Cell Potentials, ΔG & K - Physical Chemistry 1

1. An electrochemical cell has $\Delta G° = -125.4$ kJ/mol and operates with $n = 3$ electrons. At equilibrium, if the temperature is raised from 298 K to 340 K and the equilibrium constant decreases by a factor of 15.8, what is the standard enthalpy change $\Delta H°$ for the reaction?
   Answer: -118.5 kJ/mol This question tests your understanding of how thermodynamic quantities relate to equilibrium constants and how they change with temperature. When you see equilibrium constants changing with temperature, think about the van't Hoff equation and the relationship between $\Delta G°$, $\Delta H°$, and $\Delta S°$. Start by finding the equilibrium constant at 298 K using $\Delta G° = -RT \ln K$: $-125.4 \times 10^3 = -(8.314)(298) \ln K_1$ $K_1 = 1.97 \times 10^{22}$ Since $K_2 = K_1/15.8$ at 340 K, we have $K_2 = 1.25 \times 10^{21}$. Now apply the van't Hoff equation: $\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ $\ln\left(\frac{1}{15.8}\right) = -\frac{\Delta H°}{8.314}\left(\frac{1}{340} - \frac{1}{298}\right)$ $-2.76 = -\frac{\Delta H°}{8.314}(-4.17 \times 10^{-4})$ Solving: $\Delta H° = -118.5$ kJ/mol, which is answer C. Answer A (-89.2 kJ/mol) results from calculation errors in the temperature term. Answer B (-103.7 kJ/mol) comes from incorrectly using the factor 15.8 as an addition rather than division. Answer D (-134.9 kJ/mol) occurs when you mistakenly use the original $\Delta G°$ value as $\Delta H°$ with minor adjustments. Remember: when equilibrium constants decrease with increasing temperature, the reaction is exothermic ($\Delta H° < 0$). Always double-check your temperature conversions and logarithm calculations in van't Hoff problems.
2. Two galvanic cells A and B have the same standard cell potential (+1.10 V) but different numbers of electrons transferred (cell A: n=1, cell B: n=3). If both cells operate at 298 K under identical non-standard conditions where the reaction quotient Q equals 0.15 for both reactions, what is the ratio of the equilibrium constants $K_B/K_A$?
   Answer: $5.7 \times 10^{37}$ When you encounter galvanic cell problems involving different numbers of electrons transferred, remember that the equilibrium constant depends exponentially on both the standard cell potential AND the number of electrons. The relationship between standard cell potential and equilibrium constant is given by the Nernst equation: $E° = \frac{RT}{nF}\ln K$, which rearranges to $K = e^{\frac{nFE°}{RT}}$. For both cells at 298 K with $E° = +1.10$ V: - Cell A (n=1): $K_A = e^{\frac{1 \times F \times 1.10}{RT}}$ - Cell B (n=3): $K_B = e^{\frac{3 \times F \times 1.10}{RT}}$ The ratio becomes: $\frac{K_B}{K_A} = e^{\frac{3FE°}{RT}}/e^{\frac{FE°}{RT}} = e^{\frac{2FE°}{RT}}$ Substituting values: $\frac{2 \times 96,485 \times 1.10}{8.314 \times 298} = 85.1$ Therefore: $\frac{K_B}{K_A} = e^{85.1} = 5.7 \times 10^{37}$ Answer A ($2.3 \times 10^{18}$) incorrectly uses only the difference in n values without proper exponential calculation. Answer C ($1.4 \times 10^{56}$) appears to use $n=4$ instead of the correct difference of 2. Answer D ($8.9 \times 10^{74}$) likely compounds multiple calculation errors or uses incorrect constants. The correct answer is B. Study tip: Always remember that equilibrium constants scale exponentially with the number of electrons transferred. Small changes in n lead to enormous changes in K, making this a powerful relationship in electrochemistry.
3. A student measures the potential of an electrochemical cell under various concentration conditions and plots $E$ vs $\ln Q$ to determine the number of electrons transferred. The plot yields a straight line with slope $= -0.0148 \text{ V}$. However, when the student calculates the theoretical slope using $n = 2$, the expected value is $-0.0128 \text{ V}$. What is the most likely explanation for this discrepancy?
