Electrolysis is a word formed from two Latin words - 'electrum', which means electricity and 'lysis', which means to breakdown. Therefore, electrolysis is defined as the use electrical energy to breakdown a compound into its constituent elements. For instance, using electricity to decompose sodium chloride into sodium metal and chlorine gas, as depicted by their state symbols (s) and (g) in the equation below:
2NaCl(aq) ----> 2Na(s) + Cl2(g)
It is a redox process, and one needs a good understanding of redox reactions to understand it. (Kindly refer to our posts on Redox Reactions: Overview, Balancing Redox Equations and Oxidation Numbers)
Before we continue, it is imperative to understand the meanings of some terminologies in electrolysis.
Definition of Terms
Electrolyte
This is a substance which conducts electricity in the aqueous or molten state and gets decomposed in the process. An aqueous solution is formed when a substance dissolves in water as a solvent to form a solution, e.g, aqueous sodium chloride, NaCl(aq), whereas a molten state is when a solid substance is made to melt, the melted state is the molten state. The difference between aqueous and molten states is the former involves the use of water, while the latter involves the use of heat only to breakdown the force of cohesion between the solid particles.
Strong Electrolytes
These are substances which dissociate completely into ions when dissolved in water. Examples include sodium chloride, potassium iodide and strong acids like hydrochloric acid, tetraoxosulphate (VI) acid etc.
Weak Electrolytes
These are substances that undergo partial dissociation into ions when dissolved in water. Examples are weak organic acids like ethanoic acid or weak alkalis like calcium hydroxide etc.
Electrodes
These are metallic plates, rods or wires through which electric current enters and leaves the electrolyte.
Cathode
This is the negative electrode or terminal by which current leaves the electrolyte. It is the terminal where the reduction process occurs and is always connected to the negative terminal of the power supply.
Anode
This is the positive electrode or terminal by which current enters the electrolyte. It is the terminal where the oxidation process occurs and is always connected to the positive terminal of the power supply.
Electrolytic Cell
This is an assembly of the electrodes, electrolyte and power supply where the electrolysis is carried out. It is also known as voltameter. (Please refer to your textbooks for the diagram of an electrolytic cell)
Ionic Theory
According to the Arrhenius Theory, some substances undergo dissociation when dissolved in water or melted to yield electrically charged atoms or molecules, called ions. This dissociation is also known as ionization. The substances can be electrovalent or ionic in nature, e.g, sodium chloride, copper (II) tetraoxosulphate (VI), iron (II) chloride etc. or polar in nature like acids, alkalis and even water.
NaCl(aq) ---> Na+(aq) + Cl-(aq)
CuSO4(aq) ---> Cu2+(aq) + SO4--(aq)
FeCl2(aq) ---> Fe2+(aq) + 2Cl-(aq)
HCl(aq) ---> H+(aq) + Cl-(aq)
2H2O(l) <---> H3O+(aq) + OH-(aq)
The Arrhenius Theory is also known as the ionic theory and is the basis for electrolysis.
Mechanism of Electrolysis & Electrolytic Reactions
When these substances undergo the ionization in water, they form positive and negative ions which move about randomly in the solution. However, immediately a potential difference is set up by passing electric current through the solution with the help of electrodes attached to a power supply, the migration of the ions occurs. Based on the law of electrostatic attraction, i.e, unlike charges attract while like charges repel; the positively-charged ions are attracted to the negative electrode or cathode, while the negatively-charged ions move to positive electrode or anode. The positive ions that migrate to the cathode are known as the cations (cathode ions), while negative ions that migrate to the anode are called anions (anode ions).
It is important to note that it is the mobile ions in the solution and not electrons that are responsible for the conduction of electricity by electrolytes.
At the anode, the anions lose their electrons to the electron-deficient terminal and get oxidized to the non-metallic atoms. This process is known as oxidation and is illustrated below:
A-(aq) - e- ---> A …………(i)
A-(aq) ---> A + e- …………(ii) (oxidation half-reaction)
If A is a diatomic molecule, it means two of equation (ii) can be combined to give equation (iii), which is the oxidation half-equation for a diatomic gas molecule.
