Cell Potential Under Nonstandard Conditions - AP Chemistry

1. A galvanic cell is $\text{Mn}(s)\,|\,\text{Mn}^{2+}(aq)\,||\,\text{Cu}^{2+}(aq)\,|\,\text{Cu}(s)$ with overall reaction $\text{Mn}(s)+\text{Cu}^{2+}(aq)\rightarrow \text{Mn}^{2+}(aq)+\text{Cu}(s)$. The reaction is spontaneous under standard conditions. The cell is prepared with $[\text{Mn}^{2+}]$ = 1 M and $[\text{Cu}^{2+}]$ much larger than 1 M. Using qualitative Nernst reasoning about $Q$ relative to 1, what happens to $E_\text{cell}$ and spontaneity?
   Answer: $E_\text{cell}$ increases and becomes more positive (still spontaneous). 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.
2. A galvanic cell is constructed with half-cells $\text{Ni}(s)|\text{Ni}^{2+}(aq)$ and $\text{Cu}^{2+}(aq)|\text{Cu}(s)$. The spontaneous overall reaction under standard conditions is $\text{Ni}(s)+\text{Cu}^{2+}(aq)\rightarrow \text{Ni}^{2+}(aq)+\text{Cu}(s).$ Compared with standard conditions, the $\text{Cu}^{2+}$ concentration is decreased below $1.0\,\text{M}$ while all other species remain at standard conditions. Using qualitative Nernst reasoning (through the effect on $Q$), how does $E_\text{cell}$ change and does the cell remain spontaneous?
   Answer: $E_\text{cell}$ decreases but remains positive (spontaneous). 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.
3. A galvanic cell is constructed at $25^\circ\text{C}$ using the half-cells $\text{Zn}(s)|\text{Zn}^{2+}(aq)$ and $\text{Cu}^{2+}(aq)|\text{Cu}(s)$, connected by a salt bridge. Compared with standard conditions, the $\text{Zn}^{2+}$ concentration is increased above $1.0\,\text{M}$ while all other species remain at standard conditions. Using qualitative Nernst reasoning (by determining how $Q$ changes relative to standard conditions), how does the cell potential $E_\text{cell}$ change and does the cell remain spontaneous?
   Answer: $E_\text{cell}$ decreases but remains positive (spontaneous). 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.
4. Consider the cell $\text{Al}(s)\,|\,\text{Al}^{3+}(aq)\,||\,\text{Cu}^{2+}(aq)\,|\,\text{Cu}(s)$ with overall reaction $2\text{Al}(s)+3\text{Cu}^{2+}(aq)\rightarrow 2\text{Al}^{3+}(aq)+3\text{Cu}(s)$. The cell is spontaneous under standard conditions. The cell is prepared so that $[\text{Cu}^{2+}]$ is 1 M, but $[\text{Al}^{3+}]$ is much larger than 1 M. Using qualitative Nernst reasoning about the direction of the shift in $Q$, what happens to $E_\text{cell}$ and spontaneity?
   Answer: $E_\text{cell}$ decreases but remains positive (still spontaneous). This question assesses the skill of cell potential under nonstandard conditions. Increasing [Al³⁺], a product, to much larger than 1 M while keeping [Cu²⁺] at 1 M causes Q = [Al³⁺]²/[Cu²⁺]³ to be greater than 1. Since Q > 1, log Q is positive, making the Nernst correction negative. Therefore, Ecell decreases below E°cell but remains positive, keeping the cell spontaneous. A tempting distractor is that Ecell remains the same because the stoichiometric coefficients do not matter in Q, but this is incorrect because it assumes powers in Q are ignored, overlooking their impact on log Q. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.
5. A galvanic cell is $\text{Cd}(s)\,|\,\text{Cd}^{2+}(aq)\,||\,\text{Ni}^{2+}(aq)\,|\,\text{Ni}(s)$ with overall reaction $\text{Cd}(s)+\text{Ni}^{2+}(aq)\rightarrow \text{Cd}^{2+}(aq)+\text{Ni}(s)$. The reaction is spontaneous under standard conditions. The cell is prepared with $[\text{Ni}^{2+}]$ = 1 M and $[\text{Cd}^{2+}]$ much larger than 1 M. Using qualitative Nernst reasoning about $Q$ relative to 1, what happens to $E_\text{cell}$ and spontaneity?
