Recall that electrolytes are substances that conduct electricity in their aqueous or molten state, and get decomposed in the process. We are going to study the electrolysis of some electrolytes such as dilute tetraoxosulphate (VI) acid, dilute sodium chloride, concentrated sodium chloride, copper (II) tetraoxosulphate (VI) solution etc.
Electrolysis of Acidified Water Using Platinum Electrodes
The electrolysis of acidified water is carried out using the Hofmann voltameter. (Please click on the link or refer to your textbooks for a standard labelled diagram).
Acidified water, which is simply dilute tetraoxosulphate (VI) acid, H2SO4, ionizes according to the equation:
H2SO4(aq) ----> 2H+(aq) + SO4--(aq)
Meanwhile, the water molecules undergo partial dissociation to form H+ and OH-,
H2O(l) <----> H+(aq) + OH-(aq)
Therefore, the acidified water contains three ions, namely: SO4--(from the acid), OH-(from the water) and H+(from the acid and water). Since the acid is a dilute solution, its ions (H+, SO4--) are of lower concentration than those from the water (H+, OH-). Also, the platinum electrodes used are considered as inert with respect to this electrolytic process. So, being negatively-charged, the SO4-- and OH- ions migrate to the anode, while the H+ ions migrate to the cathode.
At the anode, the OH- is discharged in preference to the SO4-- because it is higher in the electrochemical series and also of higher concentration. Therefore, each OH- ion donates one electron to the anode and becomes oxidized according to the equation:
OH-(aq) ----> OH + e- ………(i)
Since a neutral OH cannot exist on its own, it combines with another OH to form a molecule of H2O and an atom of O; and for that to happen, equation (i) must occur twice as illustrated in the equations below:
OH-(aq) ----> OH + e- ....................................(x 2)
2OH-(aq) ----> 2OH + 2e- ………(ii)
but, 2OH = H2O + O because that is the only stable set of products that it can produce in electrolysis. Thus, substituting the value of 2OH in equation (ii) gives:
2OH-(aq) ----> H2O(l) + O + 2e-………(iii)
Remember that oxygen is a diatomic molecule, so equation (iii) must occur twice to enable the O atoms to combine and produce O2.
2OH-(aq) ----> H2O(l) + O + 2e- ............................ (x 2)
4OH-(aq) ----> 2H2O(l) + O2(g) + 4e-………(iv)
Equation (iv), therefore, gives the oxidation half-equation or the anodic half-reaction.
At the cathode, the H+ ions, being the only cations are readily reduced by gaining one electron each according to the equation:
H+(aq) + e- ----> H .........................(v)
being diatomic, equation (v) must occur twice, for a hydrogen molecule, H2 to be formed.
H+(aq) + e- ----> H ................................................(x 2)
2H+(aq) + 2e- ----> H2(g) ........................(vi)
Equation (vi), then, gives the reduction half-equation or the cathodic half-reaction.
If we take a closer look at the oxidation and reduction half-equations, we can observe that the 4e- were lost, while only 2e-were gained. So, to account for the electron deficit, we will multiply equation (vi) by 2 to obtain:
4H+(aq) + 4e- ----> 2H2(g) ........................(vii)
Since the number of moles of electrons transferred has been balanced, we can then combine equations (iv) and (vii) to obtain the overall equation:
4H+(aq) + 4OH-(aq) + 4e- ----> 2H2(g) + 2H2O(l) + O2(g) + 4e-................(viii)
4H2O(l) ----> 2H2(g) + O2(g) + 2H2O(l)
By netting-off the terms that appear on both sides of the equation, we obtain equation (ix) as the final overall equation.
2H2O(l) ----> 2H2(g) + O2(g) ................(ix)
From the above equation, it shows that at the end of the electrolysis of acidified water, 2 volumes of hydrogen gas are liberated at the cathode, while 1 volume of oxygen is liberated at the anode. It means that water is broken down into hydrogen and oxygen in the ratio of 2:1 respectively.
Summary:
Electrolyte - Dilute tetraoxosulphate (VI) acid, H2SO4
Ions Present - H+, SO4-- and OH-
Concentrations - High (OH-), Low (SO4--)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1
Electrolysis of Dilute Sodium Chloride Using Platinum Electrodes
Sodium chloride, NaCl, ionizes according to the equation:
NaCl(aq) ----> Na+(aq) + Cl-(aq)
being dilute, the solution will also contain the H+ and OH- ions from the water molecules
Therefore, dilute sodium chloride solution contains four ions, which are: Na+ and Cl-(from the salt), H+ and OH-(from water). Being a dilute solution, it implies that the Na+ and Cl- ions from the salt are of lower concentration than the ions from the water. The platinum electrodes used are inert because they do not partake in the electrolytic process. As expected, the Cl- and OH- ions migrate to the anode, while the Na+ and H+ ions migrate to the cathode.