2A-(aq) ---> A2 + 2e- ………(iii) (oxidation half-reaction)
From equation (i), the anion, A- loses one electron since it is univalent to form an atom A. The loss of the electron, e- is depicted by a subtraction sign before the electron on the left hand side of the equation. However, when writing chemical equations we do not use minus signs, so in applying the principle of balancing equations in mathematics, we move the electron to the right hand side and the sign changes to a plus sign. Equation (ii), therefore, is the conventional way of writing an oxidation half-equation.
On the other hand, at the cathode, the cations gain electrons from the electron-surplus terminal and get reduced to metallic atoms. This is a reduction process and is illustrated below:
M2+(aq) + 2e- ---> M(s) …………(iv) (reduction half-equation)
In equation (iv), the cation, M2+ gains two electrons because it is divalent to form a metallic atom M. The gain of the electron is indicated by an addition sign before the electron on the left hand side of the equation. Thus, equation (iv) is the conventional way of writing a reduction half-equation.
If M is a univalent metal ion, then equation (iv) will be:
M+(aq) + e- ---> M(s) …………(v) (reduction half-equation)
What do you think the reduction half-equation for a trivalent metal will be?
Note that the electrons gained by the cations are the ones that were lost by the anions, and passed through the external circuit to the cathode; and as such, the number of electrons lost must equal the number gained. So, if a univalent anion of a diatomic non-metal and a divalent cation of a monoatomic metal are involved in the reaction, then, we will make use of equations (iii) and (iv) in forming the overall redox equation:
2A-(aq) ---> A2(g) + 2e- ………(iii) (oxidation half-reaction)
M2+(aq) + 2e- ---> M(s) …………(iv) (reduction half-equation)
2A-(aq) + M2+(aq) ---> A2(g) + M(s)
or better rearranged as:
M2+(aq) + 2A-(aq) ---> M(s) + A2(g) (overall equation)
Now, let's use calcium chloride as typical example of an electrolyte to explain these jargons.
Calcium chloride undergoes ionization to give calcium ions as the cations and chloride ions as the anions as shown:
CaCl2(aq) ---> Ca2+(aq) + 2Cl-(aq)
Therefore, it can undergo electrolysis to produce calcium metal and chlorine gas as explained further. In the presence of an electric current, the chloride ions migrate to the anode, where they undergo oxidation by losing their electrons to the positive electrode to form chlorine atoms. Since chlorine atoms cannot exist on their own, they combine together to form diatomic molecules of chlorine gas, as shown:
Cl-(aq) ---> Cl + e-
Cl-(aq) ---> Cl + e-
2Cl-(aq) ---> Cl2(g) + 2e (oxidation half-equation)
On the other hand, the calcium ions migrate to the cathode, where they pick up two electrons each and get reduced to calcium metals, as represented by the equation below:
Ca2+(aq) + 2e- ---> Ca(s) (reduction half-equation)
Since the number of moles of electrons transferred in the process is balanced, then, both equations can be combined to give:
Ca2+(aq) + 2Cl-(aq) ---> Ca(s) + Cl2(g) (overall equation)
Therefore, at the end of the process, calcium is deposited at the cathode, while chlorine gas is liberated at the anode.
Nevertheless, in practice, the process is not as simple and straightforward as described above, because in a solution of any substance, there are more than two types of ions. Remember, water also undergoes dissociation, so the water molecules in the solution will also partake in the electrolytic reactions.
This implies that for the aqueous calcium chloride solution, there are four ions that will partake in the electrolytic process, namely: Ca2+, Cl- (from the CaCl2) and H+, OH- (from the H2O). So, at any of the electrodes, there will be two types of ions competing for deposition or discharge. At the anode, the Cl- and the OH- will compete for discharge, while the H+ and Ca2+ will compete for deposition or discharge at the cathode. Only one type of ions eventually gets discharged at each electrode and for that to occur, there are certain conditions that must be met. These conditions are the factors that affect the preferential discharge of ions during electrolysis.