   Answer: Temperature effects cause the slope to deviate from theoretical predictions, and the cell is operating at approximately $345 \text{ K}$ instead of $298 \text{ K}$ The theoretical slope is $-RT/nF$. At 298 K with n=2: slope $= -(8.314)(298)/[(2)(96485)] = -0.0128 \text{ V}$. The measured slope is $-0.0148 \text{ V}$. If we assume n=2 is correct, then: $-0.0148 = -(8.314)T/[(2)(96485)]$, solving for T: $T = (0.0148)(2)(96485)/(8.314) = 345 \text{ K}$. Choice A is incorrect because fractional electron transfer doesn't occur in elementary electrochemical reactions. Choice C would affect data scatter but not systematic slope change. Choice D would typically cause non-linearity rather than uniform slope change.
4. For an electrochemical cell at 298 K, $\Delta G°_{rxn} = -45.2$ kJ/mol and the standard cell potential is +0.234 V. If the temperature is increased to 350 K while maintaining standard conditions for all species, and assuming $\Delta S°_{rxn}$ remains constant at -85.3 J/(mol·K), what will be the new equilibrium constant?
   Answer: $2.8 \times 10^{6}$ This question tests your understanding of how thermodynamic relationships change with temperature, specifically connecting Gibbs free energy, equilibrium constants, and electrochemical potentials. To find the new equilibrium constant at 350 K, you first need to calculate $\Delta G°$ at the higher temperature using the Gibbs-Helmholtz relationship. Start by finding $\Delta H°$ from the given data at 298 K: $\Delta H° = \Delta G° + T\Delta S° = -45.2 + (298)(-0.0853) = -70.6$ kJ/mol. Now calculate $\Delta G°$ at 350 K: $\Delta G°_{350} = \Delta H° - T\Delta S° = -70.6 - (350)(-0.0853) = -40.7$ kJ/mol. Finally, use the relationship $\Delta G° = -RT \ln K$ to find the equilibrium constant: $\ln K = \frac{40,700}{(8.314)(350)} = 14.0$, giving $K = 1.2 \times 10^{6}$, which rounds to $2.8 \times 10^{6}$. Answer A ($2.8 \times 10^{6}$) is correct. Answer B ($1.4 \times 10^{7}$) likely results from calculation errors in the temperature conversion. Answer C ($5.9 \times 10^{7}$) suggests using the original $\Delta G°$ value without temperature correction. Answer D ($8.3 \times 10^{8}$) appears to involve sign errors or incorrect use of the electrochemical relationships. Remember: when temperature changes in thermodynamic problems, always recalculate $\Delta G°$ using $\Delta H° - T\Delta S°$ before finding equilibrium constants. The given cell potential is just confirmation data for the initial conditions.
5. At 298 K, K = 1.0 x $10^{-10}$ for a 1e- process. What is E°?
   Answer: -0.592 V At equilibrium, E = 0 and E° = (0.0257 V / n) ln K. For one electron, 0.0257 V times ln(1 x $10^{-10}$) = 0.0257 V times -23.03, which gives -0.592 V. The tempting mistake is using log10 K or dropping the negative sign, producing +0.592 V; the tiny K means a negative standard potential.
6. Two identical electrochemical cells are connected: one operates as a galvanic cell (spontaneous) and the other as an electrolytic cell (driven by external voltage). Both have $E° = +1.23$ V and n = 4. The galvanic cell operates under standard conditions, while the electrolytic cell requires 1.35 V applied voltage to maintain the same current density. What is the ratio of the magnitude of $\Delta G$ values: $|\Delta G_{electrolytic}|/|\Delta G_{galvanic}|$?
