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 $\text{Sn}(s) + \text{Pb}^{2+}(aq) \rightarrow \text{Sn}^{2+}(aq) + \text{Pb}(s)$, 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 $$ \text{Co}(s) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Co}^{2+}(aq) + 2\text{Fe}^{2+}(aq) $$, 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. 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 [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.

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.

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 $\text{Fe}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Fe}^{2+}(aq) + 2\text{Ag}(s)$, 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. 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.