4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A galvanic cell based on the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is connected to a light bulb and allowed to operate at constant temperature. How does the cell potential change as the cell operates, and why?
Answer and reasoning
AIt decreases, because Q rises as Zn²⁺ forms and Cu²⁺ is used up, nearing equilibriumCorrect As the cell operates, Q = [Zn²⁺]/[Cu²⁺] rises toward K. The closer the reaction is to equilibrium, the smaller the magnitude of the cell potential, which reaches zero at equilibrium.
BIt stays the same, because the potential is fixed by the two half-reactions in the cell A student who thinks a cell's potential is fixed by its half-reactions picks this. That is true only of E°; the actual potential depends on the concentrations, which change as the cell operates.
CIt decreases, because the zinc electrode becomes smaller as zinc atoms are oxidized A student who links the potential to electrode size picks this. The potential does fall, but because Q rises toward K, not because the electrode shrinks; electrode size does not affect potential.
DIt stays the same until the Cu²⁺ ions are used up, and it then falls to zero A student who pictures a battery working normally and then suddenly dying picks this. Q rises continuously as the cell operates, so the potential falls gradually, reaching zero when Q = K.
A galvanic cell based on Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is operating at 25°C. Solid Cu(NO₃)₂ is dissolved in the copper half-cell, increasing [Cu²⁺], and the measured cell potential increases. Which explanation of the increase is correct?
Answer and reasoning
AThe cell's equilibrium shifts toward products, as Le Châtelier's principle predicts A student who treats the operating cell as an equilibrium system picks this. A cell with a nonzero potential is not at equilibrium, so equilibrium arguments do not apply; the increase is explained by Q moving farther from K.
BMore ions are present in solution, so more charge carriers raise the potential A student who thinks the potential grows with the total number of ions picks this. The potential depends on Q, a ratio; adding Zn²⁺ ions instead would also add ions but would lower the potential.
CE° for the cell increases, because the reactant ion is now more concentrated A student who confuses E with E° picks this. E° refers to standard conditions and does not change; the measured potential E increases because Q decreases.
DQ decreases, so the cell is farther from equilibrium, and its driving force is greaterCorrect Q = [Zn²⁺]/[Cu²⁺] falls when [Cu²⁺] increases, moving the cell farther from equilibrium (Q = K). The cell potential measures this driving force, so it increases.
The diagrams represent equal volumes of the two half-cell solutions of a galvanic cell at 25°C based on Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), for which E° = +1.10 V. Nitrate ions and water molecules are not shown. How does the cell potential at this moment compare with 1.10 V, and why?
Answer and reasoning
AIt is greater than 1.10 V, because Q is lower than 1 A student who writes Q as reactants over products gets Q = 2/6 and picks this. Q = [Zn²⁺]/[Cu²⁺] = 3, so the potential is below 1.10 V.
BIt is greater than 1.10 V, because Q is greater than 1 A student who thinks a larger Q gives a larger potential picks this. In E = E° − (RT/nF) ln Q, a Q greater than 1 makes the subtracted term positive, so E < E°.
CIt is less than 1.10 V, because Q is greater than 1Correct The diagrams show three times as many Zn²⁺ ions as Cu²⁺ ions in equal volumes, so Q = [Zn²⁺]/[Cu²⁺] = 3. With Q > 1 the cell is closer to equilibrium than at standard conditions, so E < E°.
DIt equals 1.10 V, because the potential does not depend on Q A student who thinks a cell's potential always equals E° picks this. E equals E° only when Q = 1; here Q = 3.
For a galvanic cell at 25°C whose standard cell potential, E°, is known, a student wants to decide without calculation whether the actual cell potential, E, is greater or less than E°. Which information is sufficient?
Answer and reasoning
AWhether the equilibrium constant K is greater or less than 1 A student who confuses K with Q picks this. K decides the sign of E°, which is already known; how E compares with E° depends on Q.
BWhether the reaction quotient Q is greater or less than 1Correct In E = E° − (RT/nF) ln Q, ln Q is negative when Q < 1 (E > E°) and positive when Q > 1 (E < E°), so comparing Q with 1 is enough.
CWhether the reactant ion's concentration is greater or less than 1 M A student who thinks only the reactant ion's concentration matters picks this. Product concentrations are in Q too; a high product concentration can make Q > 1 even when the reactant ion is above 1 M.
DWhether any concentration is different from the standard value of 1 M A student who thinks any departure from standard conditions lowers the potential picks this. Departures can raise or lower E, depending on whether Q falls below or rises above 1.
