Unit 5: Electrochemistry — Unit Test

Assessment OF Learning
Graded
Duration: 75 min | Total: /60 | F = 96485 C/mol
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: Knowledge & Understanding [15]
Q1 [3]
Determine the oxidation number of the boldface element: (a) MnO₄⁻; (b) S in S₂O₃²⁻; (c) N in NH₄⁺.
Q2 [3]
Define galvanic cell vs electrolytic cell. State sign of E°cell for each.
Q3 [4]
Multiple choice — in cell notation Cd|Cd²⁺‖Ag⁺|Ag, the cathode is:
Q4 [5]
Given E°(Ag⁺/Ag) = +0.80, E°(Cu²⁺/Cu) = +0.34. Calculate E°cell for Cu|Cu²⁺‖Ag⁺|Ag and write the spontaneous net equation.
Part B: Thinking & Investigation [15]
Q5 [5]
Balance in acidic solution: Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂.
Q6 [5]
Calculate mass of Ag deposited from AgNO₃ when 0.500 A flows for 60.0 min. (Ag M = 107.87)
Q7 [5]
A redox titration: 25.00 mL Fe²⁺ requires 22.50 mL of 0.0200 M KMnO₄. Find [Fe²⁺]. (Use balanced equation.)
Part C: Communication [15]
Q8 [5]
Sketch a Daniell cell labelling: anode, cathode, salt bridge, e⁻ flow, ion flow, half-reactions.
Q9 [5]
Explain the role of the salt bridge in a galvanic cell. Why must charge balance be maintained?
Q10 [5]
Explain cathodic protection (sacrificial anode) and why Mg or Zn protects Fe pipelines/ships.
Part D: Application [15]
Q11 [5]
Hall-Héroult: 2 Al₂O₃ → 4 Al + 3 O₂. Calculate energy required (in C and kWh) per kg of Al. (Al³⁺ + 3e⁻ → Al; assume 5.00 V cell.)
Q12 [5]
Lead-acid car battery: write half-reactions during discharge and explain why density of H₂SO₄ decreases as battery discharges.
Q13 [5]
Compare a primary cell, secondary cell, and fuel cell — give one example of each, plus one environmental concern for each.
Complete Answer Key

Q1. (a) +7; (b) +2; (c) −3.

Q2. Galvanic = spontaneous, E°cell > 0, generates electricity. Electrolytic = non-spontaneous, requires external V, E°cell < 0.

Q3. Ag⁺/Ag (right side = cathode).

Q4. E°cell = 0.80 − 0.34 = +0.46 V. Cu + 2Ag⁺ → Cu²⁺ + 2Ag.

Q5. Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O.

Q6. Q = 0.500 × 3600 = 1800 C → n(e⁻) = 1800/96485 = 0.01866 mol; n(Ag) = 0.01866 mol → m = 2.01 g.

Q7. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. n(MnO₄⁻) = 0.0200 × 0.02250 = 4.50×10⁻⁴ mol → n(Fe²⁺) = 5 × 4.50×10⁻⁴ = 2.25×10⁻³ mol → [Fe²⁺] = 2.25×10⁻³/0.02500 = 0.0900 M.

Q8. Diagram should show Zn(s)|ZnSO₄ on left (anode, oxidation, e⁻ leave); Cu(s)|CuSO₄ on right (cathode, reduction, e⁻ enter); salt bridge with NO₃⁻/Na⁺ migrating to balance charge.

Q9. Salt bridge completes the circuit by allowing ions (cations to cathode, anions to anode) to migrate, preventing charge buildup that would halt e⁻ flow.

Q10. Mg/Zn has more negative E° than Fe → preferentially oxidized; supplies e⁻ to Fe, keeping Fe in reduced (un-corroded) state.

Q11. n(Al) = 1000/26.98 = 37.06 mol → n(e⁻) = 111.2 mol → Q = 111.2 × 96485 = 1.073×10⁷ C. E = QV = 1.073×10⁷ × 5.00 = 5.36×10⁷ J ≈ 14.9 kWh/kg Al.

Q12. Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻. Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O. SO₄²⁻ consumed → H₂SO₄ concentration (and density) drops.

Q13. Primary: Zn-C dry cell — landfill heavy metals. Secondary: Li-ion — mining + recycling. Fuel cell: H₂ — H₂ production source/footprint.