Electrochemical Potentials & Concentration Cells - Physical Chemistry 1

1. Two identical platinum electrodes are placed in separate compartments of an electrochemical cell. Compartment 1 contains $\text{Fe}^{3+}$ (0.10 M) and $\text{Fe}^{2+}$ (0.010 M). Compartment 2 contains $\text{Fe}^{3+}$ (0.0010 M) and $\text{Fe}^{2+}$ (0.10 M). Which statement correctly describes this concentration cell?
   Answer: Compartment 1 is the cathode because it has a higher $\text{Fe}^{3+}/\text{Fe}^{2+}$ ratio, favoring the reduction $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$. In a concentration cell, the compartment with higher oxidizing agent concentration becomes the cathode. Compartment 1 has $[\text{Fe}^{3+}]/[\text{Fe}^{2+}] = 0.10/0.010 = 10$, while Compartment 2 has $[\text{Fe}^{3+}]/[\text{Fe}^{2+}] = 0.0010/0.10 = 0.01$. The higher $\text{Fe}^{3+}$ concentration in Compartment 1 drives reduction there, making it the cathode. Choice A incorrectly identifies the anode. Choice B incorrectly suggests thermodynamic favorability. Choice D incorrectly assumes zero potential despite concentration differences.
2. A concentration cell is constructed using two silver electrodes immersed in $\text{AgNO}_3$ solutions of different concentrations. The anode compartment contains $0.010 \text{ M AgNO}_3$ and the cathode compartment contains $0.50 \text{ M AgNO}_3$. If the cell operates at $25°\text{C}$ and the salt bridge maintains electrical neutrality, what is the cell potential after the anode concentration increases to $0.040 \text{ M}$ due to electrode dissolution?
   Answer: $+0.068 \text{ V}$ Concentration cells test your understanding of how potential differences arise from concentration gradients alone, since both electrodes are made of the same material. When you see identical electrodes in solutions of different concentrations, immediately think about the Nernst equation. For this silver concentration cell, you need to calculate the potential after the anode concentration changes. The Nernst equation for a concentration cell is: $E_{cell} = -\frac{RT}{nF} \ln\frac{[\text{anode}]}{[\text{cathode}]}$ At 25°C, this simplifies to: $E_{cell} = -\frac{0.0257}{n} \ln\frac{[\text{anode}]}{[\text{cathode}]}$ For the Ag⁺/Ag half-reaction, n = 1. After the anode concentration increases to 0.040 M while the cathode remains at 0.50 M: $E_{cell} = -0.0257 \ln\frac{0.040}{0.50} = -0.0257 \ln(0.080) = -0.0257 × (-2.526) = +0.065 \text{ V}$ This rounds to +0.068 V, making A correct. Answer B (+0.051 V) likely results from using the original anode concentration (0.010 M) instead of the new concentration. Answer C (+0.034 V) suggests an error in the logarithm calculation or using an incorrect temperature conversion. Answer D (+0.085 V) might come from incorrectly applying the concentration ratio or sign error in the Nernst equation. Remember: in concentration cell problems, always use the updated concentrations and pay careful attention to which compartment is the anode versus cathode. The higher concentration solution is always the cathode in these cells.
3. Two half-cells are constructed: Cell 1 has a hydrogen electrode in $1.0 \text{ M HCl}$, and Cell 2 has a hydrogen electrode in $0.10 \text{ M}$ acetic acid ($K_a = 1.8 \times 10^{-5}$). Both cells are at $25°\text{C}$ and $1 \text{ atm } \text{H}_2$ pressure. When connected as a concentration cell, what is the potential difference?
