How to use this review
Use this page after the Unit 9 diagnostic and the topics it flagged. Answer all seven on paper before opening any answer; each combines two or more topics.
These are original Marlbridge practice questions, not past exam questions. Anything labelled “invented” is fictional. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Data. R = 8.314 J mol⁻¹ K⁻¹ = 0.08206 L atm mol⁻¹ K⁻¹; F = 96,485 C mol⁻¹; RT/F = 0.0257 V at 298 K. Molar masses: Co 58.93, Ni 58.69, Cu 63.55, Ag 107.87 g mol⁻¹. Standard reduction potentials (V): Ag⁺/Ag +0.80; O₂, H⁺/H₂O +1.23 (O₂ + 4H⁺ + 4e⁻ → 2H₂O); Cu²⁺/Cu +0.34; Ni²⁺/Ni −0.25; Co²⁺/Co −0.28.
Big ideas of the unit
- Entropy measures dispersal of matter (phase, moles of gas, volume) and of energy (temperature) (Topic 9.1).
- Absolute entropies give numbers. ΔS° is products minus reactants; elements are not zero (Topic 9.2).
- ΔG° weighs both drives at one temperature. Its sign decides favourability; the TΔS° term changes with T (Topic 9.3).
- ΔG° says nothing about speed. A favoured reaction with a high activation energy is under kinetic control, not at equilibrium (Topic 9.4).
- ΔG° and K carry the same information. ΔG° = −RT ln K; compare ΔG° with RT to judge how far K is from 1 (Topic 9.5).
- Dissolving is a ΔG° question too. Three particle-level factors partly cancel, so K_sp is hard to predict (Topic 9.6).
- Unfavourable changes need a push: outside energy, or a coupled reaction sharing a common intermediate (Topic 9.7).
- Cells turn ΔG° into a voltage. Galvanic cells run favoured reactions; electrolytic cells need a power supply; ΔG° = −nFE° (Topics 9.8 and 9.9).
- E tracks distance from equilibrium (E = 0 at Q = K); charge counts electrons, so amounts reacting (Topics 9.10 and 9.11).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Sign of ΔS | moles of gas, phase, volume, temperature | 9.1 |
| ΔS° | ΣnS°(products) − ΣnS°(reactants), J K⁻¹ mol⁻¹ | 9.2 |
| ΔG° | ΣnΔG°f(products) − ΣnΔG°f(reactants); or ΔH° − TΔS° (ΔS° in kJ, T in K) | 9.3 |
| Changeover temperature | T = ΔH° ÷ ΔS° when both have the same sign | 9.3, 9.5 |
| Kinetic control | ΔG° < 0 but no measurable rate | 9.4 |
| ΔG° and K | ΔG° = −RT ln K; K = e^(−ΔG°/RT) (ΔG° in J) | 9.5, 9.6 |
| Coupling | add equations and their ΔG° values; multiply K values | 9.7 |
| Cell potential | E°(cathode) − E°(anode); never scale E° by coefficients | 9.9 |
| E° and ΔG° | ΔG° = −nFE° | 9.9 |
| Nonstandard cells | E = E° − (RT/nF) ln Q; Q < 1 gives E > E°; E = 0 at Q = K | 9.10 |
| Faraday’s law | q = I × t; n(e⁻) = q ÷ F; half-reaction ratio | 9.11 |
Question 1 (multiple choice · mixed)
For an invented reaction, ΔH° = +32.0 kJ mol⁻¹ and ΔS° = +95.0 J K⁻¹ mol⁻¹. Assume both are constant. What is K at 350 K?
- (A) 0.226
- (B) 0.651
- (C) 1.54
- (D) 1.00
Answer and explanation
Answer: (C). ΔG° = 32.0 − 350(0.0950) = −1.25 kJ mol⁻¹, so K = e^(−ΔG°/RT) = e^(1250 ÷ (8.314 × 350)) = e^0.430 = 1.54. ΔG° is small next to RT, so K is near 1.
- (A) uses 298 K.
- (B) drops the minus sign in the exponent.
- (D) K = 1 only at the changeover, 32 000 ÷ 95.0 = 337 K.
Question 2 (multiple choice · mixed)
A galvanic cell has a nickel strip in 1 M Ni(NO₃)₂ and a copper strip in 1 M Cu(NO₃)₂. It delivers a steady 0.200 A for 2.00 hours. Which row is correct for the copper strip?
