The term electrolysis was first popularized in the 19th century by Michael Faraday. It was a process that helped in the study of chemical reactions in obtaining pure elements. Today, this process is commercially important as it is used widely in separating or obtaining pure elements from naturally occurring sources such as ores.
Table of Content
- What is Electrolysis?
- Electrolytic Process
- Cell Potential or Voltage
- Faraday’s Law of Electrolysis
- Product of Electrolysis
- Factors Affecting Electrolysis
- Electrolysis Applications
- Electrolysis Problems With Solutions
What is Electrolysis?
Electrolysis is defined as a process of decomposing ionic compounds into their elements by passing a direct electric current through the compound in a fluid form. The cations are reduced at cathode and anions are oxidized at the anode.
For example, acidified or salt-containing water can be decomposed by passing electric current to their original elements hydrogen and oxygen. Molten sodium chloride can be decomposed to sodium and chlorine atoms.
Electrolysis is usually done in a vessel named ‘electrolytic cell’ containing two electrodes (cathode and anode) connected to a direct current source and an electrolyte which is an ionic compound undergoing decomposition, in either molten form or in a dissolves state in a suitable solvent.
In the process of electrolysis, there is an interchange of ions and atoms due to the addition or removal of electrons from the external circuit. Basically, on passing current, cations move to the cathode, take electrons from the cathode (given by the supply source-battery), and is discharged into the neutral atom. The neutral atom, if solid, is deposited on the cathode and if gas, move upwards. This is a reduction process and the cation is, reduced at the cathode.
Also Read: Water Electrolysis
At the same time anions, give up their extra electrons to the anode and is oxidized to neutral atoms at the anode. Electrons released by the anions travel across the electrical circuit and reach the cathode completing the circuit. Electrolysis involves a simultaneous oxidation reaction at anode and a reduction reaction at the cathode.
For example, when electric current, is, passed through molten sodium chloride, the sodium ion is attracted by the cathode, from which, it takes an electrode and becomes a sodium atom.
Chloride ion reaches the anode, gives its electron, and become chlorine atom to form chlorine molecule.
Na+(in electrolyte) + e–(from cathode) → Na …. At Cathode
Cl–(from electrolyte) → e– + Cl → Cl2 …. At Anode
Electrolysis process, while useful to get elemental forms from compounds directly, it can also be used indirectly in the metallurgy of alkali and alkaline earth metals, purification of metals, deposition of metals, preparation of compounds etc.
Cell Potential or Voltage
The minimum potential needed for the electrolysis process depends on their ability of the individual ions to absorb or release electrons. This ability is, measured as an electrode potential of the ions present in the electrolytic cell. The cell potential is the sum of the potential required for the reduction and oxidation reaction. The potential involved in various redox reactions is available in literature as standard reduction potential.
Reaction with positive redox cell potentials only will be feasible as per thermodynamic Gibbs free energy (or standard potential). Generally, the electrolysis is thermodynamically controlled.
In electrolysis, a potential equal to or slightly more than that, is, applied externally. The ions, which are stable and not reacting, are made to undergo reaction in the presence of externally applied potential. External potential hence makes an unfavourable reaction to take place. In electrolysis, chemical bonds connecting atoms are either made or broken and so, electrolysis involves the conversion of electrical energy into chemical energy.
Faraday’s Law of Electrolysis
The amount of the redox reaction depends on the quantity of electricity flowing through the cell. Amount of reaction or the number of ions discharged is given by Faraday’s law of electrolysis. There are two laws.
Faraday’s first law can be summarized as;
Faraday’s second law compares the mass of different substances undergoing a change for the same current.
According to this second law,
Product of Electrolysis
Electrolysis of only two ions (cation and anion) present in a single electrolyte is direct. Electrolysis will produce products present in the compound. When more than one cation and anions are present, each ion will compete for reduction and oxidations. Reactions with more positive redox potentials will be, reduced or oxidized, in preference, to others.
Also Read: Nernst Equation
So, in spite of multiple redox couples present, only one can be reduced or oxidized. Sometimes the ions that are reduced or oxidized may depend on their relative amount. In other words, the redox reaction and electrolysis may become kinetically controlled. In such cases, the product of analysis may differ on the relative concentration of the various ions present in the electrolyte.
For example, electrolysis of aqueous sodium chloride may give different products-
- Hydrogen and chlorine,
- Hydrogen and oxygen and
- Hydrogen, oxygen and chlorine.
