Redox Potentials & Reference Electrodes - Physical Chemistry 1

1. A student constructs a concentration cell using two $\text{Ag}^+/\text{Ag}$ half-cells connected by a salt bridge. One half-cell contains 0.10 M $\text{AgNO}_3$, and the other contains 0.010 M $\text{AgNO}_3$. After measuring an initial cell potential, the student replaces the salt bridge with a direct metallic connection between the silver electrodes. What will happen to the measured potential?
   Answer: The potential will decrease to zero because both electrodes are now at the same physical potential When analyzing concentration cells, remember that the measured potential depends on maintaining separate chemical environments at each electrode. The Nernst equation shows that concentration differences drive the cell potential, but this only works when the electrodes remain in their respective solutions. With the original salt bridge setup, you have a functioning concentration cell. The $\text{Ag}^+/\text{Ag}$ electrode in 0.10 M solution acts as the cathode (reduction occurs), while the electrode in 0.010 M solution acts as the anode (oxidation occurs). The salt bridge maintains electrical contact while keeping the solutions separate, allowing the concentration difference to drive electron flow and create a measurable potential. When you replace the salt bridge with a direct metallic connection between the silver electrodes, you fundamentally change the system. Both electrodes are now made of the same material (silver) and are physically connected as one continuous piece of metal. A single piece of metal cannot have different electrical potentials at different points - this violates basic principles of electrostatics. The potential immediately drops to zero because you're essentially measuring the potential difference between two points on the same conductor. Option A incorrectly focuses on internal resistance - while resistance does decrease, this doesn't increase potential in this scenario. Option C misunderstands that the physical connection type completely changes the measurement setup, not just the electrode reactions. Option D suggests a gradual change, but the potential drop is immediate due to the physical connection creating electrical continuity. Remember: concentration cells require physical separation of the electrode environments. Direct metallic connections between identical electrodes always result in zero measured potential, regardless of solution concentrations.
2. A researcher uses a hydrogen electrode in a buffer solution with unknown pH and measures $-0.36 \text{ V}$ vs. SHE at $25°\text{C}$ under 1 atm $\text{H}_2$. However, the actual $\text{H}_2$ pressure in the electrode is later found to be 0.85 atm due to water vapor pressure. What is the corrected pH of the solution?
   Answer: pH = 6.1 This question tests your understanding of the Nernst equation and how gas pressure affects electrode potential measurements. When working with hydrogen electrodes, you need to account for the actual H₂ pressure, not just the assumed conditions. The Nernst equation for a hydrogen electrode is: $E = E° - \frac{RT}{nF}\ln\frac{a_{H^+}^2}{P_{H_2}}$. Since E° = 0 for SHE and we can convert to pH using $a_{H^+} = 10^{-pH}$, this becomes: $E = -\frac{0.0592}{2}\log\frac{10^{-2pH}}{P_{H_2}} = -0.0592pH + 0.0296\log P_{H_2}$ First, calculate the apparent pH using the measured conditions: $-0.36 = -0.0592 × pH_{apparent} + 0.0296\log(1.0)$. Since log(1) = 0: $pH_{apparent} = \frac{0.36}{0.0592} = 6.08$ Now correct for the actual pressure of 0.85 atm: $-0.36 = -0.0592 × pH_{actual} + 0.0296\log(0.85)$. Since log(0.85) = -0.071: $-0.36 = -0.0592 × pH_{actual} - 0.0021$. Solving: $pH_{actual} = \frac{0.36 - 0.0021}{0.0592} = 6.1$ Answer B (pH = 6.1) is correct. Answer A (5.8) results from incorrectly adding the pressure correction instead of subtracting it. Answer C (6.4) uses the uncorrected apparent pH value. Answer D (6.7) applies an incorrect pressure correction factor. Remember: lower H₂ pressure makes the electrode potential more negative, requiring a small upward pH correction. Always verify your pressure correction direction matches the physical expectation.
3. A student attempts to use a copper wire as a reference electrode by immersing it in 1.0 M CuSO₄ solution. After several hours of use, the measured potentials against this 'reference' electrode begin to drift significantly. What is the most fundamental reason this setup fails as a reliable reference electrode?
