Electrolysis of Some Typical Electrolytes (Part II)

In our last post: Electrolysis of Some Typical Electrolytes (Part I), we studied the electrolysis of acidified water, dilute sodium chloride and brine under different conditions. Here, we will be looking at the electrolysis of copper (II) tetraoxosulphate (VI), CuSO4, solution.

In solution, copper (II) tetraoxosulphate (VI) undergoes complete ionization to form copper (II) ions, Cu2+, and tetraoxosulphate (VI) ions, SO4--, according to the equation:

                                         CuSO4(aq) ----> Cu2+(aq) + SO4--(aq) .....................(i)

Note that the two minus signs attached to the SO4 stand for 2-

Electrolysis of Dilute Copper (II) tetraoxosulphate (VI) Using Inert (Platinum or Carbon) Electrodes

The ions present in copper (II) tetraoxosulphate (VI) solution are Cu2+, SO4-- and H+, OH-; with the latter pair coming from the dissociation of water. Expectedly, the OH- and SO4-- ions migrate to the anode, while the H+ and Cu2+ ions migrate to the cathode.



Electrolysis of copper(II) tetraoxosulphate(VI) in the laboratory

The electrolysis of dilute copper (II) tetraoxosulphate (VI) solution using inert electrodes gives the same products as the electrolysis of acidified water and dilute sodium chloride, which are 2 volumes of hydrogen gas and 1 volume of oxygen gas.

(See the electrolysis of acidified water in the  Electrolysis of Some Typical Electrolytes (Part I)  for the full anodic and cathodic half-reactions and equations).

However, in dilute copper (II) tetraoxosulphate (VI) solution, even though the Cu2+ is lower in the electrochemical series than the H+, the H+ still gets discharged in preference to the former, because of its higher concentration, which overrides their positions in the electrochemical series, especially when they are so close to each other. (See Electrolysis: Overview)

Summary:
Electrolyte - Dilute copper (II) tetraoxosulphate (VI), CuSO4
Ions Present - Cu2+, SO4-- and H+, OH-
Concentrations - High (H+, OH-), Low (Cu2+, SO4--)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1
Effect on Electrolyte - The electrolyte becomes concentrated, due to the removal of water molecules in the form of hydrogen and oxygen. This is indicated by the increased intensity of the electrolyte's colour (light blue ----> deep blue). 


Electrolysis of Concentrated Copper (II) tetraoxosulphate (VI) Using Inert (Platinum or Carbon) Electrodes

The electrolytic cell used here must be designed in a way that will enable the collection of gas at the anode and the removal of deposited metal at the cathode. (Please refer to your textbooks for a standard labelled diagram)

At the anode, despite the higher concentration of the SO4-- ions, the OH- ions are still preferentially discharged because of the wide gap separating them in the electrochemical series. So, the positions of ions in the series overrides their concentration. Hence, expectedly, they undergo oxidation to form 2H2O and O2 gas.

                                 4OH-(aq) ----> 2H2O(l) + O2(g) + 4e- .................(ii) (Anodic half-reaction)

At the cathode, the Cu2+ is preferentially discharged over H+, because it is lower in the electrochemical series and also, of higher concentration. So, each Cu2+ undergoes reduction by gaining two electrons to become deposited as metallic copper atoms:

                                  Cu2+(aq) + 2e- ----> Cu(s) ...........................(iii)

We will multiply equation (iii) by 2 to balance the number of moles of electrons transferred

                                 2Cu2+(aq) +  4e- ----> 2Cu(s) .......................(iv) (Cathodic half-equation)

Combining equations (ii) and (iv) will give the overall electrolytic reaction:

                               2Cu2+(aq) + 4OH-(aq) + 4e- ----> 2Cu(s) + 2H2O(l) + O2(g) + 4e-

                 2Cu2+(aq) + 4OH-(aq) ----> 2Cu(s) + O2(g) + 2H2O(l) .................(vi) (Overall equation)

From equation (vi) above, it can be seen that for every 2 moles of copper metal deposited at the cathode, 1 volume of oxygen is liberated at the anode. The water molecule produced in the equation is a by-product.

Note: As the Cu2+ and OH- are being removed from the solution in form of Cu metal and O2 gas, the H+ and SO4-- still remain in solution. These combine to form tetraoxosulphate (VI) acid, which makes the resulting solution after the electrolysis more acidic. The blue colour of the electrolyte also fades because of the removal of the Cu2+ from solution.

Summary:
Electrolyte - Concentrated copper (II) tetraoxosulphate (VI), CuSO4
Ions Present - Cu2+, SO4-- and H+, OH-
Concentrations - High (Cu2+, SO4--), Low (H+, OH-)
Nature of Electrodes - Inert
Product at the Cathode - Copper metal
Product at the Anode - Oxygen gas
Mole Ratio - 2:1
Effect on Electrolyte - The electrolyte becomes more acidic and faint in colour.

