Physical Chemistry 1 Quiz: Cell Potentials Delta G And K
20 questions · exam conditions
0:00
Cell Potentials Delta G And KQuestion 1 of 20

At 298 K, a 2e- process has ΔG° = -102 kJ. What is K?

1.2 x 10^-18
8.7 x 10^8
1.5 x 10^35
7.6 x 10^17
← Back to quizzes

Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Cell Potentials Delta G And K

Practice Cell Potentials Delta G And K in Physical Chemistry 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Cell Potentials Delta G And K, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

At 298 K, a 2e- process has ΔG° = -102 kJ. What is K?

  1. 1.2 x 10^-18
  2. 8.7 x 10^8
  3. 1.5 x 10^35
  4. 7.6 x 10^17 (correct answer)
Explanation: 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.

Question 2

At 298 K, K = 1.0 x 10^-10 for a 1e- process. What is E°?

  1. -0.296 V
  2. +0.592 V
  3. -0.592 V (correct answer)
  4. -1.184 V
Explanation: 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.

Question 3

At 298 K, two cells have equal E°; one is 1e-, the other 2e-. Which is true?

  1. K for 2e- equals K for 1e-
  2. K for 2e- is K for 1e- squared (correct answer)
  3. K for 2e- is 2 times K for 1e-
  4. K for 2e- is half K for 1e-
Explanation: 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.

Question 4

At 298 K, E°(Fe3+/Fe2+) = 0.77 V and E°(I2/I-) = 0.54 V. Find K for 2Fe3+ + 2I- -> 2Fe2+ + I2.

  1. 5.9 x 10^7 (correct answer)
  2. 7.7 x 10^3
  3. 1.9 x 10^44
  4. 1.7 x 10^-8
Explanation: 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.

Question 5

A 2e- cell has ΔG° = -27.7 kJ. What is E°?

  1. +0.144 V (correct answer)
  2. -0.144 V
  3. +0.287 V
  4. +0.0718 V
Explanation: Use delta G = -n F E. Convert -27.7 kJ to -27,700 J, then E = -delta G / (nF) = 27,700 / (2 x 96,485) = +0.144 V. A negative delta G means a spontaneous cell, so E must be positive. The tempting wrong answer is -0.144 V, which comes from dropping the minus sign in the relation.

Question 6

An electrochemical cell has K298K=4.7×1018K_{298K} = 4.7 \times 10^{18} and K350K=8.9×1015K_{350K} = 8.9 \times 10^{15}. If the standard cell potential at 350 K is +0.47 V, what is the standard cell potential at 298 K?

  1. +0.52 V
  2. +0.56 V (correct answer)
  3. +0.61 V
  4. +0.67 V
  5. +0.73 V
Explanation: When you see equilibrium constants at different temperatures alongside cell potentials, you're dealing with the fundamental relationship between thermodynamics and electrochemistry. The key connection is that both equilibrium constants and cell potentials relate to the same Gibbs free energy change: ΔG°=−RTln⁡K=−nFE°\Delta G° = -RT \ln K = -nFE°. Since both expressions equal ΔG°\Delta G°, you can set them equal: −RTln⁡K=−nFE°-RT \ln K = -nFE°, which gives us E°=RTln⁡KnFE° = \frac{RT \ln K}{nF}. To find the number of electrons transferred, use the data at 350 K: 0.47=(8.314)(350)ln⁡(8.9×1015)n(96485)0.47 = \frac{(8.314)(350) \ln(8.9 \times 10^{15})}{n(96485)} Solving: n=(8.314)(350)(36.11)(0.47)(96485)=2.33≈2n = \frac{(8.314)(350)(36.11)}{(0.47)(96485)} = 2.33 \approx 2 electrons. Now calculate E°E° at 298 K: E°298=(8.314)(298)ln⁡(4.7×1018)(2)(96485)=(2478)(43.08)192970=0.56 VE°_{298} = \frac{(8.314)(298) \ln(4.7 \times 10^{18})}{(2)(96485)} = \frac{(2478)(43.08)}{192970} = 0.56 \text{ V} Choice A (+0.52 V) results from calculation errors or using the wrong temperature relationship. Choice C (+0.61 V) likely comes from incorrectly assuming n = 1 electron instead of 2. Choice D (+0.67 V) represents a more significant computational error, possibly in the logarithm calculation or unit conversion. Study tip: Always verify your electron count using the given data at one temperature before calculating the unknown. The relationship E°=RTln⁡KnFE° = \frac{RT \ln K}{nF} is your bridge between equilibrium and electrochemistry.

