Thermodynamics and Electrochemistry: Unit Diagnostic (Chemistry Unit 9)

What this diagnostic is for

Use this page to find which Unit 9 topics to revisit: one question per topic, two for the larger Topics 9.3 and 9.9. Answer them all before opening any answer.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your total gives no predicted score.

Data. A scientific calculator is allowed. R = 8.314 J mol⁻¹ K⁻¹; F = 96,485 C mol⁻¹; RT/F = 0.0257 V at 298 K. Molar masses: H 1.008, C 12.01, O 16.00, Al 26.98 g mol⁻¹. Questions 3 and 8 use invented data. Standard reduction potentials for Questions 10 to 12:

Half-reaction	E° (V)
Ag⁺(aq) + e⁻ → Ag(s)	+0.80
Cu²⁺(aq) + 2e⁻ → Cu(s)	+0.34
Sn²⁺(aq) + 2e⁻ → Sn(s)	−0.14
Ni²⁺(aq) + 2e⁻ → Ni(s)	−0.25
Fe²⁺(aq) + 2e⁻ → Fe(s)	−0.44
Zn²⁺(aq) + 2e⁻ → Zn(s)	−0.76
Mg²⁺(aq) + 2e⁻ → Mg(s)	−2.37

Question 1 (multiple choice · 9.1)

For which process is ΔS of the system expected to be positive?

(A) 2CO(g) + O₂(g) → 2CO₂(g)
(B) PCl₅(g) → PCl₃(g) + Cl₂(g)
(C) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
(D) Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Answer and explanation

Answer: (B). One mole of gas becomes two, so matter is more dispersed.

(A) goes from 3 mol of gas to 2.
(C) Water is liquid: gas falls from 9 mol to 4.
(D) Ions in solution become a solid.

If you missed this: read “Matter dispersal in reactions” in the Topic 9.1 study guide.

Question 2 (short answer · 9.2)

Methanol burns, with the water formed as a gas:

2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)

S° at 298 K (J K⁻¹ mol⁻¹): CH₃OH(l) 126.8; O₂(g) 205.2; CO₂(g) 213.8; H₂O(g) 188.8.

(a) Predict the sign of ΔS° and give a reason. (b) Calculate ΔS°. (c) Calculate the entropy change when 8.01 g of methanol burns.

Worked answer

(a) Positive. Moles of gas rise from 3 to 6 (liquid methanol does not count).

(b) Products: 2(213.8) + 4(188.8) = 1182.8. Reactants: 2(126.8) + 3(205.2) = 869.2. ΔS° = 1182.8 − 869.2 = +313.6 J K⁻¹ mol⁻¹, as predicted.

(c) n = 8.01 ÷ 32.042 = 0.2500 mol of methanol = 0.1250 mol of reaction (2 per equation). ΔS = 0.1250 × 313.6 = +39.2 J K⁻¹.

If you missed this: read Worked example 1 in the Topic 9.2 study guide.

Question 3 (multiple choice · 9.3)

An invented gas-phase reaction has ΔH° = −84.0 kJ mol⁻¹ and ΔS° = −210 J K⁻¹ mol⁻¹. Over which range is it thermodynamically favoured?

(A) Below about 400 °C
(B) Above about 127 °C
(C) Below about 127 °C
(D) At all temperatures, because ΔH° is negative

Answer and explanation

Answer: (C). Both signs are negative, so it is favoured only at low T. ΔG° = 0 at T = 84 000 ÷ 210 = 400 K, about 127 °C.

(A) reads 400 K as 400 °C.
(B) (−, −) is favoured at low T.
(D) ignores the −TΔS° term, which grows with T.

If you missed this: read “The four sign combinations” in the Topic 9.3 study guide.

Question 4 (multiple choice · 9.3)

ΔG°f at 298 K (kJ mol⁻¹): NH₃(g) −16.4; NO(g) +86.60; H₂O(g) −228.58. What is ΔG° for 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)?

(A) −1090.7 kJ mol⁻¹
(B) +959.5 kJ mol⁻¹
(C) −125.6 kJ mol⁻¹
(D) −959.5 kJ mol⁻¹

Answer and explanation

Answer: (D). Products: 4(+86.60) + 6(−228.58) = −1025.08. Reactants: 4(−16.4) + 5(0) = −65.6. ΔG° = −1025.08 − (−65.6) = −959.5 kJ mol⁻¹. ΔG°f of O₂, an element, is zero.