   Answer: The Cu²⁺ concentration changes due to electrode reactions, altering the reference potential continuously A good reference electrode must maintain constant potential, requiring constant composition. In this setup, when current flows through the electrode (during measurements), Cu²⁺ + 2e⁻ → Cu or Cu → Cu²⁺ + 2e⁻ occurs, changing [Cu²⁺] and thus the potential via the Nernst equation. Choice A is wrong - wire resistance is negligible. Choice C doesn't explain the drift pattern. Choice D incorrectly attributes the effect to complex formation rather than concentration changes.
4. A galvanic cell uses a Zn/Zn²⁺ anode and a Cu/Cu²⁺ cathode, both at standard conditions initially. During operation, a current of 0.50 A flows for 2.0 hours. If the initial volume of each half-cell solution is 250 mL, what is the approximate cell potential after this period, assuming no volume change?
   Answer: +1.07 V, accounting for the modest concentration changes in both half-cell compartments during operation When you encounter galvanic cell problems involving current flow over time, you need to apply the Nernst equation to account for changing ion concentrations as the cell operates. First, calculate how the concentrations change. With 0.50 A flowing for 2.0 hours, that's 3600 coulombs of charge. Using Faraday's law: $n = \frac{Q}{nF} = \frac{3600}{2 \times 96,485} = 0.0186$ mol of electrons transferred. This means 0.0186 mol of Zn²⁺ is produced and 0.0186 mol of Cu²⁺ is consumed. In 250 mL (0.25 L), starting from 1.0 M standard conditions: - [Zn²⁺] increases from 1.0 M to 1.0 + (0.0186/0.25) = 1.074 M - [Cu²⁺] decreases from 1.0 M to 1.0 - (0.0186/0.25) = 0.926 M Apply the Nernst equation: $E = E° - \frac{0.0592}{n} \log \frac{[Zn²⁺]}{[Cu²⁺]}$ $E = 1.10 - \frac{0.0592}{2} \log \frac{1.074}{0.926} = 1.10 - 0.0296 \times 0.065 = 1.07 \text{ V}$ Option A incorrectly assumes concentrations don't affect cell potential. Option B wrongly suggests the cell potential increases significantly - the Nernst equation shows it must decrease from standard conditions. Option D overestimates the concentration change impact, suggesting an unrealistically large potential drop. Remember: whenever current flows in galvanic cells, use the Nernst equation to account for concentration changes. The cell potential always decreases from its standard value as the reaction proceeds toward equilibrium.
5. A student prepares a silver wire coated with AgBr for use as a reference electrode in a solution containing 0.010 M $\text{Br}^-$. If $K_{\text{sp}}(\text{AgBr}) = 5.0 \times 10^{-13}$ and $E^\circ(\text{Ag}^+/\text{Ag}) = +0.80 \text{ V}$, what is the potential of this electrode vs. SHE?
   Answer: +0.07 V When you encounter an electrode involving a sparingly soluble salt like AgBr, you're dealing with a system where the metal ion concentration is controlled by solubility equilibrium rather than being directly given. This requires combining solubility principles with the Nernst equation. First, find the $\text{Ag}^+$ concentration using the solubility product. For AgBr dissolving: $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$ $K_{\text{sp}} = [\text{Ag}^+][\text{Br}^-] = 5.0 \times 10^{-13}$ With $[\text{Br}^-] = 0.010 \text{ M}$: $[\text{Ag}^+] = \frac{5.0 \times 10^{-13}}{0.010} = 5.0 \times 10^{-11} \text{ M}$ Now apply the Nernst equation for the $\text{Ag}^+/\text{Ag}$ half-cell: $E = E^\circ - \frac{0.0592}{n}\log\frac{1}{[\text{Ag}^+]}$ $E = 0.80 - \frac{0.0592}{1}\log\frac{1}{5.0 \times 10^{-11}} = 0.80 - 0.0592 \times 10.3 = +0.07 \text{ V}$ This confirms answer B is correct. A (+0.11 V) likely results from calculation errors in the logarithm or using standard conditions. C (+0.03 V) and D (-0.01 V) represent more significant computational mistakes, possibly from incorrectly handling the $K_{\text{sp}}$ calculation or sign errors in the Nernst equation. Strategy tip: Always remember that sparingly soluble salts in electrochemistry problems require a two-step approach: first solve the equilibrium to find ion concentrations, then apply the Nernst equation. The extremely low $\text{Ag}^+$ concentration will significantly reduce the potential from its standard value.