   Answer: $E_\text{cell}$ decreases but remains positive (still spontaneous). 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.
6. A galvanic cell is constructed based on the overall reaction $\text{I}_2(s)+2\text{S}_2\text{O}_3^{2-}(aq)\rightarrow 2\text{I}^-(aq)+\text{S}_4\text{O}_6^{2-}(aq).$ At the same temperature as standard conditions, the thiosulfate concentration is prepared so that $[\text{S}_2\text{O}_3^{2-}]$ is much smaller than 1.0 M, while $[\text{I}^-]$ and $[\text{S}_4\text{O}_6^{2-}]$ are each 1.0 M. Using qualitative Nernst reasoning (compare $Q$ to standard conditions), what happens to $E_{\text{cell}}$ and does the reaction remain spontaneous as written?
   Answer: $E_{\text{cell}}$ decreases and becomes negative because $Q$ increases relative to standard conditions. This question tests understanding of cell potential under nonstandard conditions. For the reaction I₂(s) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq), Q = [I⁻]²[S₄O₆²⁻]/[S₂O₃²⁻]², noting that I₂(s) as a solid doesn't appear in Q. Under standard conditions Q = 1, but when [S₂O₃²⁻] is much smaller than 1.0 M while [I⁻] = [S₄O₆²⁻] = 1.0 M, Q becomes much larger than 1 because we're dividing by [S₂O₃²⁻]² which is very small. 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, and can become negative if the decrease is large enough. Choice A incorrectly states that decreasing reactant concentration decreases Q, when it actually increases Q since reactants appear in the denominator. The key is to write Q correctly excluding solids, then recognize that decreasing reactant concentration increases Q and thus decreases Ecell.
7. A galvanic cell is $\text{Zn}(s)\,|\,\text{Zn}^{2+}(aq)\,||\,\text{H}^+(aq)\,|\,\text{H}_2(g)$ with overall reaction $\text{Zn}(s)+2\text{H}^+(aq)\rightarrow \text{Zn}^{2+}(aq)+\text{H}_2(g)$. The reaction is spontaneous under standard conditions. The cell is prepared with $[\text{Zn}^{2+}]$ = 1 M and $P_{\text{H}_2}$ = 1 atm, but $[\text{H}^+]$ is much larger than 1 M. Using qualitative Nernst reasoning about whether $Q$ increases or decreases relative to 1, what happens to $E_\text{cell}$ and spontaneity?
   Answer: $E_\text{cell}$ increases and becomes more positive (still spontaneous). This question assesses the skill of cell potential under nonstandard conditions. Increasing [H⁺], a reactant, to much larger than 1 M while keeping others standard causes Q = [Zn²⁺] $P_H$₂ / [H⁺]² to be less than 1. Since Q < 1, log Q is negative, resulting in a positive correction term. Therefore, Ecell increases above E°cell, becoming more positive and remaining spontaneous. A tempting distractor is that Ecell decreases and becomes negative, but this is incorrect because it assumes more reactant always raises voltage, but here it does while keeping positive. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.
8. A galvanic cell is constructed based on the overall reaction $\text{Cl}_2(g)+2\text{Br}^-(aq)\rightarrow 2\text{Cl}^-(aq)+\text{Br}_2(l).$ At the same temperature as standard conditions, chlorine gas is introduced at a partial pressure much smaller than 1 atm, while $[\text{Br}^-]$ and $[\text{Cl}^-]$ are each 1.0 M. Using qualitative Nernst reasoning (compare $Q$ to standard conditions), how does $E_{\text{cell}}$ change and does the reaction remain spontaneous as written?
   Answer: $E_{\text{cell}}$ decreases and becomes negative because lowering reactant pressure increases $Q$ above its standard-condition value. 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.
9. A galvanic cell is constructed with the overall reaction $2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Cu}(s)$ At standard conditions, the cell is spontaneous. The cell is prepared with $[\text{Al}^{3+}]$ much smaller than $1\,\text{M}$ while $[\text{Cu}^{2+}]$ remains $1\,\text{M}$. Using qualitative Nernst reasoning (compare $Q$ to 1), how does $E_{\text{cell}}$ change and does the reaction remain spontaneous?​
   Answer: $E_{\text{cell}}$ increases and remains positive (more spontaneous). 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.
