AP - Advanced Placement Electrochemistry 1 — Questions and Answers
Question 1: Calculate ΔG° for the reaction in the Zn-Cu galvanic cell (E°cell = 1.10 V, n = 2, F = 96,485 C/mol e⁻).
- −212 kJ/mol (Correct answer)
- −106 kJ/mol
- −424 kJ/mol
- +212 kJ/mol
Correct answer: −212 kJ/mol
ΔG° = −nFE°cell = −(2)(96,485 C/mol)(1.10 V) = −212,267 J/mol ≈ −212 kJ/mol. The negative value confirms the reaction is thermodynamically spontaneous.
Question 2: A concentration cell is constructed: Cu(s) | Cu²⁺(0.0100 M) || Cu²⁺(1.00 M) | Cu(s). What is the cell potential at 25°C? (n = 2)
- 0.0592 V (Correct answer)
- 0.0296 V
- 0.118 V
- 0.00 V
Correct answer: 0.0592 V
Because both electrodes are identical, E°cell = 0. Applying the Nernst equation: E = (0.0592/2) × log([Cu²⁺]cathode / [Cu²⁺]anode) = (0.0296) × log(1.00/0.0100) = 0.0296 × 2 = 0.0592 V.
Question 3: For a spontaneous electrochemical reaction at 25°C involving n moles of electrons transferred, which expression correctly gives the equilibrium constant K from the standard cell potential E°?
- log K = nE° / 0.0592 (Correct answer)
- log K = E° / (n × 0.0592)
- log K = nFE° / R
- ln K = nE° / 0.0592
Correct answer: log K = nE° / 0.0592
Combining ΔG° = −nFE° and ΔG° = −RT ln K gives ln K = nFE°/RT. At 25°C, converting to log₁₀: log K = nE° / 0.0592. This is the standard AP formula relating E° and K.
Question 4: During the electrolysis of aqueous sodium chloride (brine), what is the product formed at the cathode?
- H₂(g) (Correct answer)
- Cl₂(g)
- Na(s)
- O₂(g)
Correct answer: H₂(g)
At the cathode, reduction occurs. Although Na⁺ is present, water is preferentially reduced (E° = −0.83 V vs. −2.71 V for Na⁺/Na), producing H₂(g) and OH⁻. Cl₂ is produced at the anode, not the cathode.
Question 5: Given standard reduction potentials — Cu²⁺/Cu = +0.34 V, Ag⁺/Ag = +0.80 V, Zn²⁺/Zn = −0.76 V, Fe²⁺/Fe = −0.44 V — which metal will NOT spontaneously dissolve in 1.0 M HCl to produce H₂(g)? (E°(H⁺/H₂) = 0.00 V)
- Cu (Correct answer)
- Zn
- Fe
- Both Zn and Fe
Correct answer: Cu
A metal displaces H₂ from acid only if its oxidation (reverse of its reduction half-reaction) yields E°cell > 0. For Cu, E°cell = 0.00 − 0.34 = −0.34 V < 0, so the reaction is non-spontaneous. Zn and Fe both have negative E° values and react readily with HCl.
Question 6: How many seconds are required to deposit 2.16 g of silver (MM = 107.87 g/mol) from an AgNO₃ solution using a constant current of 2.00 A? (Ag⁺ + e⁻ → Ag; F = 96,485 C/mol e⁻)
- 966 s (Correct answer)
- 483 s
- 1932 s
- 241 s
Correct answer: 966 s
mol Ag = 2.16 g ÷ 107.87 g/mol = 0.02003 mol. Charge needed = 0.02003 mol × 96,485 C/mol = 1932 C. Time = charge ÷ current = 1932 C ÷ 2.00 A = 966 s.
AP - Advanced Placement Electrochemistry 1
Calculate ΔG° for the reaction in the Zn-Cu galvanic cell (E°cell = 1.10 V, n = 2, F = 96,485 C/mol e⁻).