Separation of Metals from a Solution (Cu-Ni) Through

Electrolysis Karen Louise de Sousa Pesse

Supervision: Christa Sonck / christa.sonck@ugent.be

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Karen Louise de Sousa Pesse

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Introduction

Quantitative separation of Copper and Nickel

Electrolysis is a technique used in chemistry and manufacturing to separate elements from a solution by using a direct electric current, driving a non -spontaneous chemical reaction.

In this practical, Copper and Nickel were separated from an aqueous solution by their difference in equilibrium potential. This method is only possible when the difference in equilibrium potential between substances is large enough.

For Copper and Nickel, the equilibrium potentials for reduction relative to the Standard Hydrogen Electrode (SHE) are:

𝐸𝐶𝑢2+/𝐶𝑢 0 = 0,337V

𝐸𝑁𝑖+2/𝑁𝑖0 = -0,24V

These values are relatively close to each other, thus it has to be handled carefully to avoid that Nickel is deposited together with Copper on the cathode. The method of the Fixed Cathode Potential is applied in an attempt to avoid that.

Copper precipitates in acid medium at a cathode potential of -0.60 V vs Hg/Hg2SO4. Afterwards, Nickel deposits on the cathode in a strong ammonia medium.

Notice that through complexation in the form of 𝑁𝑖(𝑁𝐻3)42+ the activity of the free ions

Ni2+ is lowered, and precipitation in the form of hydroxides is prevented.

Methods

The inert Platinum electrodes are thoroughly cleaned in concentrated HNO 3 and rinsed with water. This step is fundamental to remove the traces of organic contamination from prior metal deposition. Then, the electrode which functions as cathode is dipped in alcohol and ether, and dried for about 5 minutes in the furnace with temperatures in the range of 100-110°C.

After cooling the electrode, this shall be weighted with an analytical balance. Subsequently, the electrodes are attached to a DC voltage source. The cathode is connected to a multimeter in order to measure the potential difference between cathode and the reference electrode (Hg/Hg 2SO4).

A shunt resistor converts the measured current to voltage. The measured voltage is written through an X (t) -recorder in a function of time. The paper moves forward at a speed of 30mm/s.

Karen Louise de Sousa Pesse

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Determination of Copper

Firstly 25 ml of Copper-Nickel solution was pipetted into the beaker, and then diluted until the electrode was completely immersed. Thereafter 1 ml 36N H 2SO4 by 100 ml of solution is added and heated at temperatures between 70 and 80°C.

The DC voltage source was set at a cathode potential of -0.60V with respect to the reference electrode. The movement of the electrode current recorded against the time will be measured by X (T) recorder. When all of the copper is removed from the solution, the electrode is weighed.

Determination of Nickel

The electrodes shall be cleaned as in the first step to remove the copper layer of the cathode. Regarding the solution, 12 grams of (𝑁𝐻4)2SO4 and 30 ml of 14N NH4OH were added. Also, the solution should be diluted to 300ml and heated to 70°C.

The course of the voltage is measured by using the X (t) recorder in a function of time. Afterwards, the cathode is weighed on the analytical balance.

Standard Cell Voltage

The standard cell voltage 𝐸𝑐𝑒𝑙0 can be calculated via equation:

𝐸𝑐𝑒𝑙𝑙0 = 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸𝑎𝑛𝑜𝑑𝑒 (1)

Wherein the voltage of the electrodes can be found through the Nernst equation [1]:

𝐸 = 𝐸0 +𝑅𝑇

𝑛𝐹∑ 𝜈𝑖 ln 𝑐𝑖 (2)

Take R as the ideal gas constant (3,31451 J/K mol), T the temperature in Kelvin, n the number of electrons that is exchanged, F the Faraday constant (9,6485 x 104 C/mol), 𝜈𝑖 the stoichiometric numbers and 𝑐𝑖 the concentration. At room temperature, this can be simplified to:

𝐸 = 𝐸0 + 0,059 𝑉

𝑛𝑙𝑜𝑔[𝐶] (3)

This equation is known as Nernst equation. Considering a pure solid, the activity can be equaled to 1.

Karen Louise de Sousa Pesse

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Efficiency

The efficiency is the ratio between the measured and the theoretical maximum yield. The theoretical yield is determined by : the Faraday formula

𝑚 = 𝑀𝑄

𝐹𝑛 (4)

In this formula, M is the molar mass in g/mol and Q is the total charge in Coulombs.

Results and Questions

Complete the electrolysis schedule and enter the correct polarity (+/-) on.

The set-up for the practicum is indicated in Figure 1. Since we have an Electrolytic cell, the anode should have a positive polarity and the cathode should have a negative polarity. The reaction at the anode is oxidation and in the cathode is reduction.

