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Electrolysis
Electrolysis is the process of causing the nonspontaneous reaction to occur using the external electrical source or the process of transforming the electrical energy into chemical energy. An apparatus for carrying out electrolysis is called electrolytic cell. The electrochemical principles underlying both galvanic cell and electrochemical cell are the same. An electrolytic cell, like galvanic cell, also consists of two electrodes in a molten salt or electrolyte solution that is connected to external battery or other voltage source. The external battery serves as the electron pump, which draws electrons in at one electrode (the positive electrode) and forcing them out at another (the negative electrode). Here I have shown a simplified electrolytic cell utilizing the molten NaCl as an electrolyte with two inert electrodes:
In molten sodium chloride (NaCl), there are two ions, namely, cation Na+^ and anion Cl -. At the anode, the oxidation of chloride ion takes place by depositing two electrons and forming a pure chlorine gas. At the same time, the reduction of sodium ion takes place at the cathode forming a pure sodium metal. The reactions at both electrodes are given below:
Oxidation (anode): 2Cl -^ (l) Æ Cl 2 (g) + 2e -^ E^0 ox = -1.36 V Reduction (cathode): 2Na+(l) +2e -^ Æ 2Na(l) E^0 red = -2.71 V
Overall: 2Na +(l) +2Cl -^ (l) Æ 2Na(l) + Cl 2 (g) E^0 cell = - 4.07 V
It is important to note here is that pure sodium and chlorine gas are formed at cathode and anode electrodes respectively.
The calculated E^0 cell value indicates that this reaction is a nonspontaneous one and requires an external electrical energy of equal to or greater than 4.07 V to carry out reaction. Higher voltage than 4.07 may be necessary due to inefficiencies in electrolytic process.
In the above electrolytic cell, pure sodium and pure chlorine gas are produced at the electrodes. The question is, how much of these substances are formed? That all depends
on various factors, like, type of electrolyte, duration of passing of electric current, and the amount of electric charge. The relation between all these aspects is known as quantitative aspects of electrolysis that is discussed below.
Example 1
Construct an electrolytic cell for the electrolysis of molten CaCl 2 using inert electrodes, and write half-reactions and Overall reaction.
Answer 1
Oxidation (anode) reaction: 2Cl-^ Æ Cl 2 (g) + 2e – Reduction (cathode) reaction: Ca +2^ +2 e -^ Æ Ca
Overall: Ca +2^ + 2Cl -^ Æ Ca(l) + Cl 2 (g)
Comparison between galvanic and electrolytic cells
In galvanic cell, electrons are deposited on the anode and are removed from the cathode. As a result of this, the anode carries a slight negative (-) charge and the cathode a slight positive (+) charge. This situation is reversed in electrolytic cell. In an electrolytic cell, the oxidation at the anode must be forced to remove the electrons from the reactant at the electrode making it a positive (+), and the cathode must be negative (-) to force the reactant to accept electrons at the anode.
Galvanic cell Electrolytic cell
Anode is negative (oxidation) Anode is positive (oxidation) Cathode is positive (reduction) Cathode is negative (reduction)
1 Mole electrons = 1 Faraday = 1 F = 96,490 C ≈ 96,500 C
The amount of substance produced can be related to the quantity of electricity through coulomb of charges. Thus, the coulomb is the gateway between amount and current or time:
Amount Coulomb Current, time of substance mole e -^ C = A x s
In the electrolysis of molten NaCl discussed above, the reduction of one mole of Na +^ at cathode requires one mole (1 Faraday) of electrons, i.e., 96,500 C:
Na +(l) +e -^ Æ Na(l)
On the other hand, the oxidation of two moles of Cl -^ at anode results in the formation of one mole Cl 2 by transferring two moles (2 Faraday) of electrons or 2 x 96,500 C to anode:
2Cl -^ (l) Æ Cl 2 (g) + 2e -
Similarly, 2 moles(2 Faraday) of electrons are required to reduce one mole of Ba +2^ to form one mole of Ba or 3 moles (3 Faraday) of electrons to reduce one mole of Al +3^ to one mole Al.
Ba +2^ + 2 e -^ Æ Ba Al +3^ + 3 e -^ Æ Al
In an electrolysis experiment, the current in amperes (A) (amount of electric charge per second or coulombs per second ( C/s)) is measured that passes through an electrolytic cell in a given period of time. The relation between ampere, time and charge is
Charge (Coulombs) = Current(Ampere) x Time (Second) or 1 C = 1 A x 1 s
Note that the time is expressed in seconds(s). Therefore, one Coulomb of charge is equal to passing one ampere of current for one second. The above relation involve three quantities, knowing any two of them the third one can be calculated. Once the charge is calculated, it is further related to number of moles of electrons to determine the amount of substance formed. Let us consider the following examples.
