Chapter VII: Electrowinning
Quiz Solutions
1. Electrowinning is a common way to produce pure metals. However, chemical reduction is also feasible in some instances. Consider the Eh-pH diagram below. Over what pH range would you expect to be able to reduce Co +2 to cobalt metal. Explain your answer by considering the sign of ΔE for the reaction over that pH range.
From the diagram below the pH range for reduction of Co+2 by H2 at 10 atm pressure is about 4.56 < pH < 6.46. Above pH 4.56 EhCo+2/Co > EhH+/H2. The reaction is:
Co+2 + H2 = Co + 2H+
ΔE = EhCo+2/Co – EhH+/H2 > 0 above pH 4.56.
2. After copper solvent extraction electrowinning will be practiced to produce high-purity copper metal.
(a) If the annual copper production is 45,000 tonnes per year, the PLS copper concentration is 1.9 g/L and the extent of extraction from the PLS is 91%, what will the PLS flow rate be in m3/h?
(i) PLS flow rate
Annual copper production:
45,000 t/y * 1000kg/t * y/365 days * day/24 hrs = 5,137 kg Cu/h
Copper removed per m³:
(0.91) * (1.9) = 1.71 kg/m³
PLS flow rate:
A ≈ 3004 m³/h
(b) If the rich electrolyte copper concentration is 42 g/L and the electrolyte flow rate is 570 m3/h what will the lean electrolyte copper concentration be? (Assume 365 day/year operation.)
(ii) Lean electrolyte concentration
Copper balance:
A*(CRE − CLE) = MCu
Lean electrolyte:
CLE ≈ 33.0 g/L
3. The EW plant has four banks of 28 cells each. The current efficiency is 90%. The cathodes are 1.2 m deep x 1 m wide. The current density is about 300 A/m2. The cell voltage is 2.05 V.
(a) Determine the number of cathode sheets per cell (must be an integer; round off as needed).
| Cells/bank | 28 | |||
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Cells in the plant
|
112 | |||
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Cathode sheet dimensions
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1 m X 1.2 m
|
|||
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Current efficiency
|
90 | |||
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Current density (approx.)
|
300 | A/m2 |
N = 45000 * 100 * 2 * 96485 / (300*2.4*90*63.546*0.000001*3600*24*365*112) = 59.71 = 60 sheets per cell.
(b) Calculate an accurate value for the current density given that the number of cathode sheets/cell is an integer.
J = 45000*200*96485/(2.4*60*112*90*63.546*0.000001*3600*24*365) = 298.53 A/m2
(c)
-we‘ =200*96485*2.05*1000/(90*63.546*3600000) = 1.921 kWh/kg Cu