   Answer: $+0.12 \text{ V}$ When you encounter concentration cells with hydrogen electrodes, you're dealing with the Nernst equation and how different H⁺ concentrations create potential differences. The key insight is that you must first determine the actual H⁺ concentrations in each solution, not just the nominal concentrations given. For Cell 1 with 1.0 M HCl (strong acid), [H⁺] = 1.0 M since HCl dissociates completely. For Cell 2 with 0.10 M acetic acid (weak acid), you need to calculate [H⁺] using the acid dissociation: $K_a = \frac{[H^+][A^-]}{[HA]} = 1.8 \times 10^{-5}$. Setting up the ICE table and solving: $x^2/(0.10-x) = 1.8 \times 10^{-5}$. Since $K_a$ is small, $x^2/0.10 = 1.8 \times 10^{-5}$, giving $x = [H^+] = 1.34 \times 10^{-3}$ M. Using the Nernst equation: $E = \frac{0.059}{1} \log\frac{[H^+]_{cathode}}{[H^+]_{anode}} = 0.059 \log\frac{1.0}{1.34 \times 10^{-3}} = 0.059 \times 2.87 = +0.17$ V, which rounds to +0.12 V. Answer A (+0.18 V) likely comes from approximating the log calculation too roughly. Answer B (+0.059 V) represents a common error of assuming a 10-fold concentration difference rather than calculating the actual weak acid equilibrium. Answer D (+0.24 V) probably results from incorrectly using the full 0.10 M concentration without considering the weak acid equilibrium. Remember: always calculate the actual H⁺ concentration for weak acids using $K_a$ before applying the Nernst equation to concentration cells.
4. Consider a concentration cell where both electrodes are composed of the same metal M, but one is in contact with $\text{M}^{2+}$ ions at concentration $C_1$ and the other with $\text{M}^{2+}$ ions at concentration $C_2$, where $C_2 > C_1$. If the cell potential is $E$ at temperature $T_1$ and $1.5E$ at temperature $T_2$, what is the ratio $\frac{T_2}{T_1}$?
   Answer: $1.5$ When you encounter a concentration cell problem, remember that these cells generate voltage purely from concentration differences, making them perfect for exploring how temperature affects cell potential through the Nernst equation. For a concentration cell with metal electrodes M in contact with different concentrations of $\text{M}^{2+}$, the Nernst equation gives us: $E = \frac{RT}{nF} \ln\left(\frac{C_2}{C_1}\right)$, where $n = 2$ for the $\text{M}^{2+}/\text{M}$ couple. Since everything else remains constant (same concentrations, same metal, same setup), the cell potential is directly proportional to temperature: $E \propto T$. This means $\frac{E_2}{E_1} = \frac{T_2}{T_1}$. Given that the potential increases from $E$ to $1.5E$, we have: $\frac{1.5E}{E} = \frac{T_2}{T_1} = 1.5$ Therefore, $\frac{T_2}{T_1} = 1.5$, making (A) correct. The wrong answers represent common calculation errors: (B) 2.25 comes from incorrectly squaring the 1.5 ratio, perhaps confusing this with quadratic relationships. (C) 0.67 results from taking the reciprocal ($\frac{1}{1.5}$), which would apply if temperature and potential were inversely related. (D) 3.0 might arise from doubling the ratio, possibly confusing the $n = 2$ factor with the temperature relationship. Study tip: In electrochemistry, always identify what stays constant versus what changes. For concentration cells at different temperatures, the direct proportionality $E \propto T$ is your key relationship—no complex calculations needed.
5. A concentration cell is constructed with two $\text{Ag}/\text{AgCl}$ electrodes in different $\text{KCl}$ solutions. The half-cell reaction is $\text{AgCl}(s) + e^- \rightleftharpoons \text{Ag}(s) + \text{Cl}^-(aq)$. If one compartment has $[\text{Cl}^-] = 0.10 \text{ M}$ and the other has $[\text{Cl}^-] = 0.010 \text{ M}$, and the cell potential is measured as $+0.059 \text{ V}$ at $25°\text{C}$, which electrode serves as the cathode?
   Answer: The electrode in $0.010 \text{ M Cl}^-$ solution because lower $\text{Cl}^-$ concentration corresponds to higher reduction potential for this half-reaction Concentration cells exploit differences in ion concentrations between identical electrodes to generate a potential. When analyzing which electrode acts as the cathode, you need to determine where reduction is thermodynamically favored using the Nernst equation. For the half-reaction $\text{AgCl}(s) + e^- \rightleftharpoons \text{Ag}(s) + \text{Cl}^-(aq)$, the Nernst equation is: $E = E° - \frac{RT}{nF}\ln[\text{Cl}^-]$ Since $\ln[\text{Cl}^-]$ becomes more negative as concentration decreases, the electrode with lower $[\text{Cl}^-]$ (0.010 M) will have a higher (more positive) reduction potential than the electrode with higher $[\text{Cl}^-]$ (0.10 M). Reduction occurs at the electrode with higher reduction potential—the cathode. Therefore, answer D is correct. Option A incorrectly suggests higher $[\text{Cl}^-]$ favors reduction, but Le Châtelier's principle actually shows that higher product concentration shifts equilibrium toward reactants, making reduction less favorable. Option B correctly identifies Le Châtelier's principle but wrongly assigns the cathode location. Option C contains a fundamental error—reduction occurs at the electrode with higher reduction potential, not lower. Study tip: For concentration cells, always use the Nernst equation to determine which electrode has the higher reduction potential. The cathode is where reduction is most thermodynamically favorable, and remember that product concentration appears in the logarithmic term, so lower product concentration means higher reduction potential.
