell always corresponds to a spontaneous galvanic cell reaction.

Question 2

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative. (correct answer)
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
5. The reaction is spontaneous because Ag(s) is a good conductor.

Explanation: This question tests understanding of cell potential and free energy. The written reaction shows Sn²⁺ being reduced to Sn and Ag being oxidized to Ag⁺, which is the reverse of the spontaneous direction. With Sn²⁺/Sn at E° = -0.14 V and Ag⁺/Ag at E° = +0.80 V, this reverse reaction would have E°cell = E°cathode - E°anode = (-0.14 V) - (+0.80 V) = -0.94 V. A negative E°cell indicates the reaction is nonspontaneous as written. The error would be assuming that because silver is a good conductor (option E), this affects the thermodynamics, but electrical conductivity doesn't determine reaction spontaneity. The key insight is that reversing a spontaneous galvanic reaction always produces a nonspontaneous reaction with negative E°cell.

Question 3

1. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive. (correct answer)
2. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
3. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
4. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.
5. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.

Explanation: This question assesses the skill of cell potential and free energy. In a galvanic cell, the half-cell with the more positive reduction potential serves as the cathode for reduction, while the half-cell with the less positive (or more negative) reduction potential serves as the anode for oxidation. The standard cell potential, E°cell, is determined by subtracting the standard reduction potential of the anode from that of the cathode. A positive E°cell indicates that the cell reaction is spontaneous under standard conditions, as it corresponds to a negative Gibbs free energy change. One tempting distractor is choice B, which incorrectly assumes that a spontaneous reaction must have a negative E°cell, stemming from the misconception that individual negative reduction potentials directly make the cell potential negative. A transferable strategy is to always calculate E°cell as E°cathode minus E°anode and recognize that a positive value confirms a spontaneous galvanic process.

Question 4

1. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is zero.
2. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
3. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative. (correct answer)
4. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
5. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive.

Explanation: This question assesses the skill of cell potential and free energy. When half-cells are connected with specified anode and cathode, compare their reduction potentials to find E°cell = E°cathode - E°anode. If E°cell is negative, the reaction is nonspontaneous, indicating a positive ΔG° and the need for external energy. In this assembly, Al's lower E° as cathode versus I2 as anode gives negative E°cell, making it nonspontaneous. Choice A is tempting but wrong, assuming spontaneity with negative E°cell, from the misconception that labeling reverses the natural potential without affecting the sign calculation. A transferable strategy is to always use the formula E°cell = E°red(cathode) - E°red(anode) and note that a negative value means nonspontaneity in the assembled configuration.

Question 5

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative. (correct answer)
5. The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.

Explanation: This question assesses the skill of cell potential and free energy. Labeled Zn cathode (-0.76 V), Ni anode (-0.25 V), but Ni has higher E°. E°cell = -0.76 V - (-0.25 V) = -0.51 V, negative. This means nonspontaneous as labeled. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, resulting in positive E°cell error. Use labels; a positive E°cell indicates a spontaneous galvanic process.

Question 6

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative. (correct answer)
5. The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.

Explanation: This question assesses the skill of cell potential and free energy. Claimed electron flow from Ag to Cu means Ag anode (+0.80 V), Cu cathode (+0.34 V), but Ag has higher E°. E°cell = +0.34 V - (+0.80 V) = -0.46 V, negative. This indicates nonspontaneity as described. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to positive E°cell wrongly. Base on claimed flow; a positive E°cell indicates a spontaneous galvanic process.

Question 7

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
2. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative. (correct answer)
3. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
5. The reaction is spontaneous because Mg has the more negative reduction potential.

Explanation: This question tests understanding of cell potential and free energy. The student labeled Cd as anode (oxidation) and Mg as cathode (reduction), meaning the reaction would be Cd → Cd²⁺ + 2e⁻ and Mg²⁺ + 2e⁻ → Mg. This gives E°cell = E°cathode - E°anode = (-2.37 V) - (-0.40 V) = -1.97 V. A negative E°cell indicates the reaction is nonspontaneous as labeled. The error in option E is thinking that having the more negative reduction potential automatically makes Mg the anode, but the student's labeling forces Mg to be the cathode, creating a nonspontaneous cell. The key lesson is that incorrect electrode labeling can result in a nonspontaneous cell with negative E°cell.

Question 8

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
5. The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.

Explanation: This question assesses the skill of cell potential and free energy. The standard cell potential, E°cell, is determined by subtracting the standard reduction potential of the anode from that of the cathode. Here, the Cu half-cell has a higher reduction potential (+0.34 V) compared to Zn (-0.76 V), so when Cu is the cathode and Zn is the anode, E°cell = +0.34 V - (-0.76 V) = +1.10 V, which is positive. A positive E°cell indicates that the redox reaction is spontaneous under standard conditions, as it corresponds to a negative ΔG°. A tempting distractor is choice D, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to a reversed and negative E°cell. Always identify the cathode as the half-cell with the higher reduction potential for spontaneous reactions; a positive E°cell indicates a spontaneous galvanic process.