10. A galvanic cell is $\text{F}_2(g)\,|\,\text{F}^-(aq)\,||\,\text{Fe}^{3+}(aq)\,|\,\text{Fe}^{2+}(aq)$ (inert Pt electrode) with overall reaction $\text{F}_2(g)+2\text{Fe}^{2+}(aq)\rightarrow 2\text{F}^-(aq)+2\text{Fe}^{3+}(aq)$. The reaction is spontaneous under standard conditions. The cell is prepared with $[\text{Fe}^{2+}]$ = 1 M and $[\text{Fe}^{3+}]$ = 1 M, but $[\text{F}^-]$ is much larger than 1 M while $P_{\text{F}_2}$ remains 1 atm. Using qualitative Nernst reasoning about $Q$, what happens to $E_\text{cell}$ and spontaneity?
    Answer: $E_\text{cell}$ decreases but remains positive (still spontaneous). This question assesses the skill of cell potential under nonstandard conditions. Increasing [F⁻], a product, to much larger than 1 M while keeping others standard causes Q = [F⁻]² [Fe³⁺]² / $P_F$₂ [Fe²⁺]² to be greater than 1. Since Q > 1, log Q is positive, leading to a negative Nernst correction. Therefore, Ecell decreases below E°cell but remains positive given the large E°. A tempting distractor is that Ecell decreases and becomes negative, but this is incorrect because it assumes high product concentration can flip the sign despite practical limits. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.
11. A galvanic cell is $\text{Sn}(s)\,|\,\text{Sn}^{2+}(aq)\,||\,\text{Pb}^{2+}(aq)\,|\,\text{Pb}(s)$ with overall reaction $\text{Sn}(s)+\text{Pb}^{2+}(aq)\rightarrow \text{Sn}^{2+}(aq)+\text{Pb}(s)$. The cell is spontaneous under standard conditions. The cell is prepared with $[\text{Sn}^{2+}]$ much smaller than 1 M while $[\text{Pb}^{2+}]$ is 1 M. Based on whether $Q$ is less than, equal to, or greater than 1, how does $E_\text{cell}$ change and does it remain spontaneous?
    Answer: $E_\text{cell}$ increases and becomes more positive (still spontaneous). This question assesses the skill of cell potential under nonstandard conditions. Decreasing [Sn²⁺], a product, to much smaller than 1 M while keeping [Pb²⁺] at 1 M causes Q = [Sn²⁺]/[Pb²⁺] to be less than 1. Since Q < 1, log Q is negative, leading to a positive Nernst correction term. Therefore, Ecell increases above E°cell, becoming more positive and remaining spontaneous. A tempting distractor is that Ecell decreases and becomes negative, but this is incorrect because it assumes lowering a product concentration reduces voltage, confusing it with reactant effects. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.
12. A galvanic cell is built with $\text{Co}(s)|\text{Co}^{2+}(aq)$ and $\text{Fe}^{3+}(aq),\text{Fe}^{2+}(aq)|\text{Pt}(s)$. The spontaneous overall reaction under standard conditions is $\text{Co}(s)+2\text{Fe}^{3+}(aq)\rightarrow \text{Co}^{2+}(aq)+2\text{Fe}^{2+}(aq).$ Compared with standard conditions, the ratio $\dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}$ in the platinum half-cell is increased above its standard-condition value, while all other species remain at standard conditions. Using qualitative Nernst reasoning (through the effect on $Q$), how does $E_\text{cell}$ change and does the cell remain spontaneous?
    Answer: $E_\text{cell}$ decreases but remains positive (spontaneous). 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.
13. A galvanic cell is constructed based on the overall reaction $\text{Mg}(s)+\text{Cu}^{2+}(aq)\rightarrow \text{Mg}^{2+}(aq)+\text{Cu}(s).$ At the same temperature as standard conditions, the magnesium(II) ion concentration is adjusted so that $[\text{Mg}^{2+}]$ is much smaller than 1.0 M, while $[\text{Cu}^{2+}]$ is 1.0 M. Using qualitative Nernst reasoning by comparing $Q$ to standard conditions, how does $E_{\text{cell}}$ change and does the forward reaction remain spontaneous?
