Copper Recovery Before Discharge
A printed-circuit workshop in Rubí treats 2.0 L of spent etchant containing CuSO4 0.150 M and H2SO4 0.020 M by electrodeposition onto a graphite cathode before the liquor goes to waste. The rectifier delivers a constant 3.00 A, the current efficiency for copper deposition is 94%, and a 90.0 min run is scheduled. The liquor also contains about 5 mg/L Ag⁺ and 40 mg/L Zn²⁺ as impurities. The discharge limit is [Cu²⁺] ≤ 0.100 M. The operator argues that the 3.5 V applied is excessive, since E°(Cu²⁺/Cu) = +0.34 V and E°(O2/H2O) = +1.23 V, so 'about 0.9 V should be enough'. The environmental technician must decide whether the run meets the limit, whether the voltage is justified, and what happens to the silver and zinc. F = 96485 C/mol; the cathode is flat graphite and the solution is not stirred vigorously.
1. Calculate the mass of copper deposited in 90.0 min and the final Cu²⁺ concentration. Does the batch meet the discharge limit, and if not, how much extra time is needed?
Read the worked answer
Charge passed: Q = I·t = 3.00 A × (90.0 × 60 s) = 3.00 × 5400 = 1.62 × 10⁴ C. Moles of electrons: n(e⁻) = 1.62 × 10⁴/96485 = 0.168 mol. Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 0.168/2 = 0.0840 mol before efficiency. With 94% efficiency: n(Cu) = 0.0840 × 0.94 = 0.0789 mol, m = 0.0789 × 63.55 = 5.01 g.
Initial Cu²⁺ = 0.150 M × 2.0 L = 0.300 mol. Remaining = 0.300 − 0.0789 = 0.221 mol in 2.0 L, i.e. [Cu²⁺] = 0.111 M. That is above 0.100 M, so the batch fails and must not be discharged.
Extra time: to reach 0.100 M, remove 0.300 − 0.200 = 0.100 mol Cu, requiring 0.200 mol e⁻ = 0.200 × 96485 = 1.93 × 10⁴ C. Effective current = 3.00 × 0.94 = 2.82 A, so t = 1.93 × 10⁴/2.82 = 6.84 × 10³ s ≈ 114 min. About 24 more minutes are needed (allowing for the falling Cu²⁺ concentration and mass-transport limits, a little more in practice).
2. Evaluate the operator's claim that 0.9 V should suffice. Estimate the reversible cell voltage and explain the role of overpotential and of the applied 3.5 V.
Read the worked answer
Cathode (reduction): Cu²⁺ + 2e⁻ → Cu, E = 0.34 + (0.0592/2)log[Cu²⁺] = 0.34 + 0.0296 log(0.111) = 0.34 − 0.028 = 0.312 V. Anode (oxidation): 2H2O → O2 + 4H⁺ + 4e⁻, whose reverse has E° = +1.23 V. At pH ≈ 1.4 (0.020 M H2SO4), E = 1.23 − 0.0592 × 1.4 = 1.15 V. Reversible cell voltage = 1.15 − 0.31 = 0.84 V, so the operator's 0.9 V figure is the thermodynamic minimum and is numerically reasonable.
But a real cell needs more: (i) the oxygen-evolution overpotential on graphite is typically 0.4–0.5 V; (ii) there is ohmic drop through the electrolyte and contacts; (iii) concentration overpotential grows as Cu²⁺ is depleted near the cathode. A practical cell voltage of about 1.5–2.0 V is expected, so 3.5 V is generous but not absurd - it will waste energy as heat and may accelerate side reactions such as oxygen reduction or anode corrosion. The correct judgement is that 0.9 V is a thermodynamic floor, not an operating voltage; the applied voltage should be trimmed, not justified by the Nernst calculation alone.
3. Predict the fate of the Ag⁺ and Zn²⁺ impurities during the electrolysis, and explain whether either can contaminate the copper product.
Read the worked answer
Compare reduction potentials with the cathode potential actually imposed (roughly 0.31 V, and lower as Cu²⁺ is consumed):
- Ag⁺ + e⁻ → Ag, E° = +0.80 V. Even at 5 mg/L (≈4.6 × 10⁻⁵ M), E = 0.80 + 0.0592 log(4.6 × 10⁻⁵) = 0.80 − 0.256 = 0.54 V, which is above the copper potential. Silver therefore plates before and alongside copper and will contaminate the deposit (and be lost from the recovered metal).
- Zn²⁺ + 2e⁻ → Zn, E° = −0.76 V. At 40 mg/L (≈6.1 × 10⁻⁴ M), E = −0.76 + 0.0296 log(6.1 × 10⁻⁴) = −0.76 − 0.095 = −0.85 V. This is far below the cathode potential, so zinc stays in solution.
Judgement: the copper is selectively deposited with respect to zinc but not with respect to silver. If silver-free copper is required, the liquor should be pre-treated (e.g. cemented with copper scrap or precipitated as AgCl) before electrolysis. Also note that as [Cu²⁺] falls towards the discharge limit, the cathode potential falls with it, but hydrogen evolution (E ≈ −0.06 V at pH 1.4, plus a large overpotential on copper) remains unfavourable, so current efficiency stays acceptable over the planned run.