Electrolysis of Copper (II) tetraoxosulphate (VI) Using Copper Anode and Carbon/Platinum or Copper Cathode

Here, the concentration of the electrolyte is irrelevant as the copper anode interferes with the electrolytic process and influences the ion that is discharged, because the electrolyte is a salt of copper.

So, at the anode; in as much as, the OH- and SO4-- ions migrate there, none of them is discharged. Rather, the metallic copper atoms on the surface of the electrode go into solution by losing two electrons each to form Cu2+ ions, as shown:

                                           Cu(s) ----> Cu2+(aq) + 2e- .....................(vii)

The Cu2+ ions then migrate to the cathode, through the electrolyte to pick up the two electrons that were lost at the anode and get deposited as copper metal on the cathode, neglecting the H+ and Cu2+ from the electrolyte:

                                           Cu2+(aq) + 2e- ----> Cu(s) .....................(iii)

The above reactions indicate that nothing special actually takes place than the anode being transferred to the cathode, because combining equations (vii) and (iii) to obtain the overall equation will give zero. However, the reaction can be summarized as shown:

                                      Anode                Electrolyte               Cathode
                                                     -2e-                          +2e-
                                      Cu(s)      ---->    Cu2+(aq)      ---->    Cu(s)

Over time, as the copper atoms on the anode keep dissolving in the electrolyte and get deposited on the cathode, the anode becomes smaller in size, while the cathode increases in size. The original ions from the salt and water still remain in solution, and as such, the colour, concentration and pH of the electrolyte remain unchanged until the whole anode is used up or replaced with another material.



Electrolysis of copper(II) tetraoxosulphate(VI) using copper anode

This process is employed industrially in the purification of metals (like copper, silver, gold etc.)and electroplating. Electroplating is the process of coating a substance with a metal using electricity either for beautification or protection. If the substance is coated with gold, it is known as gold-plating, if it is coated with silver, the process is silver-plating and if coated with zinc, it is said be a galvanizing process.

In purification of metals, say copper, the impure copper ore is made the anode while a strip of very pure copper is used as the cathode. The electrolyte is a solution of any soluble salt of copper (II), like copper (II) tetraoxosulphate (VI), copper (II) trioxonitrate (V), Cu(NO3)2, copper (II) chloride, CuCl2 etc. The electrodes are connected to a power supply that supplies AC current. As the process proceeds, the copper atoms from the anode are transferred to the cathode. The anode decrease in size while the cathode, which is made up of copper only increases in mass. The impurities from the ore settle at the bottom of the electrolytic tank. At the end of the process, the pure copper cathode is removed and dried.

In electroplating, say silver-plating, the substance to be silver-plated e.g. a spoon, is made the cathode, while a strip of pure silver is used as the anode. The electrolyte is a solution of a soluble salt of silver, e.g. silver trioxonitrate (V), AgNO3. The electrodes are connected to an AC power supply. As the process proceeds, the silver atoms from the anode go into solution by losing one electron each to form silver ions, Ag+ (since silver is univalent). The silver ions, then migrate to the cathode where they gain one electron each to become silver atoms, which are then deposited on the surface of the spoon, according to the equation:

                                      Anode                Electrolyte               Cathode
                                                     -e-                           +e-
                                      Ag(s)      ---->    Ag+(aq)      ---->    Ag(s)

 Depending on the thickness of the electroplating, the power supply is switched off when the desired level of electroplating has been attained, and the silver-plated spoon is removed and allowed to dry.

Summary:
In the purification of metals from their ores, the ore to be purified is the anode, while the cathode is a very pure form of the metal, whose ore is to be purified and the electrolyte is a solution of any soluble salt of the metal.

In electroplating, the substance to be electroplated is the cathode, while the metal that is to be used is the anode, and the electrolyte is a solution of any soluble salt of the anode.

Do These

Question 1
a) Draw the diagram of the electrolytic cell used in the purification of copper.
b) (i) Which of the electrodes in your diagram increases in mass during the electrolysis? Give reasons for your answers. (ii) State with reason the site of oxidation.
c) What will the electrolyte be? Explain why its colour does not change in intensity during the electrolysis.

Question 2
a) Explain why a crystal of sodium chloride is a poor conductor of electricity but its solution in water readily conducts electricity.
b) Write the equation which represents the dissociation of a salt formed from a trivalent cation A, and a divalent anion B, when dissolved in water.

Question 3
During a practical class, a student was asked to electroplate a stainless steel spoon using a silver rod as the electrode and silver trioxonitrate (V) solution as the electrolyte. The following were the results she obtained:

Mass of spoon                                                                 =       103.75g
Mass of silver-plated spoon                                            =                  g
Mass of the silver rod before electrolysis                       =       100.35g
Current                                                                            =              30A
Time taken                                                                      =         30mins

a) Draw a diagram of the set-up the student had used to electroplate the spoon. Which of the electrodes should the spoon be?
b) The student had forgotten to fill up one of the values in her results sheet. Find that value, and fill it in the empty space.
c) What was the mass of the silver rod after the electrolysis?

(To completely answer Q3, you might need to see our next post: Faraday's Laws of Electrolysis)


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