AP Electrochemistry 2 — Questions and Answers

Question 1: Calculate the standard cell potential (E°cell) for the galvanic cell: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Given: E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

+1.10 V (Correct answer)
+0.42 V
−1.10 V
+0.76 V

Correct answer: +1.10 V

E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V. Cu²⁺ is reduced at the cathode; Zn is oxidized at the anode.

A galvanic (voltaic) cell converts chemical energy to electrical energy via spontaneous redox reactions. The cell notation Zn | Zn²⁺ || Cu²⁺ | Cu indicates: - Left side = anode (oxidation): Zn → Zn²⁺ + 2e⁻, E°(Zn²⁺/Zn) = −0.76 V - Right side = cathode (reduction): Cu²⁺ + 2e⁻ → Cu, E°(Cu²⁺/Cu) = +0.34 V E°cell = E°cathode − E°anode (both as standard reduction potentials) E°cell = (+0.34 V) − (−0.76 V) = +0.34 + 0.76 = +1.10 V The positive E°cell confirms spontaneity (ΔG° = −nFE°cell = −2 × 96485 × 1.10 = −212,300 J = −212 kJ). The Daniel cell (Zn-Cu) is historically important as one of the first reliable batteries and was crucial to the development of electrochemistry in the 19th century. Its 1.10 V is large enough for practical applications.

Question 2: Using the Nernst equation, what is the cell potential at 25°C for the Zn-Cu cell when [Zn²⁺] = 2.0 M and [Cu²⁺] = 0.010 M? (E°cell = 1.10 V, n = 2)

1.04 V (Correct answer)
1.10 V
1.16 V
1.07 V

Correct answer: 1.04 V

Q = [Zn²⁺]/[Cu²⁺] = 2.0/0.010 = 200. E = E° − (0.0592/n)logQ = 1.10 − (0.0296)log(200) = 1.10 − (0.0296)(2.301) = 1.10 − 0.068 = 1.03 ≈ 1.04 V.

The Nernst equation accounts for non-standard conditions: E = E° − (RT/nF) × ln Q = E° − (0.05916/n) × log Q (at 25°C) Overall cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) Reaction quotient: Q = [Zn²⁺]/[Cu²⁺] = 2.0/0.010 = 200 (Solids and pure liquids are omitted from Q.) E = 1.10 − (0.05916/2) × log(200) E = 1.10 − 0.02958 × 2.301 E = 1.10 − 0.0681 E = 1.032 V ≈ 1.03 V (closest answer: 1.04 V) Interpreting the result: With more Zn²⁺ (product) and less Cu²⁺ (reactant) than standard conditions, Q > 1, which drives the reaction toward equilibrium, decreasing the cell potential below E°. This is consistent with Le Chatelier's principle.

Question 3: In the electrolysis of molten NaCl, which species is produced at each electrode?

Cathode: Na metal; Anode: Cl₂ gas (Correct answer)
Cathode: Cl₂ gas; Anode: Na metal
Cathode: H₂ gas; Anode: O₂ gas
Both electrodes produce NaCl

Correct answer: Cathode: Na metal; Anode: Cl₂ gas

In electrolysis, the cathode is negative and attracts Na⁺ (reduction: Na⁺ + e⁻ → Na). The anode is positive and attracts Cl⁻ (oxidation: 2Cl⁻ → Cl₂ + 2e⁻).

Electrolysis forces a non-spontaneous redox reaction using electrical energy (opposite of a galvanic cell). In molten NaCl, the only ions present are Na⁺ and Cl⁻ (no water). An external power source drives current: Cathode (negative electrode): Na⁺ + e⁻ → Na(l) Reduction occurs; liquid sodium metal is produced. This is the Downs cell process for industrial sodium production. Anode (positive electrode): 2Cl⁻ → Cl₂(g) + 2e⁻ Oxidation occurs; chlorine gas bubbles form. Overall: 2NaCl(l) → 2Na(l) + Cl₂(g) This is distinguished from electrolysis of aqueous NaCl (brine), where water competes: - Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (preferred over Na⁺ reduction) - Anode: 2Cl⁻ → Cl₂ + 2e⁻ (preferred at high [Cl⁻], overpotential dependent) The Chlor-alkali process using aqueous NaCl electrolysis produces Cl₂, H₂, and NaOH — all industrially important.