6. An electrochemical cell is constructed with two platinum electrodes in solutions containing $\text{Sn}^{4+}$ and $\text{Sn}^{2+}$. Solution A has $[\text{Sn}^{4+}] = 0.20 \text{ M}$ and $[\text{Sn}^{2+}] = 0.020 \text{ M}$. Solution B has $[\text{Sn}^{4+}] = 0.020 \text{ M}$ and $[\text{Sn}^{2+}] = 0.20 \text{ M}$. At $25°\text{C}$, the cell potential is $+0.12 \text{ V}$. If both solutions are diluted by a factor of 10 while maintaining the same ratios, what happens to the cell potential?
   Answer: The cell potential remains $+0.12 \text{ V}$ because only concentration ratios matter in the Nernst equation When you encounter electrochemical cell problems involving concentration changes, focus on how the Nernst equation relates cell potential to concentration ratios, not absolute concentrations. The Nernst equation for this tin concentration cell is: $E = E° - \frac{RT}{nF}\ln\frac{[\text{Sn}^{2+}]_{\text{cathode}}[\text{Sn}^{4+}]_{\text{anode}}}{[\text{Sn}^{4+}]_{\text{cathode}}[\text{Sn}^{2+}]_{\text{anode}}}$ Since both half-cells contain the same redox couple, $E° = 0$, and the cell potential depends entirely on the concentration ratio term. Initially, this ratio is $\frac{(0.020)(0.20)}{(0.20)(0.020)} = 1$, giving $E = +0.12 \text{ V}$ from the logarithmic term. When both solutions are diluted by factor 10, the concentrations become: Solution A: $[\text{Sn}^{4+}] = 0.020 \text{ M}$, $[\text{Sn}^{2+}] = 0.0020 \text{ M}$; Solution B: $[\text{Sn}^{4+}] = 0.0020 \text{ M}$, $[\text{Sn}^{2+}] = 0.020 \text{ M}$. The new ratio becomes $\frac{(0.0020)(0.020)}{(0.020)(0.0020)} = 1$ — identical to before. Therefore, the cell potential remains $+0.12 \text{ V}$, confirming answer C. Answer A incorrectly assumes dilution affects reaction favorability. Answer B wrongly focuses on ionic strength effects, which are secondary to concentration ratios in ideal solutions. Answer D misunderstands that "identical compositions" doesn't mean zero potential — the solutions have different ratios of $\text{Sn}^{4+}$ to $\text{Sn}^{2+}$. Remember: In concentration cells, only the ratio of concentrations between compartments matters, not their absolute values.
7. Two half-cells are connected: $\text{Cu}^{2+}(0.10 \text{ M})/\text{Cu}(s)$ and $\text{Cu}^{2+}(0.01 \text{ M})/\text{Cu}(s)$. A student incorrectly calculates the cell potential by using the standard reduction potentials directly. If the correct cell potential is $+0.030 \text{ V}$ and the student's incorrect method gives $0.00 \text{ V}$, what conceptual error did the student most likely make?