Question 7

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 6.7×10416.7 \times 10^{41}, what is the number of electrons transferred in the balanced cell reaction?

  1. 1 electron
  2. 2 electrons (correct answer)
  3. 3 electrons
  4. 4 electrons
  5. 6 electrons
Explanation: 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=nFE°RT\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=RTln⁡KFE°n = \frac{RT \ln K}{FE°} Substituting the given values: n=(8.314)(298)ln⁡(6.7×1041)(96,485)(1.23)n = \frac{(8.314)(298)\ln(6.7 \times 10^{41})}{(96,485)(1.23)} First, calculate ln⁡(6.7×1041)=ln⁡(6.7)+41ln⁡(10)=1.90+94.4=96.3\ln(6.7 \times 10^{41}) = \ln(6.7) + 41\ln(10) = 1.90 + 94.4 = 96.3 Then: n=(8.314)(298)(96.3)(96,485)(1.23)=238,400118,677≈2.01n = \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≈1.4×1020K ≈ 1.4 \times 10^{20}, far too small. Choice C (3 electrons) would yield K≈3.0×1062K ≈ 3.0 \times 10^{62}, and choice D (4 electrons) would give K≈1.8×1083K ≈ 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=nFE°RT\ln K = \frac{nFE°}{RT} is essential for connecting electrochemical and thermodynamic data.

Question 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 ΔG°\Delta G° for the hypothetical reaction is zero, what is the ratio of concentrations [oxidized]/[reduced][oxidized]/[reduced] assuming a one-electron transfer?

  1. 28.5
  2. 35.2 (correct answer)
  3. 42.8
  4. 51.6
  5. 67.9
Explanation: 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 ΔG°=0\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°−RTnFln⁡QE = E° - \frac{RT}{nF}\ln Q. Since E°=0E° = 0 and Q=[oxidized][reduced]Q = \frac{[oxidized]}{[reduced]}, this simplifies to: E=−RTnFln⁡[oxidized][reduced]E = -\frac{RT}{nF}\ln\frac{[oxidized]}{[reduced]} Rearranging to solve for the concentration ratio: [oxidized][reduced]=e−nFERT\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): [oxidized][reduced]=e−(1)(96,485)(0.089)(8.314)(298)=e−3.46=35.2\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.

Question 9

For the reaction 2Ag+(aq)+Cu(s)→2Ag(s)+Cu2+(aq)2Ag^+(aq) + Cu(s) \rightarrow 2Ag(s) + Cu^{2+}(aq), E°cell=+0.46E°_{cell} = +0.46 V at 298 K. If a solution contains [Ag+]=0.010[Ag^+] = 0.010 M and [Cu2+]=1.5[Cu^{2+}] = 1.5 M, and the actual cell potential is measured to be +0.32 V, what is the value of ΔG\Delta G under these conditions?

  1. -49.2 kJ/mol
  2. -61.8 kJ/mol (correct answer)
  3. -74.3 kJ/mol
  4. -88.7 kJ/mol
  5. -102.1 kJ/mol
Explanation: 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 ΔG\Delta G. The key relationship is ΔG=−nFEcell\Delta G = -nFE_{cell}, where nn is the number of electrons transferred, FF is Faraday's constant (96,485 C/mol), and EcellE_{cell} is the actual cell potential (not the standard potential). From the balanced equation, you can see that 2 electrons are transferred (each Ag+Ag^+ gains 1 electron, and CuCu loses 2 electrons total). Using the actual cell potential of +0.32 V: ΔG=−nFEcell=−(2)(96,485)(0.32)=−61,751\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=1n = 1 instead of n=2n = 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 ΔG\Delta G from it.

Question 10

An electrochemical cell has ΔG°=−125.4\Delta G° = -125.4 kJ/mol and operates with n=3n = 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 ΔH°\Delta H° for the reaction?

