2019 Final Exam
pdf
keyboard_arrow_up
School
University of British Columbia *
*We aren’t endorsed by this school
Course
358
Subject
Material Science
Date
Jan 9, 2024
Type
Pages
12
Uploaded by noone00
THE UNIVERSITY OF BRITISH COLUMBIA
Department of Materials Engineering
MTRL 358
–
Hydrometallurgy I
Final Examination 2019
[Total marks = 54]
Total time: 2.5 hours
Answer all questions
•
This examination has 11 pages (including supplementary material).
•
Formulas and data are included on pages 9-11.
•
Marks for each question are shown in square brackets.
1.
[6 marks]
Briefly answer the following questions:
(a) What are the two purposes of solvent extraction?
Purification and upgrading
of the PLS
(upgrading means to increase the concentration of the desired species).
(b) Indicate whether the following sentence is true or false: Leaching is the process
of dissolving a solid, in whole or in part, into an aqueous solution.
TRUE
(c) Name two hydrometallurgical process steps or operations in which the principle
of counter-current flows are applied.
Any two of counter-current leaching,
counter-current decantation and counter-current solvent extraction
(just
solvent extraction is acceptable).
(d) What happens at the cathode in an electrochemical cell?
Reduction
, or
electron transfer to effect reduction.
(e) What is the name for the aqueous stream exiting a solvent extraction loading
process?
Raffinate
This exam is
closed book and closed notes.
Read all questions carefully before beginning and work neatly.
Calculators are allowed. Cell phones and other electronic devices MUST be left at
the front of the exam room. Backpacks, bags, coats etc. MUST be left at the
front of the exam room.
(f) Silver sulfide (Ag
2
S) is highly insoluble in water (K
sp
= 6 x 10
-51
). True or false:
because of its low solubility it can’t be leached.
False
2.
[5 marks]
The mineral enargite has the formula Cu
3
AsS
4
. Write a balanced
chemical reaction for leaching of enargite by oxygen in sulfuric acid to form cupric
ion, sulfate and arsenic acid (H
3
AsO
4
; a soluble weak acid
–
assume it is
undissociated). Write the reaction in ionic and neutral forms. Include the phases in
the reaction in neutral form.
2Cu
3
AsS
4
+ 5H
2
O + 17.5O
2
= 6Cu
+2
+ 2H
3
AsO
4
+ 8SO
4
2-
+ 4H
+
2Cu
3
AsS
4 s
+ 5H
2
O
l
+ 17.5O
2 g
= 6CuSO
4 aq
+ 2H
3
AsO
4 aq
+ 2H
2
SO
4 aq
3.
[5 marks]
The mineral CuS reacts with cyanide according to the following
reaction:
2CuS + 8CN
-
+ 0.5O
2
+ H
2
O = 2[Cu(CN)
3
]
2-
+ 2SCN
-
+ 2OH
-
An ore contains 0.11% copper (1.1 kg Cu/tonne of ore) in the form of CuS. The ore
is processed at a rate of 1500 tonnes per hour. Assuming 75% of the mineral is
leached, how much sodium cyanide would be consumed by this reaction in kg/h?
3820 kg NaCN/h
4.
[8 marks]
An Eh-pH diagram for the Fe-S-H
2
O system involving only pyrite is
shown below. (Pyrite itself is not valuable, but it may contain gold.) Conditions are
indicated in the diagram. Thermodynamic data are also shown in the table below.
Species
G°
f
kJ
Species
G°
f
kJ
Fe
+2
-78.7
H
2
S
g
-33.56
Fe(OH)
2
-492
HS
-
11.44
Fe
+3
-4.6
S
8
( or S)
0
Fe(OH)
3
-704.63
SO
4
2-
-744.55
FeS
2
-166
HSO
4
-
-755.91
H
2
O
l
-237.15
(a) Is it possible to leach iron from FeS
2
at pH 8? Briefly explain your answer.
No. At pH 8 reduction of FeS
2
forms Fe
s
and HS
-
. There is no soluble Fe
species. Oxidation forms solid Fe(OH)
3
. Either way Fe is not dissolved.
(b) Write the balanced half reaction for the line numbered (1) in the diagram.