Question 9

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive.
2. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative. (correct answer)
3. The reaction is spontaneous and $E^\circ_{\text{cell}} = 0$.
4. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
5. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.

Explanation: This question assesses the skill of cell potential and free energy. Labeled Pb cathode (-0.13 V), Cu anode (+0.34 V), but Cu has higher E°. E°cell = -0.13 V - (+0.34 V) = -0.47 V, negative. This means nonspontaneous as labeled. A tempting distractor is choice A, which is incorrect because it assumes oxidation occurs at the electrode with the larger reduction potential, leading to positive E°cell incorrectly. Use labels to compute; a positive E°cell indicates a spontaneous galvanic process.

Question 10

1. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
3. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
5. The reaction is spontaneous and $E^\circ_{\text{cell}}$ equals zero.

Explanation: This question tests understanding of cell potential and free energy. When electrons flow from Fe to Ag through the external circuit, Fe is being oxidized (anode) and Ag⁺ is being reduced (cathode), which makes sense because Ag⁺/Ag has the more positive reduction potential (+0.80 V) compared to Fe²⁺/Fe (-0.44 V). The standard cell potential is E°cell = E°cathode - E°anode = +0.80 V - (-0.44 V) = +1.24 V. A positive E°cell indicates the reaction is spontaneous under standard conditions. Students might incorrectly think that because Fe has a negative reduction potential, the overall reaction must be nonspontaneous, but this confuses the individual electrode potential with the cell potential. The key strategy is to identify which electrode serves as cathode (more positive E°) and calculate E°cell = E°cathode - E°anode.

Question 11

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
5. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is zero.

Explanation: This problem examines cell potential and free energy in a Cr/Br₂ galvanic cell. With Br₂ having a much higher reduction potential (+1.07 V) than Cr³⁺ (-0.74 V), Br₂ is reduced at the cathode while Cr is oxidized at the anode. The cell potential is E°cell = E°cathode - E°anode = (+1.07 V) - (-0.74 V) = +1.81 V. A positive E°cell indicates negative ΔG° and spontaneous reaction. Students might choose option B by confusing the relationship between E°cell and spontaneity, thinking positive E°cell means nonspontaneous. Remember that for galvanic cells, positive E°cell always indicates spontaneous electron flow and energy release.

Question 12

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
2. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
5. The reaction is spontaneous and $E^\circ_{\text{cell}}$ equals zero.

Explanation: This question tests understanding of cell potential and free energy. With Pb as the cathode (E° = -0.13 V) and Cd as the anode (E° = -0.40 V), Pb²⁺ is reduced and Cd is oxidized, which makes sense because Pb²⁺/Pb has the less negative (more positive) reduction potential. The standard cell potential is E°cell = E°cathode - E°anode = -0.13 V - (-0.40 V) = +0.27 V. A positive E°cell indicates the reaction is spontaneous under standard conditions. A common misconception is assuming that negative reduction potentials for both electrodes mean the cell cannot be spontaneous, but this ignores the relative values. The strategy is to remember that the electrode with the more positive (or less negative) reduction potential serves as the cathode in a spontaneous galvanic cell.

Question 13

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
2. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
5. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is zero.

Explanation: This question tests understanding of cell potential and free energy using electron flow direction. When electrons flow from Fe to Sn through the external circuit, Fe is the anode (oxidation occurs) and Sn is the cathode (reduction occurs). The cell reaction involves Fe → Fe²⁺ + 2e⁻ at the anode and Sn²⁺ + 2e⁻ → Sn at the cathode. The standard cell potential is E°cell = E°cathode - E°anode = (-0.14 V) - (-0.44 V) = +0.30 V. A positive E°cell indicates a spontaneous reaction. A common error is assuming that because both reduction potentials are negative, the cell reaction must be nonspontaneous, but this misconception ignores that we're comparing relative potentials. Remember that electron flow from anode to cathode through the external circuit always indicates which electrode is which in a galvanic cell.

Question 14

1. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.
2. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.
3. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
4. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive. (correct answer)
5. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.

Explanation: This question assesses the skill of cell potential and free energy. The relative magnitudes of reduction potentials dictate the direction of electron flow, with oxidation occurring at the electrode with the lower reduction potential. E°cell is computed as E°cathode - E°anode, and its positive sign confirms the reaction's spontaneity, linking to a decrease in free energy. In this setup, Pb's higher E° than Ni makes Pb the cathode, yielding a positive E°cell and spontaneous reaction. Choice B is a tempting distractor, incorrectly claiming nonspontaneity with negative E°cell, arising from the misconception that both negative reduction potentials must result in a negative cell potential. A transferable strategy is to compare reduction potentials to assign anode and cathode roles, ensuring a positive E°cell indicates a spontaneous galvanic process.