    Answer: $E_{\text{cell}}$ increases but remains positive (the cell remains spontaneous). This question tests understanding of cell potential under nonstandard conditions. For the reaction Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s), Q = [Mg²⁺]/[Cu²⁺], and under standard conditions Q = 1. When [Mg²⁺] is much smaller than 1.0 M while [Cu²⁺] = 1.0 M, Q becomes much smaller than 1, indicating the reaction has not proceeded as far toward products as at standard conditions. According to the Nernst equation, when Q < 1, lnQ is negative, making -(RT/nF)lnQ positive, so Ecell = E°cell - (RT/nF)lnQ increases above the standard potential. Choice D incorrectly states that Q is smaller than 1 would decrease Ecell, confusing the inverse relationship between Q and cell potential. To solve these problems, write the Q expression, determine if Q increases or decreases from 1, then remember that smaller Q means larger Ecell.
14. A galvanic cell is $\text{Mg}(s)\,|\,\text{Mg}^{2+}(aq)\,||\,\text{Ag}^+(aq)\,|\,\text{Ag}(s)$ with overall reaction $\text{Mg}(s)+2\text{Ag}^+(aq)\rightarrow \text{Mg}^{2+}(aq)+2\text{Ag}(s)$. It is spontaneous under standard conditions. The cell is prepared with $[\text{Mg}^{2+}]$ = 1 M and $[\text{Ag}^+]$ much larger than 1 M. Using qualitative Nernst reasoning about whether $Q$ is less than or greater than 1, what happens to $E_\text{cell}$ and spontaneity?
    Answer: $E_\text{cell}$ increases and becomes more positive (still spontaneous). 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.
15. A galvanic cell is constructed with the overall reaction $\text{H}_2(g) + \text{Cu}^{2+}(aq) \rightarrow 2\text{H}^+(aq) + \text{Cu}(s)$ At standard conditions ($P_{\text{H}_2}=1\,\text{atm}$, $[\text{Cu}^{2+}]=1\,\text{M}$, and $[\text{H}^+]=1\,\text{M}$), the cell is spontaneous. The cell is prepared with $[\text{H}^+]$ much smaller than $1\,\text{M}$ while $P_{\text{H}_2}$ and $[\text{Cu}^{2+}]$ remain at their standard-state values. Using qualitative Nernst reasoning about how $Q$ changes, how does $E_{\text{cell}}$ change and does the reaction remain spontaneous?​
    Answer: $E_{\text{cell}}$ increases and remains positive (more spontaneous). This problem tests understanding of cell potential under nonstandard conditions. The reaction quotient Q = [H⁺]²/([H₂][Cu²⁺]), and when [H⁺] is much smaller than 1 M while PH₂ and [Cu²⁺] remain at 1, Q becomes much smaller than 1 (small numerator squared). 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 spontaneous, the increased Ecell makes it even more spontaneous. A misconception is that H⁺ being a product means it doesn't affect the cell (choice A), but all aqueous species in the balanced equation appear in Q and affect Ecell through the Nernst equation. The strategy is to identify all species in the balanced equation, write Q with proper exponents, then analyze how concentration changes affect Q and consequently Ecell.
16. A concentration cell uses the half-reaction $\text{Cu}^{2+}(aq)+2e^-\rightarrow \text{Cu}(s)$ at both electrodes: $\text{Cu}(s)\,|\,\text{Cu}^{2+}(aq,\,\text{dilute})\,||\,\text{Cu}^{2+}(aq,\,\text{concentrated})\,|\,\text{Cu}(s)$. Compared with the standard-state situation where both solutions are 1 M, the cell is prepared with the left half-cell having $[\text{Cu}^{2+}]$ much lower than 1 M and the right half-cell having $[\text{Cu}^{2+}]$ much higher than 1 M. Using qualitative Nernst reasoning (how $Q$ differs from 1), what is true about $E_\text{cell}$ and spontaneity?
    Answer: $E_\text{cell}$ increases above 0 and the cell is spontaneous. 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.