Factors Affecting the Preferential Discharge of Ions
These include:
a) Position of ions in the electrochemical series
b) Concentration of ions in the electrolyte
c) Nature of electrodes
Position of Ions in the Electrochemical Series (ES)
The electrochemical is a table showing the reduction potentials of various metals or non-metals arranged in a particular order as shown:
Cations
K+ - Potassium
Na+ - Sodium
Ca2+ - Calcium
Mg2+ - Magnesium
Al3+ - Aluminum
Zn2+ - Zinc
Fe2+ - Iron
Sn2+ - Tin
Pb2+ - Lead
H+ - Hydrogen
Cu2+ - Copper
Hg2+ - Mercury
Ag+ - Silver
Au+ - Gold
Pt2+ - Platinum
Anions
OH- - Hydroxide
I- - Iodide
Br- - Bromide
Cl- - Chloride
(NO3)- - Trioxonitrate (V)
SO4-- - Tetraoxosulphate (VI)
F- - Fluoride
For cations, ions lower down in the ES usually get discharged in preference to the ones higher up in the series. This is because the ions of the lower metals have the tendency of being reduced by receiving electrons more readily than the ones above, as electropositivity decreases down and increases up the series. Electropositivity is the ability to lose electrons and it is a typical metallic property.
So, in our example, H+ ions will be reduced at the cathode in preference to the Ca2+ ions because the position of the hydrogen ion is lower than that of calcium ion in the series.
Hence, at the cathode, the following reaction will take place:
H+(aq) + e- ---> H
Since hydrogen is a diatomic molecule, the final equation will be:
2H+(aq) + 2e- ---> H2(g)
The above equation illustrates that hydrogen gas will be liberated at the cathode.
Similarly, for anions, ions higher up are always preferentially discharged over the ones lower in the series. This is because the ions of the higher non-metals have the ability of being oxidized by losing electrons much readily than the ones below. This is because electronegativity, which is the ability of a substance to gain electrons decreases up and increases down the series.
Going by our illustrations, at the anode, the OH- will get discharged in preference to the Cl- because the former is higher than the latter in the ES.
So, the anodic half-reaction will be:
OH-(aq) + e- ---> OH
OH cannot exist on its own, it will have to combine with other OH radicals to form water and O atom as shown below:
OH-(aq) + e- ---> OH………(a)
OH-(aq) + e- ---> OH………(b)
Adding (a) and (b) gives:
2OH-(aq) + 2e- ---> H2O + O………(c)
However, O atom cannot exist on its own because oxygen is a diatomic molecule. Therefore, equation (c) must occur twice to make oxygen a molecule.
2OH-(aq) + 2e- ---> H2O + O………(c) x 2
4OH-(aq) + 4e- ---> 2H2O(l) + O2(g)………(d)
So, from equation (d) above, oxygen gas will be liberated at the anode.
Concentration of Ions
The ion of higher concentration in the electrolytic solution is always discharged in preference to the one of lower concentration. However, this factor is only applicable when the distance between the ions in the ES is not wide, say 3 - 4 steps apart.
So, in our example, if the concentration of the calcium chloride in the solution is high, it means that only Ca2+ and Cl- will be of high concentration, while the H+ and OH- will be of low concentration. In that case, when Cl- and OH- get to the anode, the Cl- will be preferentially discharged over the OH- because of its higher concentration. However, at the cathode, the H+ will be discharged in preference to Ca2+, despite the higher concentration of the Ca2+ in the solution. This is because the distance between them in the ES is quite wide. Hence, in the electrolysis of concentrated CaCl2 solution, chlorine gas and hydrogen gas will be liberated at the anode and cathode respectively.
Nature of Electrodes
This factor overrides the first two because of the high affinity (attraction) of certain electrodes towards some substances.There are two classes of electrodes usually involved in electrolytic reactions, namely: inert and reactive electrodes. The inert electrodes, by nature, are relatively unreactive and so do not partake in the electrolytic reactions, e.g., platinum and carbon, while the reactive electrodes affect the outcome of the electrolytic reaction because of their interference, e.g., copper electrodes used in the electrolysis of a copper salt. However, in the presence of some substances, the inert electrodes can become reactive because of their affinities for such substances.
For instance, mercury has a high affinity for sodium, and will also combine with sodium to form sodium amalgam (Na/Hg). So, in the presence of mercury cathode, sodium ion will always get discharged in preference to any other cation, irrespective of their positions in the ES. Also, platinum readily combines with chlorine to form platinum (II) chloride, and as such will preferentially discharge the chloride ions in the presence of any anion, irrespective of their positions and concentrations. A similar situation occurs between carbon and oxygen. A carbon anode will always favour the discharge of the OH- over other anions to liberate oxygen.