Fe
+2
+ 2SO
4
2-
+ 16H
+
+ 14e
-
= FeS
2
+ 8H
2
O
(c) Calculate the equation for the line in the form Eh = Slope x pH + Constant.
Eh = -0.06762pH + 0.36125
(d) (i) Write the main reactants and products for the two half reactions for leaching
of pyrite by ferric ion at pH 0.5. (The balanced half reactions are not
necessary, just the main reactants and products for each.)
Fe
+3
/Fe
+2
Fe
+2
,HSO
4
-
/FeS
2
(ii) Estimate the Eh values (including the units) for these two half reactions from
the diagram.
Eh
Fe+3/Fe+2
= 0.77 V; Eh
Fe+2/HSO4-/FeS2
= 0.32 V
(iii) Estimate the
E for the reaction.
E =0.45 V
(iv) Is the reaction favourable? Briefly explain your answer.
Yes;
E > 0
-1.0
-0.5
0.0
0.5
1.0
0
2
4
6
8
10
12
Eh
V
pH
Eh-pH diagram for the iron-sulfur-water system at 25°C and 1 atm
pressure, 0.2 m Fe solute activities and 0.5 m S solute activities.
Fe
+3
Fe
+2
HSO
4
-
SO
4
2-
Fe(OH)
3
Fe
+2
Fe
Fe(OH)
2
H
2
S
g
HS
-
FeS
2
H+
H2
O2
H2O
1
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
5.
[5 marks]
Below is a simplified flowsheet for a zinc roast-leach-electrowin
process. Zinc calcine is obtained by roasting a zinc sulfide concentrate which
contains ZnS, FeS
2
and small amounts of other minerals. The calcine contains
mainly ZnO and zinc ferrite (ZnFe
2
O
4
). Electrowinning produces most of the sulfuric
acid needed in leaching. The jarosite precipitation stage forms solid
NH
4
Fe
3
(SO
4
)
2
(OH)
6
, which is highly insoluble. Suitable conditions for jarosite
precipitation are about 95°C and pH ~1.5. (This is not sufficient to leach the zinc
ferrite.) A small fraction of the zinc calcine is used to adjust pH in jarosite
precipitation. Jarosite will not redissolve in the hot acid leach.
(a) In which step does Fe(III) enter solution?
The hot acid leach
(b) Is the jarosite precipitation step primarily a leaching process or a solution
purification process? Briefly explain your answer.
Solution purification. Fe(III)
is removed from solution as jarosite which is separated from solution by
solid-liquid separation.
(Some leaching of calcine occurs in that it acts as a base
for neutralization of some of the acid, but that is not the main point.)
(c) Indicate which of the following statements are true:
(i) This is an example of a co-current leaching process.
(ii) A disadvantage of this process is that zinc in the form of zinc ferrite added
with the calcine into jarosite precipitation is not recovered.
(iii) In this process zinc present in the calcine as zinc ferrite is leached in the hot
acid leach step.
True
(iv) In this process zinc present in the calcine as zinc ferrite is leached in the
neutral leach step.
(v) Good use of sulfuric acid in leaching is ensured by directing it back towards
the neutral leach.
True
(vi) Zinc oxide is mainly dissolved in the neutral leach step.
True
(vii) Zinc oxide is mainly dissolved in the hot acid leach step.
6.
[8 marks]
A gold ore is finely ground and then leached. The leach solution is to
be separated from the solids using CCD.
The leach slurry contains 31% solids and
the solids flow rate into the process is 930 tonnes/hour. The leach solution gold
concentration is 1.8 mg/L. The wash ratio is 1.8. The underflow slurry from each
thickener contains 40% solids. There are a total of 6 thickeners in the circuit. The
solutions have a density of about 1 t/m
3
.
(a) Calculate the feed solution flow rate in m
3
/h.
2070 m
3
/h
(b) Calculate the underflow solution flow rate and the wash water flow rate in m
3
/h.
1395 m
3
/h; 2511 m
3
/h, respectively
Neutral
leach
Zinc
calcine
L
S
Jarosite
precipitation
L
S
Hot acid
leach
L
S
Residues
(NH
4
)
2
SO
4
Zinc
calcine
Zinc dust
cementation
Electrowinning
Zinc
metal
Spent electrolyte
Spent electrolyte
Extra
H
2
SO
4
L
S
Residues
(c) Calculate the clear overflow solution flow rate from the lead thickener in m
3
/h.
