Case Study: Industrial Electrolysis
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Electrolysis reactions are the basic foundations of today's modern industry. There are various elements, chemical compounds, and organic compounds that are only produced by electrolysis including aluminum, chlorine, and NaOH. Electrolysis is the process by which an electric current spurs an otherwise non-spontaneous reaction.
Electro-refining: How it Works
The electrorefining process refines metals or compounds at a high purity for a low cost. The pure metal can coat an otherwise worthless object
Let consider the electro-refining process of copper: At the anode, there is an impure piece of copper that has other metals such as Ag, Au, Pt, Sn, Bi, Sb, As, Fe, Ni, Co, and Zn. The copper in this impure ore is oxidized to form Cu2+ at the anode, and moves through an aqueous sulfuric acid-Copper (II) sulfate solution into the cathode. When it reaches the cathode, the Cu2+ is reduced to Cu. This whole process takes place at a fairly low voltage (about .15 to .30 V), so Ag, Au, and Pt are not oxidized at the anode, as their standard oxidation electrode potentials are -.800, -1.36 and -1.20 respectively; these unoxidized impurities turn into a mixture called anode mud, a sludge at the bottom of the tank. This sludge can be recovered and used in different processes. Unlike Ag, Au and Pt, the impurities of Sb, Bi and Sn in the ore are indeed oxidized at the anode, but they are precipitated as they form hydroxides and oxides. Finally, Fe, Ni, Co and Zn are oxidized as well, but they are dissolved in water. Therefore, the only solid we are left with is the pure solid copper plate at the cathode, which has a purity level of about 99.999%. The image below gives an outline about the fate of the main components of an impure iron ore.
Electrosynthesis is the method of producing substances through electrolysis reactions. This is useful when reaction conditions must be carefully controlled. One example of Electrosynthesis is that of MnO2, Manganese dioxide. MnO2 occurs naturally in the form of the mineral pyrolusite, but this mineral is not easily used due to the nature of its size and lattice structure. However, MnO2 can be obtained a different way, through the electrolysis of MnSO4 in a sulfuric acid solution.
oxidation: Mn2+ + 2H20 --> MnO2 + 4H+ + 2e- E0MnO2/H2=-1.23
reduction: 2e- + 2H+ --> H2 -E0H+/H2= -0
overall: Mn2+ + 2H2O --> MnO2 + 2H+ +H2 E0MnO2/H2 -E0H+/H2= -1.23 - 0= -1.23
The commercial process for organic chemicals that is currently practiced on a scale comparable to that of inorganic chemicals and metals is the electrohydrodimerization of acrylonitrile to adiponitrile.
Anode: H2O ---> 2H+ + 1/2 O2 + 2e-
Cathode: 2CH2=CHCN + 2H2O + 2e- ---> NC(CH2)CN + 2OH-
Overall: (acrylonitrile)2CH2=CHCN + H2O ---> 1/2 O2 + NC(CH2)4CN(adponitrile)
The importance of adiponitrile is that it can be readily converted to other useful compounds.
The Chlor-Alkali Process
This process is the electrolysis of sodium chloride (NaCl) at an industrial level. We will begin by discussing the equation for the Chlor-Alkali process, followed by discussing three different types of the process: The diaphragm cell, the Mercury cell and the Membrane cell.
What is the equation for the electrolysis of NaCl?
We will begin the explanation of the Chlor-Alkali process by determining the reactions that occur during the electrolysis of NaCl. Because NaCl is in an aqueous solution, we also have to consider the electrolysis of water at both the anode and the cathode. Therefore, there are two possible reduction equations and two possible oxidation reactions
Na+ (aq) + 2e- ---> Na(s) E0Na+/Na= -2.71 V (1)
2 H20 (l) + 2e- ---> H2(g) +2OH- (aq) E0H20/H2= -.83 V (2)
2Cl-(aq) ---> Cl2 + 2e- -E0Cl2/Cl-= -(1.36 V) (3)
2H20(l) ---> O2 + 4H+ + 4e- -E0O2/H20= -(1.23 V) (4)
As we can see, due to the very much more negative electrode potential, the reduction of Sodium ions is much less likely to occur than the reduction of Water, so we can assume that that in the electrolysis of NaCl, the reduction that occurs is reaction (2). Therefore, we should try to determine what the oxidation reaction that occurs is. Lets say we have reaction (2) as the reduction and reaction (3) as the oxidation. We would get:
reduction: 2 H20 (l) + 2e- ---> H2(g) +2OH- (aq) E0H20/H2= -.83 V (2)
oxidation: 2Cl-(aq) ---> Cl2 + 2e- -E0Cl2/Cl-= -(1.36 V) (3)
overall: 2 H20 (l) + 2Cl-(aq) ---> H2(g) +2OH- (aq) +Cl2 E0H20/H2-E0Cl2/Cl- =-.83 + (-1.36)= -2.19 (5)
On the other hand, we could also have reaction (2) with reaction (4)
reduction: 2[2 H20 (l) + 2e- ---> H2(g) +2OH- (aq)] EH20/H20=-.83 V (2)
oxidation: 2H20(l) ---> O2 + 4H+ + 4e- -EO2/H200= -(1.23 V) (4)
overall: 2H20 (l) ---> 2H2(g) + O2 (g) E0H20/H2 - E0O2/H20= -.83 V -(1.23 V)= -2.06(6)
At first glance it would appear as though reaction (6) would occur due to the smaller (less negative) electrode potential. However, O2 actually has a fairly large overpotential, so instead Cl2 is more likely to form, making reaction (5) the most probable outcome for the electrolysis of NaCl.
Chlor-Alkali Process in a Diaphragm Cell
Depending on the method used, there can be several different products produced through the Chlor-Alkali process. The value of these products is what makes the Chlor-Alkali process so important. The name comes from the two main products of the process, chlorine and the alkali, Sodium Hydroxide (NaOH). Therefore, one of the main uses of the Chlor-Alkali process is the production of NaOH. As described earlier,the equation for the Chlor-Alkali process, that is, the electrolysis of NaCl, is as follows:
reduction: 2 H20 (l) + 2e- ---> H2(g) +2OH- (aq) EH20/H20= -.83
oxidation: 2Cl-(aq) ---> Cl2 + 2e- E0Cl2/Cl-= -1.36
Overall: 2Cl- + 2H20(l) ---> 2 OH- + H2(g) +Cl2 (g) = E0H2O/H2- E0Cl/l2-= -83V- (-1.36 V)= -2.19 V
**The Chlor-Alkali process often occurs in an apparatus called a diaphragm cell, which is illustrated below.
As you look at the cell, note the following things.
Chlor-Alkali Process in Mercury Cell
**Some of the main problems with the mercury cell are as follows:
Membrane Cell Process
One final way to make even more pure NaOH is to use a membrane cell. It is preferred over the diaphragm cell or mercury cell method because it uses the least amount of electric energy and produces the highest quality NaOH. For instance, it can produce NaOH with a degree of chlorine ion contamination of only 50 parts per million. An ion-permeable membrane is used to separate the anode and cathode.
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