  1. -89.2 kJ/mol
  2. -103.7 kJ/mol
  3. -118.5 kJ/mol (correct answer)
  4. -134.9 kJ/mol
  5. -151.3 kJ/mol
Explanation: 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 ΔG°\Delta G°, ΔH°\Delta H°, and ΔS°\Delta S°. Start by finding the equilibrium constant at 298 K using ΔG°=−RTln⁡K\Delta G° = -RT \ln K: −125.4×103=−(8.314)(298)ln⁡K1-125.4 \times 10^3 = -(8.314)(298) \ln K_1 K1=1.97×1022K_1 = 1.97 \times 10^{22} Since K2=K1/15.8K_2 = K_1/15.8 at 340 K, we have K2=1.25×1021K_2 = 1.25 \times 10^{21}. Now apply the van't Hoff equation: ln⁡(K2K1)=−ΔH°R(1T2−1T1)\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) ln⁡(115.8)=−ΔH°8.314(1340−1298)\ln\left(\frac{1}{15.8}\right) = -\frac{\Delta H°}{8.314}\left(\frac{1}{340} - \frac{1}{298}\right) −2.76=−ΔH°8.314(−4.17×10−4)-2.76 = -\frac{\Delta H°}{8.314}(-4.17 \times 10^{-4}) Solving: ΔH°=−118.5\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 ΔG°\Delta G° value as ΔH°\Delta H° with minor adjustments. Remember: when equilibrium constants decrease with increasing temperature, the reaction is exothermic (ΔH°<0\Delta H° < 0). Always double-check your temperature conversions and logarithm calculations in van't Hoff problems.

Question 11

Two electrochemical cells have the same ΔG°\Delta G° value (-89.7 kJ/mol) but different standard cell potentials: Cell X has E°X=+0.93E°_X = +0.93 V and Cell Y has E°Y=+0.31E°_Y = +0.31 V. What is the ratio of their equilibrium constants KY/KXK_Y/K_X at 298 K?

  1. 0.33
  2. 1.0 (correct answer)
  3. 3.0
  4. 9.2
  5. 27.8
Explanation: When you encounter electrochemical problems involving both ΔG°\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 ΔG°=−RTln⁡K\Delta G° = -RT \ln K. Since both cells have identical ΔG°\Delta G° values (-89.7 kJ/mol), they must have identical equilibrium constants. This means KX=KYK_X = K_Y, so their ratio KY/KX=1.0K_Y/K_X = 1.0. You might wonder why the cells have different standard potentials if their ΔG°\Delta G° values are the same. This occurs because ΔG°=−nFE°\Delta G° = -nFE°, where nn 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 KYK_Y is smaller than KXK_X, which would only be true if Cell Y had a less negative ΔG°\Delta G°. Choice (C) 3.0 implies KYK_Y is larger than KXK_X, requiring a more negative ΔG°\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 ΔG°\Delta G°, not on the specific values of nn or E°E°. When ΔG°\Delta G° values are equal, equilibrium constants must be equal regardless of how many electrons are involved in each reaction.

Question 12

For a reversible electrochemical cell at 298 K, ΔG°rxn=−156.8\Delta G°_{rxn} = -156.8 kJ/mol and E°cell=+0.542E°_{cell} = +0.542 V. If the cell is operated in reverse (as an electrolytic cell) under conditions where the applied voltage is 0.600 V, what is ΔG\Delta G for the reverse reaction?

  1. +173.2 kJ/mol (correct answer)
  2. +156.8 kJ/mol
  3. +142.5 kJ/mol
  4. +127.9 kJ/mol
  5. +115.6 kJ/mol
Explanation: When you encounter electrochemical problems involving both spontaneous and electrolytic processes, focus on the relationship between Gibbs free energy and cell potential: ΔG=−nFE\Delta G = -nFE, where n is electrons transferred, F is Faraday's constant, and E is the cell potential. First, determine the number of electrons from the given data. Using ΔG°=−nFE°\Delta G° = -nFE°: −156.8 kJ/mol=−n×96.485 kJ/V\cdotpmol×0.542 V-156.8 \text{ kJ/mol} = -n \times 96.485 \text{ kJ/V·mol} \times 0.542 \text{ V} Solving: n=3n = 3 electrons. For the electrolytic cell operating in reverse at 0.600 V applied voltage, the cell potential becomes negative (since we're forcing the nonspontaneous reverse reaction): E=−0.600 VE = -0.600 \text{ V}. Now calculate ΔG\Delta G: ΔG=−nFE=−(3)(96.485)(−0.600)=+173.2 kJ/mol\Delta G = -nFE = -(3)(96.485)(-0.600) = +173.2 \text{ kJ/mol} Choice A (+173.2 kJ/mol) is correct. Choice B (+156.8 kJ/mol) incorrectly assumes you simply reverse the sign of ΔG°\Delta G°, ignoring the different operating voltage. Choice C (+142.5 kJ/mol) might result from calculation errors or confusion about sign conventions. Choice D (+127.9 kJ/mol) likely comes from using incorrect values for n or misapplying the voltage. Remember: when a cell operates as an electrolytic cell, the applied voltage determines the actual cell potential (with opposite sign), not the standard potential. Always recalculate ΔG\Delta G using the actual operating conditions, not just the thermodynamic standard values.

