AP Chemistry Quiz: Cell Potential Under Nonstandard Conditions

What this quiz covers

This quiz focuses on Cell Potential Under Nonstandard Conditions, 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. $E_\text{cell}$ increases and remains positive (spontaneous). (correct answer)
2. $E_\text{cell}$ decreases to $0$ because increasing a reactant concentration makes the cell reach equilibrium.
3. $E_\text{cell}$ decreases but remains positive (spontaneous).
4. $E_\text{cell}$ increases but becomes negative (nonspontaneous).
5. $E_\text{cell}$ remains the same because adding more reactant does not change the driving force.

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $\text{Cd}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cd}^{2+}(aq) + 2\text{Ag}(s)$, where $Q = \frac{[\text{Cd}^{2+}]}{[\text{Ag}^{+}]^2}$. When $[\text{Ag}^{+}]$ increases above 1.0 M while $[\text{Cd}^{2+}]$ remains at 1.0 M, the denominator of $Q$ increases, so $Q$ decreases below 1. According to the Nernst equation, when $Q < 1$, $\ln Q$ is negative, making the term $-(\frac{RT}{nF})\ln Q$ positive, so $E_\text{cell} = E^\circ_\text{cell} - (\frac{RT}{nF})\ln Q$ increases above $E^\circ_\text{cell}$. Since the standard Cd/Ag cell is already spontaneous with positive $E^\circ_\text{cell}$, increasing $E_\text{cell}$ maintains positive voltage and spontaneity. Choice A incorrectly assumes that increasing a reactant concentration decreases $E_\text{cell}$, but increasing $[\text{Ag}^{+}]$ actually decreases $Q$ which increases $E_\text{cell}$. The key is recognizing that increasing reactant concentrations (in $Q$'s denominator) decreases $Q$, which increases $E_\text{cell}$.

Question 2

1. $E_\text{cell}$ decreases but remains positive (spontaneous).
2. $E_\text{cell}$ increases and remains positive (spontaneous). (correct answer)
3. $E_\text{cell}$ remains the same because solids are not in $Q$.
4. $E_\text{cell}$ decreases to $0$ because increasing an ion concentration always drives equilibrium.
5. $E_\text{cell}$ increases but becomes negative (nonspontaneous).

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $\text{Sn}(s) + \text{Pb}^{2+}(aq) \rightarrow \text{Sn}^{2+}(aq) + \text{Pb}(s)$, where $Q = \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}$. When $[\text{Pb}^{2+}]$ increases above 1.0 M while $[\text{Sn}^{2+}]$ remains at 1.0 M, the denominator of Q increases, so Q decreases below 1. According to the Nernst equation, when $Q < 1$, $\ln Q$ is negative, making the term $-(\frac{RT}{nF})\ln Q$ positive, so $E_\text{cell} = E^\circ_\text{cell} - (\frac{RT}{nF})\ln Q$ increases above $E^\circ_\text{cell}$. Since the standard cell is already spontaneous with positive $E^\circ_\text{cell}$, increasing $E_\text{cell}$ keeps it positive and spontaneous. Choice A incorrectly assumes that increasing a reactant concentration decreases $E_\text{cell}$, but increasing $[\text{Pb}^{2+}]$ actually decreases Q which increases $E_\text{cell}$. The key is recognizing that reactant concentrations appear in Q's denominator, so increasing them decreases Q and increases $E_\text{cell}$.

Question 3

1. $E_\text{cell}$ increases and remains positive (spontaneous).
2. $E_\text{cell}$ decreases but remains positive (spontaneous). (correct answer)
3. $E_\text{cell}$ remains the same because $\text{Pt}(s)$ is inert.
4. $E_\text{cell}$ decreases to $0$ because changing a ratio in one half-cell makes the system at equilibrium.
5. $E_\text{cell}$ increases but becomes negative (nonspontaneous).

