copper electrolysis, cementation and silver … electrolysis, cementation and silver sulphide...

Copper Electrolysis, Cementation and Silver Sulphide Reduction

Karen Louise de Sousa Pesse

Supervision: Christa Sonck / [email protected]

08

-12

-20

16


1

1 Introduction

In this practical class, 3 experiences are carried:

Electrolysis of copper sulphate solution containing 5 g/L Cu 2 + and 5 ml H2S04;

o With lead anode

o With copper anode

Cementation process with iron in order to obtain metallic copper retrieved from a Cu2 + 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 , because the lead will not dissolve in the inert partsolution.

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.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

12

34

56

7

Figure 1: Current x Potential curve for Electrowinning of copper with lead anode.


2

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:

𝑃𝑏 → 𝑃𝑏2+ + 2𝑒− 𝐸𝑃𝑏2+/𝑃𝑏0 = −0,126 𝑉

2𝐻2𝑂 → 4𝐻+ + 𝑂2 + 4𝑒− 𝐸𝐻+/𝐻2

0 = − 1,229 𝑉

The reaction will occur with release of oxygen gas.

In the cathode the reduction of copper will take place:

𝐶𝑢2+ + 2𝑒− → 𝐶𝑢 𝐸𝐶𝑢2+/𝐶𝑢0 = 0,337 𝑉

2𝐻+ + 2𝑒− → 𝐻2 𝐸𝐻+/𝐻2

0 = 0 𝑉

𝐸°𝑐𝑒𝑙 𝑙 = 𝐸𝑐 𝑎𝑡 ℎ𝑜𝑑 𝑒 − 𝐸𝑎 𝑛𝑜𝑑𝑒 = 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.

The reduction reactions on the cathode are:

𝐶𝑢2+ + 2𝑒− → 𝐶𝑢 𝐸𝐶𝑢2+/𝐶𝑢0 = 0,337 𝑉

2𝐻+ + 2𝑒− → 𝐻2 𝐸𝐻+/𝐻2

0 = 0 𝑉

Meanwhile on the anode, copper metal becomes ion in an oxidation reaction:

𝐶𝑢 → 𝐶𝑢2+ + 2𝑒− 𝐸𝐶𝑢2+/𝐶𝑢0 = −0,337 𝑉

0

0.5

1

1.5

2

2.5

1 2 3 4 5 6 Figure 2: Current - Potential curve for Electrorefining of copper with copper anode.


3

In this part of the experiment, E catho de and Ea no de will be the same, because it is made of the same material. Thus, de difference between each other is zero.

𝐸𝐶𝑢2+/𝐶𝑢0 − 𝐸𝐶𝑢2+ 𝐶𝑢⁄

0 = 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 uneven 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

A second set up was made, keeping the current on a constant value of 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 consisting 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.


4

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 =

63,54𝑔

𝑚𝑜𝑙

2∗

1

96500𝐶

𝑚𝑜𝑙

∫ 0,50𝑑𝑡180

0= 0,0296g

The molar mass of copper is 63,54 g/mol.

To obtain the efficiency of the electrolysis:

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦𝑐𝑎𝑡ℎ𝑜𝑑𝑒 = 𝑚𝐶𝑢

𝑚𝐶𝑢𝑖𝑑𝑒𝑎𝑙

=0,0217 𝑔

0,0296 𝑔100% = 73,31%

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦𝑎𝑛𝑜𝑑𝑒 = 𝑚𝐶𝑢

𝑚𝐶𝑢𝑖𝑑𝑒𝑎𝑙

=0,0293 𝑔

0,0296 𝑔100% = 98,98%

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.


5

3 Cementation

3.1 Reactions

On the anode, oxidation is carried:

Fe → Fe2+ + 2 e− 𝐸𝐹𝑒2+0 = −0,447 V

And on the cathode, we have the reduction of copper:

Cu2+ + 2 e− → Cu 𝐸𝐶𝑢2+0 = 0,337 V

The overall reaction is then:

Cu2+ + Fe → Cu + Fe2+

𝐸°𝑐𝑒 𝑙𝑙 = 𝐸𝑐 𝑎𝑡 ℎ𝑜𝑑 𝑒 − 𝐸𝑎 𝑛𝑜𝑑𝑒 = 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

ΔG°<0, or that 𝐸𝐹𝑒2+/𝐹𝑒0 < 𝐸𝐶𝑢2+/𝐶𝑢

0 , which is given.