6. A student measures the cell potential of $\text{Zn}|\text{Zn}^{2+}(0.01 \text{ M})||\text{Cu}^{2+}(1.0 \text{ M})|\text{Cu}$ and obtains $+1.13 \text{ V}$. Using standard potentials $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V}$ and $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}$, the student calculates an expected value of $+1.16 \text{ V}$. What accounts for the 30 mV discrepancy?
   Answer: Junction potential at the salt bridge interface is contributing approximately $-30 \text{ mV}$ to the measurement When analyzing electrochemical cell measurements that deviate from calculated values, you need to consider all sources of potential beyond the electrode reactions themselves. Real galvanic cells contain several interfaces that can contribute additional potentials. The student's calculation using the Nernst equation gives the expected cell potential based purely on the electrode half-reactions. For this cell, $E_{cell} = E°_{cell} - \frac{RT}{nF}\ln\left(\frac{[Zn^{2+}]}{[Cu^{2+}]}\right) = 1.10 - \frac{0.0257}{2}\ln\left(\frac{0.01}{1.0}\right) = +1.16 \text{ V}$. However, the measured value is 30 mV lower. Answer A is correct because junction potentials arise at the interface between the salt bridge and each half-cell solution. These potentials result from different mobilities of cations and anions diffusing across the boundary, creating a charge separation. Junction potentials typically range from a few millivolts to tens of millivolts and often reduce the measured cell potential relative to the calculated value. Answer B is incorrect because electrode passivation would affect the kinetics of the reaction, not necessarily create a systematic 30 mV offset. Answer C is wrong because temperature effects on standard potentials wouldn't selectively account for exactly 30 mV, and both electrodes experience the same temperature. Answer D is incorrect because while activity coefficients do differ from unity, the concentration differences here (0.01 M vs 1.0 M) wouldn't systematically account for this specific discrepancy. Remember: when experimental electrochemical measurements deviate from Nernst equation predictions, junction potentials at salt bridge interfaces are often the primary culprit.
7. Two different reference electrodes are used to measure the potential of the same unknown half-cell. Against a standard hydrogen electrode (SHE), the potential is +0.52 V. Against a silver/silver chloride electrode (Ag/AgCl, sat'd KCl), the measured potential is +0.32 V. What conclusion can be drawn about the silver/silver chloride electrode?
   Answer: The Ag/AgCl electrode potential vs. SHE is +0.20 V, confirming proper electrode function If the unknown half-cell shows +0.52 V vs. SHE and +0.32 V vs. Ag/AgCl, then: E_unknown - E_SHE = +0.52 V and E_unknown - E_Ag/AgCl = +0.32 V. Since E_SHE = 0, E_unknown = +0.52 V. Therefore: +0.52 - E_Ag/AgCl = +0.32, so E_Ag/AgCl = +0.20 V vs. SHE. This is close to the literature value (~0.197 V), confirming proper function. The other choices misinterpret the relationship.
8. In a research application, two identical platinum electrodes are placed in solutions containing different concentrations of Ce⁴⁺/Ce³⁺ redox couples. Solution A has [Ce⁴⁺] = 0.10 M and [Ce³⁺] = 0.010 M, while Solution B has [Ce⁴⁺] = 0.010 M and [Ce³⁺] = 0.10 M. What is the cell potential when these half-cells are connected?