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $\text{Co}(s) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Co}^{2+}(aq) + 2\text{Fe}^{2+}(aq)$, where $Q = \frac{[\text{Co}^{2+}][\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}$. When the ratio $\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}$ increases while $[\text{Co}^{2+}]$ remains at 1.0 M, this means either $[\text{Fe}^{2+}]$ increased or $[\text{Fe}^{3+}]$ decreased (or both). Either way, Q increases because $[\text{Fe}^{2+}]^2$ is in the numerator and $[\text{Fe}^{3+}]^2$ is in the denominator. According to the Nernst equation, when Q increases above 1, lnQ becomes positive, making $-(\frac{RT}{nF})\ln Q$ negative, so $E_\text{cell} = E^\circ_\text{cell} - (\frac{RT}{nF})\ln Q$ decreases. Since the standard Co/Fe cell has positive $E^\circ_\text{cell}$, the decrease still leaves $E_\text{cell}$ positive and spontaneous. Choice A incorrectly predicts an increase, failing to recognize that increasing the product/reactant ratio increases Q. The strategy is to express Q in terms of all species, then determine how the given ratio change affects Q.

Question 4

1. $E_\text{cell}$ increases and remains positive (spontaneous).
2. $E_\text{cell}$ decreases but remains positive (spontaneous). (correct answer)
3. $E_\text{cell}$ remains the same because reactant concentration does not appear in $Q$.
4. $E_\text{cell}$ decreases to $0$ because any decrease in a reactant concentration makes the cell reach equilibrium.
5. $E_\text{cell}$ increases but becomes negative (nonspontaneous).

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $\text{Ni}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Ni}^{2+}(aq) + \text{Cu}(s)$, where $Q = \frac{[\text{Ni}^{2+}]}{[\text{Cu}^{2+}]}$. When $[\text{Cu}^{2+}]$ decreases below 1.0 M while $[\text{Ni}^{2+}]$ remains at 1.0 M, the denominator of $Q$ decreases, so $Q$ increases above 1. According to the Nernst equation, when $Q > 1$, $\ln Q$ is positive, making the term $-\frac{RT}{nF} \ln Q$ negative, so $E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q$ decreases below $E^\circ_\text{cell}$. Since the standard Ni/Cu cell has positive $E^\circ_\text{cell}$, the decrease still leaves $E_\text{cell}$ positive, maintaining spontaneity. Choice A would be correct if a product concentration decreased, but here a reactant concentration decreased, which has the opposite effect on $Q$. The strategy is to identify whether the changed species is a reactant (in $Q$'s denominator) or product (in $Q$'s numerator), then determine $Q$'s change.

Question 5

1. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
2. $E_\text{cell}$ remains the same because $Q$ only depends on reactants, not products.
3. $E_\text{cell}$ increases and becomes more positive (still spontaneous).
4. $E_\text{cell}$ decreases but remains positive (still spontaneous). (correct answer)
5. $E_\text{cell}$ becomes zero because adding product ions forces equilibrium.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Increasing [Cd²⁺], a product, to much larger than 1 M while keeping [Ni²⁺] at 1 M causes Q = [Cd²⁺]/[Ni²⁺] to be greater than 1. Since Q > 1, log Q is positive, making the Nernst term negative. Therefore, Ecell decreases below E°cell but remains positive, keeping the cell spontaneous. A tempting distractor is that Ecell becomes zero because adding product ions forces equilibrium, but this is incorrect because it assumes Q=1 without calculation. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 6

1. $E_{\text{cell}}$ decreases but remains positive (still spontaneous).
2. $E_{\text{cell}}$ decreases to zero because lowering a product concentration forces equilibrium.
3. $E_{\text{cell}}$ increases and remains positive (more spontaneous). (correct answer)
4. $E_{\text{cell}}$ remains the same because only reactant concentrations affect $E_{\text{cell}}$.
5. $E_{\text{cell}}$ increases but becomes negative (no longer spontaneous).

Explanation: This problem tests understanding of cell potential under nonstandard conditions. The reaction quotient Q = [Al³⁺]²/[Cu²⁺]³, and when [Al³⁺] is much smaller than 1 M while [Cu²⁺] remains 1 M, Q becomes much smaller than 1 (the numerator is squared and small). According to the Nernst equation, when Q < 1, the term -(RT/nF)lnQ becomes positive (since lnQ is negative), which increases Ecell above its standard value. Since the standard cell is already spontaneous, the increased Ecell makes it even more spontaneous. A common misconception is that only reactant concentrations affect Ecell (choice D), but the Nernst equation includes all aqueous species in Q, both reactants and products. To solve these problems, write Q with all concentration terms raised to their stoichiometric powers, evaluate whether Q is greater or less than 1, then determine the direction of Ecell change.