At the anode, the OH- is preferentially discharged over the Cl- because it is higher in the electrochemical series and also of higher concentration, and the oxidation process and the products will be similar to the one discussed in the electrolysis of acidified water above.
At the cathode, the H+ is discharged in preference to the Na+ because its position is lower in the electrochemical series than the Na+. So, it undergoes reduction as described in the electrolysis of acidified water to produce hydrogen gas.
Just like the electrolysis of acidified water, the end-products of the electrolysis of dilute sodium chloride are 2 volumes of hydrogen gas at the cathode and 1 volume of oxygen gas at the anode.
Summary:
Electrolyte - Dilute sodium chloride, NaCl
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (H+, OH-), Low (Na+, Cl-)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1
Electrolysis of Brine Using Graphite (Carbon) Electrodes
Brine is simply concentrated sodium chloride solution, and contains the same types of ions as the dilute solution, except that the Na+ and Cl- ions will be of higher concentrations than the H+ and OH- ions.
At the anode, the Cl- is preferentially discharged over the OH- because of its higher concentration. Here, the concentration of ions override their positions in the electrochemical series because their distance apart is not wide. So, the chloride ions get oxidized by donating their electrons to the anode to become chlorine atoms, which then combine in pairs to form molecules of chlorine gas, according to the equations:
Cl-(aq) ----> Cl + e- .............................(x)
Cl-(aq) ----> Cl + e-
2Cl-(aq) ----> Cl2(g) + 2e- ..................(xi) (Oxidation half-equation)
At the cathode, the H+ is still discharged in preference to the Na+, despite the higher concentration of the Na+ in the solution. This is because their positions in the electrochemical series are very wide apart, and this overrides the concentration of ions in solution. So, it undergoes reduction to produce hydrogen gas.
2H+(aq) + 2e- ----> H2(g) ................(vi) (Reduction half-equation)
Since the number of moles of electrons transferred in the electrolytic process is balanced, we can combine the anodic and cathodic half-reactions to obtain the overall reaction.
2H+(aq) + 2Cl-(aq) + 2e- ----> H2(g) + Cl2(g) + 2e- ............................(xii)
2H+(aq) + 2Cl-(aq) ----> H2(g) + Cl2(g) ............................(xiii)
2HCl(aq) ----> H2(g) + Cl2(g) ............................(xiv)
From equation (xiv), we can see that the electrolysis of brine produces equal volumes of hydrogen and chlorine gases liberated at the cathode and anode respectively.
Summary:
Electrolyte - Concentrated sodium chloride, NaCl (Brine)
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (Na+, Cl-), Low (H+, OH-)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Chlorine gas
Gas Volume Ratio - 1:1
Electrolysis of Brine Using Mercury Cathode and Graphite (Carbon) Anode
At the anode, just like the electrolysis of brine using carbon electrodes, Cl- will be discharged preferentially, and chlorine gas will be liberated at the anode.
2Cl-(aq) ----> Cl2(g) + 2e- ..................(x) (Oxidation half-equation)
At the cathode, the mercury electrode will influence the discharge of ions. Because of its strong affinity to sodium, the mercury cathode combines with the sodium ion, Na+, to form sodium amalgam (Na/Hg), thereby making the energy required for the discharge of Na+ to be lower than that required for the H+, according to the equation:
Na+(aq) + Hg(l) + e- ----> Na/Hg(l) ................(xv)
To balance the number of moles of electrons transferred in the process, we multiply equation (xv) by 2, to obtain:
2Na+(aq) + 2Hg(l) + 2e- ----> 2Na/Hg(l) ................(xvi) (Reduction half-equation)
Combining the balanced oxidation and reduction half-equations, the overall reaction becomes:
2Na+(aq) + 2Hg(l) + 2Cl-(aq) ----> 2Na/Hg(l) + Cl2(g) ...........(xvii)
but, the sodium amalgam is not quite useful, and to obtain a more useful product from it, it is treated with water, to form sodium hydroxide, hydrogen gas and mercury according to the equation:
2Na/Hg(l) + 2H2O(l) ----> 2NaOH(aq) + H2(g) + 2Hg(l) .........(xviii)
At the end of the process, the mercury regenerated in equation (xviii) can be reused if the need arises. The sodium hydroxide is produced as a by-product, while the hydrogen gas is liberated at the cathode.