3186 m
3
/h
(d) How many wash stages are there in the circuit?
5
(e) Calculate the wash efficiency and the recovery of gold in % in the clear overflow
solution from the lead thickener.
97.6%, 99.0%, respectively
(f) Calculate the gold concentration in the clear overflow solution from the lead
thickener in mg/L.
1.16 mg/L
7.
[8 marks]
A copper SX plant has a 2E 1S configuration. The stripping isotherm
for copper using 7% Lix 984N and a lean electrolyte [Cu
+2
] of 35 g/L is shown
below (next page).
(a) The rich electrolyte contains 47 g/L Cu
+2
. Estimate the [Cu
+2
] in the barren
organic. From the stripping isotherm:
1.40 g/L
(See diagram)
(b) The PLS contains 1.8 g/L Cu
+2
and has a pH of 1.8. The copper extraction from
the PLS is 90%. What is the raffinate copper concentration in g/L?
0.18 g/L
(c) The loaded organic copper concentration is 3.3 g/L. What is the A/O ratio?
Show
your calculation.
A/O =
y/
x = (3.3-1.40)/(1.8
–
0.18) =
1.173
(d) Two copies of a loading curve are provided below (pages 7 and 8). Draw a
McCabe-Thiele diagram for the solvent extraction loading process. Indicate the
stage numbers on the diagram. Indicate the 4 streams involved, e.g. the PLS, etc.
Detach your final diagram from the exam paper and include it with your answer
booklet. Make sure your name is on the diagram.
See diagram below
(e) If the copper production rate is 50,000 tonnes per year what is the electrolyte
flow rate in m
3
/h? (Assume 365 days per year operation; no down time.)
475.6 m
3
/h
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Stripping isotherm for Question 7:
30
35
40
45
50
55
1.0
1.1
1.2
1.3
1.4
1.5
1.6
Electrolyte [Cu
+2
] g/L
Organic [Cu
+2
] g/L
Stripping isotherm for 7% lix 984N with 180 g/L H
2
SO
4
, 35 g/L Cu
+2
0
1
2
3
4
0
0.5
1
1.5
2
2.5
Organic [Cu
+2
] g/L
Aqueous [Cu
+2
] g/L
Loading curve for 7% Lix 984N
Name:
Barren
organic
Loaded
organic
PLS
Raffinate
1
2
Question 7 McCabe-Thiele diagram
8.
[9 marks]
A copper electrowinning process consists of 172 cells, each
containing 60 cathodes. The cathode immersed dimensions are 1.1 m x 1 m. The
current efficiency is 90% and the energy consumption is 1940 kWh/tonne Cu. The
annual production rate is 64,300 tonnes of copper. (Assume 365 days/year
operation.)
(a) What is the applied voltage, including the units?
2.07 V
(b) What is the current density in A/m
2
?
303 A/m
2
(c) In a different plant the current efficiency is 91% and the current density is 247
A/m
2
. The immersed cathode dimensions are 1.1 m x 1.0 m. The density of copper
is 8.94 g/cm
3
. How long (
in days
) will it take to plate the cathodes to a thickness of
0.50 cm?
6.99 days
(d) What is the bare minimum applied cell voltage required for copper
electrowinning?
0.89 V
(positive
–
an applied voltage)
(e) Explain why the actual applied cell voltage is much greater than the minimum
cell voltage.