- (A) Gains 0.474 g; ΔG° = −56.9 kJ mol⁻¹
- (B) Gains 0.948 g; ΔG° = −114 kJ mol⁻¹
- (C) Loses 0.474 g; ΔG° = +114 kJ mol⁻¹
- (D) Gains 0.474 g; ΔG° = −114 kJ mol⁻¹
Answer and explanation
Answer: (D). Cu²⁺/Cu is higher, so copper is the cathode. E° = 0.34 − (−0.25) = 0.59 V and ΔG° = −(2)(96,485)(0.59) = −114 kJ mol⁻¹. Faraday: q = 0.200 × 7200 = 1440 C, n(e⁻) = 0.01492 mol, n(Cu) = 0.007462 mol, mass gained = 0.474 g.
- (A) uses n = 1 in ΔG° = −nFE°.
- (B) uses one electron per Cu.
- (C) makes copper the anode; a galvanic cell runs its favoured reaction.
Question 3 (multiple choice · mixed)
An invented galvanic cell has E° = +0.020 V and n = 2 at 298 K. It is set up with Q = 10. Which statement is correct?
- (A) K is about 4.7. Because Q > 1, E is positive but a little smaller than E°.
- (B) K is about 4.7. Because Q > K, E is negative and the net reaction runs in reverse.
- (C) K is about 2.2, and E is positive because E° is positive.
- (D) K equals 1, because E° is so close to zero, so E = 0.
Answer and explanation
Answer: (B). E° = (RT/nF) ln K, so ln K = 2(0.020) ÷ 0.0257 = 1.56 and K = 4.7. Q = 10 is past equilibrium, so the cell runs backwards: E = 0.020 − (0.0257/2) ln 10 = −0.0096 V.
- (A) compares Q only with 1; once Q passes K, E changes sign.
- (C) uses n = 1 and ignores Q.
- (D) A small E° gives K close to 1, not equal to 1.
Question 4 (constructed response · mixed)
An invented salt hydrate loses its water when heated:
D·2H₂O(s) → D(s) + 2H₂O(g) ΔH° = +110.0 kJ mol⁻¹
S° at 298 K (J K⁻¹ mol⁻¹): D·2H₂O(s) 180.0; D(s) 95.0; H₂O(g) 188.8.
(a) Predict the sign of ΔS° and justify it. (b) Calculate ΔS°. (c) Calculate ΔG° and K at 298 K. Describe what K tells you. (d) Calculate the temperature above which the dehydration is favoured. (e) A student says: “The hydrate is stable on the shelf because the dehydration is under kinetic control.” Evaluate this claim. (f) An invented drying agent Z absorbs water vapour: Z(s) + H₂O(g) → Z·H₂O(s), ΔG° = −30.0 kJ mol⁻¹. Show that coupling the dehydration with this reaction makes the overall process favoured at 298 K, and name the common intermediate.
Worked solution and suggested Marlbridge rubric
(a) Positive. A solid gives 2 mol of gas from no gas, so matter is far more dispersed.
(b) ΔS° = 95.0 + 2(188.8) − 180.0 = +292.6 J K⁻¹ mol⁻¹.
(c) ΔG° = 110.0 − 298(0.2926) = +22.8 kJ mol⁻¹. K = e^(−22 810 ÷ 2477.6) = e^(−9.21) = 1.0 × 10⁻⁴: far below 1 (reactant-favoured): very little water vapour forms at 298 K.
(d) Both terms are positive, so it is favoured at high T: T = 110 000 ÷ 292.6 = 376 K (about 103 °C).
(e) Wrong. Kinetic control means a favoured process (ΔG° < 0) that is too slow to see. Here ΔG° > 0: the hydrate is stable for thermodynamic reasons, and a catalyst would not make it lose water.
(f) Add the dehydration to twice the drying reaction. The 2H₂O(g) made is used up, so water vapour is the common intermediate: D·2H₂O(s) + 2Z(s) → D(s) + 2Z·H₂O(s) ΔG° = +22.8 + 2(−30.0) = −37.2 kJ mol⁻¹ < 0, so favoured.
| Point | What earns it |
|---|---|
| 1 | (a) Positive, with 0 → 2 mol of gas |
| 1 | (b) +292.6 J K⁻¹ mol⁻¹, including the coefficient 2 |
| 1 | (c) +22.8 kJ mol⁻¹, with ΔS° converted to kJ |
| 1 | (c) K = 1.0 × 10⁻⁴ and “reactant-favoured” |
| 1 | (d) 376 K from ΔH° ÷ ΔS° |
| 1 | (e) Rejects it: ΔG° > 0; kinetic control needs ΔG° < 0 |
| 1 | (f) H₂O(g) as common intermediate; drying reaction doubled |
| 1 | (f) −37.2 kJ mol⁻¹ and “favoured” |
Total: 8 points.