This process is applied industrially in the electrolysis of brine to produce chlorine gas, hydrogen gas and sodium hydroxide.
Summary:
Electrolyte - Concentrated sodium chloride, NaCl (Brine)
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (Na+, Cl-), Low (H+, OH-)
Nature of Electrodes - Anode: Graphite (Inert); Cathode: Mercury (Reactive)
Product at the Cathode - Hydrogen gas and Sodium hydroxide
Product at the Anode - Chlorine gas
Gas Volume Ratio - 1:1
From the above, does it mean it is not possible to achieve the deposition of sodium metal only at the cathode?
The answer is: Yes, it is possible!
The deposition of sodium metal only at the cathode can be achieved by using molten sodium chloride (melted sodium chloride without water) and graphite (carbon) electrodes. The anodic half-reaction will remain the same as equation (x) above, while the cathodic half-reaction will be:
2Na+(aq) + 2e- ----> 2Na(s) ...............(xix)
Combining both half-reactions will give us the overall equation:
2Na+(aq) + 2Cl-(aq) + 2e- ----> 2Na(s) + Cl2(g) + 2e-
2Na+(aq) + 2Cl-(aq) ----> 2Na(s) + Cl2(g) ...............(xx)
2NaCl(l) ----> 2Na(s) + Cl2(g) ...........(xxi)
This principle is employed industrially in the extraction of sodium metal using molten sodium chloride.
We will study the electrolysis of copper (II) tetraoxosulphate (VI), CuSO4, solution in our next post: Electrolysis of Some Typical Electrolytes (Part II).
Do These:
Question 1
a) Define the term electrolysis.
b) State three factors that affect the discharge of ions during electrolysis, and discuss any two.
c) Briefly describe what happens when a concentrated solution of sodium chlorine is electrolyzed using carbon electrodes. What will the outcome be if mercury is used as the cathode?
Question 2
a) Explain briefly each of the following terms: i) anode ii) inert electrode iii) voltameter
b) Sodium and aluminium are extracted by the electrolysis of molten sodium chloride and alumina, Al2O3, respectively. Write balanced equations for the reactions at the anode and cathode during the extraction of: i) sodium ii) aluminium
(Hint: Alumina ionizes according to the equation - Al203(aq) ----> 2Al3+(aq) + 3O2-(aq). There are no water molecules and each of the aluminium ions will gain three electrons to become deposited as aluminium metal atoms, while the oxides will donate two electrons each to become oxygen atoms. Remember, oxygen is a diatomic molecule)
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Electrolysis of Acidified Water Using Platinum Electrodes
The electrolysis of acidified water is carried out using the Hofmann voltameter. (Please click on the link or refer to your textbooks for a standard labelled diagram).
 |
| Hofmann Voltameter |
H2SO4(aq) ----> 2H+(aq) + SO4--(aq)
Meanwhile, the water molecules undergo partial dissociation to form H+ and OH-,
H2O(l) <----> H+(aq) + OH-(aq)
Therefore, the acidified water contains three ions, namely: SO4--(from the acid), OH-(from the water) and H+(from the acid and water). Since the acid is a dilute solution, its ions (H+, SO4--) are of lower concentration than those from the water (H+, OH-). Also, the platinum electrodes used are considered as inert with respect to this electrolytic process. So, being negatively-charged, the SO4-- and OH- ions migrate to the anode, while the H+ ions migrate to the cathode.
At the anode, the OH- is discharged in preference to the SO4-- because it is higher in the electrochemical series and also of higher concentration. Therefore, each OH- ion donates one electron to the anode and becomes oxidized according to the equation:
OH-(aq) ----> OH + e- ………(i)
Since a neutral OH cannot exist on its own, it combines with another OH to form a molecule of H2O and an atom of O; and for that to happen, equation (i) must occur twice as illustrated in the equations below:
OH-(aq) ----> OH + e- ....................................(x 2)
2OH-(aq) ----> 2OH + 2e- ………(ii)
but, 2OH = H2O + O because that is the only stable set of products that it can produce in electrolysis. Thus, substituting the value of 2OH in equation (ii) gives:
2OH-(aq) ----> H2O(l) + O + 2e-………(iii)
Remember that oxygen is a diatomic molecule, so equation (iii) must occur twice to enable the O atoms to combine and produce O2.
2OH-(aq) ----> H2O(l) + O + 2e- ............................ (x 2)
4OH-(aq) ----> 2H2O(l) + O2(g) + 4e-………(iv)
Equation (iv), therefore, gives the oxidation half-equation or the anodic half-reaction.
