The problems are on the electrochemistry worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electrochemistry links redox reactions to electrical energy. These worked answers correct the 20 problems of the AP Chemistry worksheet on electrochemistry, from reading a lead-copper cell to two cells in series: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the final result with its unit in bold. Check the sign of E°cell first, then n, because most lost marks come from those two steps.
Level: AP Chemistry · Curriculum: AP Chem Unit 9
Problem 1 Reading a lead-copper cell
1. Lead has the lower E° (−0.13 V against +0.34 V), so lead is oxidised: electrode X is the anode, Pb → Pb²⁺ + 2e⁻.
2. Electrons leave the anode and flow through the wire and voltmeter to the copper cathode: from X to Y.
3. Pb²⁺ builds up in the anode beaker, so nitrate anions move toward the lead half-cell to keep it neutral.
Problem 2 Magnesium and silver cell
1. Silver has the higher E°, so silver is the cathode.
2. E°cell = 0.80 − (−2.37) = +3.17 V.
3. Mg → Mg²⁺ + 2e⁻ and 2Ag⁺ + 2e⁻ → 2Ag give Mg + 2Ag⁺ → Mg²⁺ + 2Ag. The silver potential is not doubled.
Problem 3 Will the iron nail react?
1. Zn²⁺ is the species reduced: E° = −0.76 − (−0.44) = −0.32 V, not spontaneous.
2. E° = 0.34 − (−0.44) = +0.78 V, spontaneous.
3. Only the nail in copper(II) sulfate reacts: it becomes coated with reddish-brown copper and the blue solution fades toward pale green.
Problem 4 From current to electrons
1. t = 2.00 × 3600 = 7200 s.
2. Q = It = 0.250 × 7200 = 1800 C.
3. n(e⁻) = 1800 / 96 485 = 0.0187 mol.
4. N = 0.018656 × 6.022 × 10²³ = 1.12 × 10²² electrons.
Problem 5 Balancing aluminium and copper
1. Anode: Al → Al³⁺ + 3e⁻. Cathode: Cu²⁺ + 2e⁻ → Cu.
2. Multiply the anode by 2 and the cathode by 3 so each has 6 electrons: 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu.
3. n = 6. E°cell = 0.34 − (−1.66) = +2.00 V.
Problem 6 The hydrogen electrode
1. 1.0 mol/dm³ H⁺(aq), H₂ gas at 1 atm, inert platinum electrode, 25 °C; E° is defined as 0.00 V.
2. Copper has the higher E°, so the copper half-cell is the cathode and reads +0.34 V.
3. Zn + 2H⁺ → Zn²⁺ + H₂ has E° = 0.00 − (−0.76) = +0.76 V, so zinc does react and bubbles of hydrogen form.
Problem 7 Electrolysis of molten potassium bromide
1. A is joined to the positive terminal, so A is the anode (+); B is the cathode (−).
2. Anode: 2Br⁻ → Br₂ + 2e⁻ (brown vapour). Cathode: K⁺ + e⁻ → K (liquid potassium).
3. The reaction 2K + Br₂ → 2KBr is spontaneous, so the reverse is not; the battery supplies the energy that forces it.
Problem 8 Free energy of a copper-silver cell
1. E°cell = 0.80 − 0.34 = +0.46 V.
2. Two electrons move: n = 2.
3. ΔG° = −nFE° = −2 × 96 485 × 0.46 = −88 766 J/mol = −88.8 kJ/mol; negative, so spontaneous.
Problem 9 Equilibrium constant for iron and copper
1. E°cell = 0.34 − (−0.44) = 0.78 V; n = 2.
2. log K = 2 × 0.78 / 0.0592 = 26.35, so K ≈ 2.2 × 10²⁶.
3. K is enormous, so at equilibrium essentially all the Cu²⁺ has been converted.
Problem 10 Nickel plating a bracket
1. t = 1500 s; Q = 1.50 × 1500 = 2250 C.