Question 5 (constructed response · mixed)
A galvanic cell has a cobalt strip in 1.0 M Co(NO₃)₂ and a silver strip in 1.0 M AgNO₃, joined by a wire and a KNO₃ salt bridge.
(a) Identify the anode, write both half-reactions and the overall equation, and describe how electrons, K⁺ and NO₃⁻ move. (b) Calculate E° and ΔG°. Use E° to explain whether K is greater or less than 1. (c) The cell delivers 0.150 A for 45.0 minutes. Calculate the change in mass of each electrode. (d) A second cell is identical except that [Co²⁺] = 0.010 M. Predict whether its starting E is greater or less than E°, and justify your answer in terms of Q.
Worked solution and suggested Marlbridge rubric
(a) Ag⁺/Ag (+0.80 V) is above Co²⁺/Co (−0.28 V), so Ag⁺ is reduced. Cobalt is the anode. Anode: Co(s) → Co²⁺(aq) + 2e⁻. Cathode: Ag⁺(aq) + e⁻ → Ag(s). Overall: Co(s) + 2Ag⁺(aq) → Co²⁺(aq) + 2Ag(s). Electrons flow through the wire from cobalt to silver. NO₃⁻ moves towards the cobalt half-cell (balancing Co²⁺ made); K⁺ moves towards the silver half-cell.
(b) E° = 0.80 − (−0.28) = +1.08 V. ΔG° = −(2)(96,485)(1.08) = −208 kJ mol⁻¹. A positive E° means ΔG° < 0, so K > 1 (ln K = 2(1.08) ÷ 0.0257 = 84; K is about 10³⁶).
(c) q = 0.150 × 2700 = 405 C; n(e⁻) = 405 ÷ 96,485 = 0.004198 mol. Silver: 0.004198 mol × 107.87 = 0.453 g gained. Cobalt: 0.004198 ÷ 2 = 0.002099 mol × 58.93 = 0.124 g lost.
(d) Q = [Co²⁺] ÷ [Ag⁺]² = 0.010 ÷ 1.0² = 0.010. Q < 1, so the cell starts further from equilibrium than at standard conditions: E > E°. (The Nernst equation gives about 1.14 V.)
| Point | What earns it |
|---|---|
| 1 | (a) Cobalt anode, half-reactions and overall equation |
| 1 | (a) Electrons Co → Ag in the wire; NO₃⁻ towards Co, K⁺ towards Ag |
| 1 | (b) +1.08 V and −208 kJ mol⁻¹ with n = 2 |
| 1 | (b) K > 1 because E° > 0 (ΔG° < 0) |
| 1 | (c) 0.453 g gained by silver |
| 1 | (c) 0.124 g lost by cobalt, using 2 electrons per Co |
| 1 | (d) E > E°, because Q < 1 puts the cell further from equilibrium |
Total: 7 points. No point in (d) for a Le Châtelier argument: a working cell is not at equilibrium.
Question 6 (constructed response · mixed)
An invented 1:1 salt MY dissolves endothermically: MY(s) ⇌ M⁺(aq) + Y⁻(aq). K_sp = 5.09 × 10⁻³ at 298 K and 2.17 × 10⁻² at 348 K.
(a) Calculate ΔG° of dissolution at each temperature. (b) Assume ΔH° and ΔS° are constant. Calculate both. Predict whether MY is more or less soluble at 5 °C. (c) Explain the signs of ΔH° and ΔS° in terms of the three particle-level factors in dissolving. (d) A student says: “K_sp is less than 1 at 298 K, so dissolving is not favoured and no MY dissolves.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) ΔG° = −RT ln K_sp. 298 K: −(8.314)(298) ln(5.09 × 10⁻³) = +13.1 kJ mol⁻¹. 348 K: −(8.314)(348) ln(2.17 × 10⁻²) = +11.1 kJ mol⁻¹.
(b) Write ΔG° = ΔH° − TΔS° twice: 13.1 = ΔH° − 298ΔS° and 11.1 = ΔH° − 348ΔS°. Subtract: 2.0 = 50ΔS°, so ΔS° = 0.040 kJ K⁻¹ mol⁻¹ = +40 J K⁻¹ mol⁻¹. Then ΔH° = 13.1 + 298(0.040) = +25.0 kJ mol⁻¹. At 278 K (5 °C), ΔG° = 25.0 − 278(0.040) = +13.9 kJ mol⁻¹, larger than at 298 K, so K_sp is smaller (about 2.5 × 10⁻³): MY is less soluble.
(c) ΔH° > 0: separating the ions and breaking some water–water attractions needs more energy than the new ion–water attractions release. ΔS° > 0 but modest: ions leaving the lattice gain entropy, partly cancelled by water ordering around the ions.