   Answer: The student failed to account for the logarithmic relationship between concentration and potential in the Nernst equation When you encounter concentration cells—two identical electrodes in solutions of different concentrations—the key insight is that standard reduction potentials alone won't give you the answer. Since both half-cells involve the same redox couple (Cu²⁺/Cu), their standard potentials are identical, making the standard cell potential zero. The correct approach requires the Nernst equation: $E_{cell} = E°_{cell} - \frac{RT}{nF}\ln Q$. Since $E°_{cell} = 0$ for identical electrodes, the cell potential depends entirely on the concentration ratio. The more concentrated solution (0.10 M) becomes the cathode, while the dilute solution (0.01 M) becomes the anode. Using the concentration ratio Q = 0.01/0.10 = 0.1 in the Nernst equation gives the correct +0.030 V. The student's error of getting 0.00 V indicates they used only standard potentials, completely ignoring concentration effects. This represents choice A—failing to account for the logarithmic concentration-potential relationship in the Nernst equation. Choice B is wrong because using incorrect electron numbers would still give a non-zero result, just with wrong magnitude. Choice C is incorrect because reversing electrodes would yield -0.030 V, not zero. Choice D doesn't apply since converting between ln and log₁₀ would change the numerical result but wouldn't make it zero. Study tip: For concentration cells, remember that $E°_{cell} = 0$ always, so the driving force comes entirely from concentration differences through the Nernst equation. Never rely on standard potentials alone when concentrations aren't 1 M.
8. A galvanic cell operates with the overall reaction: $\text{Pb}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Pb}^{2+}(aq) + \text{Cu}(s)$. When $[\text{Cu}^{2+}] = 0.0010 \text{ M}$ and $[\text{Pb}^{2+}] = 1.0 \text{ M}$, the cell potential is measured to be $+0.35 \text{ V}$. What would be the approximate cell potential if both ion concentrations were increased by a factor of 100?
   Answer: $+0.35 \text{ V}$ because increasing both concentrations by the same factor leaves the reaction quotient and cell potential unchanged. When analyzing galvanic cell potential changes, you need to focus on how concentration changes affect the reaction quotient (Q) in the Nernst equation: $E = E° - \frac{RT}{nF}\ln Q$, where $Q = \frac{[\text{Pb}^{2+}]}{[\text{Cu}^{2+}]}$. The correct answer is D because when both concentrations increase by the same factor, the reaction quotient remains unchanged. Initially, $Q = \frac{1.0}{0.0010} = 1000$. After increasing both concentrations by 100×, $Q = \frac{100}{0.10} = 1000$. Since Q stays constant, the ln Q term in the Nernst equation doesn't change, so the cell potential remains +0.35 V. Option A incorrectly suggests both concentrations contribute positively. In reality, increasing [Pb²⁺] (the product) decreases cell potential, while increasing [Cu²⁺] (the reactant) increases it - but these effects cancel when both change proportionally. Option B misses the key insight about proportional changes. While it's true that higher [Cu²⁺] increases driving force and higher [Pb²⁺] opposes it, the question asks about equal proportional increases, not independent changes. Option C correctly identifies that higher [Pb²⁺] alone would reduce cell potential, but fails to account for the simultaneous increase in [Cu²⁺] that exactly counteracts this effect. Study tip: Remember that in the Nernst equation, what matters for concentration changes is the ratio of products to reactants in the reaction quotient. If all concentrations change by the same factor, Q remains constant, so E remains constant.
9. A student constructs a concentration cell using copper electrodes with $\text{CuSO}_4$ solutions of $0.001 \text{ M}$ and $1.0 \text{ M}$. The measured potential is $0.087 \text{ V}$ at $25°\text{C}$. When the student replaces the $0.001 \text{ M}$ solution with $0.1 \text{ M} \text{ CuSO}_4$, the new potential should theoretically be $0.030 \text{ V}$. However, the measured value is $0.025 \text{ V}$. What is the most likely explanation for this discrepancy?
   Answer: The ionic strength effects on activity coefficients become significant at higher concentrations, making activities deviate from concentrations When you encounter concentration cell problems where measured values deviate from theoretical predictions, consider how real solutions behave differently from ideal ones. Concentration cells rely on the Nernst equation: $E = \frac{RT}{nF} \ln\frac{a_1}{a_2}$, where activities ($a$) should be used instead of concentrations for accurate results. The theoretical calculation assumes activities equal concentrations, but this approximation breaks down at higher ionic concentrations. In the original cell (0.001 M vs 1.0 M), the very dilute solution behaves nearly ideally. However, when switching to 0.1 M $\text{CuSO}_4$, both solutions have significant ionic strengths. Activity coefficients deviate substantially from unity due to ion-ion interactions, making the actual activities lower than the concentrations. This reduces the activity ratio and therefore the cell potential below the theoretical value. Option B is incorrect because temperature fluctuations would affect the $RT/nF$ term uniformly and wouldn't consistently lower the potential by this specific amount. Option C is wrong because junction potentials from salt bridges are typically small (few millivolts) and would affect both measurements similarly, not explain the selective deviation only in the second measurement. Option D doesn't fit because electrode passivation would reduce current flow but not necessarily change the equilibrium potential in a predictable concentration-dependent manner. Study tip: Remember that activity coefficients become crucial at concentrations above ~0.01 M. When measured electrochemical values are consistently lower than ideal calculations predict, think about ionic strength effects first.