We will see how to apply all these factors when we look at the electrolysis of some typical electrolytes like sodium chloride solution, dilute tetraoxosulphate (VI) acid, copper (II) tetraoxosulphate (VI) solution etc. in our next post.
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2NaCl(aq) ----> 2Na(s) + Cl2(g)
It is a redox process, and one needs a good understanding of redox reactions to understand it. (Kindly refer to our posts on Redox Reactions: Overview, Balancing Redox Equations and Oxidation Numbers)
Before we continue, it is imperative to understand the meanings of some terminologies in electrolysis.
Definition of Terms
Electrolyte
This is a substance which conducts electricity in the aqueous or molten state and gets decomposed in the process. An aqueous solution is formed when a substance dissolves in water as a solvent to form a solution, e.g, aqueous sodium chloride, NaCl(aq), whereas a molten state is when a solid substance is made to melt, the melted state is the molten state. The difference between aqueous and molten states is the former involves the use of water, while the latter involves the use of heat only to breakdown the force of cohesion between the solid particles.
Strong Electrolytes
These are substances which dissociate completely into ions when dissolved in water. Examples include sodium chloride, potassium iodide and strong acids like hydrochloric acid, tetraoxosulphate (VI) acid etc.
Weak Electrolytes
These are substances that undergo partial dissociation into ions when dissolved in water. Examples are weak organic acids like ethanoic acid or weak alkalis like calcium hydroxide etc.
Electrodes
These are metallic plates, rods or wires through which electric current enters and leaves the electrolyte.
Cathode
This is the negative electrode or terminal by which current leaves the electrolyte. It is the terminal where the reduction process occurs and is always connected to the negative terminal of the power supply.
Anode
This is the positive electrode or terminal by which current enters the electrolyte. It is the terminal where the oxidation process occurs and is always connected to the positive terminal of the power supply.
Electrolytic Cell
This is an assembly of the electrodes, electrolyte and power supply where the electrolysis is carried out. It is also known as voltameter. (Please refer to your textbooks for the diagram of an electrolytic cell)
Ionic Theory
According to the Arrhenius Theory, some substances undergo dissociation when dissolved in water or melted to yield electrically charged atoms or molecules, called ions. This dissociation is also known as ionization. The substances can be electrovalent or ionic in nature, e.g, sodium chloride, copper (II) tetraoxosulphate (VI), iron (II) chloride etc. or polar in nature like acids, alkalis and even water.
NaCl(aq) ---> Na+(aq) + Cl-(aq)
CuSO4(aq) ---> Cu2+(aq) + SO4--(aq)
FeCl2(aq) ---> Fe2+(aq) + 2Cl-(aq)
HCl(aq) ---> H+(aq) + Cl-(aq)
2H2O(l) <---> H3O+(aq) + OH-(aq)
The Arrhenius Theory is also known as the ionic theory and is the basis for electrolysis.
Mechanism of Electrolysis & Electrolytic Reactions
When these substances undergo the ionization in water, they form positive and negative ions which move about randomly in the solution. However, immediately a potential difference is set up by passing electric current through the solution with the help of electrodes attached to a power supply, the migration of the ions occurs. Based on the law of electrostatic attraction, i.e, unlike charges attract while like charges repel; the positively-charged ions are attracted to the negative electrode or cathode, while the negatively-charged ions move to positive electrode or anode. The positive ions that migrate to the cathode are known as the cations (cathode ions), while negative ions that migrate to the anode are called anions (anode ions).
It is important to note that it is the mobile ions in the solution and not electrons that are responsible for the conduction of electricity by electrolytes.
At the anode, the anions lose their electrons to the electron-deficient terminal and get oxidized to the non-metallic atoms. This process is known as oxidation and is illustrated below:
A-(aq) - e- ---> A …………(i)
A-(aq) ---> A + e- …………(ii) (oxidation half-reaction)
If A is a diatomic molecule, it means two of equation (ii) can be combined to give equation (iii), which is the oxidation half-equation for a diatomic gas molecule.
2A-(aq) ---> A2 + 2e- ………(iii) (oxidation half-reaction)
From equation (i), the anion, A- loses one electron since it is univalent to form an atom A. The loss of the electron, e- is depicted by a subtraction sign before the electron on the left hand side of the equation. However, when writing chemical equations we do not use minus signs, so in applying the principle of balancing equations in mathematics, we move the electron to the right hand side and the sign changes to a plus sign. Equation (ii), therefore, is the conventional way of writing an oxidation half-equation.