Question 7

1. $E_\text{cell}$ increases and remains positive (spontaneous).
2. $E_\text{cell}$ decreases but remains positive (spontaneous). (correct answer)
3. $E_\text{cell}$ remains the same because the electrodes are solids.
4. $E_\text{cell}$ decreases to $0$ because nonstandard conditions imply equilibrium.
5. $E_\text{cell}$ increases but becomes negative (nonspontaneous).

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), where Q = [Zn²⁺]/[Cu²⁺]. When [Zn²⁺] increases above 1.0 M while [Cu²⁺] remains at 1.0 M, Q increases above 1 (its standard value). According to the Nernst equation, when Q > 1, the term -(RT/nF)lnQ becomes more negative, so Ecell = E°cell - (RT/nF)lnQ decreases from its standard value. Since the standard Zn/Cu cell has E°cell = +1.10 V, even with the decrease, Ecell remains positive and the reaction stays spontaneous. Choice C incorrectly assumes that solid electrodes make concentrations irrelevant, but ion concentrations still affect Q and thus Ecell. The key strategy is to determine whether Q increases or decreases relative to standard conditions, then apply that Ecell moves opposite to Q.

Question 8

1. $E_\text{cell}$ increases but becomes negative (nonspontaneous).
2. $E_\text{cell}$ increases and remains positive (spontaneous). (correct answer)
3. $E_\text{cell}$ remains the same because changing product concentration only affects mass, not voltage.
4. $E_\text{cell}$ decreases but remains positive (spontaneous).
5. $E_\text{cell}$ decreases to $0$ because lowering a product concentration forces equilibrium.

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $\text{Fe}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Fe}^{2+}(aq) + 2\text{Ag}(s)$, where $Q = \frac{[\text{Fe}^{2+}]}{[\text{Ag}^{+}]^2}$. When $[\text{Fe}^{2+}]$ decreases below 1.0 M while $[\text{Ag}^{+}]$ remains at 1.0 M, the numerator of Q decreases, so Q decreases below 1. According to the Nernst equation, when Q < 1, $\ln Q$ is negative, making the term $-\frac{RT}{nF} \ln Q$ positive, so $E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q$ increases above $E^\circ_\text{cell}$. Since the standard Fe/Ag cell is already spontaneous with positive $E^\circ_\text{cell}$, increasing $E_\text{cell}$ maintains positive voltage and spontaneity. Choice A incorrectly assumes that decreasing a product concentration decreases $E_\text{cell}$, but decreasing $[\text{Fe}^{2+}]$ actually decreases Q which increases $E_\text{cell}$. The key insight is that product concentrations appear in Q's numerator, so decreasing them decreases Q and increases $E_\text{cell}$.

Question 9

1. $E_\text{cell}$ remains the same as standard because the electrodes are identical.
2. $E_\text{cell}$ increases above the standard value and remains positive.
3. $E_\text{cell}$ is positive but smaller than the standard value for a $\text{Cu}/\text{Cu}^{2+}$ cell. (correct answer)
4. $E_\text{cell}$ becomes zero because concentration cells cannot be spontaneous.
5. $E_\text{cell}$ becomes negative because higher ion concentration always increases $Q$.

Explanation: This question tests understanding of cell potential under nonstandard conditions. In a concentration cell, both half-cells have the same electrode (Cu/Cu²⁺), so E°cell = 0. The cell operates because of the concentration difference: electrons flow from the dilute side (lower [Cu²⁺]) to the concentrated side (higher [Cu²⁺]). With one side at 1.0 M and the other above 1.0 M, the reaction is Cu(s) + Cu²⁺(conc) → Cu²⁺(dil) + Cu(s), where Q = [Cu²⁺]dil/[Cu²⁺]conc < 1. Since Q < 1, lnQ is negative, making Ecell = 0 - (RT/nF)lnQ positive. However, this Ecell is smaller than a standard Cu/Cu²⁺ cell paired with a different metal because concentration cells rely only on concentration differences, not standard potential differences. Choice D incorrectly claims concentration cells cannot be spontaneous, but they are spontaneous until concentrations equalize. The strategy for concentration cells is recognizing that E°cell = 0 and Ecell depends entirely on the concentration ratio through the Nernst equation.

