3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
For the reaction Ni(s) + Zn²⁺(aq) → Ni²⁺(aq) + Zn(s), E°cell = −0.51 V at 25°C. Which statement about this reaction under standard conditions is correct?
Answer and reasoning
AIt is thermodynamically favored, because a negative potential means energy is released A student who carries the sign rule for ΔH° over to cell potentials picks this. Because ΔG° = −nFE°, a negative E° gives a positive ΔG°: the reaction is unfavored.
BIt is thermodynamically unfavored, so it proceeds only with an external potentialCorrect A negative cell potential means ΔG° = −nFE° is positive: the reaction is thermodynamically unfavored. It proceeds only in an electrolytic cell, where an applied potential greater than 0.51 V drives it.
CIt is thermodynamically unfavored, so it occurs on its own but only very slowly A student who equates 'unfavored' with 'slow' picks this. Unfavored describes ΔG° > 0, not a rate: the reaction does not proceed to a significant extent on its own at any speed.
DIt is thermodynamically unfavored, so it cannot be made to occur under any conditions A student who equates 'unfavored' with 'impossible' picks this. An external potential larger than 0.51 V, applied in an electrolytic cell, drives the reaction.
A galvanic cell at 25°C is built from a standard Al³⁺/Al half-cell and a standard Cu²⁺/Cu half-cell. The cell reaction is 2 Al(s) + 3 Cu²⁺(aq) → 2 Al³⁺(aq) + 3 Cu(s). Using the standard reduction potentials in the table, what is E°cell?
Answer and reasoning
A+4.34 V A student who multiplies each E° by its coefficient picks this: 3(0.34 V) − 2(−1.66 V) = 4.34 V. E° is intensive and is not multiplied when a half-reaction is multiplied.
B−1.32 V A student who adds the two tabulated reduction potentials picks this: 0.34 V + (−1.66 V) = −1.32 V. Al is oxidized, so its reduction potential must be subtracted: 0.34 − (−1.66) = +2.00 V.
C+1.32 V A student who subtracts the sizes of the two potentials and ignores their signs picks this: 1.66 − 0.34 = 1.32 V. With signs, E°cell = 0.34 − (−1.66) = +2.00 V.
D+2.00 VCorrect Cu²⁺ is reduced at the cathode and Al is oxidized at the anode, so E°cell = E°cathode − E°anode = 0.34 V − (−1.66 V) = +2.00 V. The coefficients balance the electrons but do not multiply the potentials.
Working Cu²⁺ is reduced (cathode) and Al is oxidized (anode). E°cell = E°cathode − E°anode = (+0.34 V) − (−1.66 V) = +2.00 V. The coefficients 2 and 3 balance the electrons (6 e⁻) but do not change the potentials.
The standard reduction potentials at 25°C are +0.34 V for Cu²⁺(aq) + 2 e⁻ → Cu(s) and −0.74 V for Cr³⁺(aq) + 3 e⁻ → Cr(s). What is ΔG° for the reaction 2 Cr(s) + 3 Cu²⁺(aq) → 2 Cr³⁺(aq) + 3 Cu(s)? (F = 96,485 C/mol e⁻)
Answer and reasoning
A−208 kJ/molrxn A student who takes n = 2 from the half-reaction Cu²⁺ + 2 e⁻ → Cu as written in the table picks this. In the balanced overall equation three Cu²⁺ ions each gain two electrons, so n = 6.
B−521 kJ/molrxn A student who adds the 3 electrons in the Cr³⁺/Cr half-reaction to the 2 electrons in the Cu²⁺/Cu half-reaction, taking n = 5, picks this. The electrons lost by Cr are the same electrons gained by Cu²⁺, and in the balanced equation six are transferred, so n = 6.
C−625 kJ/molrxnCorrect E°cell = 0.34 − (−0.74) = 1.08 V and n = 6 (six electrons pass from the two Cr atoms to the three Cu²⁺ ions). ΔG° = −nFE° = −6(96,485)(1.08) J = −625 kJ/molrxn.
D−232 kJ/molrxn A student who subtracts the sizes of the two potentials without their signs picks this: E° = 0.74 − 0.34 = 0.40 V. Each value is used with its sign, so E°cell = 0.34 − (−0.74) = 1.08 V.