On the other hand, at the cathode, the cations gain electrons from the electron-surplus terminal and get reduced to metallic atoms. This is a reduction process and is illustrated below:
M2+(aq) + 2e- ---> M(s) …………(iv) (reduction half-equation)
In equation (iv), the cation, M2+ gains two electrons because it is divalent to form a metallic atom M. The gain of the electron is indicated by an addition sign before the electron on the left hand side of the equation. Thus, equation (iv) is the conventional way of writing a reduction half-equation.
If M is a univalent metal ion, then equation (iv) will be:
M+(aq) + e- ---> M(s) …………(v) (reduction half-equation)
What do you think the reduction half-equation for a trivalent metal will be?
Note that the electrons gained by the cations are the ones that were lost by the anions, and passed through the external circuit to the cathode; and as such, the number of electrons lost must equal the number gained. So, if a univalent anion of a diatomic non-metal and a divalent cation of a monoatomic metal are involved in the reaction, then, we will make use of equations (iii) and (iv) in forming the overall redox equation:
2A-(aq) ---> A2(g) + 2e- ………(iii) (oxidation half-reaction)
M2+(aq) + 2e- ---> M(s) …………(iv) (reduction half-equation)
2A-(aq) + M2+(aq) ---> A2(g) + M(s)
or better rearranged as:
M2+(aq) + 2A-(aq) ---> M(s) + A2(g) (overall equation)
Now, let's use calcium chloride as typical example of an electrolyte to explain these jargons.
Calcium chloride undergoes ionization to give calcium ions as the cations and chloride ions as the anions as shown:
CaCl2(aq) ---> Ca2+(aq) + 2Cl-(aq)
Therefore, it can undergo electrolysis to produce calcium metal and chlorine gas as explained further. In the presence of an electric current, the chloride ions migrate to the anode, where they undergo oxidation by losing their electrons to the positive electrode to form chlorine atoms. Since chlorine atoms cannot exist on their own, they combine together to form diatomic molecules of chlorine gas, as shown:
Cl-(aq) ---> Cl + e-
Cl-(aq) ---> Cl + e-
2Cl-(aq) ---> Cl2(g) + 2e (oxidation half-equation)
On the other hand, the calcium ions migrate to the cathode, where they pick up two electrons each and get reduced to calcium metals, as represented by the equation below:
Ca2+(aq) + 2e- ---> Ca(s) (reduction half-equation)
Since the number of moles of electrons transferred in the process is balanced, then, both equations can be combined to give:
Ca2+(aq) + 2Cl-(aq) ---> Ca(s) + Cl2(g) (overall equation)
Therefore, at the end of the process, calcium is deposited at the cathode, while chlorine gas is liberated at the anode.
Nevertheless, in practice, the process is not as simple and straightforward as described above, because in a solution of any substance, there are more than two types of ions. Remember, water also undergoes dissociation, so the water molecules in the solution will also partake in the electrolytic reactions.
This implies that for the aqueous calcium chloride solution, there are four ions that will partake in the electrolytic process, namely: Ca2+, Cl- (from the CaCl2) and H+, OH- (from the H2O). So, at any of the electrodes, there will be two types of ions competing for deposition or discharge. At the anode, the Cl- and the OH- will compete for discharge, while the H+ and Ca2+ will compete for deposition or discharge at the cathode. Only one type of ions eventually gets discharged at each electrode and for that to occur, there are certain conditions that must be met. These conditions are the factors that affect the preferential discharge of ions during electrolysis.
Factors Affecting the Preferential Discharge of Ions
These include:
a) Position of ions in the electrochemical series
b) Concentration of ions in the electrolyte
c) Nature of electrodes
Position of Ions in the Electrochemical Series (ES)
The electrochemical is a table showing the reduction potentials of various metals or non-metals arranged in a particular order as shown:
Cations
K+ - Potassium
Na+ - Sodium
Ca2+ - Calcium
Mg2+ - Magnesium
Al3+ - Aluminum
Zn2+ - Zinc
Fe2+ - Iron
Sn2+ - Tin
Pb2+ - Lead
H+ - Hydrogen
Cu2+ - Copper
Hg2+ - Mercury
Ag+ - Silver
Au+ - Gold
Pt2+ - Platinum
Anions
OH- - Hydroxide
I- - Iodide
Br- - Bromide
Cl- - Chloride
(NO3)- - Trioxonitrate (V)
SO4-- - Tetraoxosulphate (VI)
F- - Fluoride
For cations, ions lower down in the ES usually get discharged in preference to the ones higher up in the series. This is because the ions of the lower metals have the tendency of being reduced by receiving electrons more readily than the ones above, as electropositivity decreases down and increases up the series. Electropositivity is the ability to lose electrons and it is a typical metallic property.