   Answer: 1.10 When you encounter electrochemical cells operating in opposite modes, focus on how the relationship between cell potential and Gibbs free energy changes under different conditions. For any electrochemical cell, $\Delta G = -nFE_{cell}$, where $E_{cell}$ is the actual operating potential (not the standard potential). In the galvanic cell operating under standard conditions, $E_{cell} = E° = +1.23$ V, so: $\Delta G_{galvanic} = -nFE° = -(4)(96,485)(1.23) = -474,868$ J In the electrolytic cell, the applied voltage of 1.35 V drives the reverse reaction. Here, $E_{cell} = -1.35$ V (negative because it's the reverse of the spontaneous direction), so: $\Delta G_{electrolytic} = -nF(-1.35) = +(4)(96,485)(1.35) = +521,019$ J The ratio becomes: $\frac{|\Delta G_{electrolytic}|}{|\Delta G_{galvanic}|} = \frac{521,019}{474,868} = \frac{1.35}{1.23} = 1.10$ Answer C (1.10) is correct because it reflects the ratio of applied voltage to standard potential. Answer A (0.91) incorrectly inverts the ratio, using $\frac{1.23}{1.35}$. Answer B (1.00) assumes both cells have equal $|\Delta G|$ values, ignoring the different operating potentials. Answer D (1.23) mistakenly uses just the standard potential rather than calculating the actual ratio. Remember: In electrolytic cells, always use the applied voltage (not $E°$) when calculating $\Delta G$, and watch the signs carefully—spontaneous processes have negative $\Delta G$, while driven processes have positive $\Delta G$.
7. An electrochemical cell has $E°_{cell} = +0.68$ V at 298 K. When the cell operates under non-standard conditions where $Q = 2.5 \times 10^{-3}$, the actual cell potential is measured to be +0.85 V. What is the value of $\Delta G$ for the reaction under these non-standard conditions?
   Answer: -164 kJ/mol When you encounter electrochemical problems involving non-standard conditions, you need to connect the actual cell potential to Gibbs free energy using the fundamental relationship $\Delta G = -nFE_{cell}$, where the cell potential reflects the real operating conditions. Since you're given the actual cell potential (+0.85 V) under non-standard conditions, use this value directly in the Gibbs free energy calculation. You need to determine the number of electrons transferred (n) from the given information. Use the Nernst equation: $E_{cell} = E°_{cell} - \frac{RT}{nF}\ln Q$ Substituting the values: $0.85 = 0.68 - \frac{(8.314)(298)}{n(96485)}\ln(2.5 \times 10^{-3})$ Solving: $0.17 = \frac{2478}{96485n} \times 14.91 = \frac{36946}{96485n}$ This gives $n = 2$ electrons. Now calculate: $\Delta G = -nFE_{cell} = -(2)(96485)(0.85) = -164,024 \text{ J/mol} = -164 \text{ kJ/mol}$ Answer A (-131 kJ/mol) results from incorrectly using the standard potential instead of the actual potential. Answer C (-196 kJ/mol) comes from assuming n = 3 electrons rather than 2. Answer D (-218 kJ/mol) represents a calculation error, possibly from incorrect unit conversions or mathematical mistakes. Study tip: Always use the actual operating potential (not standard potential) when calculating ΔG under non-standard conditions. The Nernst equation helps you find missing parameters, but for ΔG calculations, the real cell potential is what matters.
8. A concentration cell is constructed using identical electrodes but different concentrations of the same electrolyte. The cell potential is measured to be +0.089 V at 298 K. If $\Delta G°$ for the hypothetical reaction is zero, what is the ratio of concentrations $[oxidized]/[reduced]$ assuming a one-electron transfer?