Working From E = E° − (RT/nF) ln Q, the sign of ln Q decides the comparison: if Q < 1, ln Q < 0 and E > E°; if Q > 1, ln Q > 0 and E < E°. So it is sufficient to know whether Q is greater or less than 1.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.10.A.1 Nonstandard conditions Fix
Nonstandard conditions
Conditions in which the concentrations (or partial pressures) of the species in the cell reaction are not all at their standard values (1 M, 1 atm). The cell potential E then differs from E° whenever Q ≠ 1.
Cell potential as a driving force
The cell potential measures how far the cell reaction is from equilibrium and in which direction it will proceed: the farther from equilibrium, the greater the magnitude of E.
Students often think A cell's potential is fixed by its two half-reactions, so it equals E° whatever the concentrations of the solutions. In fact No. E equals E° only when Q = 1. Under nonstandard conditions the potential depends on the concentrations of the species in the cell reaction, through Q.
Students often think A cell's potential stays at a constant value while it operates and drops to zero only when a reactant has been completely used up. In fact No. As the cell operates, products build up and reactants are used up, so Q rises toward K and the potential falls gradually, reaching zero at equilibrium.
9.10.A.2 Operating cell is not at equilibrium Fix
Operating cell is not at equilibrium
While a cell has a nonzero potential, its reaction is not at equilibrium, so equilibrium-shift arguments (Le Châtelier's principle) are not used to explain its potential; changes are explained by how Q compares with 1 and with K.
Students often think An operating galvanic cell is at equilibrium, so changes in its potential are explained by Le Châtelier's principle (equilibrium shifts). In fact No. A cell with a nonzero potential is not at equilibrium; its potential measures how far it is from equilibrium. Equilibrium arguments such as Le Châtelier's principle do not apply; changes are explained by how a change moves Q relative to K.
9.10.A.3 E° and Q = 1 Fix
E° and Q = 1
The standard cell potential E° is the cell potential when Q = 1 (all species at standard concentrations or pressures).
E = 0 at equilibrium
As a cell operates, Q approaches K and the magnitude of the cell potential decreases, reaching zero when Q = K; a cell at equilibrium can do no electrical work.
Concentration cell
A cell made from two half-cells with the same half-reaction but different concentrations (for example, Cu in 0.010 M Cu²⁺ and Cu in 1.0 M Cu²⁺). E° = 0, and electrons flow spontaneously in the direction that makes the two concentrations more nearly equal. In a cell of a metal and its ions, such as this one, that means oxidation in the dilute half-cell and reduction in the concentrated half-cell.
Students often think A cell's potential is greatest at standard conditions, so any departure from 1 M concentrations lowers it below E°. In fact No. Departures that move Q below 1 (farther from equilibrium for a favored cell) raise the potential above E°; departures that move Q above 1 lower it.
Students often think A cell with two electrodes of the same metal produces no potential, because there is no difference between the half-reactions. In fact No. A concentration cell has E° = 0, but if the two solutions have different concentrations, Q ≠ 1 and the cell has a nonzero potential that drives electrons until the concentrations are equal.
9.10.A.4 Nernst equation Fix
Nernst equation
E = E° − (RT/nF) ln Q. Used qualitatively: if Q < 1, ln Q < 0 and E > E°; if Q > 1, E < E°; at Q = K, E = 0.
Students often think If K for the cell reaction is greater than 1, the cell potential is greater than E°. In fact No. K (fixed at a given temperature) determines the sign of E°; whether E is above or below E° depends on Q, the actual concentrations: E > E° when Q < 1 and E < E° when Q > 1.
Students often think A cell's potential increases as Q increases, because Q is in the numerator of the Nernst term. In fact No. In E = E° − (RT/nF) ln Q the term with ln Q is subtracted, so as Q increases (more products, fewer reactants) E decreases.
4 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 4
A student measures the potential of a cell made from a Zn electrode in 1.0 M Zn(NO₃)₂(aq) and a Cu electrode in 1.0 M Cu(NO₃)₂(aq) at 25°C; the cell reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). The student repeats the measurement after replacing the copper half-cell's solution with 0.010 M Cu(NO₃)₂(aq). How does the second reading compare with the first, and why?
Answer and reasoning
AIt is higher, because the smaller [Cu²⁺] increases Q and a larger Q raises the potential A student who thinks a cell's potential increases as Q increases picks this. Q does rise from 1 to 100, but the ln Q term is subtracted in E = E° − (RT/nF) ln Q, so E falls.