So, in our example, H+ ions will be reduced at the cathode in preference to the Ca2+ ions because the position of the hydrogen ion is lower than that of calcium ion in the series.
Hence, at the cathode, the following reaction will take place:
H+(aq) + e- ---> H
Since hydrogen is a diatomic molecule, the final equation will be:
2H+(aq) + 2e- ---> H2(g)
The above equation illustrates that hydrogen gas will be liberated at the cathode.
Similarly, for anions, ions higher up are always preferentially discharged over the ones lower in the series. This is because the ions of the higher non-metals have the ability of being oxidized by losing electrons much readily than the ones below. This is because electronegativity, which is the ability of a substance to gain electrons decreases up and increases down the series.
Going by our illustrations, at the anode, the OH- will get discharged in preference to the Cl- because the former is higher than the latter in the ES.
So, the anodic half-reaction will be:
OH-(aq) + e- ---> OH
OH cannot exist on its own, it will have to combine with other OH radicals to form water and O atom as shown below:
OH-(aq) + e- ---> OH………(a)
OH-(aq) + e- ---> OH………(b)
Adding (a) and (b) gives:
2OH-(aq) + 2e- ---> H2O + O………(c)
However, O atom cannot exist on its own because oxygen is a diatomic molecule. Therefore, equation (c) must occur twice to make oxygen a molecule.
2OH-(aq) + 2e- ---> H2O + O………(c) x 2
4OH-(aq) + 4e- ---> 2H2O(l) + O2(g)………(d)
So, from equation (d) above, oxygen gas will be liberated at the anode.
Concentration of Ions
The ion of higher concentration in the electrolytic solution is always discharged in preference to the one of lower concentration. However, this factor is only applicable when the distance between the ions in the ES is not wide, say 3 - 4 steps apart.
So, in our example, if the concentration of the calcium chloride in the solution is high, it means that only Ca2+ and Cl- will be of high concentration, while the H+ and OH- will be of low concentration. In that case, when Cl- and OH- get to the anode, the Cl- will be preferentially discharged over the OH- because of its higher concentration. However, at the cathode, the H+ will be discharged in preference to Ca2+, despite the higher concentration of the Ca2+ in the solution. This is because the distance between them in the ES is quite wide. Hence, in the electrolysis of concentrated CaCl2 solution, chlorine gas and hydrogen gas will be liberated at the anode and cathode respectively.
Nature of Electrodes
This factor overrides the first two because of the high affinity (attraction) of certain electrodes towards some substances.There are two classes of electrodes usually involved in electrolytic reactions, namely: inert and reactive electrodes. The inert electrodes, by nature, are relatively unreactive and so do not partake in the electrolytic reactions, e.g., platinum and carbon, while the reactive electrodes affect the outcome of the electrolytic reaction because of their interference, e.g., copper electrodes used in the electrolysis of a copper salt. However, in the presence of some substances, the inert electrodes can become reactive because of their affinities for such substances.
For instance, mercury has a high affinity for sodium, and will also combine with sodium to form sodium amalgam (Na/Hg). So, in the presence of mercury cathode, sodium ion will always get discharged in preference to any other cation, irrespective of their positions in the ES. Also, platinum readily combines with chlorine to form platinum (II) chloride, and as such will preferentially discharge the chloride ions in the presence of any anion, irrespective of their positions and concentrations. A similar situation occurs between carbon and oxygen. A carbon anode will always favour the discharge of the OH- over other anions to liberate oxygen.
We will see how to apply all these factors when we look at the electrolysis of some typical electrolytes like sodium chloride solution, dilute tetraoxosulphate (VI) acid, copper (II) tetraoxosulphate (VI) solution etc. in our next post.
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good work. you have shared a great deal of information. keep it up
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