The questions are on the electrochemistry test. Take it under real conditions before reading the mark scheme.
Quick answer: Electrochemistry links redox reactions to electrical energy. This mark scheme gives the model answers and the marks for the 8 questions of the AP Chemistry test on electrochemistry, from anode and cathode to free energy to potential, with the equivalent answers to accept.
Model answers and mark allocations are shown below. Award error carried forward where a correct method follows an earlier slip, and withhold a mark when a final answer has no unit.
Level: AP Chemistry · Curriculum: AP Chem Unit 9
Question 1 Anode and cathode (2 marks)
- Oxidation (loss of electrons) occurs at the anode.
- The anode is negative in a galvanic cell and positive in an electrolytic cell.
Marking: 1 mark for oxidation; 1 mark for both signs correct.
Question 2 Magnesium and nickel cell (3 marks)
- Nickel has the higher E°: cathode, Ni²⁺ + 2e⁻ → Ni.
- E°cell = −0.25 − (−2.37) = +2.12 V.
- Mg + Ni²⁺ → Mg²⁺ + Ni.
Marking: 1 mark per part; award part (b) for the correct sign and unit.
Question 3 Free energy and K for tin and copper (3 marks)
- E°cell = 0.34 − (−0.14) = 0.48 V.
- n = 2: ΔG° = −2 × 96 485 × 0.48 = −92 626 J/mol = −92.6 kJ/mol.
- log K = 2 × 0.48 / 0.0592 = 16.2, so K ≈ 1.6 × 10¹⁶.
Marking: 1 mark per part; accept ΔG° = −92.6 or −93 kJ/mol.
Question 4 Voltage away from standard (2 marks)
- Q = [Zn²⁺]/[Cu²⁺] = 0.010/1.0 = 0.010.
- E = 1.10 − (0.0592/2) log(0.010) = 1.10 + 0.0592 = 1.16 V.
Marking: 1 mark for Q; 1 mark for E (accept 1.159 V).
Question 5 Silver plating (3 marks)
- Q = 0.500 × 1800 = 900 C.
- n(e⁻) = 900 / 96 485 = 9.33 × 10⁻³ mol.
- One electron per atom, so m = 9.33 × 10⁻³ × 107.87 = 1.01 g.
Marking: 1 mark per part; allow error carried forward.
Question 6 Brine electrolysis (2 marks)
Water is reduced at a less negative potential (−0.83 V) than Na⁺ (−2.71 V), so 2H₂O + 2e⁻ → H₂ + 2OH⁻ occurs and sodium stays dissolved as Na⁺.
Marking: 1 mark for comparing the two potentials; 1 mark for the conclusion with the half-reaction.
Question 7 Reading a nickel-silver cell (3 marks)
- Nickel (E° −0.25 V) is the anode: electrons flow from X to Y.
- Ni → Ni²⁺ + 2e⁻ so electrode X loses mass; Ag⁺ + e⁻ → Ag so electrode Y gains mass.
- Q = [Ni²⁺]/[Ag⁺]² rises, so E falls below 1.05 V.
Marking: 1 mark per part.
Question 8 Free energy to potential (2 marks)
- E° = −ΔG°/(nF) = 193 000 / (2 × 96 485) = 1.00 V.
- K is greater than 1 because ΔG° is negative and E° is positive.
Marking: 1 mark per part.