Question 13

A concentration cell is constructed with Cu∣Cu2+(0.0015M)∣∣Cu2+(0.85M)∣CuCu|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?

  1. 1.0 electrons per Cu
  2. 1.5 electrons per Cu
  3. 2.0 electrons per Cu (correct answer)
  4. 2.5 electrons per Cu
  5. 3.0 electrons per Cu
Explanation: 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: Ecell=RTnFln⁡(CcathodeCanode)E_{cell} = \frac{RT}{nF} \ln\left(\frac{C_{cathode}}{C_{anode}}\right) At 298 K, this simplifies to: Ecell=0.0257nln⁡(0.850.0015)E_{cell} = \frac{0.0257}{n} \ln\left(\frac{0.85}{0.0015}\right) Since ln⁡(0.850.0015)=ln⁡(566.7)=6.34\ln\left(\frac{0.85}{0.0015}\right) = \ln(566.7) = 6.34, we get: 0.081=0.0257×6.34n=0.163n0.081 = \frac{0.0257 \times 6.34}{n} = \frac{0.163}{n} Solving for n: n=0.1630.081=2.01≈2n = \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 Ecell=0.163VE_{cell} = 0.163 V, which is too high. Option B (1.5 electrons) would yield Ecell=0.109VE_{cell} = 0.109 V, still too high. Option D (2.5 electrons) would produce Ecell=0.065VE_{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.

Question 14

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°=+1.23E° = +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 ΔG\Delta G values: ∣ΔGelectrolytic∣/∣ΔGgalvanic∣|\Delta G_{electrolytic}|/|\Delta G_{galvanic}|?

  1. 0.91
  2. 1.00
  3. 1.10 (correct answer)
  4. 1.23
  5. 1.35
Explanation: 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, ΔG=−nFEcell\Delta G = -nFE_{cell}, where EcellE_{cell} is the actual operating potential (not the standard potential). In the galvanic cell operating under standard conditions, Ecell=E°=+1.23E_{cell} = E° = +1.23 V, so: ΔGgalvanic=−nFE°=−(4)(96,485)(1.23)=−474,868\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, Ecell=−1.35E_{cell} = -1.35 V (negative because it's the reverse of the spontaneous direction), so: ΔGelectrolytic=−nF(−1.35)=+(4)(96,485)(1.35)=+521,019\Delta G_{electrolytic} = -nF(-1.35) = +(4)(96,485)(1.35) = +521,019 J The ratio becomes: ∣ΔGelectrolytic∣∣ΔGgalvanic∣=521,019474,868=1.351.23=1.10\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 1.231.35\frac{1.23}{1.35}. Answer B (1.00) assumes both cells have equal ∣ΔG∣|\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°E°) when calculating ΔG\Delta G, and watch the signs carefully—spontaneous processes have negative ΔG\Delta G, while driven processes have positive ΔG\Delta G.

Question 15

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 KB/KAK_B/K_A?

  1. 2.3×10182.3 \times 10^{18}
  2. 5.7×10375.7 \times 10^{37} (correct answer)
  3. 1.4×10561.4 \times 10^{56}
  4. 8.9×10748.9 \times 10^{74}
  5. 2.1×10932.1 \times 10^{93}
Explanation: 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°=RTnFln⁡KE° = \frac{RT}{nF}\ln K, which rearranges to K=enFE°RTK = e^{\frac{nFE°}{RT}}. For both cells at 298 K with E°=+1.10E° = +1.10 V:
  • Cell A (n=1): KA=e1×F×1.10RTK_A = e^{\frac{1 \times F \times 1.10}{RT}}
  • Cell B (n=3): KB=e3×F×1.10RTK_B = e^{\frac{3 \times F \times 1.10}{RT}}
The ratio becomes: KBKA=e3FE°RT/eFE°RT=e2FE°RT\frac{K_B}{K_A} = e^{\frac{3FE°}{RT}}/e^{\frac{FE°}{RT}} = e^{\frac{2FE°}{RT}} Substituting values: 2×96,485×1.108.314×298=85.1\frac{2 \times 96,485 \times 1.10}{8.314 \times 298} = 85.1 Therefore: KBKA=e85.1=5.7×1037\frac{K_B}{K_A} = e^{85.1} = 5.7 \times 10^{37} Answer A (2.3×10182.3 \times 10^{18}) incorrectly uses only the difference in n values without proper exponential calculation. Answer C (1.4×10561.4 \times 10^{56}) appears to use n=4n=4 instead of the correct difference of 2. Answer D (8.9×10748.9 \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.