Question 10

1. $E_\text{cell}$ decreases but remains positive (spontaneous). (correct answer)
2. $E_\text{cell}$ increases and remains positive (spontaneous).
3. $E_\text{cell}$ remains the same because changing concentration only affects reaction rate.
4. $E_\text{cell}$ increases but becomes negative (nonspontaneous).
5. $E_\text{cell}$ decreases to $0$ because any concentration change forces equilibrium.

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $2\text{Ag}^{+}(aq) + \text{Cu}(s) \rightarrow 2\text{Ag}(s) + \text{Cu}^{2+}(aq)$, where $Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2}$. When $[\text{Ag}^{+}]$ decreases below 1.0 M while $[\text{Cu}^{2+}]$ remains at 1.0 M, the denominator of Q becomes smaller, so Q increases above 1. According to the Nernst equation, when Q increases, $E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q$ decreases because the logarithmic term becomes more positive. Since the standard Ag/Cu cell has a positive $E^\circ_\text{cell}$, the decrease still leaves $E_\text{cell}$ positive, maintaining spontaneity. Choice B incorrectly assumes that decreasing a reactant concentration increases $E_\text{cell}$, but this actually increases Q which decreases $E_\text{cell}$. The strategy is to write Q for the reaction, determine how concentration changes affect Q, then infer that $E_\text{cell}$ moves opposite to Q.

Question 11

1. $E_\text{cell}$ decreases but remains positive (spontaneous).
2. $E_\text{cell}$ increases and remains positive (spontaneous). (correct answer)
3. $E_\text{cell}$ remains the same because only reactant concentrations affect $Q$.
4. $E_\text{cell}$ decreases to $0$ because lowering a product concentration makes the cell stop.
5. $E_\text{cell}$ increases but becomes negative (nonspontaneous).

Explanation: This question tests understanding of cell potential under nonstandard conditions. The reaction is $\text{Pb}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Pb}^{2+}(aq) + \text{Cu}(s)$, where $Q = \frac{[\text{Pb}^{2+}]}{[\text{Cu}^{2+}]}$. When $[\text{Pb}^{2+}]$ decreases below 1.0 M while $[\text{Cu}^{2+}]$ remains at 1.0 M, the numerator of Q decreases, so Q decreases below 1. According to the Nernst equation, when Q < 1, lnQ is negative, making the term $-(\frac{RT}{nF})\ln Q$ positive, so $E_\text{cell} = E^\circ_\text{cell} - (\frac{RT}{nF})\ln Q$ increases above $E^\circ_\text{cell}$. Since the standard Pb/Cu cell is already spontaneous with positive $E^\circ_\text{cell}$, increasing $E_\text{cell}$ maintains positive voltage and spontaneity. Choice A incorrectly predicts a decrease, which would occur if a reactant concentration decreased instead of a product. The strategy is to identify the changed species as a product (numerator of Q), recognize that decreasing it decreases Q, then apply that $E_\text{cell}$ moves opposite to Q.

Question 12

1. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
2. $E_\text{cell}$ remains the same because changing pressure only changes the amount of gas present, not $Q$.
3. $E_\text{cell}$ decreases but remains positive (still spontaneous).
4. $E_\text{cell}$ becomes zero because increasing pressure drives the reaction to equilibrium.
5. $E_\text{cell}$ increases and becomes more positive (still spontaneous). (correct answer)

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Increasing P_H₂, a reactant, to much larger than 1 atm while keeping ions at 1 M causes Q = [H⁺]²/[Ag⁺]² P_H₂ to be less than 1. Since Q < 1, log Q is negative, leading to a positive Nernst correction. Therefore, Ecell increases above E°cell, becoming more positive and remaining spontaneous. A tempting distractor is that Ecell decreases but remains positive, but this is incorrect because it assumes increasing pressure raises Q, misapplying Le Chatelier to potential. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 13