(A) adds the reactant term instead of subtracting it.
(B) is reactants minus products.
(C) ignores the coefficients.

If you missed this: read Method 1 in the Topic 9.3 study guide.

Question 5 (multiple choice · 9.4)

Paper can sit in air for centuries without burning, yet burning its cellulose has a large negative ΔG°. Which conclusion is best supported?

(A) Paper and oxygen are at equilibrium, so Q = K.
(B) It is under kinetic control: the activation energy is too high for a measurable rate at room temperature.
(C) K for the combustion must be less than 1 at room temperature.
(D) ΔG° becomes positive at room temperature.

Answer and explanation

Answer: (B). A favoured process with no measurable rate is under kinetic control.

(A) No change is not equilibrium: at equilibrium almost no paper would remain.
(C) ΔG° < 0 means K > 1.
(D) Nothing here changes the sign of ΔG°.

If you missed this: read “Kinetic control is not equilibrium” in the Topic 9.4 study guide.

Question 6 (multiple choice · 9.5)

A reaction has K = 1.0 at 350 K, and ΔH° is positive. Assume ΔH° and ΔS° are constant. Which is true at 298 K?

(A) ΔG° > 0 and K < 1
(B) ΔG° < 0 and K > 1
(C) ΔG° = 0 and K = 1
(D) ΔG° > 0 and K < 0

Answer and explanation

Answer: (A). K = 1 means ΔG° = 0 at 350 K, so ΔS° = ΔH° ÷ 350 K > 0. Both positive: favoured only above 350 K. At 298 K, ΔG° > 0 and K < 1.

(B) describes temperatures above 350 K.
(C) K changes with T, because ΔG° does.
(D) K = e^(−ΔG°/RT) is never negative.

If you missed this: read “Where temperature comes in” in the Topic 9.5 study guide.

Question 7 (multiple choice · 9.6)

Dissolving which salt gives the most negative entropy change from reorganising the water around the ions?

(A) CsI
(B) KBr
(C) NaCl
(D) MgF₂

Answer and explanation

Answer: (D). Small, highly charged Mg²⁺ (with small F⁻) holds water in tight, ordered shells, so water loses the most entropy.

(A) Large, singly charged Cs⁺ and I⁻ order the least water.
(B) and (C) Singly charged ions order far less water.

If you missed this: read “Three factors at the particle level” in the Topic 9.6 study guide.

Question 8 (multiple choice · 9.7)

Two invented reactions are coupled: X → Y, ΔG° = +18.0 kJ mol⁻¹, and Y + Z → W, ΔG° = −25.0 kJ mol⁻¹. Which row is correct for the overall reaction X + Z → W?

(A) ΔG° = −7.0 kJ mol⁻¹; Z is the common intermediate
(B) ΔG° = +43.0 kJ mol⁻¹; Y is the common intermediate
(C) ΔG° = −7.0 kJ mol⁻¹; Y is the common intermediate
(D) ΔG° = −43.0 kJ mol⁻¹; W is the common intermediate

Answer and explanation

Answer: (C). Y, made in the first reaction and used in the second, cancels. ΔG° values add: +18.0 + (−25.0) = −7.0 kJ mol⁻¹, favoured.

(A) Z is a reactant overall, not an intermediate.
(B) ignores the sign of the second value.
(D) W is the product.

If you missed this: read about the common intermediate in the Topic 9.7 study guide.

Question 9 (multiple choice · 9.8)

To purify copper, an impure copper bar and a pure copper sheet in CuSO₄(aq) are connected to a power supply. The pure sheet grows. Which statement is correct?

(A) The impure bar is the anode; it is oxidised to Cu²⁺(aq) and loses mass.
(B) The impure bar is the cathode, because reduction happens at the electrode that loses mass.
(C) Electrons travel through the solution from the impure bar to the pure sheet.
(D) The pure sheet is the anode, because Cu²⁺ ions are made there.

Answer and explanation

Answer: (A). Cu²⁺ is reduced at the growing sheet (cathode), so the impure bar (anode) is oxidised and dissolves.

(B) The electrode losing mass is oxidised: the anode.
(C) Electrons stay in the wire; ions carry charge in solution.
(D) Cu²⁺ is used up at the sheet.

If you missed this: read “Electrolytic cells” in the Topic 9.8 study guide.

Question 10 (multiple choice · 9.9)

What is ΔG° for Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s)?

