Unit 9 practice test: Thermodynamics and Electrochemistry
12 exam-style questions from the unit, free, with an explanation for every option. Four answer choices each, and data sets that share one table, graph or particle diagram, as on the real exam.
Covers: Introduction to Entropy; Absolute Entropy and Entropy Change; Gibbs Free Energy and Thermodynamic Favorability; Thermodynamic and Kinetic Control; Free Energy and Equilibrium; Free Energy of Dissolution; Coupled Reactions; Galvanic (Voltaic) and Electrolytic Cells; Cell Potential and Free Energy; Cell Potential Under Nonstandard Conditions; Electrolysis and Faraday's Law.
Particle view
Ions in water
A student draws one cation (M⁺) and one anion (X⁻) surrounded by water molecules after a salt dissolves. In each water molecule the oxygen end is partly negative and the hydrogen ends are partly positive.
Key: dark circle with label, an ion; mid-size orange circle, oxygen atom; small white circle, hydrogen atom.
1. Is the orientation of the water molecules in the drawing correct?
- No: the oxygen should face the anion, because both are electron-rich.
- No: the hydrogens should face the cation, because hydrogen bonds form to cations.
- Yes: δ− oxygen faces the cation and a δ+ hydrogen faces the anion.
- Yes, because water molecules point the same way around any ion.
Show the answer
In ion-dipole attraction, oxygen (δ−) points toward cations and hydrogen (δ+) points toward anions.
- No: the oxygen should face the anion, because both are electron-rich.: Like charges repel; the partly negative oxygen is attracted to the cation, not the anion.
- No: the hydrogens should face the cation, because hydrogen bonds form to cations.: Partly positive hydrogens are repelled by a cation. They point toward anions.
- Correct: Yes: δ− oxygen faces the cation and a δ+ hydrogen faces the anion.: Right: opposite partial charges attract, so this is the ion-dipole orientation.
- Yes, because water molecules point the same way around any ion.: The drawing shows different orientations around the two ions, which is what makes it right.
2. Forming the attractions shown in the drawing has which energy effect?
- It absorbs energy, because water has to be pulled toward the ions.
- It has no energy effect, because no chemical bonds form.
- It releases energy, as opposite partial charges come together.
- It absorbs energy, because the ions break apart from the crystal at that moment.
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Dissolving has three steps: separating solute particles (absorbs), separating solvent particles (absorbs) and forming solute-solvent attractions (releases).
- It absorbs energy, because water has to be pulled toward the ions.: Attractions forming release energy; separating particles is what absorbs it.
- It has no energy effect, because no chemical bonds form.: Ion-dipole attractions are intermolecular, but forming them still releases energy.
- Correct: It releases energy, as opposite partial charges come together.: Right: forming any attraction lowers the potential energy and releases energy (the third step of dissolving).
- It absorbs energy, because the ions break apart from the crystal at that moment.: Separating the ions from the crystal is a different step, and that one absorbs energy.
3. The ion-dipole attractions in the drawing are stronger for Mg²⁺ than for Na⁺. Which is the best explanation?
- Mg²⁺ has a larger charge and smaller radius, so its Coulombic attraction is greater.
- Mg²⁺ has a larger molar mass, so it attracts water more strongly.
- Mg²⁺ has more electrons than Na⁺, so it is more polarizable.
- Mg²⁺ forms covalent bonds with water, while Na⁺ forms ionic bonds.
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Coulombic attraction grows with charge and falls with distance. Mg²⁺ (2+, smaller) beats Na⁺ (1+, larger).
- Correct: Mg²⁺ has a larger charge and smaller radius, so its Coulombic attraction is greater.: Right: both the charge and the distance favor Mg²⁺.
- Mg²⁺ has a larger molar mass, so it attracts water more strongly.: Mass does not set the strength of an electrostatic attraction; charge and distance do.
- Mg²⁺ has more electrons than Na⁺, so it is more polarizable.: Mg²⁺ and Na⁺ have the same number of electrons (10), and the key is charge and size.
- Mg²⁺ forms covalent bonds with water, while Na⁺ forms ionic bonds.: Both form ion-dipole attractions with water; the strength differs because of charge and size.
4. How does the ordering of water shown in the drawing affect ΔS°soln?
- It raises ΔS°soln, because water molecules move in closer to the ions.