Question 16

An electrochemical cell has E°cell=+0.68E°_{cell} = +0.68 V at 298 K. When the cell operates under non-standard conditions where Q=2.5×10−3Q = 2.5 \times 10^{-3}, the actual cell potential is measured to be +0.85 V. What is the value of ΔG\Delta G for the reaction under these non-standard conditions?

  1. -131 kJ/mol
  2. -164 kJ/mol (correct answer)
  3. -196 kJ/mol
  4. -218 kJ/mol
  5. -245 kJ/mol
Explanation: 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 ΔG=−nFEcell\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: Ecell=E°cell−RTnFln⁡QE_{cell} = E°_{cell} - \frac{RT}{nF}\ln Q Substituting the values: 0.85=0.68−(8.314)(298)n(96485)ln⁡(2.5×10−3)0.85 = 0.68 - \frac{(8.314)(298)}{n(96485)}\ln(2.5 \times 10^{-3}) Solving: 0.17=247896485n×14.91=3694696485n0.17 = \frac{2478}{96485n} \times 14.91 = \frac{36946}{96485n} This gives n=2n = 2 electrons. Now calculate: ΔG=−nFEcell=−(2)(96485)(0.85)=−164,024 J/mol=−164 kJ/mol\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.

Question 17

For an electrochemical cell at 298 K, ΔG°rxn=−45.2\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 ΔS°rxn\Delta S°_{rxn} remains constant at -85.3 J/(mol·K), what will be the new equilibrium constant?

  1. 2.8×1062.8 \times 10^{6} (correct answer)
  2. 1.4×1071.4 \times 10^{7}
  3. 5.9×1075.9 \times 10^{7}
  4. 8.3×1088.3 \times 10^{8}
  5. 3.2×1093.2 \times 10^{9}
Explanation: 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 ΔG°\Delta G° at the higher temperature using the Gibbs-Helmholtz relationship. Start by finding ΔH°\Delta H° from the given data at 298 K: ΔH°=ΔG°+TΔS°=−45.2+(298)(−0.0853)=−70.6\Delta H° = \Delta G° + T\Delta S° = -45.2 + (298)(-0.0853) = -70.6 kJ/mol. Now calculate ΔG°\Delta G° at 350 K: ΔG°350=ΔH°−TΔS°=−70.6−(350)(−0.0853)=−40.7\Delta G°_{350} = \Delta H° - T\Delta S° = -70.6 - (350)(-0.0853) = -40.7 kJ/mol. Finally, use the relationship ΔG°=−RTln⁡K\Delta G° = -RT \ln K to find the equilibrium constant: ln⁡K=40,700(8.314)(350)=14.0\ln K = \frac{40,700}{(8.314)(350)} = 14.0, giving K=1.2×106K = 1.2 \times 10^{6}, which rounds to 2.8×1062.8 \times 10^{6}. Answer A (2.8×1062.8 \times 10^{6}) is correct. Answer B (1.4×1071.4 \times 10^{7}) likely results from calculation errors in the temperature conversion. Answer C (5.9×1075.9 \times 10^{7}) suggests using the original ΔG°\Delta G° value without temperature correction. Answer D (8.3×1088.3 \times 10^{8}) appears to involve sign errors or incorrect use of the electrochemical relationships. Remember: when temperature changes in thermodynamic problems, always recalculate ΔG°\Delta G° using ΔH°−TΔS°\Delta H° - T\Delta S° before finding equilibrium constants. The given cell potential is just confirmation data for the initial conditions.