1. $E_\text{cell}$ remains the same because the electrodes are identical.
2. $E_\text{cell}$ becomes zero because the same half-reaction occurs at both electrodes.
3. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
4. $E_\text{cell}$ increases above 0 and the cell is spontaneous. (correct answer)
5. $E_\text{cell}$ decreases but remains positive because dilution always lowers voltage.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Making [Cu²⁺] dilute much lower than 1 M on the left and concentrated much higher than 1 M on the right causes Q = [Cu²⁺ dilute]/[Cu²⁺ conc] to be much less than 1. Since Q < 1, log Q is negative, making the Nernst correction term positive. Therefore, Ecell increases above the standard 0 V, making the cell spontaneous. A tempting distractor is that Ecell becomes zero because the same half-reaction occurs at both electrodes, but this is incorrect because it assumes no driving force in concentration cells, ignoring the entropy-driven potential from concentration differences. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 14

1. $E_\text{cell}$ decreases but remains positive (still spontaneous).
2. $E_\text{cell}$ increases and becomes more positive (still spontaneous). (correct answer)
3. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
4. $E_\text{cell}$ remains the same because only $E^\circ$ determines spontaneity.
5. $E_\text{cell}$ becomes zero because increasing reactant concentration makes $Q=1$.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Increasing [Ag⁺], a reactant, to much larger than 1 M while keeping [Mg²⁺] at 1 M causes Q = [Mg²⁺]/[Ag⁺]² to be less than 1. Since Q < 1, log Q is negative, leading to a positive Nernst correction. Therefore, Ecell increases above E°cell, becoming more positive and remaining spontaneous. A tempting distractor is that Ecell decreases but remains positive, but this is incorrect because it assumes increasing reactant raises Q, confusing reactant and product roles. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 15

1. $E_\text{cell}$ increases and becomes more positive (still spontaneous).
2. $E_\text{cell}$ remains the same because chlorine is a reactant and reactants do not appear in $Q$.
3. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
4. $E_\text{cell}$ decreases but remains positive (still spontaneous). (correct answer)
5. $E_\text{cell}$ becomes zero because lowering gas pressure forces equilibrium.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Decreasing P_Cl₂, a reactant, to much smaller than 1 atm while keeping ions at 1 M causes Q = [Cl⁻]² [Zn²⁺]/P_Cl₂ to be greater than 1. Since Q > 1, log Q is positive, making the correction negative. Therefore, Ecell decreases below E°cell but remains positive due to the large E°. A tempting distractor is that Ecell decreases and becomes negative, but this is incorrect because it assumes low pressure can realistically flip the sign given the high standard potential. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 16

1. $E_\text{cell}$ increases and becomes more positive (still spontaneous).
2. $E_\text{cell}$ decreases but remains positive (still spontaneous). (correct answer)
3. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
4. $E_\text{cell}$ becomes zero because the cell is now at equilibrium.
5. $E_\text{cell}$ remains the same because solids do not appear in $Q$.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Increasing the concentration of Zn²⁺, a product, to much larger than 1 M while keeping Cu²⁺ at 1 M causes the reaction quotient Q = [Zn²⁺]/[Cu²⁺] to be greater than 1. Since Q > 1, log Q is positive, leading to a negative correction in the Nernst equation. Therefore, Ecell decreases compared to E°cell but remains positive, keeping the reaction spontaneous. A tempting distractor is that Ecell remains the same because solids do not appear in Q, but this is incorrect because it assumes concentrations of ions do not influence Q. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 17

1. $E_\text{cell}$ increases and becomes more positive (still spontaneous).
2. $E_\text{cell}$ decreases but remains positive (still spontaneous). (correct answer)
3. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
4. $E_\text{cell}$ remains the same because product ions do not affect a galvanic cell.
5. $E_\text{cell}$ becomes zero because adding product forces $Q=1$.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Increasing [Pb²⁺], a product, to much larger than 1 M while keeping [Ag⁺] at 1 M causes Q = [Pb²⁺]/[Ag⁺]² to be greater than 1. Since Q > 1, log Q is positive, making the Nernst term negative. Therefore, Ecell decreases below E°cell but remains positive, keeping the cell spontaneous. A tempting distractor is that Ecell remains the same because product ions do not affect a galvanic cell, but this is incorrect because it assumes products are absent from Q. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 18