2. n(e⁻) = 2250 / 96 485 = 0.02332 mol; n(Ni) = 0.01166 mol = 1.17 × 10⁻² mol.
3. m = 0.01166 × 58.69 = 0.684 g.
Problem 11 How long to deposit silver?
1. n(Ag) = 5.00 / 107.87 = 0.04635 mol.
2. One electron per atom, so Q = 0.04635 × 96 485 = 4472 C.
3. t = Q / I = 4472 / 2.00 = 2236 s = 37.3 min.
Problem 12 Cell voltage at other concentrations
1. Q = [Zn²⁺] / [Cu²⁺].
2. Q = 0.050 / 0.80 = 0.0625, log Q = −1.204.
3. E = 1.10 − (0.0592 / 2)(−1.204) = 1.10 + 0.0356 = 1.14 V. Q is below 1, so E is above E°.
Problem 13 Reading a Nernst graph
1. E° is the value at log Q = 0: 1.10 V.
2. Slope = (1.04 − 1.16) / (2 − (−2)) = −0.12 / 4 = −0.030 V per unit of log Q.
3. n = 0.0592 / 0.030 = 1.97, so n = 2.
Problem 14 Galvanic against electrolytic
Galvanic: chemical energy becomes electrical energy; anode negative; ΔG° negative (E°cell positive); electrons come from the oxidation at the anode itself.
Electrolytic: electrical energy drives a chemical change; anode positive; ΔG° positive (E°cell negative); electrons are pumped by the external power supply. Oxidation is at the anode in both.
Problem 15 Galvanised and tin-plated steel
1. Zinc has a lower E° than iron, so in the scratch zinc is oxidised in preference: it acts as a sacrificial anode and the iron stays the cathode.
2. Iron has a lower E° than tin, so iron becomes the anode and corrodes faster than it would bare.
Problem 16 Permanganate and iron(II) titration cell
1. Neither half-cell has a solid metal that takes part, so an inert electrode is needed to carry electrons.
2. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, with Fe²⁺ → Fe³⁺ + e⁻ ×5: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ (charge +17 on each side).
3. E°cell = 1.51 − 0.77 = +0.74 V; n = 5; ΔG° = −5 × 96 485 × 0.74 = −356 995 J/mol = −357 kJ/mol.
Problem 17 Aluminium smelting
1. Q = 5.00 × 10⁴ × 3600 = 1.80 × 10⁸ C; n(e⁻) = 1.80 × 10⁸ / 96 485 = 1865.6 mol; n(Al) = 1865.6 / 3 = 621.9 mol; m = 621.9 × 26.98 = 16 778 g = 16.8 kg.
2. n(O₂) = 1865.6 / 4 = 466 mol.
3. Water is reduced at −0.83 V, a less negative potential than −1.66 V, so hydrogen forms at the cathode before aluminium does.
Problem 18 A concentration cell
1. Both half-cells are the same couple, so E°cell = 0; the voltage comes from Q ≠ 1.
2. Oxidation raises the concentration of the dilute side, so the 0.0010 mol/dm³ half-cell is the anode.
3. Q = 0.0010 / 1.0 = 10⁻³, so E = 0 − (0.0592 / 2)(−3) = +0.089 V.
4. The two concentrations become equal, Q = 1, so E → 0.
Problem 19 Spot the errors
1. The student doubled the silver potential and used n = 1 although two electrons are transferred.
2. E°cell = 0.80 − (−0.76) = 1.56 V.
3. ΔG° = −2 × 96 485 × 1.56 = −301 033 J/mol = −301 kJ/mol.
Problem 20 Two cells in series
1. n(Cu) = 0.640 / 63.55 = 0.010071 mol; two electrons each, so n(e⁻) = 0.0201 mol.
2. Ag⁺ + e⁻ → Ag, so n(Ag) = 0.020142 mol and m = 0.020142 × 107.87 = 2.17 g.
3. Q = 0.020142 × 96 485 = 1943 C; I = 1943 / 1800 = 1.08 A.