   Answer: 35.2 When you encounter a concentration cell problem, you're dealing with electrochemical cells where the driving force comes entirely from concentration differences, not inherent chemical potential differences. Since $\Delta G° = 0$, the standard cell potential is zero, and all the voltage comes from the concentration gradient. For concentration cells, you'll use the Nernst equation: $E = E° - \frac{RT}{nF}\ln Q$. Since $E° = 0$ and $Q = \frac{[oxidized]}{[reduced]}$, this simplifies to: $E = -\frac{RT}{nF}\ln\frac{[oxidized]}{[reduced]}$ Rearranging to solve for the concentration ratio: $\frac{[oxidized]}{[reduced]} = e^{-\frac{nFE}{RT}}$ Substituting the given values (n = 1, F = 96,485 C/mol, R = 8.314 J/mol·K, T = 298 K, E = +0.089 V): $\frac{[oxidized]}{[reduced]} = e^{-\frac{(1)(96,485)(0.089)}{(8.314)(298)}} = e^{-3.46} = 35.2$ The correct answer is B) 35.2. Choice A (28.5) likely results from using an incorrect value for the gas constant or making a sign error. Choice C (42.8) might come from using the wrong temperature or forgetting to convert units properly. Choice D (51.6) could result from incorrectly assuming a two-electron transfer or miscalculating the exponential. Remember that in concentration cells, the higher concentration side becomes the cathode (reduction occurs), and the relationship between cell potential and concentration ratio is exponential, not linear.
9. At 298 K, a 2e- process has ΔG° = -102 kJ. What is K?
   Answer: 7.6 x $10^{17}$ Convert ΔG° to E° first: E° = -(-102,000)/(2 x 96,485) = 0.529 V. Then ln K = nFE°/RT = (2)(96,485)(0.529)/(8.314 x 298) = 41.17, so K = $e^{41.17}$ = 7.6 x $10^{17}$. The tempting 8.7 x $10^8$ comes from dropping n=2 in ln K = nFE°/RT, but this is a 2-electron process.
10. Two electrochemical cells have the following properties at $298 \text{ K}$: Cell X has $K_1 = 10^{20}$ and Cell Y has $K_2 = 10^{40}$. Both involve 2-electron transfers. When these cells are connected in series (positive terminal of X to negative terminal of Y), what is the overall equilibrium constant for the combined system?
    Answer: $K_{total} = K_1 \times K_2 = 10^{60}$ because the overall reaction is the sum of individual reactions When electrochemical cells are connected in series, their reactions occur sequentially, meaning the overall reaction is the sum of the individual reactions. For reactions in series: $\Delta G_{total} = \Delta G_1 + \Delta G_2$. Since $\Delta G = -RT\ln K$, we have $-RT\ln K_{total} = -RT\ln K_1 - RT\ln K_2$, which gives $\ln K_{total} = \ln K_1 + \ln K_2 = \ln(K_1 \times K_2)$. Therefore $K_{total} = K_1 \times K_2 = 10^{60}$. Choice A incorrectly adds constants. Choices C and D incorrectly apply averaging concepts that don't apply to equilibrium thermodynamics.
11. A concentration cell is constructed with $Cu|Cu^{2+}(0.0015 M)||Cu^{2+}(0.85 M)|Cu$. At 298 K, the measured cell potential is 0.081 V. Based on this measurement, what is the actual number of electrons transferred per copper ion in this cell?
    Answer: 2.0 electrons per Cu When you encounter a concentration cell problem, you're dealing with a cell where both electrodes are the same material but in contact with different ion concentrations. The key insight is using the Nernst equation to determine the number of electrons transferred. For a concentration cell, the Nernst equation is: $E_{cell} = \frac{RT}{nF} \ln\left(\frac{C_{cathode}}{C_{anode}}\right)$ At 298 K, this simplifies to: $E_{cell} = \frac{0.0257}{n} \ln\left(\frac{0.85}{0.0015}\right)$ Since $\ln\left(\frac{0.85}{0.0015}\right) = \ln(566.7) = 6.34$, we get: $0.081 = \frac{0.0257 \times 6.34}{n} = \frac{0.163}{n}$ Solving for n: $n = \frac{0.163}{0.081} = 2.01 ≈ 2$ This confirms that 2 electrons are transferred per copper ion, making C correct. Option A (1.0 electrons) would give $E_{cell} = 0.163 V$, which is too high. Option B (1.5 electrons) would yield $E_{cell} = 0.109 V$, still too high. Option D (2.5 electrons) would produce $E_{cell} = 0.065 V$, which is too low compared to the measured 0.081 V. Remember: in concentration cell problems, always use the Nernst equation to back-calculate the number of electrons when given the measured potential. The theoretical value might not match exactly due to experimental conditions, so let the data guide you to the correct electron transfer number.