Working E°cell = E°cathode − E°anode = 0.34 V − (−0.74 V) = 1.08 V. Six electrons are transferred (2 Cr → 2 Cr³⁺ + 6 e⁻; 3 Cu²⁺ + 6 e⁻ → 3 Cu), so n = 6. ΔG° = −nFE° = −(6 mol e⁻)(96,485 C/mol e⁻)(1.08 J/C) = −6.25 × 10⁵ J = −625 kJ/molrxn.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.9.A.1 Electrochemistry Fix
Electrochemistry
The study of redox reactions that occur in electrochemical cells, where the oxidation and reduction half-reactions occur at separate electrodes and electrons pass between them through an external circuit.
Sign of the cell potential
A thermodynamically favored cell reaction gives a positive cell potential (a galvanic cell); a thermodynamically unfavored reaction gives a negative cell potential and proceeds only if an external potential is applied (an electrolytic cell).
Students often think A negative cell potential means that energy is released, so a reaction with a negative E° is thermodynamically favored, and more negative potentials release more energy. In fact No. For cell potentials the sign convention is the opposite of that for ΔH° and ΔG°: a positive E° corresponds to a negative ΔG°, a thermodynamically favored reaction. A negative E° means the reaction is unfavored.
Students often think A thermodynamically unfavored reaction does occur on its own, just very slowly. In fact No. Unfavored describes ΔG° > 0 (E° < 0): reactants are favored at equilibrium (K < 1), so on its own the reaction stops at a reactant-favored equilibrium however long it is left; when E° is strongly negative almost no product forms. How fast a reaction occurs is a separate, kinetic question.
9.9.A.2 Standard reduction potential, E° Fix
Standard reduction potential, E°
The potential, under standard conditions, of a half-reaction written as a reduction. A more positive E° means the species on the left is more readily reduced.
Standard cell potential, E°cell
The potential of a cell under standard conditions, found from the standard reduction potentials of its two half-reactions: E°cell = E°(reduction half-reaction, cathode) − E°(oxidation half-reaction written as a reduction, anode).
Cell potential as an intensive quantity
A cell potential does not depend on the amounts of substances or on how many times a half-reaction is used, so E° values are not multiplied by coefficients when half-reactions are balanced.
Students often think When a half-reaction is multiplied by a coefficient to balance the electrons, its E° is multiplied by the same coefficient. In fact No. E° is an intensive property: it does not depend on how many times the half-reaction occurs, so it is used unchanged however the half-reaction is multiplied.
Students often think E°cell is the difference between the sizes of the two tabulated reduction potentials, with their signs ignored. In fact No. The signs matter: E°cell = E°cathode − E°anode with each value taken with its sign, so a cathode at +0.34 V and an anode at −1.66 V give 0.34 − (−1.66) = +2.00 V.
9.9.A.3 n in ΔG° = −nFE° Fix
n in ΔG° = −nFE°
The number of moles of electrons transferred in the balanced overall redox equation as written (electrons lost by the species oxidized = electrons gained by the species reduced).
Faraday constant, F
The charge on one mole of electrons, F = 96,485 C/mol e⁻; with E° in volts (J/C), nFE° gives an energy in joules per mole of reaction.
ΔG° = −nFE°
The relationship between the standard free energy change of a redox reaction and its standard cell potential: a positive E° corresponds to a negative ΔG° (thermodynamically favored), and a negative E° to a positive ΔG°. ΔG° scales with the coefficients (through n) while E° does not.
Students often think n is the number of electrons shown in one of the half-reactions as written in a table of reduction potentials. In fact No. n is the number of moles of electrons transferred in the balanced overall equation; for Zn(s) + 2 Ag⁺(aq) → Zn²⁺(aq) + 2 Ag(s), n = 2, although Ag⁺ + e⁻ → Ag shows one electron.
Students often think n is found by adding the electrons in the oxidation half-reaction to the electrons in the reduction half-reaction. In fact No. The electrons lost in the oxidation are the same electrons gained in the reduction, so they are counted once: n = 2 for Zn(s) + 2 Ag⁺(aq) → Zn²⁺(aq) + 2 Ag(s).
6 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 6
A student builds a galvanic cell at 25°C from a Cu electrode in 1.0 M Cu(NO₃)₂(aq) and an Ag electrode in 1.0 M AgNO₃(aq) and records the voltmeter reading. The student then builds a second cell with electrodes of twice the surface area, dipping into twice the volume of each solution at the same concentrations and temperature. How does the reading for the second cell compare with the first, and why?