1. $E_\text{cell}$ decreases but remains positive (still spontaneous).
2. $E_\text{cell}$ decreases and becomes negative (no longer spontaneous).
3. $E_\text{cell}$ increases and becomes more positive (still spontaneous). (correct answer)
4. $E_\text{cell}$ becomes zero because high $[\text{Cu}^{2+}]$ means the reaction goes to completion.
5. $E_\text{cell}$ remains the same because the identity of the electrodes determines the voltage, not concentrations.

Explanation: This question assesses the skill of cell potential under nonstandard conditions. Increasing [Cu²⁺], a reactant, to much larger than 1 M while keeping [Mn²⁺] at 1 M causes Q = [Mn²⁺]/[Cu²⁺] to be less than 1. Since Q < 1, log Q is negative, leading to a positive correction. Therefore, Ecell increases above E°cell, becoming more positive and remaining spontaneous. A tempting distractor is that Ecell becomes zero because high [Cu²⁺] means the reaction goes to completion, but this is incorrect because it assumes initial conditions set equilibrium immediately. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.

Question 19

1. $E_{\text{cell}}$ remains the same because the same species appear in both half-cells.
2. $E_{\text{cell}}$ decreases to zero because identical electrodes always give $E_{\text{cell}}=0$.
3. $E_{\text{cell}}$ increases above zero and is spontaneous. (correct answer)
4. $E_{\text{cell}}$ becomes negative because increasing $[\text{Ag}^+]$ on one side reverses electron flow.
5. $E_{\text{cell}}$ remains zero because solids do not appear in $Q$.

Explanation: This problem tests understanding of cell potential under nonstandard conditions. In a concentration cell, the net reaction is Ag⁺(high conc) → Ag⁺(low conc), where electrons flow to reduce Ag⁺ at higher concentration. With one compartment at [Ag⁺] = 1 M and the other much larger than 1 M, Q = [Ag⁺]low/[Ag⁺]high < 1. According to the Nernst equation, when Q < 1, the term -(RT/nF)lnQ becomes positive (since lnQ is negative), which increases Ecell above zero. The cell becomes spontaneous in the direction that equalizes concentrations. A common misconception is that identical electrodes always give Ecell = 0 (choice B), but this is only true when concentrations are equal; concentration differences drive the cell. The strategy for concentration cells is to identify which direction equalizes concentrations, then recognize that Q ≠ 1 creates a nonzero, positive Ecell.

Question 20

1. $E_{\text{cell}}$ decreases and becomes negative because lowering reactant pressure increases $Q$ above its standard-condition value. (correct answer)
2. $E_{\text{cell}}$ increases but remains positive (the cell remains spontaneous).
3. $E_{\text{cell}}$ remains the same because liquids and gases do not affect $Q$.
4. $E_{\text{cell}}$ decreases but remains positive because $Q$ decreases below its standard-condition value.
5. $E_{\text{cell}}$ increases and becomes zero because changing pressure forces equilibrium.

Explanation: This question tests understanding of cell potential under nonstandard conditions. For the reaction Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l), Q = [Cl⁻]²/[P(Cl₂)][Br⁻]², noting that Br₂(l) as a pure liquid doesn't appear in Q. Under standard conditions Q = 1, but when P(Cl₂) is much smaller than 1 atm while [Br⁻] = [Cl⁻] = 1.0 M, Q becomes much larger than 1 because we're dividing by a small pressure. According to the Nernst equation, when Q > 1, lnQ is positive, making -(RT/nF)lnQ negative, so Ecell = E°cell - (RT/nF)lnQ decreases below the standard potential. Since this decrease can be substantial when Q >> 1, and if E°cell is not very large, Ecell can become negative, making the forward reaction nonspontaneous. Choice D incorrectly claims that Q decreases when reactant pressure decreases, missing that lower reactant concentration or pressure increases Q. To solve these problems, write Q including gases but excluding pure liquids and solids, then analyze the concentration/pressure effects.