Physical Chemistry 1 Quiz: Redox Potentials And Reference Electrodes

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Redox Potentials And Reference ElectrodesQuestion 1 of 20

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?

The

\text{Zn}^{2+}

concentration is exactly 1.0 M as expected under standard conditions

The

\text{Zn}^{2+}

concentration is greater than 1.0 M due to the more negative potential observed

The

\text{Zn}^{2+}

concentration is less than 1.0 M due to the more negative potential observed

The

\text{Zn}^{2+}

concentration cannot be determined without knowing the temperature of the measurement

The measurement contains systematic error since the observed potential differs from the standard value

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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Redox Potentials And Reference Electrodes

Practice Redox Potentials And Reference Electrodes 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 Redox Potentials And Reference Electrodes, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

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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

The $\text{Zn}^{2+}$ concentration is exactly 1.0 M as expected under standard conditions
The $\text{Zn}^{2+}$ concentration is greater than 1.0 M due to the more negative potential observed
The $\text{Zn}^{2+}$ concentration is less than 1.0 M due to the more negative potential observed (correct answer)
The $\text{Zn}^{2+}$ concentration cannot be determined without knowing the temperature of the measurement
The measurement contains systematic error since the observed potential differs from the standard value

Explanation: 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.

Question 2

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?

The $\text{Cu}^{2+}/\text{Cu}$ half-cell potential will be more positive than its initial value due to increased copper ion concentration
The $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell potential will be more negative than its initial value due to decreased iron(III) concentration (correct answer)
Both half-cell potentials will return to their standard values since equilibrium represents standard conditions
The potential difference between half-cells will be zero, but individual half-cell potentials remain unchanged from initial values
The system cannot reach true equilibrium due to the presence of the salt bridge creating a junction potential

Explanation: 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.

Question 3

In a potentiometric titration of $\text{Fe}^{2+}$ with $\text{Ce}^{4+}$ , the potential is monitored using a platinum indicator electrode and a calomel reference electrode. Near the equivalence point, the potential changes from $+0.85 \text{ V}$ to $+1.25 \text{ V}$ vs. SCE. Given: $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \text{ V}$ and $E^\circ(\text{Ce}^{4+}/\text{Ce}^{3+}) = +1.44 \text{ V}$ vs. SHE, what controls the potential in this region?

The $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple controls the potential throughout this entire region since iron is the analyte
The $\text{Ce}^{4+}/\text{Ce}^{3+}$ couple controls the potential throughout this entire region since cerium is the titrant
The potential transitions from being controlled by the $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple to the $\text{Ce}^{4+}/\text{Ce}^{3+}$ couple (correct answer)
Both couples contribute equally to the potential throughout this region due to their similar standard potentials
The platinum electrode potential determines the measured values since it acts as an inert indicator electrode

Explanation: When analyzing potentiometric titrations involving redox couples, you need to understand that the electrode potential is governed by whichever redox couple can most readily establish equilibrium at the electrode surface. This depends on the relative concentrations of the oxidized and reduced forms present in solution. Before the equivalence point, excess

\text{Fe}^{2+}

remains unreacted, so both

\text{Fe}^{2+}

and

\text{Fe}^{3+}

(produced from the reaction) are present in significant concentrations. The

\text{Fe}^{3+}/\text{Fe}^{2+}

couple dominates the potential because these species can readily exchange electrons with the platinum electrode. After the equivalence point, essentially all iron has been oxidized, and excess

\text{Ce}^{4+}

is present alongside

\text{Ce}^{3+}

(produced from the reaction). Now the

\text{Ce}^{4+}/\text{Ce}^{3+}

couple controls the potential. The dramatic potential jump from +0.85 V to +1.25 V signals this transition at the equivalence point. Option A is wrong because the analyte doesn't automatically control potential throughout—it's about which species are present in measurable concentrations. Option B is incorrect for the same reason regarding the titrant. Option D misses the point entirely; having similar standard potentials doesn't mean both couples contribute equally—the controlling couple depends on which oxidized and reduced forms are both present in significant amounts. Remember: In redox titrations, the potential is controlled by whichever couple has both its oxidized and reduced forms present in appreciable concentrations. The equivalence point marks the transition between these controlling couples.

Question 4

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?

