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 $E_\text{cell}$.

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.

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 $E_\text{cell}$.

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 $E_\text{cell}$.

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.

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.

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.

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.

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.

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.