Karen Louise de Sousa Pesse
Supervision: Christa Sonck / [email protected]
08-12-2016
Copper Electrolysis, Cementation
and Silver Sulphide Reduction
1 Introduction
In this practical class, 3 experiences are carried:

Electrolysis of copper sulphate solution containing 5 g/L Cu 2 + and 5 ml H 2 S0 4 ;
o
With lead anode
o
With copper anode

Cementation process with iron in order to obtain metallic copper retrieved from a
Cu 2 + solution;

Reduction of Silver Sulphide;
1.1 Copper Electrolysis
1.1.1
With Lead Anode
On the first part, electrolysis was done using as an electrode a lead anode and a copper
cathode. Lead anode will be the inert part , because the lead will no t dissolve in the
solution.
Electrowinning, also called electroextraction, is the electrodeposition of metal that will be
happening by taking out ions from the solution. A similar process called Electrorefining is
also carried, using Copper-Copper electrodes, in which it can be observed the formation of a
layer on the cathode.
During winning, an increase of voltage will display a behaviour of the current in the solution
(Figure 1). The minimum potential, i.e. the voltage in which the system starts having
current, is defined experimentally as 1,8 V, and increases according to the potential
difference.
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
1
2
3
4
5
6
7
Figure 1: Current x Potential curve for Electrowinning of copper with lead anode.
Karen Louise de Sousa Pesse
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The metallic copper layer formed on the cathode is not homogeneous, and present sort of
bubbles. This layer is formed by copper ions which were compelled to reduce from the
solution because of the potential that was applied.
In the anode we will have the following oxidative half – reaction:
0
𝑃𝑏 → 𝑃𝑏 2+ + 2𝑒 − 𝐸𝑃𝑏
2+ /𝑃𝑏 = −0,126 𝑉
𝐸𝐻0 + /𝐻2 = − 1,229 𝑉
2𝐻2 𝑂 → 4𝐻 + + 𝑂2 + 4𝑒 −
The reaction will occur with release of oxygen gas.
In the cathode the reduction of copper will take place:
0
𝐶𝑢2+ + 2𝑒 − → 𝐶𝑢 𝐸𝐶𝑢
2+ /𝐶𝑢 = 0,337 𝑉
2𝐻 + + 2𝑒 − → 𝐻2
𝐸𝐻0 + /𝐻2 = 0 𝑉
𝐸° 𝑐 𝑒 𝑙 𝑙 = 𝐸 𝑐 𝑎𝑡 ℎ 𝑜 𝑑 𝑒 − 𝐸 𝑎 𝑛𝑜 𝑑 𝑒 = 0,337 + 0,126 + 1,229 = 1,7 V at 25°C
In this part, it is necessary to overcome a certain potential so that the reacti ons happen at
the same time at the cathode and anode. This has a very negative effect on the efficiency of
the electrolysis.
1.1.2
With Copper Anode
On the second part, copper anode and copper cathode are used.
2
The reduction reactions on the cathode are:
0
𝐶𝑢2+ + 2𝑒 − → 𝐶𝑢 𝐸𝐶𝑢
2+ /𝐶𝑢 = 0,337 𝑉
2𝐻 + + 2𝑒 − → 𝐻2 𝐸𝐻0 + /𝐻2 = 0 𝑉
Meanwhile on the anode, copper metal becomes ion in an oxidation reaction:
0
𝐶𝑢 → 𝐶𝑢2+ + 2𝑒 − 𝐸𝐶𝑢
2+ /𝐶𝑢 = −0,337 𝑉
2.5
2
1.5
1
0.5
0
1
2
3
4
5
6
Figure 2: Current - Potential curve
for Electrorefining of copper with
copper anode.
Karen Louise de Sousa Pesse
In this part of the experiment, E c a t ho de and E a no de will be the same, because it is made of the
same material. Thus, de difference between each other is zero.
0
0
𝐸𝐶𝑢
2+ /𝐶𝑢 − 𝐸𝐶𝑢2+ ⁄𝐶𝑢 = 0,337 V − 0,337 V = 0 V
This means that they do not have to overcome a potential like in the first experiment.