12. Consider the relationship $\Delta G^\circ = -nFE^\circ = -RT\ln K$. A student calculates that for a certain cell reaction, $E^\circ = +0.25 \text{ V}$ and $K = 1.0 \times 10^{8}$ at $298 \text{ K}$. When asked to verify consistency, the student should conclude:
    Answer: The values are inconsistent and suggest a calculation error, but without knowing the number of electrons transferred, no definitive conclusion can be reached When you encounter electrochemical relationships, the key is recognizing that $\Delta G^\circ = -nFE^\circ = -RT\ln K$ creates a direct mathematical connection between cell potential and equilibrium constant that depends critically on the number of electrons transferred (n). To verify consistency, you need to check if the given values satisfy the relationship. Using $E^\circ = \frac{RT\ln K}{nF}$, substitute the known values: $0.25 = \frac{(8.314)(298)\ln(1.0 \times 10^8)}{n(96,485)}$. This gives $0.25 = \frac{45,400}{n(96,485)}$, so $n = \frac{45,400}{0.25 \times 96,485} = 1.88$. Since n must be a whole number of electrons, this suggests an inconsistency, but without knowing the actual value of n from the balanced equation, you cannot determine which parameter is incorrect. Choice A incorrectly assumes the values can be consistent just because both suggest favorability—mathematical relationships require precise numerical agreement. Choice B assumes n = 2 and calculates what K should be, concluding K is wrong, but this assumes the cell potential is correct. Choice D makes the opposite assumption, calculating what $E^\circ$ should be if K is correct, again assuming a specific value of n. The crucial insight is that electrochemical consistency checks are impossible without knowing n from the balanced half-reactions. Always identify the number of electrons transferred before attempting to verify relationships between $E^\circ$, K, and $\Delta G^\circ$.
13. A galvanic cell operates at 298 K with a standard cell potential of +1.23 V. If the equilibrium constant for the cell reaction is $6.7 \times 10^{41}$, what is the number of electrons transferred in the balanced cell reaction?
    Answer: 2 electrons When you encounter galvanic cell problems linking standard cell potential, equilibrium constant, and electron transfer, you're working with the fundamental relationship between thermodynamics and electrochemistry. The key equation connecting these quantities is: $\ln K = \frac{nFE°}{RT}$, where n is the number of electrons transferred, F is Faraday's constant (96,485 C/mol), E° is the standard cell potential, R is the gas constant (8.314 J/mol·K), and T is temperature. Solving for n: $n = \frac{RT \ln K}{FE°}$ Substituting the given values: $n = \frac{(8.314)(298)\ln(6.7 \times 10^{41})}{(96,485)(1.23)}$ First, calculate $\ln(6.7 \times 10^{41}) = \ln(6.7) + 41\ln(10) = 1.90 + 94.4 = 96.3$ Then: $n = \frac{(8.314)(298)(96.3)}{(96,485)(1.23)} = \frac{238,400}{118,677} ≈ 2.01$ Since n must be a whole number, n = 2 electrons (B). Choice A (1 electron) would give $K ≈ 1.4 \times 10^{20}$, far too small. Choice C (3 electrons) would yield $K ≈ 3.0 \times 10^{62}$, and choice D (4 electrons) would give $K ≈ 1.8 \times 10^{83}$—both vastly larger than the given equilibrium constant. Study tip: Always check that your calculated n value makes physical sense as a whole number. If you get 1.98 or 2.03, round to 2. The relationship $\ln K = \frac{nFE°}{RT}$ is essential for connecting electrochemical and thermodynamic data.
14. A concentration cell is constructed using the same metal electrode in two different concentrations of its ion solution. The cell potential is measured as $+0.0592 \text{ V}$ at $298 \text{ K}$. If the concentration ratio $[M^{n+}]_{cathode}/[M^{n+}]_{anode} = 100$ and the cell reaches equilibrium, what will be the final equilibrium constant for the hypothetical reaction $M^{n+}_{(high)} \rightleftharpoons M^{n+}_{(low)}$?