Explain the principle of the method of fixed cathode potential based on the

given standard potentials.

As explained before, the main idea of this method is to avoid that Nickel is deposited together with Copper on the cathode. The standard redu ction potentials of Copper and Nickel are relatively close to each other: 0.337 V for Copper and -0.24 V for Nickel, relative to the Standard Hydrogen Electrode (SHE). Using a constant potential of -0.60 V relative to an Hg/Hg2SO4 reference electrode will allow the copper to precipitate separately.

The cathode is then cleaned and set up for the deposit of Nickel in strong ammonia

environment, in which the free Ni 2+ ions are complexed into 𝑁𝑖(𝑁𝐻3)42+, having their

activity reduced and preventing hydroxide precipitation. In this way, Nickel will precipitate on the cathode in its metallic form.

Figure 1: Completed scheme of separation of metal by means of electrolysis indicating the polarity.

Anode Cathode

Karen Louise de Sousa Pesse

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The determination of Copper

Which Reactions occur at the cathode and anode? Calculate the standard cell

voltage at equilibrium.

At the cathode the reduction of Copper will happen as follow:

𝐶𝑢2+ + 2𝑒− → 𝐶𝑢 E° = 0,337 V

And at the anode the oxidation of water will be carried:

2𝐻2𝑂 → 𝑂2 + 4𝐻+ + 4𝑒− E° = 1.229 V

Thus, the balanced global reaction is:

2𝐶𝑢2+ + 2𝐻2𝑂 → 2𝐶𝑢 + 𝑂2 + 4𝐻+

And the Standard Cell Voltage at Equilibrium according to (1) is:

𝐸𝑐𝑒𝑙𝑙0 = 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒

0 − 𝐸𝑎𝑛𝑜𝑑𝑒0 = 0,337 − 1,229 = −0,892 𝑉

Compare the standard cathode potential with the specified electrolysis

potential.

The standard cathode potential is the likelihood that a species will be reduced.

During the experiment there is an attempt to av oid formation of Hydrogen gas. When H 2 is formed, an amorphous decomposition of copper can jeopardize the experiment.

Thus the cathode potential of -0.60 V and the reference electrode Hg/Hg 2SO4 (E0=0,657 V) are combined forming the same potential as 0,057 V relative to the Standard Hydrogen Electrode (SHE).

In this way the potential is kept as low as possible, but still higher than 0 volts relative to the Standard Hydrogen Electrode.

Karen Louise de Sousa Pesse

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Calculate and determine the amount of deposited copper and the concentration

of Cu2+ in the original solution. Calculate the equation potential of Cu2+/Cu in

the original solution.

During the electrolysis Oxygen gas is formed at the cathode. The copper deposition is promptly visible, being amorphous and easy to crumble. The initial weigh of the Platinum cathode was 35,4760 grams and the final weigh was 35,5994 grams .

The amount of deposited Copper is then:

𝑚𝐶𝑢 = 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒𝑎𝑓𝑡𝑒𝑟− 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒𝑏𝑒𝑓𝑜𝑟𝑒

= 35,5994𝑔 − 35,4760𝑔 = 0,12 𝑔𝑟𝑎𝑚𝑠

With this information, and the molar mass of copper being 63,54 g/mol , the concentration of copper ions in 200 ml of solution is determined by:

𝐶𝐶𝑢2+ =𝑚𝐶𝑢

𝑀𝑀𝐶𝑢

1=

0,12𝑔

63,54𝑔/𝑚𝑜𝑙

1

0,2𝑙= 0,00944

𝑚𝑜𝑙

𝑙

𝑉

To calculate the equation potential of Cu 2+/Cu in the original solution, the Nernst equati on (3) shall be used:

𝑬 = 𝑬𝟎 + 𝟎,𝟎𝟓𝟗 𝑽

𝒏𝒍𝒐𝒈[𝑪]

𝐸𝑐𝑎𝑡 = 𝐸𝐶𝑢2+/𝐶𝑢0 +

0,059 V

2log(𝐶𝐶𝑢2+) = 0,337 V +

0,059 V

2log (

35,5994𝑔 − 35,4760𝑔

63,54𝑔/𝑚𝑜𝑙

1

0.2𝑙) = 0,278 𝑉

To calculate the Nernst equation for anode potential, the concentration of Hydrogen ions shall be determined. This value can be obtained from the addition of 2 ml of H 2SO4 (18M) to 200 ml of solution:

𝐶𝐻+ =(2𝐶𝐻2𝑆𝑂4

𝑉𝑎𝑑𝑑𝑒𝑑)

𝑉𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛=

(2 ∗ 18 𝑀 ∗ 0,002 𝑙)

0,2 𝑙= 0,36 𝑀

It is important not to forget about the factor 2 in the previous equation, since there are two ions H+ for each H2SO4.