In the refinement of the copper, one electrode is contaminated with silver, cobalt, nickel,
etc. and those ions are resting on the anode. The copper then flows from one electrode to
the other.
2 Results and Conclusion
During Electrowinning, the hydrogen forming in the cathode in the solution influences the
copper layer, making it not homogeneous. This happens because the reduction potential is
lower than the hydrogen’s potential. The layer present bubbles and is une ven due to the
current density being higher at the corner of your sample.
The system involving Lead anode will require more energy because of the cell potential
different than zero. This can be observed on the graph: during winning, the first values are
zero, and only at 1,8 volts the current starts increasing. During refining, the current starts
increasing together with the potential since the beginning
2.1.1
Electrolysis with Constant Current
3
A second set up was made, keeping the c urrent on a constant value o f 0,5 A and room
temperature. The time was set for 3 minutes (180 seconds). There is no evolution of V in a
function of time: the current remained stable during the experiment thus it was not
necessary to vary the potential.
The electrochemical cell consis ting of 2 copper electrodes had its potential as a variable in
order to keep the current constant . On table 1, the mass of copper was determined before
and after the experiment in order to further obtain the efficiency of the operation.
Electrode
Mass (g) before Electrolysis
Mass (g) after Electrolysis
Deposit in Mass (g)
Anode
14,4715
14,4422
0,0293
Cathode
14,6995
14,7212
0,0217
Table 1 : Mass of copper was determined before and after the experiment.
The efficiency is the ratio between the measured and the theoretical maximum yield. The
theoretical yield is determined by the Faraday formula:
𝑚𝑚𝑎𝑥 =
𝑀𝑄
𝑛𝐹
In this formula, M is the molar mass in g/mol and Q is the total charge in Coulombs.
Karen Louise de Sousa Pesse
Applying the Faraday formula to obtain the ideal mass of copper that should be deposited in
the cathode during a constant current and well defined time:
𝑚𝑚𝑎𝑥 =
𝑀𝑄
𝑛𝐹
=
63,54
𝑔
𝑚𝑜𝑙
2
∗
180
1
96500
𝐶
𝑚𝑜𝑙
∫0
0,50𝑑𝑡 = 0,0296g
The molar mass of copper is 63,54 g/mol.
To obtain the efficiency of the electrolysis:
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦𝑐𝑎𝑡ℎ𝑜𝑑𝑒 =
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦𝑎𝑛𝑜𝑑𝑒 =
𝑚𝐶𝑢
0,0217 𝑔
=
100% = 73,31%
𝑚𝐶𝑢𝑖𝑑𝑒𝑎𝑙 0,0296 𝑔
𝑚𝐶𝑢
0,0293 𝑔
=
100% = 98,98%
𝑚𝐶𝑢𝑖𝑑𝑒𝑎𝑙 0,0296 𝑔
The efficiency of the anode, which corresponds for the copper that is gone, is higher than
the efficiency of the cathode, which is the amount of deposited copper. Theoretically, bot
efficiencies should have been 100%, since the same amount of electrons that are generated
should be consumed.
However it is possible that copper stays in the solution instead of depositing , not adhering
properly on the cathode and falling off of it.
2.1.2
Scanning Electron Microscope Imaging
The SEM-microscope shows that the copper on the electrode has a non -homogeneous
appearance, cause both by the release of H 2 gas during electrolysis and is uneven due to the
current density being higher at the corner of the sample.
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3 Cementation
3.1 Reactions
On the anode, oxidation is carried:
0
𝐸𝐹𝑒
2+ = −0,447 V
Fe → Fe2+ + 2 e−
And on the cathode, we have the reduction of copper:
0
𝐸𝐶𝑢
2+ = 0,337 V
Cu2+ + 2 e− → Cu
The overall reaction is then:
Cu2+ + Fe → Cu + Fe2+
𝐸° 𝑐 𝑒 𝑙𝑙 = 𝐸 𝑐 𝑎𝑡 ℎ 𝑜 𝑑 𝑒 − 𝐸 𝑎 𝑛𝑜 𝑑 𝑒 = 0,784 V
3.2 Determine the condition in which case the cementation takes place
This process is a heterogeneous way to obtain precipitation, in which the copper ions in a
solution precipitate out of it in the presence of a solid iron. During the reaction; the copper
ions reduce through transfer of electrons, meanwhile the iron oxidizes. No potential was
imposed to this experiment, thus the reactions shall be spontaneous, which means that
0
0
ΔG°<0, or that 𝐸𝐹𝑒
2+ /𝐹𝑒 < 𝐸𝐶𝑢2+ /𝐶𝑢 , which is given.