In addition to overcoming the thermodynamic voltage (-
E) extra driving
force is needed to force the oxygen evolution (anodic) half reaction (
O2
)
and the copper reduction (cathodic) half reaction (
Cu
). There are also
voltage drops due to resistances such as the electrical conductors and the
solution (I
R
i
). In total these make
E
appl
significantly greater than
–
E.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Formulas and data
R =
F/U + x
–
1
where x = O/U
F/U + x -
x
n+1
–
x
x
n+1
–
1
For a generalized half reaction:
aA + bB + mH
+
+ ne
-
= cC + dD
Eh = -
G°
- 2.303RT
log
a
C
c
a
D
d
- 2.303RTm
pH
nF
nF
a
A
a
a
B
b
nF
For a generalized hydrolysis reaction:
aA+ bB + mH
+
= cC + dD
pH
=
-
G°
- 1
log
a
C
c
a
D
d
2.303RTm
m
a
A
a
a
B
b
E =
E° - 2.303RT
logQ
nF
Faraday’s law relationships:
q = n
M
nF
M = It AW
nF
Constants:
CE = 100nFM
R = 8.3145 J/mol K = 0.08206 L atm/mol K
ItAW
EE =
E
rev
CE
F = 96485 C/mol e
-
E
appl
-w
e
’ = 100nF
E
appl
Conversions:
CE
1 kWh = 3.6
10
6
J
dn
M
=
jA
c
NCE
dt
100nF
0°C = 273.15 K
2007 standard atomic weights
Atomic
Atomic
Atomic
weight
weight
weight
actinium
Ac
227 hafnium
Hf
178.49 promethium
Pm
145
aluminium
Al
26.981538 hassium
Hs
269 protactinium
Pa
231.03588
americium
Am
243 helium
He
4.002602 radium
Ra
226
antimony
Sb
121.76 holmium
Ho
164.93032 radon
Rn
222
argon
Ar
39.948 hydrogen
H
1.00794 rhenium
Re
186.207
arsenic
As
74.9216 indium
In
114.818 rhodium
Rh
102.9055
astatine
At
210 iodine
I
126.90447 rubidium
Rb
85.4678
barium
Ba
137.327 iridium
Ir
192.217 ruthenium
Ru
101.07
berkelium
Bk
247 iron
Fe
55.845 rutherfordium
Rf
261
beryllium
Be
9.012182 krypton
Kr
83.8 samarium
Sm
150.36
bismuth
Bi
208.98038 lanthanum
La
138.9055 scandium
Sc
44.95591
bohrium
Bh
264 lawrencium
Lr
262 seaborgium
Sg
266
boron
B
10.811 lead
Pb
207.2 selenium
Se
78.96
bromine
Br
79.904 lithium
Li
6.941 silicon
Si
28.0855
cadmium
Cd
112.411 lutetium
Lu
174.9668 silver
Ag
107.8682
caesium
Cs
132.90545 magnesium
Mg
24.305 sodium
Na
22.98977
calcium
Ca
40.078 manganese
Mn
54.938049 strontium
Sr
87.62
californium
Cf
251 meitnerium
Mt
268 sulphur
S
32.066
carbon
C
12.0107 mendelevium
Md
258 tantalum
Ta
180.9479
cerium
Ce
140.116 mercury
Hg
200.59 technetium
Tc
98
chlorine
Cl
35.4527 molybdenum
Mo
95.96 tellurium
Te
127.6
chromium
Cr
51.9961 neodymium
Nd
144.24 terbium
Tb
158.92534
cobalt
Co
58.9332 neon
Ne
20.1797 thallium
Tl
204.3833
copper
Cu
63.546 neptunium
Np
237 thorium
Th
232.0381
curium
Cm
247 nickel
Ni
58.6934 thulium
Tm
168.93421
dubnium
Db
262 niobium
Nb
92.90638 tin
Sn
118.71
dysprosium
Dy
162.5 nitrogen
N
14.00674 titanium
Ti
47.867
einsteinium
Es
252 nobelium
No
259 tungsten
W
183.84
erbium
Er
167.26 osmium
Os
190.23 copernicium
Cn
277
europium
Eu
151.964 oxygen
O
15.9994 darmstadtium
Ds
269
fermium
Fm
257 palladium
Pd
106.42 roentgenium
Rg
272
fluorine
F
18.9984032 phosphorus
P
30.973762 uranium
U
238.0289
francium
Fr
223 platinum
Pt
195.078 vanadium
V
50.9415
gadolinium
Gd
157.25 plutonium
Pu
244 xenon
Xe
131.29
gallium
Ga
69.723 polonium
Po
210 ytterbium
Yb
173.04
germanium
Ge
72.61 potassium
K
39.0983 yttrium
Y
88.90585
gold
Au
196.96655 praseodymium
Pr
140.90765 zinc
Zn
65.38
zirconium
Zr
91.224
Name and
symbol
Name and
symbol
Name and
symbol
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help