Answer and reasoning
AIt is the same, because cell potential does not depend on the amounts usedCorrect The cell potential is set by the half-reactions, the concentrations and the temperature, all unchanged. Larger electrodes and volumes let the cell deliver more charge in total, but at the same potential.
BIt is twice as large, because twice as many electrons can be transferred by the cell A student who thinks potential grows with the number of electrons a cell can transfer picks this. The second cell can transfer twice the charge, but its potential (energy per coulomb) is unchanged.
CIt is larger, because larger electrodes allow a greater current to flow in the circuit A student who treats voltage and current as the same thing picks this. Larger electrodes can allow a larger current, but the voltmeter measures potential, which does not change.
DIt is twice as large, because the larger cell stores twice as much chemical energy A student who thinks a cell's potential measures the total energy it stores picks this. The total energy available doubles, but it is delivered at the same potential.
For a redox reaction at 25°C, E° = +0.40 V and ΔG° = −77 kJ/molrxn. All the coefficients in the balanced equation are then doubled. How do E° and ΔG° for the new equation compare with the original values?
Answer and reasoning
AE° is doubled, and ΔG° is doubled A student who thinks E° is multiplied along with the equation picks this. E° is intensive; only n, and with it ΔG°, doubles.
BE° is unchanged, and ΔG° is unchanged A student who thinks ΔG° is fixed whatever the coefficients picks this. ΔG° is per mole of reaction as written, so it doubles when the coefficients double.
CE° is unchanged, and ΔG° is doubledCorrect E° depends only on the half-reactions and conditions, so it stays +0.40 V. n doubles, so ΔG° = −nFE° doubles to −154 kJ/molrxn.
DE° is halved, and ΔG° is unchanged A student who holds ΔG° fixed and doubles n in E° = −ΔG°/(nF) picks this. ΔG° doubles along with n, so E° stays at +0.40 V.
Working Doubling the coefficients doubles n, the moles of electrons transferred per mole of reaction as written. E° is intensive and does not change: +0.40 V. ΔG° = −nFE° doubles: −154 kJ/molrxn.
The diagram shows a galvanic cell at 25°C made from a standard Ag⁺/Ag half-cell (E° = +0.80 V for Ag⁺ + e⁻ → Ag) and a standard half-cell of an unknown metal, M, in M(NO₃)₂(aq). The arrows show the direction of electron flow, and the voltmeter shows the cell potential. What is E° for M²⁺(aq) + 2 e⁻ → M(s)?
Answer and reasoning
A−0.25 VCorrect Electrons flow out of M, so M is the anode and Ag⁺ is reduced at the cathode. 1.05 V = 0.80 V − E°(M²⁺/M), giving E°(M²⁺/M) = −0.25 V.
B+1.85 V A student who thinks electrons flow toward the anode, and so takes M²⁺ as the species reduced, picks this: 1.05 = E°(M) − 0.80 gives +1.85 V. Electrons leave the anode, so M is oxidized.
C−1.05 V A student who takes the voltmeter reading as the unknown half-cell's potential, made negative because M is oxidized, picks this. The reading is the difference between the two half-cell potentials, and the Ag⁺/Ag half-cell is at +0.80 V, not 0 V.
D+0.55 V A student who doubles the Ag⁺/Ag potential because two Ag⁺ ions are reduced per M atom picks this: 1.05 = 2(0.80) − E°(M) gives +0.55 V. E° is not multiplied by coefficients.
Working Electrons leave the M electrode, so M is oxidized (anode) and Ag⁺ is reduced (cathode). E°cell = E°cathode − E°anode: 1.05 V = 0.80 V − E°(M²⁺/M), so E°(M²⁺/M) = 0.80 V − 1.05 V = −0.25 V.
A student will build a galvanic cell at 25°C from two of the standard half-cells whose reduction potentials are listed in the table. Which two half-cells give the cell with the largest standard cell potential?
Answer and reasoning
ACu and Al half-cells A student who multiplies each E° by its coefficient in the balanced equation picks this: 3(0.34) + 2(1.66) = 4.34 V looks largest. Without multiplying, this cell gives 0.34 − (−1.66) = +2.00 V, less than Ag/Mg.
BAg and Cu half-cells A student who adds the tabulated reduction potentials picks the two most positive values: 0.80 + 0.34 = 1.14 V. The anode's potential is subtracted, so this cell gives only 0.80 − 0.34 = +0.46 V.