Calculation of Charge
Suppose a current of 1.50 A is passed through an electrolytic solution for 2.5 hr, what is total amount of charge passed through?
Answer
Charge is measured in coulombs, and hence
C = 1.50 A x 2.5 hr x ( 3600 s/hr) = 1.50 C/s x 2.5 hr x (3600 s /hr) = 13,500 C
Calculation of Current
Suppose a coulomb of charge of 5000 is passed through an electrolytic solution for 5 minutes, what is the current in amperes?
Answer
A = C / s = 5000 C / (5 min x 60 s/min) = 16.67 C/s = 16.67 A
Calculation of Time
How much time is required to produce 6,500 C of charges using a current of 15 A?
Answer
s = C /A = 6500 C / 15 A = 6500 C / (15 C/s) = 433.33 s or 7.22 min
Example 2
A current of 10 A is passed through the molten sodium chloride for 2.5 hours. What is the amount of metallic sodium at the cathode? How much chlorine gas is produced at the anode?
Answer
First calculate the coulombs using given current and time
C = 10 A x 2.5 hr = 10 (C/s) x 2.5 hr (3600 s /hr) = 90,000 C
Convert 90,000C to mole of electrons using the relation, 1 mole electrons = 96,500 C. Thus
Example 4
How many minutes are needed to deposit 25g of copper from a Cu+2^ solution, using 4.5 A of current?
Answer
The time is calculated known coulomb of charges ( C) and current (A) through the following equation:
s = C /A = C/ 4.5 A
Now we calculate the coulombs. One mole of Cu +2^ requires 2 moles of electrons to deposit one mole of Cu (63.55 g), that is,
2 mole e -^ = 63.55 g Cu
One mole of electrons equal to 1 Faraday or 96,500 C, that is, 1 F = 1 mol e -^ = 96,500 C. Hence,
C = 25 g Cu x (2 mole e -^ / 63.55 g Cu) x (96,500 C / 1 mole e-) = 7.59 x 10^4 C.
Therefore the time required in seconds is
s = 7.59 x 10^4 C / 4.5 A = 16,872.10 s 0r 281.20 min or 4.69 hrs.
Example 5
Three cells are setup in series, and a current of 1.5 A is passed through them for 30 minutes. The cell A contains an molten salt of sodium chloride, cell B contains molten copper (I) chloride, and cell C contains molten aluminum chloride as shown below:
What is the amount in grams of sodium, copper, and aluminum produced in cell A, cell B, and cell C respectively?
Answer
First, let us calculate the amount of charge when 1.5 A current is passed for 30 minutes.
C = A x s = 1.5 (C/s) x ( 30 min x 60 s /min) = 2,700 C
Then convert this coulomb into moles of electrons.
Mol of e-^ = 2,700 C x (1 mol e-/ 96,500 C) = 0.028 mol of e-
Now we can calculate the amount of each substance deposited using 0.028 moles of electrons in each cell. First we write the reduction of ion to metal reactions as
Cell A: Na+^ + e -^ Æ Na Cell B: Cu+2^ +2e -^ Æ Cu Cell C: Al +3^ + 3e -^ Æ Al
To produce one mole or 22.99 g of sodium, it requires one mole electrons, to produce one mole or 63.55g of copper, it requires 2 moles of electrons, and to produce one mole or 26.98g of aluminum, it requires 3 moles of electrons. Then calculate the grams of each substance produced by 0.028 moles of electrons.
g of Na = 0.028 mol e-^ x (22.99 g Na/ 1 mol e-) = 0.64 g Na g of Cu = 0.028 mol e-^ x (63.55 g Cu/ 2 mol e-) = 0.89 g Cu g of Al = 0.028 mol e-^ x ( 26.98 g Al / 3 mol e-) = 0.25 g Al
Example 6
A particular object with a surface area of 200 cm^2 is silver-plated using a solution containing Ag(CN) - 2. How many minutes does it take to cover an object with silver to a thickness of 0.0050 mm by using a current of 2.5 A? The density of silver is 10.5 g/cm^3.
Answer
First we need to calculate the amount of silver required to cover surface area of 200 cm^2 to a thickness of 0.0050 mm (0.00050 cm). When we multiply the surface area by the thickness, we get the volume of the metal.
Volume of metal = 200 cm^2 x 0.00050 cm = 0.100 cm^3
Now using the density, we can calculate the amount in grams to cover this volume.
g Ag = 0.100 cm^3 x 10.5 g /cm^3 = 1.05 g Ag
When the silver is deposited from Ag(CN)- 2 , the following reduction reaction takes place.
Ag +^ + e -^ Æ Ag