   Answer: +0.118 V, with Solution A serving as the cathode due to its higher Ce⁴⁺/Ce³⁺ ratio When you encounter electrochemical cells with identical redox couples at different concentrations, you're dealing with a concentration cell. The key insight is that even though both half-cells use the same Ce⁴⁺/Ce³⁺ couple, different concentration ratios create different electrode potentials. To find the cell potential, use the Nernst equation for each half-cell: $E = E° - \frac{0.059}{n}\log\frac{[Ce³⁺]}{[Ce⁴⁺]}$. Since both use the same redox couple, the standard potentials cancel out, leaving only the concentration effects. For Solution A: $E_A = E° - 0.059\log\frac{0.010}{0.10} = E° - 0.059(-1) = E° + 0.059 \text{ V}$ For Solution B: $E_B = E° - 0.059\log\frac{0.10}{0.010} = E° - 0.059(+1) = E° - 0.059 \text{ V}$ The cell potential is $E_{cell} = E_A - E_B = +0.118 \text{ V}$, with Solution A as the cathode since it has the higher potential. Option A incorrectly assumes concentration differences don't matter when the redox couples are identical. Option B gets the wrong sign and magnitude—the potential doesn't oppose anything, and the value should be 0.118 V. Option D correctly identifies the tenfold concentration difference but miscalculates the result by forgetting that the cell potential is the difference between two half-cell potentials, each contributing 0.059 V. Remember: In concentration cells, the half-cell with the higher oxidized-to-reduced species ratio always serves as the cathode, and concentration differences of 10-fold increments create 0.059 V differences per electron transferred.
9. Two identical platinum electrodes are placed in solutions containing: (1) $1.0 \text{ mM}$ $\text{Fe}^{3+}$ and $10 \text{ mM}$ $\text{Fe}^{2+}$, and (2) $10 \text{ mM}$ $\text{Fe}^{3+}$ and $1.0 \text{ mM}$ $\text{Fe}^{2+}$. When connected through a salt bridge, which electrode acts as the cathode and what is the approximate cell potential?
   Answer: Electrode 2 is the cathode; cell potential is approximately $+0.118 \text{ V}$ When you encounter electrochemical cells with concentration differences, you're dealing with concentration cells where the Nernst equation determines both electrode potentials and electron flow direction. First, calculate each electrode's potential using the Nernst equation: $E = E° + \frac{0.059}{n}\log\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}$. Since both electrodes use the same Fe³⁺/Fe²⁺ couple, the standard potential $E°$ cancels out when finding the cell potential. For electrode 1: $E_1 = \frac{0.059}{1}\log\frac{1.0}{10} = 0.059 \times (-1) = -0.059 \text{ V}$ For electrode 2: $E_2 = \frac{0.059}{1}\log\frac{10}{1.0} = 0.059 \times (+1) = +0.059 \text{ V}$ Since electrode 2 has the higher potential, it acts as the cathode (reduction occurs here). The cell potential is $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.059 - (-0.059) = 0.118 \text{ V}$. Answer A incorrectly identifies electrode 1 as the cathode—this electrode has lower potential and will be the anode. Answer C makes the same electrode error but also uses the wrong cell potential (just the difference from standard conditions rather than the full potential difference). Answer D correctly calculates one electrode's potential relative to standard conditions but fails to account for the full potential difference between the two electrodes. Remember: in concentration cells, the electrode with higher oxidized-to-reduced species ratio has higher potential and becomes the cathode. Always calculate the full potential difference between electrodes, not just individual deviations from standard conditions.
10. An electrochemical cell contains a platinum electrode in a solution with Fe³⁺ and Fe²⁺ ions, with [Fe³⁺] = 0.020 M and [Fe²⁺] = 0.080 M. This half-cell is connected to a standard hydrogen electrode. If E°(Fe³⁺/Fe²⁺) = +0.77 V, what is the measured cell potential and which electrode serves as the cathode?
    Answer: +0.73 V with the platinum electrode as cathode, accounting for the non-standard concentration ratio Using the Nernst equation: E = 0.77 + (0.0257/1)ln(0.020/0.080) = 0.77 + 0.0257 × ln(0.25) = 0.77 - 0.036 = +0.73 V. Since +0.73 V > 0 V (SHE), the Fe³⁺/Fe²⁺ couple has higher potential and serves as cathode. Choice A ignores concentration effects. Choice B incorrectly identifies the cathode. Choice C miscalculates the concentration effect direction.