(d) Only partly right. ΔG° > 0 means K_sp < 1: the ions cannot reach 1 M. But K_sp is not zero: [M⁺] = [Y⁻] = √(5.09 × 10⁻³) = 0.071 M at equilibrium. MY is slightly soluble, not insoluble.
| Point | What earns it |
|---|---|
| 1 | (a) +13.1 and +11.1 kJ mol⁻¹ (ΔG° in J converted to kJ) |
| 1 | (b) Two equations in ΔH° and ΔS° set up correctly |
| 1 | (b) ΔS° = +40 J K⁻¹ mol⁻¹ and ΔH° = +25.0 kJ mol⁻¹ |
| 1 | (b) Less soluble at 5 °C, with a reason |
| 1 | (c) ΔH°: energy in versus energy out for the factors |
| 1 | (c) ΔS°: lattice gain partly cancelled by water ordering |
| 1 | (d) Small but not zero solubility (0.071 M) |
Total: 7 points.
Question 7 (constructed response · mixed)
A steel hinge is nickel-plated from NiSO₄(aq) with an inert platinum anode. Nickel deposits on the hinge and oxygen forms at the platinum; assume all the current goes to these two reactions.
(a) Write the half-reaction at each electrode and the overall equation. Explain why the cell needs a power supply. (b) Calculate E° and ΔG° for the overall reaction. State the smallest voltage the supply must provide under standard conditions. (c) A current of 2.50 A flows for 25.0 minutes. Calculate the mass of nickel deposited and the volume of oxygen at 298 K and 1.00 atm. (d) The platinum anode is replaced by a nickel bar. Predict, with a reason, how the smallest voltage needed changes, and what happens to the bar’s mass.
Worked solution and suggested Marlbridge rubric
(a) Cathode: Ni²⁺(aq) + 2e⁻ → Ni(s). Anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻. Overall (4 electrons): 2Ni²⁺(aq) + 2H₂O(l) → 2Ni(s) + O₂(g) + 4H⁺(aq). This direction is unfavoured (the reverse is favoured), so electrical energy from the supply must drive it.
(b) E° = −0.25 − 1.23 = −1.48 V. ΔG° = −(4)(96,485)(−1.48) = +571 kJ mol⁻¹. The supply must provide more than 1.48 V.
(c) q = 2.50 × 1500 = 3750 C; n(e⁻) = 0.03887 mol. Ni: 0.03887 ÷ 2 = 0.01943 mol × 58.69 = 1.14 g. O₂: 0.03887 ÷ 4 = 0.009717 mol; V = nRT/P = 0.009717 × 0.08206 × 298 ÷ 1.00 = 0.238 L (238 mL).
(d) Now nickel is oxidised at the anode, Ni(s) → Ni²⁺(aq) + 2e⁻. The overall change just moves nickel from bar to hinge, so E° = 0 and almost no voltage is needed (a little drives the current). The anode loses mass: 1.14 g, the mass the hinge gains.
| Point | What earns it |
|---|---|
| 1 | (a) Both half-reactions and the balanced overall equation |
| 1 | (a) Unfavoured reaction driven by electrical energy |
| 1 | (b) −1.48 V with E°(cathode) − E°(anode) |
| 1 | (b) +571 kJ mol⁻¹ with n = 4, and a minimum of 1.48 V |
| 1 | (c) 1.14 g of nickel |
| 1 | (c) 0.238 L of O₂, using 4 electrons per O₂ |
| 1 | (d) Nickel oxidised at the anode, so E° = 0 and almost no voltage is needed |
| 1 | (d) Anode loses 1.14 g, matching the hinge’s gain |
Total: 8 points.
How did you do?
Total: 33 points (3 multiple choice, 30 constructed response). Where you lost points matters most.
- ΔS° (4(a)–(b), 6(c)): revisit Topics 9.1 and 9.2.
- ΔG°, K and temperature (1, 4(c)–(d), 6(a)–(b)): revisit Topics 9.3 and 9.5.
- Kinetic control (4(e)), dissolving (6) or coupling (4(f), 7(a)): revisit Topics 9.4, 9.6 and 9.7.
- Cells, E° and Faraday’s law (2, 3, 5, 7): revisit Topics 9.8–9.11.
- Explanations marked down: name the quantity (ΔH°, ΔS°, ΔG°, Q or E), its sign and which way it pushes.
If several topics were weak, return to the Unit 9 diagnostic, then tick off each topic checklist: 9.1, 9.2, 9.3, 9.4, 9.5, 9.6, 9.7, 9.8, 9.9, 9.10, 9.11.