The electrode is responding normally since $3 \times 59.2 = 177.6 \text{ mV}$ , within experimental error (correct answer)
The electrode has sub-Nernstian response indicating surface contamination or electrode aging
The electrode has super-Nernstian response suggesting interference from other ionic species
The reference electrode is drifting and contributing additional potential changes during measurement
The temperature of the system is higher than $25°\text{C}$ based on the observed response slope

Explanation: 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.

Question 5

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?

Water evaporation from the SCE is increasing the $\text{KCl}$ concentration and making the electrode potential more positive (correct answer)
Temperature fluctuations in the laboratory are causing the SCE potential to vary according to its temperature coefficient
The SCE salt bridge is becoming clogged, increasing the junction potential between reference and sample
Mercury oxidation within the SCE is gradually changing the $\text{Hg}_2\text{Cl}_2$ concentration and electrode composition
Chloride leakage from the SCE into the sample solution is creating a concentration gradient at the junction

Explanation: 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.

Question 6

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?

The double-junction electrode has a higher internal resistance, causing a systematic voltage drop during measurement
Sulfide ions react with the outer electrolyte of the double-junction electrode, altering its composition
The double-junction design eliminates junction potential effects that were present with the single-junction electrode
Silver sulfide precipitation at the single-junction electrode created additional potential contributions (correct answer)
The double-junction electrode contains different electrolyte concentrations, changing its reference potential

Explanation: 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.

Question 7

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?

Junction potential at the salt bridge interface is contributing approximately $-30 \text{ mV}$ to the measurement (correct answer)
The zinc electrode surface is passivated, reducing its effective activity and making it less negative
Temperature effects have shifted both standard potentials in opposite directions by equal amounts
Activity coefficients for the ionic species differ significantly from unity at these concentrations
Internal resistance of the cell is causing a voltage drop during the potential measurement

Explanation: 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.

Question 8

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?

Electrode 1 is the cathode; cell potential is approximately $+0.118 \text{ V}$
Electrode 2 is the cathode; cell potential is approximately $+0.118 \text{ V}$ (correct answer)
Electrode 1 is the cathode; cell potential is approximately $+0.059 \text{ V}$
Electrode 2 is the cathode; cell potential is approximately $+0.059 \text{ V}$
Neither electrode acts as cathode since both contain the same redox couple

Explanation: 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.

Question 9

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?

pH = 5.8
pH = 6.1 (correct answer)
pH = 6.4
pH = 6.7
pH = 7.0

Explanation: 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.

Question 10

An ion-selective electrode for $\text{Ca}^{2+}$ shows the following response in calibration solutions: 10⁻² M Ca²⁺ gives -58 mV, 10⁻³ M Ca²⁺ gives -28 mV, and 10⁻⁴ M Ca²⁺ gives +2 mV (all vs. constant reference). What can be concluded about electrode performance?

The electrode shows perfect Nernstian response with the expected slope of +29.6 mV per decade
The electrode shows perfect Nernstian response with the expected slope of -29.6 mV per decade (correct answer)
The electrode has super-Nernstian response indicating interference from other divalent cations
The electrode has sub-Nernstian response suggesting membrane degradation or poor selectivity
The electrode response is non-linear and cannot be described by simple Nernstian behavior

Explanation: When you encounter ion-selective electrode questions, focus on the Nernst equation relationship between potential and concentration. For a divalent cation like

\text{Ca}^{2+}

, the theoretical Nernstian slope should be

\frac{RT}{nF} \times 2.303 = \frac{59.2}{n}

mV per decade at 25°C, where n = 2 for the charge. This gives an expected slope of +29.6 mV per decade when plotting potential vs. log[Ca²⁺]. Let's calculate the actual slope from the data. From 10⁻⁴ M to 10⁻³ M (one decade increase in concentration): the potential changes from +2 mV to -28 mV, giving a slope of (-28 - 2)/(log 10⁻³ - log 10⁻⁴) = -30 mV per decade. From 10⁻³ M to 10⁻² M: (-58 - (-28))/(log 10⁻² - log 10⁻³) = -30 mV per decade. The electrode shows a consistent -30 mV per decade slope, very close to the theoretical -29.6 mV. Answer A is wrong because the slope is negative, not positive. The sign depends on electrode construction and reference electrode polarity. Answer C is incorrect because super-Nernstian response would show slopes greater than 29.6 mV in magnitude, but we observe exactly the expected magnitude. Answer D is wrong because sub-Nernstian response would show smaller slopes than expected, indicating poor performance, but this electrode performs ideally. Remember: "Nernstian" refers to the slope magnitude matching theory (29.6 mV for divalent ions), while the sign depends on experimental setup. Focus on calculating the actual slope and comparing its magnitude to theoretical predictions.