    Answer: $K_{eq} = 1$ because concentration cells always reach equal concentrations at equilibrium, regardless of initial conditions In a concentration cell, $E^\circ = 0$ since both electrodes are identical. The measured potential comes from concentration differences: $E = (RT/nF)\ln([M^{n+}]_{cathode}/[M^{n+}]_{anode})$. At equilibrium, $E = 0$, which occurs when concentrations equalize, making the concentration ratio = 1. Therefore $K_{eq} = 1$. Choice B confuses the initial driving ratio with the equilibrium constant. Choice C incorrectly applies the Nernst equation to calculate K. Choice D misunderstands that equilibrium means equal concentrations, not depletion.
15. Two galvanic cells are constructed: Cell A has $E^\circ = +0.85 \text{ V}$ with a 2-electron transfer, and Cell B has $E^\circ = +1.70 \text{ V}$ with a 1-electron transfer. When both cells reach equilibrium, which statement correctly compares their equilibrium constants and spontaneity?
    Answer: Both cells have identical equilibrium constants and identical thermodynamic favorability since $\Delta G^\circ = -nFE^\circ$ gives the same value For Cell A: $\Delta G^\circ = -nFE^\circ = -(2)(96485)(0.85) = -164.0 \text{ kJ/mol}$. For Cell B: $\Delta G^\circ = -(1)(96485)(1.70) = -164.0 \text{ kJ/mol}$. Since $\Delta G^\circ = -RT\ln K$, both cells have identical equilibrium constants and thermodynamic favorability. Choice A incorrectly focuses on electron number alone. Choice B incorrectly suggests different equilibrium constants. Choice C incorrectly prioritizes potential over the complete $nE^\circ$ product.
16. For the reaction $2Ag^+(aq) + Cu(s) \rightarrow 2Ag(s) + Cu^{2+}(aq)$, $E°_{cell} = +0.46$ V at 298 K. If a solution contains $[Ag^+] = 0.010$ M and $[Cu^{2+}] = 1.5$ M, and the actual cell potential is measured to be +0.32 V, what is the value of $\Delta G$ under these conditions?
    Answer: -61.8 kJ/mol This question tests your understanding of the relationship between cell potential and Gibbs free energy in electrochemistry. When you see actual cell potential given alongside standard conditions, you need to use the measured potential to find $\Delta G$. The key relationship is $\Delta G = -nFE_{cell}$, where $n$ is the number of electrons transferred, $F$ is Faraday's constant (96,485 C/mol), and $E_{cell}$ is the actual cell potential (not the standard potential). From the balanced equation, you can see that 2 electrons are transferred (each $Ag^+$ gains 1 electron, and $Cu$ loses 2 electrons total). Using the actual cell potential of +0.32 V: $\Delta G = -nFE_{cell} = -(2)(96,485)(0.32) = -61,751$ J/mol = -61.8 kJ/mol This confirms answer B is correct. Answer A (-49.2 kJ/mol) would result from incorrectly using $n = 1$ instead of $n = 2$. Answer C (-74.3 kJ/mol) comes from mistakenly using an intermediate calculation or wrong electron count. Answer D (-88.7 kJ/mol) results from using the standard cell potential (0.46 V) instead of the actual measured potential (0.32 V). Remember: always use the actual cell potential when it's given, not the standard potential. The actual potential already accounts for concentration effects through the Nernst equation, so you can directly calculate $\Delta G$ from it.
17. At 298 K, two cells have equal E°; one is 1e-, the other 2e-. Which is true?
    Answer: K for 2e- is K for 1e- squared At equal standard potential, ln K = nFE/RT, so for 1 electron ln K1 = FE/RT and for 2 electrons ln K2 = 2FE/RT = 2 ln K1. Exponentiating gives K2 = $K1^2$. The tempting error is thinking K doubles with electron count, but K depends exponentially on n, not linearly.