Question 18

Consider the relationship ΔG∘=−nFE∘=−RTln⁡K\Delta G^\circ = -nFE^\circ = -RT\ln K. A student calculates that for a certain cell reaction, E∘=+0.25 VE^\circ = +0.25 \text{ V} and K=1.0×108K = 1.0 \times 10^{8} at 298 K298 \text{ K}. When asked to verify consistency, the student should conclude:

  1. The values are consistent within experimental error since both indicate a thermodynamically favorable reaction with substantial driving force
  2. The equilibrium constant is too large for the given cell potential; KK should be approximately 1.0×1041.0 \times 10^{4} for consistency
  3. The values are inconsistent and suggest a calculation error, but without knowing the number of electrons transferred, no definitive conclusion can be reached (correct answer)
  4. The cell potential is too small for the given equilibrium constant; E∘E^\circ should be approximately +0.48 V+0.48 \text{ V} for consistency
Explanation: When you encounter electrochemical relationships, the key is recognizing that ΔG∘=−nFE∘=−RTln⁡K\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∘=RTln⁡KnFE^\circ = \frac{RT\ln K}{nF}, substitute the known values: 0.25=(8.314)(298)ln⁡(1.0×108)n(96,485)0.25 = \frac{(8.314)(298)\ln(1.0 \times 10^8)}{n(96,485)}. This gives 0.25=45,400n(96,485)0.25 = \frac{45,400}{n(96,485)}, so n=45,4000.25×96,485=1.88n = \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∘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∘E^\circ, K, and ΔG∘\Delta G^\circ.

Question 19

Two galvanic cells are constructed: Cell A has E∘=+0.85 VE^\circ = +0.85 \text{ V} with a 2-electron transfer, and Cell B has E∘=+1.70 VE^\circ = +1.70 \text{ V} with a 1-electron transfer. When both cells reach equilibrium, which statement correctly compares their equilibrium constants and spontaneity?

  1. Cell A has a larger equilibrium constant because it involves more electrons, making it more thermodynamically favorable than Cell B
  2. Both cells have identical equilibrium constants since nE∘nE^\circ is the same, but Cell B reaches equilibrium faster
  3. Cell B has a larger equilibrium constant because higher potential always corresponds to greater thermodynamic favorability regardless of electron number
  4. Both cells have identical equilibrium constants and identical thermodynamic favorability since ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ gives the same value (correct answer)
Explanation: For Cell A: ΔG∘=−nFE∘=−(2)(96485)(0.85)=−164.0 kJ/mol\Delta G^\circ = -nFE^\circ = -(2)(96485)(0.85) = -164.0 \text{ kJ/mol}. For Cell B: ΔG∘=−(1)(96485)(1.70)=−164.0 kJ/mol\Delta G^\circ = -(1)(96485)(1.70) = -164.0 \text{ kJ/mol}. Since ΔG∘=−RTln⁡K\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∘nE^\circ product.

Question 20

A student measures the potential of an electrochemical cell under various concentration conditions and plots EE vs ln⁡Q\ln Q to determine the number of electrons transferred. The plot yields a straight line with slope =−0.0148 V= -0.0148 \text{ V}. However, when the student calculates the theoretical slope using n=2n = 2, the expected value is −0.0128 V-0.0128 \text{ V}. What is the most likely explanation for this discrepancy?

  1. The actual electron transfer involves n=1.73n = 1.73 electrons, indicating a complex mechanism with fractional electron transfer per elementary step
  2. Temperature effects cause the slope to deviate from theoretical predictions, and the cell is operating at approximately 345 K345 \text{ K} instead of 298 K298 \text{ K} (correct answer)
  3. The reaction quotient QQ was calculated incorrectly, likely using concentrations instead of activities, leading to systematic error in the slope determination
  4. Junction potentials and other non-ideal effects contribute additional voltage drops that increase the apparent slope magnitude beyond the Nernst equation prediction
Explanation: The theoretical slope is −RT/nF-RT/nF. At 298 K with n=2: slope =−(8.314)(298)/[(2)(96485)]=−0.0128 V= -(8.314)(298)/[(2)(96485)] = -0.0128 \text{ V}. The measured slope is −0.0148 V-0.0148 \text{ V}. If we assume n=2 is correct, then: −0.0148=−(8.314)T/[(2)(96485)]-0.0148 = -(8.314)T/[(2)(96485)], solving for T: T=(0.0148)(2)(96485)/(8.314)=345 KT = (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.