Question 11

A student constructs a galvanic cell: $\text{Pt}|\text{H}_2(1 \text{ atm})|\text{HCl}(0.1 \text{ M})||\text{AgCl}(\text{sat}),\text{KCl}(\text{sat})|\text{Ag}$ . The measured cell potential is +0.352 V at 25°C. Using $E^\circ(\text{AgCl}/\text{Ag}) = +0.222 \text{ V}$ , what is the mean activity coefficient of HCl in the 0.1 M solution?

$\gamma_{\pm} = 0.65$
$\gamma_{\pm} = 0.78$ (correct answer)
$\gamma_{\pm} = 0.85$
$\gamma_{\pm} = 0.92$
$\gamma_{\pm} = 1.00$

Explanation: This galvanic cell problem tests your understanding of how activity coefficients relate real solution behavior to ideal behavior in electrochemistry. When you see a cell with known standard potentials but unknown activity coefficients, you'll need to use the Nernst equation to bridge measured and theoretical values. Start by identifying the half-reactions. The anode is the standard hydrogen electrode:

\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-

(

E° = 0.000 \text{ V}

). The cathode involves the silver chloride electrode:

\text{AgCl} + e^- \rightarrow \text{Ag} + \text{Cl}^-

(

E° = +0.222 \text{ V}

). The cell potential equation becomes:

E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF}\ln Q

, where

E°_{\text{cell}} = 0.222 - 0.000 = 0.222 \text{ V}

. For the reaction quotient,

Q = \frac{a_{\text{H}^+} \cdot a_{\text{Cl}^-}}{P_{\text{H}_2}} = a_{\text{H}^+} \cdot a_{\text{Cl}^-}

since

P_{\text{H}_2} = 1

. Since

a_{\text{H}^+} = a_{\text{Cl}^-} = \gamma_{\pm} \cdot 0.1

for the 1:1 electrolyte HCl, we get

Q = (\gamma_{\pm} \cdot 0.1)^2

. Substituting:

0.352 = 0.222 - \frac{0.0592}{2}\log[(\gamma_{\pm} \cdot 0.1)^2]

. Solving:

0.130 = -0.0592\log(\gamma_{\pm}^2 \cdot 0.01)

, which gives

\gamma_{\pm} = 0.78

. Choice A (0.65) would result from calculation errors in the logarithmic terms. Choice C (0.85) and D (0.92) represent values too close to ideal behavior for a 0.1 M electrolyte solution. Choice B (0.78) correctly accounts for the significant deviation from ideality expected at this concentration. Remember: activity coefficients for electrolytes are always less than 1 due to interionic attractions, and they decrease with increasing concentration.

Question 12

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?

$\text{Fe}^{2+}$ is oxidized first because it has the higher standard potential and is more easily oxidized
$\text{Ti}^{3+}$ is oxidized first because it has the lower standard potential and is more easily oxidized (correct answer)
$\text{Fe}^{2+}$ is oxidized first because the $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple has a larger potential difference with $\text{Ce}^{4+}/\text{Ce}^{3+}$
$\text{Ti}^{3+}$ is oxidized first because the $\text{Ti}^{4+}/\text{Ti}^{3+}$ couple has a larger potential difference with $\text{Ce}^{4+}/\text{Ce}^{3+}$
Both species are oxidized simultaneously since $\text{Ce}^{4+}$ can oxidize either species thermodynamically

Explanation: 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.

Question 13

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?

+0.11 V
+0.07 V (correct answer)
+0.03 V
-0.01 V
-0.05 V

Explanation: 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.

Question 14

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?