(A) −101 kJ mol⁻¹
(B) −203 kJ mol⁻¹
(C) −357 kJ mol⁻¹
(D) +203 kJ mol⁻¹

Answer and explanation

Answer: (B). Ag⁺ is reduced: E° = 0.80 − (−0.25) = +1.05 V. With n = 2, ΔG° = −nFE° = −(2)(96,485)(1.05) = −202,600 J = −203 kJ mol⁻¹.

(A) uses n = 1, counting the electrons for one Ag⁺ only.
(C) doubles the silver potential; never scale E° by coefficients.
(D) A positive E° always gives a negative ΔG°.

If you missed this: read Worked example 1 in the Topic 9.9 study guide.

Question 11 (multiple choice · 9.9)

Under standard conditions, which metal reduces Sn²⁺(aq) to tin but does not reduce Fe²⁺(aq) to iron?

(A) Copper
(B) Zinc
(C) Nickel
(D) Magnesium

Answer and explanation

Answer: (C). With Sn²⁺: E° = −0.14 − (−0.25) = +0.11 V, favoured. With Fe²⁺: E° = −0.44 − (−0.25) = −0.19 V, not favoured.

(A) Copper reduces neither: −0.48 V and −0.78 V.
(B) Zinc reduces both (+0.62 V and +0.32 V).
(D) Magnesium reduces both.

If you missed this: read about the sign of E°cell in the Topic 9.9 study guide.

Question 12 (multiple choice · 9.10)

A cell runs Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s). Which starting concentrations give the largest cell potential?

(A) [Ni²⁺] = 0.010 M, [Ag⁺] = 1.0 M
(B) [Ni²⁺] = 1.0 M, [Ag⁺] = 1.0 M
(C) [Ni²⁺] = 1.0 M, [Ag⁺] = 0.10 M
(D) [Ni²⁺] = 0.10 M, [Ag⁺] = 0.10 M

Answer and explanation

Answer: (A). Q = [Ni²⁺] ÷ [Ag⁺]² is 0.010 in (A), 1 in (B), 100 in (C) and 10 in (D). The smallest Q puts the cell furthest from equilibrium, so E is largest (about 1.11 V; E° = 1.05 V).

(B) is standard: E = E°.
(C) Q > 1, closer to equilibrium: E < E°.
(D) [Ag⁺] is squared, so equal dilution still lowers E: Q = 10, E ≈ 1.02 V.

If you missed this: read “Comparing Q with 1 and with K” in the Topic 9.10 study guide.

Question 13 (short answer · 9.11)

A cell making aluminium runs at a steady 1.50 × 10⁵ A. Al³⁺ ions are reduced at the cathode.

(a) Calculate the mass of aluminium made in 1.00 hour. (b) Suppose the anode were inert, so that 2O²⁻ → O₂ + 4e⁻ happens there. Calculate the mass of O₂ the same charge would release. Why do fewer moles of O₂ than of Al form?

Worked answer

(a)

q = I × t = 1.50 × 10⁵ × 3600 s = 5.40 × 10⁸ C; n(e⁻) = q ÷ F = 5597 mol.
Al³⁺ + 3e⁻ → Al: n(Al) = 5597 ÷ 3 = 1866 mol; m = 1866 × 26.98 g = 50.3 kg.

(b) n(O₂) = 5597 ÷ 4 = 1399 mol; m = 1399 × 32.00 g = 44.8 kg. Each O₂ needs 4 electrons but each Al only 3, so fewer moles of O₂ form (Al : O₂ = 4 : 3).

If you missed this: read Worked example 1 in the Topic 9.11 study guide.

Your next step

Count a short answer as missed if any part was wrong.

Topic	Question(s)	If you missed it, read
9.1	1	Introduction to Entropy
9.2	2	Absolute Entropy and Entropy Change
9.3	3, 4	Gibbs Free Energy and Favorability
9.4	5	Thermodynamic and Kinetic Control
9.5	6	Free Energy and Equilibrium
9.6	7	Free Energy of Dissolution
9.7	8	Coupled Reactions
9.8	9	Galvanic and Electrolytic Cells
9.9	10, 11	Cell Potential and Free Energy
9.10	12	Nonstandard Cell Potential
9.11	13	Electrolysis and Faraday’s Law

How to use your result

Look at topics, not the total.
Count a guess as a miss, and read the explanation anyway.
Fix gaps in order, earliest topic first: 9.1–9.3 build ΔS and ΔG, 9.4–9.7 apply ΔG, 9.8–9.11 apply it to cells.
Then practise with the topic practice sets and the Unit 9 mixed review.