3.3 Determine the Equilibrium concentration of Fe2+ and Cu2+ The equilibrium concentration can be achieved from the Nernst equations:

𝐸𝐶𝑢 = 𝐸𝐶𝑢2+0 +

𝑅𝑇

𝑛𝐹ln

𝑎𝐶𝑢2+

𝑎𝐶𝑢= 0,337 𝑉 +

0,059

2log 𝐶𝑢2+

𝐸𝐹𝑒 = 𝐸𝐹𝑒2+0 +

𝑅𝑇

𝑛𝐹ln

𝑎𝐹𝑒2+

𝑎𝐹𝑒= −0.447 𝑉 +

0,059

2log 𝐹𝑒2+

Because we want cementation to happen: 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 > 𝐸𝑎𝑛𝑜𝑑𝑒:

0,337 𝑉 +0,059

2log 𝐶𝑢2+ > −0.447 𝑉 +

0,059

2log 𝐹𝑒2+

(0,337 𝑉 + 0,447 𝑉) ∗2

0,059> log

𝐹𝑒2+

𝐶𝑢2+

−26,5763 < log𝐹𝑒2+

𝐶𝑢2+ → 10−26,5763 <

𝐶𝑢2+

𝐹𝑒2+

2,6528−27 <𝐶𝑢2+

𝐹𝑒2+

𝐹𝑒2+

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*1026 of iron ions.

So at equilibrium there is almost no copper left. This means that:

𝐹𝑒2+𝑖𝑛𝑖𝑡𝑖𝑎𝑙 → 𝐹𝑒2+𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 = 𝐶𝑢2+𝑖𝑛𝑖𝑡𝑖𝑎𝑙 → 𝐶𝑢2+𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚


6

2,6528−27 = 𝐶𝑢2+𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

𝐶𝑢2+𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐶𝑢2+𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

2,6528−27𝐶𝑢2+𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐶𝑢2+𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

𝐶𝑢2+𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 < 𝐶𝑢2+𝑖𝑛𝑖𝑡𝑖𝑎𝑙

𝐹𝑒2+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:

𝛥𝐺𝑟𝑒𝑑0 = −𝑛𝐹𝐸0

Thus:

𝛥𝐺𝑟𝑒𝑑0 = −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 induced 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.


7

4 Silver Sulphide Reduction

Silver is a noble metal, however susceptible to the effect s of sulphide radicals. Silver sulphide (Ag2S) 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 2S 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− + 4𝐻2𝑂 + 6𝑒−

Cathode 3𝐴𝑔2𝑆 + 6𝑒− → 6𝐴𝑔 + 3𝑆2−

Table 3: The reduction on the cathode and oxidation on the anode are summarized.

The general reaction will then be:

3𝐴𝑔2𝑆 + 2𝐴𝑙 + 8𝑂𝐻− → 2𝐴𝑙𝑂2− + 4𝐻2𝑂 + 6𝐴𝑔 + 3𝑆2−

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 = ∑ 𝑁𝑖𝜇𝑖

𝛥𝐺red = 3𝜇𝑆2 −

0 + 6𝜇𝐴𝑔0 − 3𝜇𝐴𝑔2𝑆

0 = 92.533𝐽

𝑚𝑜𝑙+ 0 + 34.056

𝐽

𝑚𝑜𝑙= 126.589

𝐽

𝑚𝑜𝑙

𝛥𝐺oxi = 8𝜇𝑂𝐻−0 − 4𝜇𝐻2𝑂

0 − 2𝜇𝐴𝑙𝑂2−

0 = 4(−157.410)𝐽

𝑚𝑜𝑙 + 2(236.817)

𝐽

𝑚𝑜𝑙 + 831,161

𝐽

𝑚𝑜𝑙= −675.145

𝐽

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:

𝛥𝐺𝑟𝑒𝑑0 = −𝑛𝐹𝐸0

𝐸0 = −𝛥𝐺

𝑛𝐹

𝐸10 = −

126589𝐽

𝑚𝑜𝑙

(2 ∗ 96485𝐶

𝑚𝑜𝑙)

= −0.656 𝑉


8

𝐸20 −

675145𝐽

𝑚𝑜𝑙

(2 ∗ 96485𝐶

𝑚𝑜𝑙)

= −2.33 𝑉

It can be pointed out that the equilibrium potential is h igher than zero, which reinforces the reaction as a spontaneous one:

𝐸𝑐𝑒𝑙𝑙0 = 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸𝑎𝑛𝑜𝑑𝑒 = −0,656 𝑉 + 2,33𝑉 = 1,674 𝑉

This can be endorsed by the previous determination of the total 𝛥G:

𝛥𝐺𝑡𝑜𝑡𝑎𝑙 = − 970.523𝐽

𝑚𝑜𝑙

After obtaining these values, the equilibrium of the cell can be recalculated:

𝐸0 = −𝛥𝐺

𝑛𝐹

𝐸𝑐𝑒𝑙𝑙0 = −

−970.523 𝐽

𝑚𝑜𝑙⁄

6 ∗ 96.485 𝐶𝑚𝑜𝑙⁄

= 1,676 𝑉

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

AlO2

- 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.

copper electrolysis, cementation and silver … electrolysis, cementation and silver sulphide...

Documents