The potential of the anode will then be:

𝐸𝑎𝑛𝑜𝑑𝑒 = 𝐸𝑂2/𝐻2𝑂0 +

0,059

4log(𝐶𝐻+

4 ) = 1,229 + 0,059

4log {

(2 ∗ 18 𝑀 ∗ 0,002 𝑙)

0,2 𝑙 }

4

= 1,2028 𝑉

The experimental cell potential stands for:

𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸𝑎𝑛𝑜𝑑𝑒 = 0,278 − 1,2030 𝑉 = −0,925 𝑉

Karen Louise de Sousa Pesse

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Why does the measured cathode potential differ from this value?

The measured value for E catho de was 0,837 V. This value is obtained in the beginning of the experiment, before the electrolysis starts. The system voltage is 0,180 V and Hg/Hg2SO4 which is used as reference electrode has a potential of 0,657 V. The measured cathode potential is then 0,180 + 0,657 = 0,837 V.

The difference in values can be blamed to the fact that a Platinum electrode was used. The potential of Platinum during reduction reaction equals 1,2 volts . This creates a discrepancy caused by the measurement of the reduction of platinum as well.

Calculate the efficiency of the copper electrolysis based on the course of the

electrolysis current in function of the time.

After the end of the experiment, the area underneath the I (t) curve mad e was calculated. A value of 517,68 C was obtained. This value is then used in the Faraday law (4) in order to calculate the amount of copper deposited in an ideal case:

𝑚𝑐𝑜𝑝𝑝𝑒𝑟 = 𝑀𝑄

𝐹𝑛=

𝑀𝑀𝑐𝑜𝑝𝑝𝑒𝑟

2

1

96500 𝐶/𝑚𝑜𝑙∗ ∫ 𝐼(𝑡) =

63,54𝑔

𝑚𝑜𝑙2

1

96500𝐶

𝑚𝑜𝑙

517,68 𝐶 = 0,170 𝑔

To obtain the efficiency of the electrolysis:

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜼 =𝑚𝑐𝑜𝑝𝑝𝑒𝑟𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑

𝑚𝑐𝑜𝑝𝑝𝑒𝑟𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑𝑖𝑑𝑒𝑎𝑙

∗ 100% =0,12 𝑔

0,170𝑔∗ 100% = 70%

The determination of Nickel

Which reactions occur at the cathode and anode? Calculate the standard-cell

voltage at equilibrium.

At the cathode, the reduction of Nickel will happen:

𝑁𝑖(𝑁𝐻3)4 + 2𝑒− → 𝑁𝑖 + 4𝑁𝐻3 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒0 = −0,24 𝑉

Meanwhile at the anode there is the oxidation of OH-:

4 𝑂𝐻− → 𝑂2 + 2𝐻2𝑂 + 4 𝑒− 𝐸𝑎𝑛𝑜𝑑𝑒0 = 0,401 𝑉

The balanced global reaction corresponds to:

2 𝑁𝑖(𝑁𝐻3)4 + 4 𝑂𝐻− → 2𝑁𝑖 + 8 𝑁𝐻3 + 𝑂2 + 2 𝐻2𝑂

Thus, the Standard-Cell Voltage at Equilibrium is:

𝐸𝑐𝑒𝑙𝑙0 = 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒

0 − 𝐸𝑎𝑛𝑜𝑑𝑒0 = −0,24𝑉 − 0,401 𝑉 = −0,641 𝑉

Karen Louise de Sousa Pesse

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Calculate the amount of deposited nickel and the concentration of Ni2+ in the

original solution. Determine, based on that concentration, the equilibrium

potential of Ni2+/Ni in the original solution.

The cathode was weighted before and after the experiment, presenting an initia l mass of 35,4759 grams and a final mass of 35,5342 grams .

The amount of deposited Nickel is then:

𝑚𝑁𝑖 = 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒𝑎𝑓𝑡𝑒𝑟− 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒𝑏𝑒𝑓𝑜𝑟𝑒

= 35,5342𝑔 − 35,4759𝑔 = 0,0583 𝑔𝑟𝑎𝑚𝑠

It is then possible to calculate the concentration of Nickel from a solution of 300 ml, knowing that the molar mass of Nickel is 58,71 g/mol:

𝐶𝑁𝑖2+ =𝑚𝑁𝑖

𝑀𝑀𝑁𝑖∗

1

𝑉𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛=

0,0583 𝑔

58,71𝑔

𝑚𝑜𝑙

∗1

0,3𝑙 = 3,31 ∗ 10−3

𝑚𝑜𝑙

𝑙

Subsequently, the cathode potential shall be calculate from the Nernst equation:

𝑬 = 𝑬𝟎 + 𝟎,𝟎𝟓𝟗 𝑽

𝒏𝒍𝒐𝒈[𝑪]

𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 = 𝐸𝑁𝑖+2/𝑁𝑖0 +

0.059

2log(𝐶𝑁𝑖2+) = −0.24 𝑉 +

0.059

2log(3,31 10−3) = −0,313 𝑉

To calculate the anode potential, the concentration of OH- shall be determined. Taking

into consideration 30 ml of 14M NH 4OH into a diluted solution of 300 ml, this solution being saturated with H + from previous experiment:

𝐶𝑂𝐻− = 𝐶𝑁𝐻4𝑂𝐻𝑉𝑎𝑑𝑑𝑒𝑑

𝑉𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛= 14 𝑀 ∗

0,03𝑙

0,30𝑙= 1,4M

(𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑐𝑖𝑑)

𝐶𝐻+ = 0,36 𝑀 → 200 𝑚𝑙 = 0,26 𝑀 ∗0,20𝑙

0,30𝑙 → 300 𝑚𝑙 = 0.24 𝑀 𝑖𝑛 300 𝑚𝑙

𝐶𝑂𝐻− = 1.4 𝑀 − 0.24 𝑀 = 1.16 𝑀

(𝑤𝑖𝑡ℎ 𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑐𝑖𝑑)

𝐸𝑎𝑛𝑜𝑑𝑒 = 𝐸𝑂2/𝑂𝐻−0 −

0.059

4log(𝐶𝑂𝐻−

4 ) = 0.401 𝑉 − 0.059 log (14 𝑀 ∗0,03𝑙

0,30𝑙− 0,26 𝑀 ∗

0,20𝑙

0,30𝑙) = 0,3972 𝑉

𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸𝑎𝑛𝑜𝑑𝑒 = −0,313 𝑉 − 0,3972 𝑉 = −0,7102 𝑉

Karen Louise de Sousa Pesse

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Calculate the efficiency of the nickel electrolysis based on the course of the

electrolysis current in function of time.

After the end of the experiment, the area underneath the I (t) curve mad e was calculated. A value of 711,232 C was obtained. This value is then used in the Faraday law (4) in order to calculate the amount of nickel deposited in an ideal case:

𝑚𝑁𝑖𝑐𝑘𝑒𝑙 = 𝑀𝑄

𝐹𝑛=

𝑀𝑀𝑁𝑖𝑐𝑘𝑒𝑙

2

1

96500 𝐶/𝑚𝑜𝑙∗ ∫ 𝐼(𝑡) =

58,71𝑔

𝑚𝑜𝑙2

1

96500𝐶

𝑚𝑜𝑙

711,232 𝐶 = 0,2163 𝑔

The efficiency will then be:

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜼 =𝑚𝑛𝑖𝑐𝑘𝑒𝑙𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑

𝑚𝑛𝑖𝑐𝑘𝑒𝑙𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑𝑖𝑑𝑒𝑎𝑙

∗ 100% =0.0726 𝑔

0.2163𝑔∗ 100% = 33,56%

Karen Louise de Sousa Pesse

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Conclusion

Compare the performance of the copper electrolysis and explain the difference.

Declare in the same connection, also the difference in view of the current

curves in both cases.

Electrolysis is a suitable method to purify Copper, and it is often applied in industry.

Comparing with the copper electrolysis one can clearly see that nickel electrolysis is a lot less efficient.

This is visible on the I(t) curve and can be explained by the fact that durin g the Nickel electrolysis more ions can be found in the solution, resulting in a higher rate of side reactions if compared to the copper electrolysis. In a real environment, ammonium sulfate is a strong acidic salt ammonium hydroxide is a weak base, making room for ionizations with many by-products causing higher viscosity during the nickel deposition, which may affect the reaction negatively.

The reason you can notice this behavior in the I(t) curve is because the high amount of ions in the solution containing Nickel will increase the transfer of electrons, which affects the Faraday equation:

𝑚𝐸𝑙𝑒𝑚𝑒𝑛𝑡 = 𝑀𝑄

𝐹𝑛

This all leads to a low efficiency and higher residual current, culminating in a curve that ends further from 0 amperes during the nickel electrolysis.

References

[1] Christopher M. A. Brett. Electrochemistry: principles, methods, and applications .

Oxford science publications. Oxford University Press, 1993. isbn: 0198553897.

Annex - I (t) curves

Due to the shunt resistance, the voltage of 20 mV is associated with a current of 8 amperes. Thus, on the graph, the range of current will vary between -4 A and 4A.