11. When switching from a hydrogen electrode reference to a silver/silver chloride reference electrode ($E^\circ = +0.197 \text{ V}$ vs. SHE) in a potentiometric measurement, a researcher observes that all previously measured potentials shift by $-0.203 \text{ V}$. What is the most likely explanation for this discrepancy?
    Answer: The silver/silver chloride electrode contains a different chloride concentration than the standard electrode When you encounter potentiometric measurements with reference electrode changes, focus on the Nernst equation and how concentration affects electrode potential. Reference electrodes must maintain constant, known potentials, but this depends on maintaining standard conditions. The expected potential shift when switching from SHE to Ag/AgCl should be exactly $+0.197 \text{ V}$ (the standard potential difference). However, you observe a shift of $-0.203 \text{ V}$, which differs by $0.400 \text{ V}$ from the expected value. This large discrepancy points to a non-standard Ag/AgCl electrode. The Ag/AgCl electrode potential follows: $E = E° - \frac{RT}{F}\ln[\text{Cl}^-]$. When the chloride concentration differs from the standard (typically 1 M or saturated KCl), the electrode potential shifts significantly. A lower chloride concentration would make the potential more positive than standard, explaining why your measured shift is more negative than expected. Answer B correctly identifies this concentration effect. Answer A suggests only a 6 mV contamination effect, but your discrepancy is 400 mV—far too large for simple contamination. Answer C proposes junction potential changes, but these typically cause much smaller shifts (few millivolts) and wouldn't account for such a systematic difference. Answer D mentions temperature drift, but again, thermal effects alone wouldn't produce a 400 mV shift from standard conditions. Study tip: Always check whether reference electrodes match standard conditions. Non-standard concentrations in reference electrodes are common sources of large, systematic errors in potentiometric measurements.
12. An electrochemical cell contains a glass pH electrode and a reference electrode. When the pH of the solution changes from 7.0 to 4.0, the measured potential changes by $+177 \text{ mV}$. If the theoretical Nernstian response is $59.2 \text{ mV/pH}$ unit at $25°\text{C}$, what can be concluded about the electrode system?
    Answer: The electrode is responding normally since $3 \times 59.2 = 177.6 \text{ mV}$, within experimental error When analyzing pH electrode performance, you need to compare the observed potential change to the theoretical Nernstian response. The Nernst equation predicts that at 25°C, a pH electrode should respond with 59.2 mV per pH unit change. Let's calculate what should happen when pH changes from 7.0 to 4.0. This is a change of 3 pH units (7.0 - 4.0 = 3.0). Since the solution becomes more acidic, the glass electrode potential should increase (become more positive). The theoretical response would be: $3.0 \times 59.2 \text{ mV/pH} = 177.6 \text{ mV}$ The measured change of +177 mV matches this theoretical value within experimental error (only 0.6 mV difference), indicating normal electrode function. Choice A is correct because it properly calculates the expected Nernstian response and recognizes that the small difference falls within typical experimental uncertainty. Choice B is wrong because sub-Nernstian response would mean a smaller potential change than expected, not the near-perfect match observed here. Choice C is incorrect because super-Nernstian response would produce a larger potential change than the theoretical 177.6 mV, which didn't occur. Choice D is wrong because reference electrode drift would typically cause erratic or non-linear responses over time, not a response that matches Nernstian theory so closely. Study tip: Always calculate the expected Nernstian response first (ΔpH × 59.2 mV), then compare to the observed value. Responses within ±5% typically indicate proper electrode function.
13. A quinhydrone electrode is prepared by placing a platinum wire in a solution containing equimolar amounts of quinone and hydroquinone. When this electrode is used to measure the pH of an unknown solution, a potential of +0.45 V vs. SHE is observed at 25°C. Given that $E^\circ(\text{quinone}/\text{hydroquinone}) = +0.699 \text{ V}$, what is the pH of the unknown solution?