3.3 Determine the Equilibrium concentration of Fe2+ and Cu2+
The equilibrium concentration can be achieved from the Nernst equations:
0
𝐸𝐶𝑢 = 𝐸𝐶𝑢
2+ +
0
𝐸𝐹𝑒 = 𝐸𝐹𝑒
2+ +
𝑅𝑇 𝑎𝐶𝑢2+
0,059
ln
= 0,337 𝑉 +
log 𝐶𝑢2+ 
𝑛𝐹
𝑎𝐶𝑢
2
𝑅𝑇 𝑎𝐹𝑒 2+
0,059
ln
= −0.447 𝑉 +
log 𝐹𝑒 2+ 
𝑛𝐹
𝑎𝐹𝑒
2
Because we want cementation to happen : 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 > 𝐸𝑎𝑛𝑜𝑑𝑒 :
0,337 𝑉 +
0,059
0,059
log 𝐶𝑢2+  > −0.447 𝑉 +
log 𝐹𝑒 2+ 
2
2
(0,337 𝑉 + 0,447 𝑉) ∗
−26,5763 < log
2
𝐹𝑒 2+ 
> log
0,059
𝐶𝑢2+ 
𝐹𝑒 2+ 
𝐶𝑢2+ 
−26,5763
→
10
<
𝐶𝑢2+ 
𝐹𝑒 2+ 
𝐶𝑢2+ 
𝐹𝑒 2+ 
2,6528−27 < 𝐹𝑒 2+  𝐶𝑢2+ > 3,77 ∗ 1026
This means that the cementation process is carried until 1 ion of copper is available in a
solution of 3,77*10 26 of iron ions.
So at equilibrium there is almost no copper left. This means that:
𝐹𝑒 2+ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 → 𝐹𝑒 2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 = 𝐶𝑢2+ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 → 𝐶𝑢2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
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2,6528−27 =
𝐶𝑢2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
𝐶𝑢2+ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐶𝑢2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
2,6528−27 𝐶𝑢2+ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐶𝑢2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
𝐶𝑢2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 < 𝐶𝑢2+ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝐹𝑒 2+ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐶𝑢2+ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑖𝑟𝑢𝑚
3.4 Determine ΔG. Does the reaction occur spontaneously?
It is known that to calculate the Gibbs free energy, one have to use:
0
𝛥𝐺𝑟𝑒𝑑
= −𝑛𝐹𝐸0
Thus:
0
𝛥𝐺𝑟𝑒𝑑
= −𝑛𝐹𝐸0 = -2*(0,337+0,447)*96.500 = -151.312 J/mol
The reaction can be considered spontaneous, since the Gibbs free energy is negative.
3.5 How to do this in practice?
In practice, in an industry is not viable to have the frequent stoppage of the reaction after
the surface is entirely occupied. The most obvious solution would be to have someone
replacing the surface frequently, or stir the system. If a Foucault current is added, i.e. a
loop of electrical current ind uced within conductors by a changing magnetic field, and
combined with scrapping the copper off the electrode by a mechanical scrapper , copper
powder can be obtained. To filter i t, a centrifuge can be used.
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4 Silver Sulphide Reduction
Silver is a noble metal, however susceptible to the effect s of sulphide radicals. Silver
sulphide (Ag 2 S) is a commonly found compound in either aerobic or anaerobic environment,
and can be easily reduced in electrolysis. This can be considered a cleaning method to
eliminate the layer of Ag 2 S that gives the silver an anaesthetic appearance.
4.1 Method
The silver material being analysed is immersed in sodium carbonate and water alkali
solution. Afterwards, the material is involved in aluminium foil, connecting aluminium and
silver will enable electron transfer between metals.