17. A galvanic cell uses $\text{Ni}(s)\,|\,\text{Ni}^{2+}(aq)\,||\,\text{Co}^{2+}(aq)\,|\,\text{Co}(s)$ with overall reaction $\text{Ni}(s)+\text{Co}^{2+}(aq)\rightarrow \text{Ni}^{2+}(aq)+\text{Co}(s)$. The reaction is spontaneous under standard conditions. The cell is assembled with $[\text{Co}^{2+}]$ much larger than 1 M while $[\text{Ni}^{2+}]$ is 1 M. Using qualitative Nernst reasoning about $Q$ relative to 1, what happens to $E_\text{cell}$ and spontaneity?
    Answer: $E_\text{cell}$ increases and becomes more positive (still spontaneous). This question assesses the skill of cell potential under nonstandard conditions. Increasing [Co²⁺], a reactant, to much larger than 1 M while keeping [Ni²⁺] at 1 M causes Q = [Ni²⁺]/[Co²⁺] to be less than 1. Since Q < 1, log Q is negative, resulting in a positive correction term in the Nernst equation. Therefore, Ecell increases above E°cell, becoming more positive and remaining spontaneous. A tempting distractor is that Ecell remains the same because increasing a reactant concentration does not change Q, but this is incorrect because it assumes reactants are not in the denominator of Q. To analyze similar problems, determine whether Q increases or decreases relative to 1, then infer how Ecell changes using Nernst logic.
18. A concentration cell is built with two $\text{Cu}(s)|\text{Cu}^{2+}(aq)$ half-cells connected by a salt bridge. Initially, one beaker contains $\text{Cu}^{2+}$ at $1.0\,\text{M}$ and the other contains $\text{Cu}^{2+}$ at a concentration greater than $1.0\,\text{M}$. Without using numerical calculations, apply qualitative Nernst reasoning to determine how the cell potential compares to standard conditions.
    Answer: $E_\text{cell}$ is positive but smaller than the standard value for a $\text{Cu}/\text{Cu}^{2+}$ cell. 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.
19. A galvanic cell runs with the overall reaction $\text{Pb}(s) + \text{Cl}_2(g) \rightarrow \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)$ At standard conditions ($P_{\text{Cl}_2}=1\,\text{atm}$ and aqueous ions at $1\,\text{M}$), the cell is spontaneous. The cell is prepared with $P_{\text{Cl}_2}$ much smaller than $1\,\text{atm}$ while $[\text{Pb}^{2+}]$ and $[\text{Cl}^-]$ are each $1\,\text{M}$. Using qualitative Nernst reasoning about $Q$ relative to 1, how does $E_{\text{cell}}$ change and does the reaction remain spontaneous?​
    Answer: $E_{\text{cell}}$ decreases but remains positive (still spontaneous). This problem tests understanding of cell potential under nonstandard conditions. The reaction quotient Q = [Pb²⁺][Cl⁻]²/PCl₂, and when PCl₂ is much smaller than 1 atm while ion concentrations remain at 1 M, Q becomes much greater than 1 (small denominator). According to the Nernst equation, when Q > 1, the term -(RT/nF)lnQ becomes negative, which decreases Ecell from its standard value. Since the standard cell is spontaneous (positive Ecell), and the decrease is finite for finite Q, the cell remains spontaneous but with reduced voltage. A misconception is that gases don't appear in Q (choice C), but gaseous species are included using their partial pressures in atm. The strategy is to write Q including all species (gases use pressure, aqueous use molarity), determine if Q increases or decreases from 1, then predict the corresponding change in Ecell.
20. A galvanic cell is made from $\text{Sn}(s)|\text{Sn}^{2+}(aq)$ and $\text{Pb}^{2+}(aq)|\text{Pb}(s)$. The net ionic equation for the spontaneous reaction under standard conditions is $\text{Sn}(s)+\text{Pb}^{2+}(aq)\rightarrow \text{Sn}^{2+}(aq)+\text{Pb}(s).$ Compared with standard conditions, the $\text{Pb}^{2+}$ concentration is increased above $1.0\,\text{M}$ while all other species remain at standard conditions. Using qualitative Nernst reasoning (via the change in $Q$), how does $E_\text{cell}$ change and does the cell remain spontaneous?
    Answer: $E_\text{cell}$ increases and remains positive (spontaneous). 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 $E_\text{cell}$.

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