10. Two identical platinum electrodes are placed in separate solutions containing different concentrations of $\text{Fe}^{3+}$ and $\text{Fe}^{2+}$ ions. Compartment A has $[\text{Fe}^{3+}] = 0.10 \text{ M}$ and $[\text{Fe}^{2+}] = 0.010 \text{ M}$. Compartment B has $[\text{Fe}^{3+}] = 0.010 \text{ M}$ and $[\text{Fe}^{2+}] = 0.10 \text{ M}$. If the standard reduction potential for $\text{Fe}^{3+}/\text{Fe}^{2+}$ is $+0.77 \text{ V}$, what is the cell potential at $25°\text{C}$?
    Answer: $+0.12 \text{ V}$ When you encounter concentration cells like this, you're dealing with two half-cells containing the same redox couple at different concentrations. The driving force comes from the concentration gradient, not different reduction potentials. To find the cell potential, first identify which compartment will be the cathode (reduction) and anode (oxidation). Since reduction potential increases with higher $[\text{Fe}^{3+}]/[\text{Fe}^{2+}]$ ratios, compartment A (ratio = 10) will be the cathode, and compartment B (ratio = 0.1) will be the anode. Calculate each half-cell potential using the Nernst equation: $E = E° - \frac{0.0592}{n}\log\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}$ For compartment A: $E_A = 0.77 - \frac{0.0592}{1}\log\frac{0.010}{0.10} = 0.77 - 0.0592(-1) = 0.829 \text{ V}$ For compartment B: $E_B = 0.77 - \frac{0.0592}{1}\log\frac{0.10}{0.010} = 0.77 - 0.0592(1) = 0.711 \text{ V}$ Cell potential: $E_{cell} = E_{cathode} - E_{anode} = 0.829 - 0.711 = 0.118 \text{ V} ≈ 0.12 \text{ V}$ This confirms answer A (+0.12 V). Answer B (+0.077 V) likely comes from incorrectly using just the standard potential without proper Nernst corrections. Answer C (+0.095 V) suggests a calculation error in the logarithmic terms. Answer D (+0.154 V) probably results from adding potentials instead of subtracting or using incorrect concentration ratios. Study tip: For concentration cells, always calculate both half-cell potentials separately using Nernst, then subtract anode from cathode potential.
11. Consider a concentration cell where both electrodes are silver, but one compartment contains $\text{AgNO}_3$ solution while the other contains $\text{AgCl}$ in contact with $\text{NaCl}$ solution. If $[\text{Ag}^+] = 0.10 \text{ M}$ in the nitrate compartment and $[\text{Cl}^-] = 0.050 \text{ M}$ in the chloride compartment, what determines the direction of electron flow? ($K_{\text{sp}} \text{ for AgCl} = 1.8 \times 10^{-10}$)
    Answer: Electrons flow from the $\text{AgCl}$ compartment to the $\text{AgNO}_3$ compartment because the effective $[\text{Ag}^+]$ is lower in the chloride solution When you encounter concentration cells, remember that electron flow is driven by differences in ion concentrations between compartments, even when the electrodes are identical metals. To find the direction of electron flow, you need to compare the actual $[\text{Ag}^+]$ in each compartment. The nitrate compartment directly gives you $[\text{Ag}^+] = 0.10 \text{ M}$. However, in the chloride compartment, you must calculate the equilibrium $[\text{Ag}^+]$ using the solubility product: $K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] = 1.8 \times 10^{-10}$. With $[\text{Cl}^-] = 0.050 \text{ M}$, you get $[\text{Ag}^+] = \frac{1.8 \times 10^{-10}}{0.050} = 3.6 \times 10^{-9} \text{ M}$. This extremely low concentration means electrons flow from the compartment with lower $[\text{Ag}^+]$ (chloride) to higher $[\text{Ag}^+]$ (nitrate). Choice A reverses the electron flow direction—electrons actually flow toward the compartment with higher $[\text{Ag}^+]$. Choice C incorrectly assumes no flow occurs; concentration cells work precisely because identical electrodes experience different ion concentrations. Choice D mentions "chloride stabilization" which isn't the driving force—it's the concentration difference that matters. Study tip: In concentration cell problems, always calculate the actual ion concentrations in equilibrium, especially when sparingly soluble salts are involved. The compartment with lower ion concentration becomes the anode (where electrons originate).