    Answer: pH = 4.2 When you encounter a quinhydrone electrode problem, you're dealing with a pH-sensitive redox system where both quinone and hydroquinone concentrations affect the electrode potential through the Nernst equation. The quinhydrone electrode follows the half-reaction: $\text{quinone} + 2\text{H}^+ + 2e^- \rightarrow \text{hydroquinone}$ Since the solution contains equimolar amounts of quinone and hydroquinone, their concentration ratio equals 1, simplifying the Nernst equation to: $E = E° - \frac{0.0592}{2} \log\left(\frac{1}{[\text{H}^+]^2}\right)$ This becomes: $E = E° - 0.0592 \times \text{pH}$ Substituting the given values: $0.45 = 0.699 - 0.0592 \times \text{pH}$ Solving for pH: $0.0592 \times \text{pH} = 0.699 - 0.45 = 0.249$ $\text{pH} = \frac{0.249}{0.0592} = 4.2$ This confirms answer (B) pH = 4.2 is correct. Answer (A) pH = 2.1 would result from doubling the pH calculation error or mishandling the 2-electron transfer. Answer (C) pH = 6.3 likely comes from incorrectly adding rather than subtracting the Nernst correction. Answer (D) pH = 8.4 represents a sign error in the Nernst equation setup. Remember that quinhydrone electrodes become unreliable above pH 8 due to side reactions, and always check that your final pH makes chemical sense given the measured potential relative to the standard potential.
14. A student measures the potential of a $\text{Zn}^{2+}/\text{Zn}$ half-cell against a saturated calomel electrode (SCE) and obtains $-1.02 \text{ V}$. Given that $E^\circ(\text{SCE}) = +0.24 \text{ V}$ vs. SHE and $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V}$ vs. SHE, what can be concluded about the $\text{Zn}^{2+}$ concentration in the half-cell?
    Answer: The $\text{Zn}^{2+}$ concentration is less than 1.0 M due to the more negative potential observed When you encounter electrochemical cell potential measurements that deviate from standard values, you're dealing with the Nernst equation and concentration effects on cell potentials. First, let's determine what the measured potential tells us. The cell potential measured is -1.02 V with the Zn²⁺/Zn half-cell as the anode and SCE as the cathode. Under standard conditions, this cell potential would be: $E°_{cell} = E°_{cathode} - E°_{anode} = 0.24 - (-0.76) = 1.00 \text{ V}$ Since the measured potential (-1.02 V) is more negative than the standard potential (-1.00 V), the Zn²⁺/Zn half-cell potential must be more negative than its standard value of -0.76 V. The Nernst equation shows that $E = E° - \frac{RT}{nF}\ln Q$, where Q is the reaction quotient. For the Zn²⁺/Zn half-cell, a lower Zn²⁺ concentration makes Q smaller, which makes the ln Q term more negative, resulting in a more negative overall potential. Option A is incorrect because the observed potential differs from the standard condition prediction. Option B incorrectly suggests higher concentration causes more negative potential—this is backwards. Option D is wrong because while temperature affects the calculation quantitatively, we can still determine the concentration trend qualitatively from the potential direction. Study tip: Remember that for reduction reactions, lower ion concentrations lead to more negative (less favorable) reduction potentials. Always compare your measured cell potential to the calculated standard potential to determine concentration effects.
15. In a potentiometric titration of a mixture containing both $\text{Fe}^{2+}$ and $\text{Ti}^{3+}$ with $\text{Ce}^{4+}$, two distinct equivalence points are observed. Given the standard potentials: $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \text{ V}$, $E^\circ(\text{Ti}^{4+}/\text{Ti}^{3+}) = +0.04 \text{ V}$, and $E^\circ(\text{Ce}^{4+}/\text{Ce}^{3+}) = +1.44 \text{ V}$, which species is oxidized first and why?