- It has no effect on ΔS°soln, because the water was already a liquid.
- It raises ΔS°soln, because ion-dipole attractions release heat.
- It lowers ΔS°soln: water held around ions has fewer arrangements.
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Hydration orders the water (S down) while the ions disperse (S up). The net sign depends on which is larger.
- It raises ΔS°soln, because water molecules move in closer to the ions.: Being held close to an ion restricts the water molecules, which lowers their entropy.
- It has no effect on ΔS°soln, because the water was already a liquid.: The water molecules in the shells lose freedom compared with bulk liquid, so the entropy changes.
- It raises ΔS°soln, because ion-dipole attractions release heat.: Releasing heat affects ΔH°, and through the surroundings, not the ordering effect asked about here.
- Correct: It lowers ΔS°soln: water held around ions has fewer arrangements.: Right: hydration shells reduce the freedom of the water, an entropy decrease that offsets the ions dispersing.
Data table
Metal strips in solutions
A student places a clean metal strip in each of six solutions and records whether a coating forms on the strip after 10 minutes.
| Trial | Metal | Solution | Observation |
|---|---|---|---|
| 1 | Zn | Cu(NO₃)₂ | red-brown coating of Cu forms |
| 2 | Cu | Zn(NO₃)₂ | no change |
| 3 | Cu | AgNO₃ | gray coating of Ag forms |
| 4 | Ag | Cu(NO₃)₂ | no change |
| 5 | Zn | AgNO₃ | gray coating of Ag forms |
| 6 | Ag | Zn(NO₃)₂ | no change |
5. In Trial 1, which species is oxidized?
- Zn(s)
- Cu²⁺(aq)
- NO₃⁻(aq)
- Cu(s)
Show the answer
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s): zinc goes from 0 to +2 (oxidized).
- Correct: Zn(s): Right: Zn atoms lose electrons and become Zn²⁺(aq), while Cu²⁺ gains them and plates out.
- Cu²⁺(aq): Cu²⁺ gains electrons to become the copper coating: it is reduced.
- NO₃⁻(aq): Nitrate is a spectator ion here; its oxidation number does not change.
- Cu(s): Copper metal is a product in Trial 1, not something that reacts.
6. A galvanic cell is built from a Cu strip in Cu(NO₃)₂(aq) and an Ag strip in AgNO₃(aq). Based on the trials, which electrode is the anode?
- Silver, because Trial 4 shows silver does not react with Cu²⁺
- Silver, because silver is the more valuable metal
- Copper, because copper has the larger mass of the two metals
- Copper, because in Trial 3 Cu(s) gives electrons to Ag⁺
Show the answer
The cell runs the favored reaction from Trial 3. Copper is oxidized, so the copper electrode is the anode.
- Silver, because Trial 4 shows silver does not react with Cu²⁺: Trial 4 shows that silver is not oxidized by Cu²⁺; silver ions are reduced instead, at the cathode.
- Silver, because silver is the more valuable metal: The cost of a metal has nothing to do with which electrode is the anode.
- Copper, because copper has the larger mass of the two metals: Atomic mass does not decide which metal is oxidized; the observed reaction does.
- Correct: Copper, because in Trial 3 Cu(s) gives electrons to Ag⁺: Right: the favored reaction is Cu + 2 Ag⁺ → Cu²⁺ + 2 Ag, so copper is oxidized at the anode.
7. Which ranks the metals from the one most easily oxidized to the one least easily oxidized?
- Ag > Cu > Zn
- Zn > Cu > Ag
- Cu > Zn > Ag
- Zn > Ag > Cu
Show the answer
A metal that coats with another metal from solution is being oxidized by those ions. Zn reacts with both ions, Cu with one, Ag with none.
- Ag > Cu > Zn: This is reversed: silver never reacts, so it is the least easily oxidized.
- Correct: Zn > Cu > Ag: Right: Zn gives electrons to Cu²⁺ and Ag⁺; Cu gives electrons to Ag⁺ but not to Zn²⁺; Ag gives electrons to neither.
- Cu > Zn > Ag: Copper does not react with Zn²⁺ (Trial 2), so zinc is more easily oxidized than copper.
- Zn > Ag > Cu: Silver does not react with Cu²⁺ (Trial 4), so copper is more easily oxidized than silver.