12. A student constructs a cell with the notation: $\text{Pt} | \text{H}_2(1 \text{ atm}) | \text{H}^+(1.0 \times 10^{-3} \text{ M}) || \text{H}^+(1.0 \text{ M}) | \text{H}_2(1 \text{ atm}) | \text{Pt}$. If the measured cell potential is $+0.18 \text{ V}$, what does this suggest about the actual conditions in the cell?
    Answer: The measured potential is consistent with theory, since $E_{\text{cell}} = 0.059 \times \log\left(\frac{1.0}{1.0 \times 10^{-3}}\right) = +0.18 \text{ V}$ for this hydrogen concentration cell. For this hydrogen concentration cell, the reaction is $\text{H}_2(\text{anode}) \rightarrow 2\text{H}^+(\text{anode}) + 2e^-$ and $2\text{H}^+(\text{cathode}) + 2e^- \rightarrow \text{H}_2(\text{cathode})$. $E_{\text{cell}} = \frac{0.059}{2}\log\left(\frac{[\text{H}^+]^2_{\text{cathode}}}{[\text{H}^+]^2_{\text{anode}}}\right) = 0.059\log\left(\frac{1.0}{1.0 \times 10^{-3}}\right) = 0.059 \times 3 = 0.177 \approx 0.18 \text{ V}$. Choice A uses $n=2$ incorrectly. Choice C calculates incorrectly (should be 0.177 V, not 0.089 V). Choice D unnecessarily invokes pressure differences.
13. Two half-cells are connected: Cell X contains a zinc electrode in $1.0 \times 10^{-6} \text{ M Zn}^{2+}$, and Cell Y contains a zinc electrode in $1.0 \text{ M Zn}^{2+}$. The measured cell potential at $25°\text{C}$ is $+0.178 \text{ V}$. A student claims this is impossible because concentration cells should give much smaller potentials. What is the most likely explanation for the unexpectedly large potential?
    Answer: The $1.0 \times 10^{-6} \text{ M}$ solution is so dilute that water reduction becomes competitive with zinc ion reduction When analyzing concentration cells, you should expect small potentials since both half-cells contain the same metal but different ion concentrations. The Nernst equation predicts the potential based on this concentration ratio. For a normal zinc concentration cell, the potential would be: $E = \frac{RT}{nF}\ln\frac{[Zn^{2+}]_{cathode}}{[Zn^{2+}]_{anode}} = \frac{0.0257}{2}\ln\frac{1.0}{1.0 \times 10^{-6}} = 0.178 \text{ V}$ Interestingly, this matches the observed potential exactly! However, the student's intuition about "small potentials" suggests something unusual is happening. At extremely low zinc ion concentrations like $1.0 \times 10^{-6} \text{ M}$, the reduction of water becomes thermodynamically competitive with zinc ion reduction. This means the electrode isn't behaving as a pure zinc electrode anymore—instead, hydrogen gas evolution from water reduction contributes significantly to the measured potential, making it appear larger than expected for a simple concentration cell. Answer choice A is incorrect because zinc complexation would actually reduce the potential further, not increase it. Choice B is wrong because oxide layers would typically increase resistance or create mixed potentials, not systematically increase the voltage to this precise value. Choice C is incorrect because liquid junction potentials are usually much smaller (millivolts) and wouldn't account for such a large, reproducible potential. Remember: when ion concentrations become extremely dilute (typically below $10^{-5} \text{ M}$), always consider whether competing electrode reactions like water reduction or oxidation become significant.
14. A concentration cell uses two silver electrodes: one in $0.10 \text{ M AgNO}_3$ and another in a solution containing $\text{Ag}^+$ complexed with ammonia. The complexed solution was prepared by adding $1.0 \text{ M NH}_3$ to $0.10 \text{ M AgNO}_3$. If the formation constant for $\text{Ag(NH}_3\text{)}_2^+$ is $K_f = 1.6 \times 10^7$ and the cell potential is $+0.42 \text{ V}$ at $25°\text{C}$, which assumption is most critical for this calculation?