    Answer: $\text{Ti}^{3+}$ is oxidized first because it has the lower standard potential and is more easily oxidized When analyzing potentiometric titrations involving multiple species, you need to determine which species will be oxidized first based on thermodynamic favorability. The key insight is that species with lower standard reduction potentials are more easily oxidized. In redox reactions, the species that gets oxidized must have a lower standard potential than the oxidizing agent. Here, $\text{Ce}^{4+}$ ($E° = +1.44 \text{ V}$) can oxidize both $\text{Fe}^{2+}$ ($E° = +0.77 \text{ V}$) and $\text{Ti}^{3+}$ ($E° = +0.04 \text{ V}$) since both have lower potentials. However, $\text{Ti}^{3+}$ will be oxidized first because it has the significantly lower standard potential, making it the more thermodynamically favorable reaction. The larger potential difference between $\text{Ce}^{4+}/\text{Ce}^{3+}$ and $\text{Ti}^{4+}/\text{Ti}^{3+}$ (1.40 V vs. 0.67 V) means this oxidation occurs more readily. Answer A incorrectly suggests that higher potential means easier oxidation - it's actually the opposite. Answer C correctly identifies that $\text{Fe}^{2+}$ has a smaller potential difference with cerium, but incorrectly concludes this means it oxidizes first. Answer D correctly identifies that titanium has the larger potential difference but incorrectly states the conclusion about which oxidizes first. Remember: in competitive oxidation reactions, the species with the lowest standard reduction potential gets oxidized first because it represents the most thermodynamically favorable reaction with the oxidizing agent.
16. In a laboratory setting, a student measures the potential of a hydrogen electrode against a saturated calomel electrode (SCE) and obtains a reading of -0.48 V. Given that E°(SCE) = +0.241 V vs. SHE, what can be concluded about the hydrogen electrode conditions?
    Answer: The pH is approximately 12.2, indicating a strongly basic solution with standard H₂ pressure The potential of the H electrode vs. SHE = -0.48 - 0.241 = -0.721 V. Using the Nernst equation for H⁺/H₂: E = 0 - (0.0592)pH (assuming 1 atm H₂ at 25°C). Therefore: -0.721 = -0.0592 × pH, so pH = 0.721/0.0592 ≈ 12.2. This indicates a strongly basic solution. Choice B is incorrect because low H₂ pressure would give a more positive potential. Choice C is wrong as negative potentials are possible in basic solutions. Choice D is unsupported by the data.
17. In a potentiometric measurement using a saturated calomel electrode (SCE) as reference, a student observes that the potential readings drift slowly over time, becoming approximately 2 mV more positive each hour. The sample solution and indicator electrode are stable. What is the most probable cause of this drift?
    Answer: Water evaporation from the SCE is increasing the $\text{KCl}$ concentration and making the electrode potential more positive When you encounter potentiometric drift problems, focus on the systematic changes that can affect reference electrode stability over time. The key is identifying which factors cause predictable, unidirectional potential shifts. A saturated calomel electrode (SCE) maintains its potential through the equilibrium: $\text{Hg}_2\text{Cl}_2(s) + 2e^- \rightleftharpoons 2\text{Hg}(l) + 2\text{Cl}^-$. The electrode potential depends on chloride ion activity, which is controlled by the saturated KCl solution. When water evaporates from the SCE, the KCl concentration increases beyond saturation, raising the chloride activity. According to the Nernst equation, higher $\text{Cl}^-$ activity makes the electrode potential more positive—exactly matching the observed 2 mV/hour positive drift. This makes A correct. B is wrong because temperature fluctuations would cause random, bidirectional changes, not the steady positive drift described. Temperature effects are typically reversible and don't show this consistent trend. C is incorrect because a clogged salt bridge would cause erratic, unstable readings and poor electrical contact, not a smooth, predictable drift. Junction potential changes from clogging are usually irregular and accompanied by noisy measurements. D is wrong because mercury oxidation would consume $\text{Hg}_2\text{Cl}_2$ and destabilize the electrode permanently, causing large, irreversible potential changes rather than the small, steady drift observed. Study tip: For reference electrode problems, always consider water evaporation first when you see steady, unidirectional drift. It's the most common cause of SCE instability and produces predictable potential changes through concentration effects.
18. A galvanic cell is constructed with a $\text{Cu}^{2+}/\text{Cu}$ half-cell ($E^\circ = +0.34 \text{ V}$) and a $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell ($E^\circ = +0.77 \text{ V}$) connected by a salt bridge. If both half-cells initially contain 0.10 M concentrations of all ionic species, which statement correctly describes the system at equilibrium?