The potential will increase because the internal resistance of the cell decreases significantly
The potential will decrease to zero because both electrodes are now at the same physical potential (correct answer)
The potential will remain unchanged since the electrode reactions are unaffected by the connection type
The potential will initially remain the same but gradually decrease to zero as the solutions mix
The potential cannot be measured accurately due to the short-circuit created by the metallic connection

Explanation: 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.

Question 15

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?

pH = 2.1
pH = 4.2 (correct answer)
pH = 6.3
pH = 8.4
pH = 10.5

Explanation: 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.

Question 16

A combination pH electrode (glass electrode + Ag/AgCl reference) is calibrated using pH 4.00 and pH 7.00 buffer solutions, giving potentials of +180 mV and +2 mV respectively. When measuring an unknown sample, the electrode reads +95 mV. After recalibrating with fresh pH 7.00 buffer, the same electrode now reads -8 mV for the pH 7.00 solution. What is the most likely explanation for this shift?

The glass membrane has developed microcracks, allowing direct contact between sample and reference electrolyte
The internal Ag/AgCl reference has shifted due to chloride concentration changes in the internal filling solution (correct answer)
Temperature fluctuations have caused the electrode slope to change between calibration and measurement
The junction potential at the reference electrode salt bridge has changed due to contamination
The glass electrode has developed alkaline error due to high sodium concentration in the unknown sample

Explanation: When troubleshooting pH electrode problems, you need to distinguish between issues affecting the glass membrane versus the reference electrode system. The key diagnostic clue here is that the electrode reading for the same pH 7.00 buffer changed from +2 mV to -8 mV after recalibration. This 10 mV shift in the reference point indicates the reference electrode potential has drifted. In combination pH electrodes, the internal Ag/AgCl reference electrode maintains a stable potential based on the equilibrium:

\text{AgCl(s) + e^- \rightleftharpoons Ag(s) + Cl^-}

. When the chloride concentration in the internal filling solution changes (due to evaporation, leakage, or contamination), this equilibrium shifts, causing the reference potential to drift. This is exactly what answer B describes. Looking at the wrong answers: A is incorrect because microcracks in the glass membrane would cause erratic, unstable readings and likely complete electrode failure, not a systematic 10 mV shift. C doesn't explain the issue since temperature effects would be corrected during recalibration with fresh buffer at the same temperature. D is wrong because junction potential changes typically cause gradual drift during measurements, not a sudden shift between calibrations with the same buffer. The systematic shift in the reference reading is the smoking gun that points to internal reference electrode problems rather than glass membrane or junction issues. Study tip: When pH electrodes show consistent potential shifts for the same buffer solution, suspect reference electrode drift first. Glass membrane problems usually cause instability, while reference issues cause systematic shifts.

Question 17

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?

The silver/silver chloride electrode is contaminated and reading 6 mV more positive than expected
The silver/silver chloride electrode contains a different chloride concentration than the standard electrode (correct answer)
The junction potential between the reference and sample has changed significantly between measurements
Temperature effects have caused the silver/silver chloride potential to drift from its standard value
The hydrogen electrode was not properly calibrated and was reading more negative than expected

Explanation: 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.

Question 18

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?

The pH is approximately 12.2, indicating a strongly basic solution with standard H₂ pressure (correct answer)
The H₂ pressure is significantly below 1 atm while maintaining standard pH = 0 conditions
The electrode is malfunctioning since hydrogen electrodes cannot show negative potentials against SCE
The solution contains impurities that shift the reference potential by approximately 0.5 V from standard

Explanation: 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.

Question 19

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?

The Ag/AgCl electrode potential vs. SHE is +0.20 V, confirming proper electrode function (correct answer)
The Ag/AgCl electrode potential vs. SHE is -0.20 V, indicating an inverted measurement setup
The 0.20 V difference suggests significant junction potential effects between the reference electrodes
The measurements indicate that the Ag/AgCl electrode is contaminated or improperly prepared

Explanation: 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.

Question 20

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?

+1.10 V, since the standard cell potential remains constant during current flow under these conditions
+1.13 V, because the increased Zn²⁺ concentration enhances the overall cell driving force significantly
+1.07 V, accounting for the modest concentration changes in both half-cell compartments during operation (correct answer)
+0.98 V, reflecting the substantial decrease in Cu²⁺ concentration after prolonged current flow

Explanation: 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.