Electrode
Reaction
Anode
2𝐴𝑙 + 8𝑂𝐻 − → 2𝐴𝑙𝑂2− + 4𝐻2 𝑂 + 6𝑒 −
Cathode
3𝐴𝑔2 𝑆 + 6𝑒 − → 6𝐴𝑔 + 3𝑆 2−
Table 3: The reduction on the cathode and oxidation on the anode are summarized.
The general reaction will then be:
3𝐴𝑔2 𝑆 + 2𝐴𝑙 + 8𝑂𝐻 − → 2𝐴𝑙𝑂2− + 4𝐻2 𝑂 + 6𝐴𝑔 + 3𝑆 2−
A kinetically favored redox reaction will take place since their potentials will be equal in
value. This can be obtained with the following calculations of the Gibbs free energy:
𝛥G = ∑ 𝑁𝑖 𝜇𝑖
0
0
𝛥𝐺red = 3𝜇𝑆02− + 6𝜇𝐴𝑔
− 3𝜇𝐴𝑔
= 92.533
2𝑆
0 −
0
0 −
𝛥𝐺oxi = 8𝜇𝑂𝐻
− 4𝜇𝐻
− 2𝜇𝐴𝑙𝑂
= 4(−157.410)
2𝑂
2
𝐽
𝐽
𝐽
+ 0 + 34.056
= 126.589
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝐽
𝐽
𝐽
𝐽
+ 2(236.817)
+ 831,161
= −675.145
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
To obtain the total Gibbs free energy, one should take into account the balance of the
reaction. The anodic reaction can be divided by 2, and the cathodic reaction can be divided
by 3. Thus, the general Gibbs free energy is defined as:
𝛥𝐺𝑡𝑜𝑡𝑎𝑙 = 3𝛥𝐺𝑟𝑒𝑑 + 2𝛥𝐺oxi = 3 ∗ 126589 − 2 ∗ 675145 = −970.523 𝐽/𝑚𝑜𝑙
This is definitely a very exothermic and spontaneous reaction.
Afterwards, E° reactions can be calculated:
0
𝛥𝐺𝑟𝑒𝑑
= −𝑛𝐹𝐸0
𝐸0 = −
𝛥𝐺
𝑛𝐹
𝐽
𝑚𝑜𝑙
𝐸10 = −
= −0.656 𝑉
𝐶
(2 ∗ 96485
)
𝑚𝑜𝑙
126589
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𝐽
𝑚𝑜𝑙
𝐸20 −
= −2.33 𝑉
𝐶
(2 ∗ 96485
)
𝑚𝑜𝑙
675145
It can be pointed out that the equilibrium potential is h igher than zero, which reinforces the
reaction as a spontaneous one :
0
𝐸𝑐𝑒𝑙𝑙
= 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸𝑎𝑛𝑜𝑑𝑒 = −0,656 𝑉 + 2,33𝑉 = 1,674 𝑉
This can be endorsed by the previous determination of the total 𝛥G:
𝛥𝐺𝑡𝑜𝑡𝑎𝑙 = − 970.523
𝐽
𝑚𝑜𝑙
After obtaining these values, the equilibrium of the cell can be recalculated:
𝐸0 = −
0
𝐸𝑐𝑒𝑙𝑙
𝛥𝐺
𝑛𝐹
𝐽
−970.523 ⁄𝑚𝑜𝑙
= −
= 1,676 𝑉
6 ∗ 96.485 𝐶⁄𝑚𝑜𝑙
The values are substantially equal between potentials, which confirm our experiment.
As an additional step, the Pourbaix
diagram can be consulted in order to
define which pH values are not
recommended during the experience ,
regarding the aluminum foil. The
passivation of the material happens
between pH 4,0 and 8,5. For this
reason, the addition of sodium
carbonate is essential, so that the ions
AlO 2 can be formed with the higher
pH. These ions do not form a layer on
the material and the experience
happens as expected.
Figure 3: Pourbaix Diagram of Aluminum.
Karen Louise de Sousa Pesse
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