    Answer: The $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell potential will be more negative than its initial value due to decreased iron(III) concentration When analyzing galvanic cells at equilibrium, you need to understand how concentration changes affect individual half-cell potentials through the Nernst equation, and recognize that equilibrium means zero net current flow. Initially, the $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell has a higher standard potential (+0.77 V) than the $\text{Cu}^{2+}/\text{Cu}$ half-cell (+0.34 V), so electrons flow from copper to iron. This means copper gets oxidized ($\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$) and iron(III) gets reduced ($\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$). As the cell operates, $[\text{Fe}^{3+}]$ decreases while $[\text{Fe}^{2+}]$ increases. According to the Nernst equation, when the ratio $[\text{Fe}^{2+}]/[\text{Fe}^{3+}]$ increases, the iron half-cell potential becomes more negative than its initial value. This continues until both half-cells reach the same potential, establishing equilibrium with zero potential difference. Answer choice A is wrong because copper ion concentration increases, which would make the copper half-cell potential more positive than standard, but this doesn't match the iron half-cell description. Answer choice C incorrectly assumes equilibrium represents standard conditions – it doesn't, since concentrations have changed from 1 M. Answer choice D is partially correct about zero potential difference but wrong about individual potentials remaining unchanged; concentrations have shifted, so potentials must change according to Nernst. Remember: at galvanic cell equilibrium, the potential difference is zero, but individual half-cell potentials have shifted from their initial values due to concentration changes during cell operation.
19. A double-junction reference electrode is used instead of a single-junction Ag/AgCl electrode for measuring the potential of a solution containing high concentrations of sulfide ions. The measured potential differs by $+15 \text{ mV}$ from the value obtained with the single-junction electrode. What explains this difference?
    Answer: Silver sulfide precipitation at the single-junction electrode created additional potential contributions When you encounter electrochemistry problems involving reference electrodes and interfering ions, focus on how specific ions can disrupt electrode function through chemical reactions. The key issue here is that sulfide ions (S²⁻) readily react with silver to form silver sulfide (Ag₂S), a highly insoluble precipitate. In a single-junction Ag/AgCl electrode, the AgCl coating comes into direct contact with the sample solution through the junction. When sulfide ions diffuse to the electrode surface, they react with the silver, forming Ag₂S and creating an additional electrochemical potential that wasn't part of the original Ag/AgCl system. This extra potential contribution explains the +15 mV difference you observe. Answer D correctly identifies this silver sulfide precipitation as the source of the potential difference. The Ag₂S formation creates a mixed potential system that shifts the measured voltage. Answer A incorrectly suggests resistance effects, but the difference is due to chemical reaction, not electrical resistance. Answer B misidentifies the location of the problem—sulfide doesn't significantly react with the outer electrolyte but rather with the silver electrode itself. Answer C incorrectly explains the double-junction's role; while double-junction electrodes do minimize junction potential issues, the primary benefit here is preventing direct contact between sulfide and the Ag/AgCl electrode. Study tip: Remember that Ag/AgCl electrodes are vulnerable to sulfide, protein, and other ions that can react with silver. When you see these interfering species mentioned, immediately consider whether they form precipitates or complexes that would alter electrode potentials.
20. A concentration cell is constructed using two silver electrodes in AgNO₃ solutions of different concentrations. The measured cell potential is +0.089 V at 25°C. If one compartment contains 0.10 M AgNO₃, what is the approximate concentration of AgNO₃ in the other compartment?
    Answer: 0.0030 M AgNO₃, with the dilute solution serving as the anode in this configuration For a concentration cell: E = (RT/nF)ln(C₁/C₂). With E = 0.089 V and n = 1 for Ag⁺/Ag: 0.089 = 0.0257 × ln(0.10/C₂), so ln(0.10/C₂) = 3.46, giving 0.10/C₂ = 32, thus C₂ = 0.0031 M ≈ 0.0030 M. The dilute solution is the anode since electrons flow from low to high concentration. Choice A gives wrong direction, C and D give incorrect concentrations.

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