Question 4: How many grams of copper (MM = 63.55 g/mol) are deposited during the electrolysis of a CuSO₄ solution when a current of 2.50 A flows for 30.0 minutes? (F = 96,485 C/mol e⁻)

1.48 g (Correct answer)
2.96 g
0.740 g
0.370 g

Correct answer: 1.48 g

Charge = I × t = 2.50 × (30.0 × 60) = 4500 C. mol e⁻ = 4500/96485 = 0.04663 mol. Cu²⁺ + 2e⁻ → Cu, so mol Cu = 0.04663/2 = 0.02332 mol. Mass = 0.02332 × 63.55 = 1.48 g.

This problem applies Faraday's laws of electrolysis: Step 1: Calculate total charge. Q = I × t = 2.50 A × (30.0 min × 60 s/min) = 2.50 × 1800 = 4500 C. Step 2: Convert charge to moles of electrons. mol e⁻ = Q / F = 4500 C / 96,485 C/mol = 0.04663 mol e⁻. Step 3: Apply stoichiometry. Half-reaction: Cu²⁺ + 2e⁻ → Cu (requires 2 electrons per Cu atom) mol Cu = 0.04663 / 2 = 0.02332 mol. Step 4: Convert to grams. mass Cu = 0.02332 mol × 63.55 g/mol = 1.48 g. Faraday's laws: (1) The amount of substance deposited is proportional to the charge passed. (2) For different substances, equal charges deposit amounts proportional to their equivalent masses (molar mass / charge). Applications include electroplating, copper refining, and the Hall-Héroult process for aluminum production.

Question 5: Which of the following best describes the role of the salt bridge in a galvanic cell?

It allows ions to flow between half-cells to maintain electrical neutrality without mixing the solutions (Correct answer)
It transfers electrons directly between the anode and cathode
It prevents any ion migration so the two half-cell solutions remain pure
It provides additional electromotive force to the cell

Correct answer: It allows ions to flow between half-cells to maintain electrical neutrality without mixing the solutions

As the cell operates, charge builds up in each half-cell (Zn²⁺ accumulates at the anode; Cu²⁺ depletes at the cathode). The salt bridge allows ions (typically K⁺ and NO₃⁻) to migrate to neutralize these charges without mixing the electrode solutions.

A galvanic cell requires both electron flow (through the external circuit) and ion flow (through the solution) to operate continuously. At the anode: Zn → Zn²⁺ + 2e⁻. Positive charge (Zn²⁺) accumulates in the anode compartment. At the cathode: Cu²⁺ + 2e⁻ → Cu. Positive ions are removed from the cathode compartment. Without a salt bridge, positive charge would accumulate in the anode and deplete in the cathode, creating a potential that opposes further current flow — the cell would quickly stop. The salt bridge (typically a gel of KCl or KNO₃) allows: - Anions (Cl⁻ or NO₃⁻) to migrate into the anode compartment, neutralizing the excess positive charge - Cations (K⁺) to migrate into the cathode compartment, replenishing lost positive charge Ions are chosen to be chemically inert (not react with either electrode solution). This ion migration maintains electrical neutrality and allows continuous current flow, but the solutions themselves do not mix significantly.

Question 6: For the half-reaction: Fe³⁺ + e⁻ → Fe²⁺, E° = +0.77 V. What is the value of ΔG° for this process when n = 1? (F = 96,485 C/mol)

−74.3 kJ/mol (Correct answer)
+74.3 kJ/mol
−148.6 kJ/mol
+148.6 kJ/mol

Correct answer: −74.3 kJ/mol

ΔG° = −nFE° = −(1)(96,485 C/mol)(+0.77 V) = −74,300 J/mol = −74.3 kJ/mol. The negative ΔG° confirms this half-reaction is thermodynamically favorable as written.

The relationship between free energy and cell potential is: ΔG° = −nFE° where: - n = number of moles of electrons transferred = 1 - F = Faraday's constant = 96,485 C/mol (or J/(V·mol)) - E° = standard reduction potential = +0.77 V ΔG° = −(1 mol e⁻)(96,485 J/V·mol)(+0.77 V) = −74,293 J/mol ≈ −74.3 kJ/mol. The negative ΔG° means the reduction of Fe³⁺ to Fe²⁺ is thermodynamically spontaneous under standard conditions — Fe³⁺ is a moderately strong oxidizing agent. This relationship connects three key thermodynamic/electrochemical quantities: - ΔG° = −nFE° (free energy ↔ cell potential) - ΔG° = −RT ln K (free energy ↔ equilibrium constant) - E° = (RT/nF) ln K (cell potential ↔ equilibrium constant) Knowing any one of ΔG°, E°, or K allows calculation of the other two — a unifying concept in thermodynamics.

AP Electrochemistry 2

Calculate the standard cell potential (E°cell) for the galvanic cell: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
Given: E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.