At the cathode, the H+ ions, being the only cations are readily reduced by gaining one electron each according to the equation:
H+(aq) + e- ----> H .........................(v)
being diatomic, equation (v) must occur twice, for a hydrogen molecule, H2 to be formed.
H+(aq) + e- ----> H ................................................(x 2)
2H+(aq) + 2e- ----> H2(g) ........................(vi)
Equation (vi), then, gives the reduction half-equation or the cathodic half-reaction.
If we take a closer look at the oxidation and reduction half-equations, we can observe that the 4e- were lost, while only 2e-were gained. So, to account for the electron deficit, we will multiply equation (vi) by 2 to obtain:
4H+(aq) + 4e- ----> 2H2(g) ........................(vii)
Since the number of moles of electrons transferred has been balanced, we can then combine equations (iv) and (vii) to obtain the overall equation:
4H+(aq) + 4OH-(aq) + 4e- ----> 2H2(g) + 2H2O(l) + O2(g) + 4e-................(viii)
4H2O(l) ----> 2H2(g) + O2(g) + 2H2O(l)
By netting-off the terms that appear on both sides of the equation, we obtain equation (ix) as the final overall equation.
2H2O(l) ----> 2H2(g) + O2(g) ................(ix)
From the above equation, it shows that at the end of the electrolysis of acidified water, 2 volumes of hydrogen gas are liberated at the cathode, while 1 volume of oxygen is liberated at the anode. It means that water is broken down into hydrogen and oxygen in the ratio of 2:1 respectively.
Summary:
Electrolyte - Dilute tetraoxosulphate (VI) acid, H2SO4
Ions Present - H+, SO4-- and OH-
Concentrations - High (OH-), Low (SO4--)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1
Electrolysis of Dilute Sodium Chloride Using Platinum Electrodes
Sodium chloride, NaCl, ionizes according to the equation:
NaCl(aq) ----> Na+(aq) + Cl-(aq)
being dilute, the solution will also contain the H+ and OH- ions from the water molecules
Therefore, dilute sodium chloride solution contains four ions, which are: Na+ and Cl-(from the salt), H+ and OH-(from water). Being a dilute solution, it implies that the Na+ and Cl- ions from the salt are of lower concentration than the ions from the water. The platinum electrodes used are inert because they do not partake in the electrolytic process. As expected, the Cl- and OH- ions migrate to the anode, while the Na+ and H+ ions migrate to the cathode.
At the anode, the OH- is preferentially discharged over the Cl- because it is higher in the electrochemical series and also of higher concentration, and the oxidation process and the products will be similar to the one discussed in the electrolysis of acidified water above.
At the cathode, the H+ is discharged in preference to the Na+ because its position is lower in the electrochemical series than the Na+. So, it undergoes reduction as described in the electrolysis of acidified water to produce hydrogen gas.
Just like the electrolysis of acidified water, the end-products of the electrolysis of dilute sodium chloride are 2 volumes of hydrogen gas at the cathode and 1 volume of oxygen gas at the anode.
Summary:
Electrolyte - Dilute sodium chloride, NaCl
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (H+, OH-), Low (Na+, Cl-)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1
Electrolysis of Brine Using Graphite (Carbon) Electrodes
Brine is simply concentrated sodium chloride solution, and contains the same types of ions as the dilute solution, except that the Na+ and Cl- ions will be of higher concentrations than the H+ and OH- ions.
At the anode, the Cl- is preferentially discharged over the OH- because of its higher concentration. Here, the concentration of ions override their positions in the electrochemical series because their distance apart is not wide. So, the chloride ions get oxidized by donating their electrons to the anode to become chlorine atoms, which then combine in pairs to form molecules of chlorine gas, according to the equations:
Cl-(aq) ----> Cl + e- .............................(x)
Cl-(aq) ----> Cl + e-
2Cl-(aq) ----> Cl2(g) + 2e- ..................(xi) (Oxidation half-equation)
At the cathode, the H+ is still discharged in preference to the Na+, despite the higher concentration of the Na+ in the solution. This is because their positions in the electrochemical series are very wide apart, and this overrides the concentration of ions in solution. So, it undergoes reduction to produce hydrogen gas.
2H+(aq) + 2e- ----> H2(g) ................(vi) (Reduction half-equation)
Since the number of moles of electrons transferred in the electrolytic process is balanced, we can combine the anodic and cathodic half-reactions to obtain the overall reaction.