Question 15

1. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
2. The reaction is spontaneous because both half-reactions have reduction potentials.
3. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
4. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
5. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)

Explanation: This question tests understanding of cell potential and free energy. In a galvanic cell, the species with the more positive reduction potential (Cu²⁺/Cu at +0.34 V) undergoes reduction at the cathode, while the species with the more negative reduction potential (Zn²⁺/Zn at -0.76 V) undergoes oxidation at the anode. The standard cell potential is calculated as E°cell = E°cathode - E°anode = (+0.34 V) - (-0.76 V) = +1.10 V. Since E°cell is positive, the reaction is spontaneous under standard conditions. A common misconception is thinking that a negative reduction potential for one half-cell makes the overall reaction nonspontaneous, but what matters is the difference between the two potentials. Remember that for galvanic cells, a positive E°cell always indicates a spontaneous reaction with negative ΔG°.

Question 16

1. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
2. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.
3. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
4. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive. (correct answer)
5. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.

Explanation: This question assesses the skill of cell potential and free energy. The cathode is where the species with the highest reduction potential is reduced, and the anode is where oxidation occurs for the lower potential species. E°cell is positive when this natural order is followed, indicating spontaneity and a favorable free energy change. With F2's higher E° than Au's, the setup yields positive E°cell and spontaneous reaction. A distractor like choice B wrongly states spontaneity with negative E°cell, arising from the misconception that the highest E° should be oxidized at the anode. A transferable strategy is to select the higher reduction potential as cathode and ensure E°cell positivity denotes a spontaneous galvanic cell reaction.

Question 17

1. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is negative.
2. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is positive.
3. The reaction is spontaneous and $E^\circ_{\text{cell}}$ is positive. (correct answer)
4. The reaction is nonspontaneous and $E^\circ_{\text{cell}}$ is negative.
5. The reaction is spontaneous because both $E^\circ$ values are negative.

Explanation: This question tests understanding of cell potential and free energy. In the reaction, Ni is oxidized (reverse of Ni²⁺ + 2e⁻ → Ni) and Pb²⁺ is reduced to Pb. Since Pb²⁺/Pb has E° = -0.13 V and Ni²⁺/Ni has E° = -0.25 V, the cell potential is E°cell = E°cathode - E°anode = (-0.13 V) - (-0.25 V) = +0.12 V. A positive E°cell indicates the reaction is spontaneous. The misconception in option E assumes that because both reduction potentials are negative, the reaction must be spontaneous, but what matters is which species has the more positive (less negative) reduction potential. Remember that even with two negative reduction potentials, the cell can be spontaneous if E°cell is positive.

Question 18

1. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
2. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive. (correct answer)
3. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.
4. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
5. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.

Explanation: This question assesses the skill of cell potential and free energy. The half-cell with the higher reduction potential naturally becomes the cathode, while the lower one is the anode, determining the direction of spontaneous electron flow. E°cell, calculated as the difference in reduction potentials, is positive for spontaneous reactions, correlating with exothermic free energy changes. With Mn as anode and Cl2 as cathode, the large positive E°cell confirms spontaneity. A distractor like choice C incorrectly claims nonspontaneity with negative E°cell, stemming from the misconception that the highly negative E° for Mn makes the overall cell potential negative. A transferable strategy is to assign the anode as the half-cell with the smaller reduction potential and verify a positive E°cell for spontaneous galvanic reactions.

Question 19

1. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative. (correct answer)
2. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive.
3. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.
4. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
5. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.

Explanation: This question assesses the skill of cell potential and free energy. Relative reduction potentials indicate natural roles, but if labeled oppositely, E°cell may be negative, signaling nonspontaneity. The sign of E°cell determines if the reaction proceeds without input, with negative values meaning it's nonspontaneous as assembled. Here, labeling Cd as anode and Mg as cathode (against natural potentials) yields negative E°cell and nonspontaneity. Choice E is a tempting distractor, suggesting spontaneity with positive E°cell, based on the misconception that the more negative individual E° always drives spontaneity regardless of labeling. A transferable strategy is to compute E°cell using the given anode and cathode assignments and recognize that a negative E°cell indicates a nonspontaneous process in that setup.

Question 20

1. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is zero.
2. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is positive. (correct answer)
3. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is negative.
4. The cell reaction is nonspontaneous, and $E^\circ_{\text{cell}}$ is positive.
5. The cell reaction is spontaneous, and $E^\circ_{\text{cell}}$ is negative.

Explanation: This question assesses the skill of cell potential and free energy. Electron flow from one electrode to another identifies the anode (oxidation) and cathode (reduction), with relative reduction potentials confirming the setup. A positive E°cell, from E°cathode - E°anode, ensures the reaction is spontaneous, tied to negative free energy. Observation of flow from Cr to Cu aligns with Cu's higher E° , giving positive E°cell and spontaneity. Choice D is incorrect but tempting, claiming nonspontaneity with positive E°cell, from the misconception that differing electron counts in half-reactions affect the potential sign. A transferable strategy is to use observed electron flow to assign roles and calculate a positive E°cell for confirming spontaneous galvanic processes.