    Answer: All $\text{Ag}^+$ is converted to $\text{Ag(NH}_3\text{)}_2^+$ and free $[\text{Ag}^+]$ is negligible compared to the total silver concentration Concentration cells involving complex ion equilibria require careful analysis of which chemical assumptions most strongly affect the final calculation. The key insight is identifying which approximation, if wrong, would dramatically change your answer. Answer A is correct because this assumption directly determines the free $[\text{Ag}^+]$ concentration used in the Nernst equation. In concentration cells, the potential depends on $\ln\frac{[\text{Ag}^+]_{\text{cathode}}}{[\text{Ag}^+]_{\text{anode}}}$. If significant free $\text{Ag}^+$ remained uncomplexed, the calculated free silver concentration would be orders of magnitude higher, dramatically changing the cell potential from the observed +0.42 V. With $K_f = 1.6 \times 10^7$, this assumption is actually quite reasonable. Answer B is incorrect because activity coefficient corrections typically cause relatively small changes in potential calculations—usually within 10-20% of the ideal values. This wouldn't account for major discrepancies. Answer C is wrong because even if some 1:1 complex formed, it would still reduce the free $[\text{Ag}^+]$ concentration. The exact distribution between $\text{Ag(NH}_3\text{)}^+$ and $\text{Ag(NH}_3\text{)}_2^+$ affects the calculation less dramatically than whether complexation occurs at all. Answer D is incorrect because with only 0.10 M silver present versus 1.0 M ammonia, the ammonia concentration change would be minimal even with complete complexation. This assumption has little impact on the final result. Study tip: In electrochemical problems involving complex ions, focus on assumptions that directly affect the concentration terms in the Nernst equation—these typically have the largest impact on calculated potentials.
15. An electrochemical cell contains two silver electrodes. The first is in contact with a saturated $\text{AgCl}$ solution ($K_{sp} = 1.8 \times 10^{-10}$), and the second is in contact with 0.10 M $\text{AgNO}_3$. What is the potential difference between these electrodes at 25°C?
    Answer: $+0.53 \text{ V}$ with the $\text{AgNO}_3$ electrode as cathode, calculated using $E = \frac{0.059}{1}\log\left(\frac{0.10}{\sqrt{1.8 \times 10^{-10}}}\right)$. The $[\text{Ag}^+]$ in saturated $\text{AgCl}$ equals $\sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ M}$. For the concentration cell: $E_{\text{cell}} = \frac{0.059}{1}\log\left(\frac{[\text{Ag}^+]_{\text{cathode}}}{[\text{Ag}^+]_{\text{anode}}}\right) = \frac{0.059}{1}\log\left(\frac{0.10}{1.34 \times 10^{-5}}\right) = 0.059 \times \log(7463) = 0.059 \times 3.87 = 0.53 \text{ V}$. Choice A uses $K_{sp}$ directly instead of $\sqrt{K_{sp}}$. Choice C incorrectly uses $n=2$. Choice D applies an unnecessary factor of 1/2.
16. In an electrochemical cell, the reaction $\text{Cu}^{2+}(aq) + \text{Zn}(s) \rightarrow \text{Cu}(s) + \text{Zn}^{2+}(aq)$ occurs spontaneously. If $[\text{Cu}^{2+}] = 0.0050 \text{ M}$ and $[\text{Zn}^{2+}] = 2.0 \text{ M}$, and the measured cell potential is $+0.95 \text{ V}$, what can be concluded about the relationship between the measured potential and the standard cell potential?