2H+(aq) + 2Cl-(aq) + 2e- ----> H2(g) + Cl2(g) + 2e- ............................(xii)
2H+(aq) + 2Cl-(aq) ----> H2(g) + Cl2(g) ............................(xiii)
2HCl(aq) ----> H2(g) + Cl2(g) ............................(xiv)
From equation (xiv), we can see that the electrolysis of brine produces equal volumes of hydrogen and chlorine gases liberated at the cathode and anode respectively.
Summary:
Electrolyte - Concentrated sodium chloride, NaCl (Brine)
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (Na+, Cl-), Low (H+, OH-)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Chlorine gas
Gas Volume Ratio - 1:1
Electrolysis of Brine Using Mercury Cathode and Graphite (Carbon) Anode
At the anode, just like the electrolysis of brine using carbon electrodes, Cl- will be discharged preferentially, and chlorine gas will be liberated at the anode.
2Cl-(aq) ----> Cl2(g) + 2e- ..................(x) (Oxidation half-equation)
At the cathode, the mercury electrode will influence the discharge of ions. Because of its strong affinity to sodium, the mercury cathode combines with the sodium ion, Na+, to form sodium amalgam (Na/Hg), thereby making the energy required for the discharge of Na+ to be lower than that required for the H+, according to the equation:
Na+(aq) + Hg(l) + e- ----> Na/Hg(l) ................(xv)
To balance the number of moles of electrons transferred in the process, we multiply equation (xv) by 2, to obtain:
2Na+(aq) + 2Hg(l) + 2e- ----> 2Na/Hg(l) ................(xvi) (Reduction half-equation)
Combining the balanced oxidation and reduction half-equations, the overall reaction becomes:
2Na+(aq) + 2Hg(l) + 2Cl-(aq) ----> 2Na/Hg(l) + Cl2(g) ...........(xvii)
but, the sodium amalgam is not quite useful, and to obtain a more useful product from it, it is treated with water, to form sodium hydroxide, hydrogen gas and mercury according to the equation:
2Na/Hg(l) + 2H2O(l) ----> 2NaOH(aq) + H2(g) + 2Hg(l) .........(xviii)
At the end of the process, the mercury regenerated in equation (xviii) can be reused if the need arises. The sodium hydroxide is produced as a by-product, while the hydrogen gas is liberated at the cathode.
This process is applied industrially in the electrolysis of brine to produce chlorine gas, hydrogen gas and sodium hydroxide.
Summary:
Electrolyte - Concentrated sodium chloride, NaCl (Brine)
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (Na+, Cl-), Low (H+, OH-)
Nature of Electrodes - Anode: Graphite (Inert); Cathode: Mercury (Reactive)
Product at the Cathode - Hydrogen gas and Sodium hydroxide
Product at the Anode - Chlorine gas
Gas Volume Ratio - 1:1
From the above, does it mean it is not possible to achieve the deposition of sodium metal only at the cathode?
The answer is: Yes, it is possible!
The deposition of sodium metal only at the cathode can be achieved by using molten sodium chloride (melted sodium chloride without water) and graphite (carbon) electrodes. The anodic half-reaction will remain the same as equation (x) above, while the cathodic half-reaction will be:
2Na+(aq) + 2e- ----> 2Na(s) ...............(xix)
Combining both half-reactions will give us the overall equation:
2Na+(aq) + 2Cl-(aq) + 2e- ----> 2Na(s) + Cl2(g) + 2e-
2Na+(aq) + 2Cl-(aq) ----> 2Na(s) + Cl2(g) ...............(xx)
2NaCl(l) ----> 2Na(s) + Cl2(g) ...........(xxi)
This principle is employed industrially in the extraction of sodium metal using molten sodium chloride.
We will study the electrolysis of copper (II) tetraoxosulphate (VI), CuSO4, solution in our next post: Electrolysis of Some Typical Electrolytes (Part II).
Do These:
Question 1
a) Define the term electrolysis.
b) State three factors that affect the discharge of ions during electrolysis, and discuss any two.
c) Briefly describe what happens when a concentrated solution of sodium chlorine is electrolyzed using carbon electrodes. What will the outcome be if mercury is used as the cathode?
Question 2
a) Explain briefly each of the following terms: i) anode ii) inert electrode iii) voltameter
b) Sodium and aluminium are extracted by the electrolysis of molten sodium chloride and alumina, Al2O3, respectively. Write balanced equations for the reactions at the anode and cathode during the extraction of: i) sodium ii) aluminium
(Hint: Alumina ionizes according to the equation - Al203(aq) ----> 2Al3+(aq) + 3O2-(aq). There are no water molecules and each of the aluminium ions will gain three electrons to become deposited as aluminium metal atoms, while the oxides will donate two electrons each to become oxygen atoms. Remember, oxygen is a diatomic molecule)
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