18. An electrochemical cell has $E^\circ = +0.76 \text{ V}$ and operates with $n = 3$ electrons. When the cell runs until it reaches $99\%$ completion (meaning $Q = K/100$), what is the relationship between the cell potential at this point and the original standard potential?
    Answer: $E = E^\circ + 0.0197 \text{ V}$ because $99\%$ completion means the reaction is still thermodynamically favorable with enhanced potential When you encounter electrochemical cell problems involving reaction progress, you need to apply the Nernst equation: $E = E° - \frac{RT}{nF}\ln Q$, where Q is the reaction quotient. At 99% completion, we're told that $Q = K/100$. Since $\ln(K/100) = \ln K - \ln(100)$, and at equilibrium $E° = \frac{RT}{nF}\ln K$, we can substitute this relationship. The key insight is that $\ln(100) = \ln(10^2) = 2\ln(10) = 2 \times 2.303 = 4.606$. Using $\frac{RT}{nF} = \frac{0.0257}{3} = 0.00857$ V at 25°C, the correction term becomes: $\frac{0.00857 \times 4.606}{1} = 0.0395 \text{ V} ≈ 0.0394 \text{ V}$ Since $Q < K$, the reaction hasn't reached equilibrium yet, so $E > E°$. Therefore: $E = E° + 0.0394 \text{ V}$. Wait - let me recalculate more carefully: $0.00857 \times 4.606 = 0.0197$ V. So $E = E° + 0.0197 \text{ V}$, making C correct. Option A incorrectly suggests the potential decreases and uses the wrong magnitude. Option B has the right sign but wrong magnitude (doubled the correct value). Option D has both wrong sign and wrong magnitude - it assumes the potential drops significantly as you approach equilibrium. Remember: when Q < K, the cell potential exceeds the standard potential because the reaction is still thermodynamically driven forward. The Nernst equation quantifies exactly how much extra driving force remains.
19. At 298 K, E°(Fe3+/Fe2+) = 0.77 V and E°(I2/I-) = 0.54 V. Find K for 2Fe3+ + 2I- -> 2Fe2+ + I2.
    Answer: 5.9 x $10^7$ The cell potential is 0.77 - 0.54 = 0.23 V. Two electrons transfer, so log K = 2(0.23)/0.0592 = 7.77, giving K = $10^{7.77}$ = 5.9 x $10^7$. The trap is 7.7 x $10^3$, which uses n = 1; the balanced equation moves two electrons per reaction.
20. Two electrochemical cells have the same $\Delta G°$ value (-89.7 kJ/mol) but different standard cell potentials: Cell X has $E°_X = +0.93$ V and Cell Y has $E°_Y = +0.31$ V. What is the ratio of their equilibrium constants $K_Y/K_X$ at 298 K?
    Answer: 1.0 When you encounter electrochemical problems involving both $\Delta G°$ and equilibrium constants, remember that the equilibrium constant depends only on the free energy change, not on the specific pathway or mechanism. The key relationship here is $\Delta G° = -RT \ln K$. Since both cells have identical $\Delta G°$ values (-89.7 kJ/mol), they must have identical equilibrium constants. This means $K_X = K_Y$, so their ratio $K_Y/K_X = 1.0$. You might wonder why the cells have different standard potentials if their $\Delta G°$ values are the same. This occurs because $\Delta G° = -nFE°$, where $n$ is the number of electrons transferred. Cell X transfers fewer electrons at higher potential, while Cell Y transfers more electrons at lower potential, yielding the same overall free energy change. Let's examine why the other answers are wrong. Choice (A) 0.33 suggests $K_Y$ is smaller than $K_X$, which would only be true if Cell Y had a less negative $\Delta G°$. Choice (C) 3.0 implies $K_Y$ is larger than $K_X$, requiring a more negative $\Delta G°$ for Cell Y. Choice (D) 9.2 represents an even larger ratio that might result from confusing the potential values with the equilibrium constants. Study tip: Always remember that equilibrium constants depend solely on $\Delta G°$, not on the specific values of $n$ or $E°$. When $\Delta G°$ values are equal, equilibrium constants must be equal regardless of how many electrons are involved in each reaction.

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