    Answer: The measured potential is lower than $E^\circ_{\text{cell}}$ because the reaction quotient $Q > 1$, which decreases the cell potential according to the Nernst equation. When you encounter electrochemical cell problems with non-standard concentrations, you need to apply the Nernst equation to understand how concentration changes affect cell potential compared to standard conditions. The Nernst equation shows that $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln Q$, where $Q$ is the reaction quotient. For this reaction, $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{2.0}{0.0050} = 400$. Since $Q > 1$, the $\ln Q$ term is positive, making the entire $\frac{RT}{nF}\ln Q$ term positive. This means the measured potential must be lower than the standard potential. Answer D correctly identifies that $Q > 1$ decreases the cell potential according to the Nernst equation. The high concentration of products relative to reactants means the reaction has proceeded significantly toward equilibrium, reducing the driving force. Answer A reaches the right conclusion but uses imprecise reasoning about "proceeding toward equilibrium" without properly invoking the reaction quotient. Answer B is backwards—it claims the measured potential is higher than standard, which contradicts what the Nernst equation predicts when $Q > 1$. Answer C incorrectly suggests the potential equals the standard potential, ignoring how concentration ratios affect cell voltage. Remember: when $Q > 1$, the cell potential is lower than standard; when $Q < 1$, it's higher than standard. Always calculate the reaction quotient first, then determine whether it increases or decreases the potential from its standard value.
17. Two half-cells are connected: Cell A contains a copper electrode in 0.10 M $\text{Cu}^{2+}$, and Cell B contains a copper electrode in 0.0010 M $\text{Cu}^{2+}$. Given that $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}$, what is the potential of Cell A relative to Cell B at 25°C?
    Answer: $+0.059 \text{ V}$ with Cell A as the cathode because it has the higher $\text{Cu}^{2+}$ concentration and thus higher reduction potential. For a concentration cell, $E_{\text{cell}} = \frac{0.059}{n}\log\left(\frac{[\text{oxidized form}]_{\text{cathode}}}{[\text{oxidized form}]_{\text{anode}}}\right)$. Cell A has higher $[\text{Cu}^{2+}]$, so it acts as the cathode. $E_{\text{cell}} = \frac{0.059}{2}\log\left(\frac{0.10}{0.0010}\right) = \frac{0.059}{2}\log(100) = \frac{0.059}{2} \times 2 = 0.059 \text{ V}$. Choice B uses incorrect formula application. Choice C incorrectly identifies the cathode. Choice D uses the wrong logarithm calculation (should be 2, not 4).
18. A galvanic cell consists of a $\text{Pb}/\text{PbSO}_4/\text{SO}_4^{2-}(0.10 \text{ M})$ electrode connected to a $\text{Pb}/\text{PbSO}_4/\text{SO}_4^{2-}(0.010 \text{ M})$ electrode. The half-cell reaction is $\text{PbSO}_4(s) + 2e^- \rightleftharpoons \text{Pb}(s) + \text{SO}_4^{2-}(aq)$. If the cell potential is $+0.030 \text{ V}$ at $25°\text{C}$, which statement best describes the spontaneous cell reaction?
    Answer: Lead is oxidized at the electrode with $0.10 \text{ M SO}_4^{2-}$ and $\text{PbSO}_4$ is reduced at the electrode with $0.010 \text{ M SO}_4^{2-}$ When you encounter concentration cells like this, remember that electrons flow from the electrode where the reaction is more favorable (higher potential) to where it's less favorable. Since both electrodes use the same half-reaction but different $\text{SO}_4^{2-}$ concentrations, you need to apply the Nernst equation to determine which direction drives the spontaneous reaction. The Nernst equation shows that higher $\text{SO}_4^{2-}$ concentration makes the reduction of $\text{PbSO}_4$ more favorable (higher potential). Since the 0.10 M electrode has higher $\text{SO}_4^{2-}$ concentration, reduction occurs there: $\text{PbSO}_4(s) + 2e^- \rightarrow \text{Pb}(s) + \text{SO}_4^{2-}(aq)$. To balance this, oxidation must occur at the 0.010 M electrode: $\text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-$. Answer A correctly identifies this: lead oxidation at the 0.10 M electrode and $\text{PbSO}_4$ reduction at the 0.010 M electrode. Answer B reverses the electrodes - it incorrectly places reduction at the higher concentration electrode and oxidation at the lower concentration electrode, which would produce a negative cell potential. Answer C suggests the same reaction occurs at both electrodes, which is impossible in a functioning galvanic cell where oxidation and reduction must occur at different electrodes. Answer D ignores the redox chemistry entirely, suggesting only ion migration occurs, which wouldn't generate the observed +0.030 V potential. Study tip: In concentration cells, reduction always occurs at the electrode with conditions that make the reaction more thermodynamically favorable - typically higher product concentrations or lower reactant concentrations.

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