CAl and Mg half-cells A student who thinks more negative potentials release more energy picks the two most negative values. This pair gives only −1.66 − (−2.37) = +0.71 V.
DAg and Mg half-cellsCorrect E°cell = E°cathode − E°anode is largest when the half-reaction with the most positive E° (Ag⁺/Ag, +0.80 V) is paired with the one with the most negative E° (Mg²⁺/Mg, −2.37 V): E°cell = 0.80 − (−2.37) = +3.17 V. No other pair exceeds this (for example, Cu/Mg gives +2.71 V and Ag/Al gives +2.46 V).
The standard reduction potentials at 25°C are +0.80 V for Ag⁺(aq) + e⁻ → Ag(s) and −0.14 V for Sn²⁺(aq) + 2 e⁻ → Sn(s). What is E° for the reaction 2 Ag(s) + Sn²⁺(aq) → 2 Ag⁺(aq) + Sn(s)?
Answer and reasoning
A−0.94 VCorrect As written, Sn²⁺ is reduced and Ag is oxidized, so E° = −0.14 V − (+0.80 V) = −0.94 V. The coefficient 2 does not change the potential of the Ag⁺/Ag half-reaction, and the negative sign shows the reaction as written is thermodynamically unfavored.
B−1.74 V A student who multiplies the potential of the Ag⁺/Ag half-reaction by its coefficient picks this: −0.14 − 2(0.80). A standard reduction potential does not change when a half-reaction is multiplied, so E° = −0.14 − 0.80 = −0.94 V.
C+0.66 V A student who adds the two reduction potentials as they appear in the table picks this: 0.80 + (−0.14). Ag is oxidized in this reaction, so its reduction potential is subtracted: −0.14 − 0.80 = −0.94 V.
D+0.94 V A student who always subtracts the less positive reduction potential from the more positive one picks this: 0.80 − (−0.14). That is E° for the reverse reaction, in which Sn is oxidized; here Ag is oxidized and Sn²⁺ is reduced, so E° = −0.94 V.
Working In the equation as written, Sn²⁺ gains electrons (reduction) and Ag loses electrons (oxidation). E° = E°(reduction half-reaction) − E°(oxidation half-reaction), both as reduction potentials = −0.14 V − (+0.80 V) = −0.94 V. The potential of Ag⁺/Ag is not multiplied by 2. The negative value shows that the reaction as written is thermodynamically unfavored.
The particulate diagrams represent the two half-cells of a galvanic cell before and after the cell operates for a short time. Each diagram shows part of an electrode and the solution next to it; water molecules and spectator ions are not shown. Based on the diagrams, how many electrons moved through the wire between the two electrodes during this time?
Answer and reasoning
A1 A student who takes the electrons transferred to be the number shown in one tabulated half-reaction, Ag⁺ + e⁻ → Ag, picks this. That is the number gained by one Ag⁺ ion; six Ag⁺ ions are reduced in the diagrams, so 6 electrons are transferred.
B6Correct Two Cr atoms leave the electrode as Cr³⁺ ions, each losing 3 electrons, so 2 × 3 = 6 electrons are released. Six Ag⁺ ions become Ag atoms, each gaining 1 electron, which accounts for the same 6 electrons.
C4 A student who adds the electrons in the two half-reactions, 3 for Cr → Cr³⁺ + 3 e⁻ and 1 for Ag⁺ + e⁻ → Ag, picks this. The electrons lost in the oxidation are the same electrons gained in the reduction, and here 2 Cr atoms lose 6 electrons in total.
D8 A student who counts one electron for each particle that reacts picks this: 2 Cr atoms plus 6 Ag⁺ ions. Each Cr atom loses 3 electrons, so the 2 Cr atoms release 6 electrons, the same 6 that the six Ag⁺ ions gain.
Working Cr half-cell: the electrode goes from 5 to 3 Cr atoms and the solution from 1 to 3 Cr³⁺ ions, so 2 Cr atoms are oxidized, Cr → Cr³⁺ + 3 e⁻, releasing 2 × 3 = 6 electrons. Ag half-cell: the solution goes from 7 to 1 Ag⁺ ions and the electrode from 2 to 8 Ag atoms, so 6 Ag⁺ ions are reduced, Ag⁺ + e⁻ → Ag, gaining 6 × 1 = 6 electrons. Electrons transferred = 6.
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