BIt is lower, because the smaller [Cu²⁺] increases Q and brings the cell closer to equilibriumCorrect Q = [Zn²⁺]/[Cu²⁺] rises from 1 to 100. The cell is then closer to equilibrium (Q nearer to K), so the cell potential is smaller than E°.
CIt is higher, because the smaller [Cu²⁺] decreases Q and takes the cell farther from equilibrium A student who writes Q as reactants over products picks this. Q = [Zn²⁺]/[Cu²⁺] (products over reactants), so lowering [Cu²⁺] increases Q and lowers the potential.
DIt is lower, because the dilute solution conducts too poorly for the cell to work well A student who links the voltmeter reading to how well the solution conducts picks this. The reading is lower, but because Q has increased; a voltmeter measures potential, which does not depend on the solution's conductivity.
The diagram shows a concentration cell at 25°C. In which direction do electrons flow through the external circuit, and why?
Answer and reasoning
AFrom right to left, because electrons move away from the more concentrated Cu²⁺ solution A student who expects electrons to move from high to low concentration, as in diffusion, picks this. That direction would make the concentrations more different; electrons flow from the dilute half-cell (anode) to the concentrated one (cathode).
BNo electrons flow, because both electrodes are made of the same metal A student who thinks identical electrodes give no potential picks this. E° = 0 for a concentration cell, but Q ≠ 1 here, so the cell has a potential and electrons flow until the concentrations are equal.
CNo electrons flow, because the cell's potential is equal to its E° value, which is zero A student who thinks a cell's potential always equals E° picks this. E° = 0, but with unequal concentrations Q = 0.010 ≠ 1, so E = −(RT/2F) ln Q is positive.
DFrom left to right, because this brings the two Cu²⁺ concentrations closer togetherCorrect The cell moves toward equilibrium, equal concentrations. Cu is oxidized in the dilute left half-cell (adding Cu²⁺) and Cu²⁺ is reduced in the concentrated right half-cell (removing Cu²⁺), so electrons flow from left to right.
The graph shows how the potential, E, of a galvanic cell at 25°C depends on log Q for the cell reaction. Which statement is supported by the graph?
Answer and reasoning
AThe cell reaction is at equilibrium when Q = 1 A student who thinks standard conditions are equilibrium conditions picks this. At Q = 1 (log Q = 0) the graph shows E = E° ≈ 0.30 V, not zero, so the reaction is not at equilibrium.
BBoth forward and reverse reactions stop when E falls to 0 V A student who thinks reactions stop at equilibrium picks this. Where the line reaches E = 0 the cell is at equilibrium: the forward and reverse reactions continue at equal rates, so there is no net current.
CThe cell reaction is at equilibrium when Q = 1 × 10¹⁰Correct The potential falls to zero at log Q = 10. E = 0 means there is no driving force: the reaction is at equilibrium, so K = 1 × 10¹⁰.
DThe cell potential would be below E° if Q were less than 1 A student who thinks any departure from standard conditions lowers the potential picks this. The line rises toward smaller log Q, so for Q < 1 (log Q < 0) E would be greater than E°.
Working E = 0 where the line crosses the horizontal axis, at log Q = 10. The cell has no driving force there, so the reaction is at equilibrium: Q = K = 1 × 10¹⁰. At log Q = 0 (Q = 1), E = E° ≈ 0.30 V; for log Q < 0 the line would lie above 0.30 V.
A student measures the potentials of four cells at 25°C, each based on Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with the concentrations shown in the table. Which claim is supported by the data?
Answer and reasoning
AE increases as the total concentration of ions in the two solutions increases A student who thinks more ions give a higher potential picks this. Cells 1 and 4 have very different total ion concentrations but the same potential, and cell 3 has fewer ions than cell 1 but a higher potential.
BE depends on [Cu²⁺], the reactant ion's concentration, and not on [Zn²⁺] A student who thinks only the reactant ion's concentration matters picks this. Cells 1 and 3 have the same [Cu²⁺] but different potentials, so [Zn²⁺] affects E.
CE is greatest when both solutions are at the standard concentration of 1.0 M A student who thinks standard conditions give the largest potential picks this. Cell 3, with [Zn²⁺] = 0.10 M, has a higher potential (1.13 V) than cell 1 (1.10 V).
DE depends on the ratio of [Zn²⁺] to [Cu²⁺], and not on either concentration aloneCorrect Cells 1 and 4 have the same ratio, [Zn²⁺]/[Cu²⁺] = 1, and the same potential, although cell 4 is ten times more dilute. A larger ratio (cell 2, ratio 10) gives a lower potential and a smaller ratio (cell 3, ratio 0.1) a higher one, as expected as Q changes.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