8. If a strip of Ag is placed in Zn(NO₃)₂(aq), as in Trial 6, and connected to a power supply that forces zinc metal to plate onto it, what kind of cell is this?
- An electrolytic cell: outside energy drives an unfavored reaction
- A galvanic cell, because a metal plates onto an electrode
- A galvanic cell, because the power supply produces a current
- Neither, because a reaction that is not favored will not happen
Show the answer
An electrolytic cell uses an external energy source to drive a thermodynamically unfavored redox reaction.
- Correct: An electrolytic cell: outside energy drives an unfavored reaction: Right: Trial 6 shows Zn²⁺ is not reduced by Ag on its own, so the power supply must drive it.
- A galvanic cell, because a metal plates onto an electrode: Plating happens in both kinds of cell. What decides the type is whether the reaction is favored or driven.
- A galvanic cell, because the power supply produces a current: In a galvanic cell the reaction itself produces the current; here the current is supplied from outside.
- Neither, because a reaction that is not favored will not happen: Outside energy can drive an unfavored reaction (topic 9.7); that is what an electrolytic cell does.
9. On a cold morning, water vapor condenses into droplets on a window. Which other process has the same sign of change in entropy for the substance involved?
- Liquid bromine freezing to a solid
- Dry ice, solid CO₂, turning straight to gas
- Sugar dissolving in a cup of tea
- A puddle of water evaporating
Show the answer
Condensation (gas → liquid) lowers the entropy of the water. Freezing (liquid → solid) is the only other change here that lowers entropy.
- Correct: Liquid bromine freezing to a solid: Right: freezing, like condensing, moves the particles to a phase where they have fewer arrangements, so ΔS < 0.
- Dry ice, solid CO₂, turning straight to gas: Solid to gas (sublimation) spreads the particles out, so ΔS > 0, the opposite sign.
- Sugar dissolving in a cup of tea: Dissolving spreads the sugar molecules through the liquid, so ΔS > 0.
- A puddle of water evaporating: Liquid to gas spreads the molecules out, so ΔS > 0, the opposite of condensing.
10. Without a table, which substance has the largest standard molar entropy at 298 K?
- C₂H₆(g)
- C(s, diamond)
- H₂O(l)
- CH₄(g)
Show the answer
Gases beat liquids and solids. Between two gases, the molecule with more atoms has more ways to rotate and vibrate.
- Correct: C₂H₆(g): Right: a gas, and a molecule with eight atoms, so it has the most ways to store energy.
- C(s, diamond): A rigid network solid has very few microstates: its S° is among the smallest of all substances.
- H₂O(l): A liquid has more entropy than a solid but much less than a gas.
- CH₄(g): A gas, but with fewer atoms per molecule than C₂H₆, so fewer ways to store energy.
11. For N₂(g) + 3 H₂(g) → 2 NH₃(g), ΔH° = −92.2 kJ/mol and ΔS° = −198.1 J/(mol·K). Calculate ΔG° at 298 K.
Type a number and its unit.
Show the answer
ΔG° = ΔH° − TΔS° = −92.2 kJ/mol − (298 K)(−0.1981 kJ/(mol·K)) = −92.2 + 59.03 = −33.2 kJ/mol. Rounded to tenths, matching ΔH°.
- Answer: -33.2 kJ/mol
12. Adding a catalyst to a reaction that is under kinetic control has which effect?
- It makes ΔG° more negative, so the reaction becomes more favored.
- It lowers Ea so the reaction can run at a measurable rate; ΔG° stays the same.
- It makes a reaction with ΔG° > 0 favored.
- It raises the activation energy of the reverse reaction alone, so products build up.
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A catalyst changes the pathway and Ea, so it changes the rate. It cannot change ΔG°, ΔH° or K.
- It makes ΔG° more negative, so the reaction becomes more favored.: ΔG° depends on the reactants and products, which a catalyst does not change.
- Correct: It lowers Ea so the reaction can run at a measurable rate; ΔG° stays the same.: Right: a catalyst provides a lower-barrier pathway but does not change the starting or ending free energies.
- It makes a reaction with ΔG° > 0 favored.: A catalyst cannot make an unfavored reaction favored; it speeds both directions equally.
- It raises the activation energy of the reverse reaction alone, so products build up.: A catalyst lowers the barrier in both directions.
Keep going
Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8 · Unit 9.