Factors Affecting Electrolysis
The factors that may affect the electrolysis are;
i) The nature of the electrode
ii) Nature and state of the electrolyte
iii) Nature and electrode potential of ions present in the electrolyte and
iv) Overvoltage at the electrodes.
i) Nature and State of the Electrolyte
Electrolysis involves the movement of ions towards the oppositely charged electrodes. Naturally, the electrolyte should have mobile ions. In solids, ions are in specific positions and cannot move at ordinary temperatures. Hence, solids are unsuitable for electrolysis.
For electrolysis, electrolyte should be in the liquid form- molten or in solution with a suitable polar solvent. Sodium chloride will undergo electrolysis in the molten state or in aqueous solution.
ii) Nature and Electrode Potential of Ions Present in the Electrolyte
- Electrolysis of electrolytes of two elemental ions is straight forward giving the two elements on electrolysis. Molten sodium chloride gives sodium atoms and chlorine molecule.
- Electrolysis of radical ions does not give the elemental atoms.
- Electrolytes containing more than one ionic compound depends on the relative redox potentials.
- Electrolysis of aqueous solutions of electrolytes. Water molecules also can undergo redox reactions and will compete with redox reactions of the electrolyte ions.
- Electrolysis of molten sodium chloride gives sodium and chlorine. But electrolysis of aqueous sodium chloride gives hydrogen and chlorine and not sodium.
Also Check ⇒ Electrolytes
iii) Nature of the Electrode
For the same electrolyte, the nature of the electrolyte may give different products. When aqueous copper sulphate solution is, electrolyzed, the following redox reactions are possible.
At cathode: Reduction at pH =7
Cu2+ (aq) + 2e– →Cu (s) E° = 0.34V and 2H2O + 2e–→H2 + 2OH– E° = -1.02V
At anode: Oxidation at pH = 7
Cu(s) →Cu2+ (aq) + 2e– E° = – 0.34V and 2H2O → O2(g) + 4H+ + 4e– E° = +1.4 V
At the cathode, out of the two electrodes reduction potential of copper ions is more positive than the reduction of water. So, irrespective of electrode, copper ions from the electrolyte will be reduced and deposited on the cathode, increasing its mass. But the reaction at anode depends on the electrode.
Also Read: Electrochemical Cells
Electrolysis with inert electrodes like platinum, graphite, etc. Inert electrodes do not react with the electrolyte or the products and so does not undergo any changes. Since oxidation of water has more positive potential, oxygen will be evolved at the anode.
But, if the copper is used as an anode, it will react with the sulphate ion to retain the electrolyte concentration. So, there will not be any gas evolution. Instead, the anode mass slowly decreases going into the solution.
iv) Overvoltage at the Electrodes.
The redox potential of electrolyte ions decides the electrolysis reactions and products. Sometimes, redox potentials of some half-reactions during the electrolysis is more than the thermodynamic potentials. This excess voltage (over-voltage) of the half-reaction may make the reaction unfavourable and change the product of electrolysis.
In the hydrolysis of aqueous sodium chloride, at the anode, two oxidation reactions can take place. The reduction potential of water and chloride is +0.82V and 0.1.36V respectively.
2H2O→O2(g) + 4H+ + 4e– E° = -0.82 V
2Cl– → Cl2 + 2e– E = – 1.36V
Oxidation of water being more positive is more feasible and so, the evolution of oxygen gas should happen at the anode. But, the evolution of oxygen from water has an overvoltage of -0.6V making the voltage for the oxidation of water as -1.42V. Chloride oxidation is more positive than the net voltage of water oxidation. Chloride is oxidized to chlorine at the anode. Chlorine is liberated and not oxygen because of overvoltage.
Electrolysis, as stated above, is a process of converting the ions of a compound in a liquid state into their reduced or oxidized state by passing an electric current through the compound. Thus, electrolysis finds many applications both in experimental and industrial products. Some of the important ones are:
1) Determination of equivalent eight of substances.
2) Metallurgy of alkali and alkaline earth metals.
3) Purification of metals.
4) Manufacture of pure gases.
5) Manufacture of compounds like sodium hydroxide, sodium carbonate, potassium chlorate etc.
6) Electroplating for corrosion resistance, ornaments etc.
We shall discuss the different applications of electrolysis in detail below.
Determination of Equivalent Eight of Substances
We know Faraday’s Second Law states that the mass of substances deposited is proportional to their equivalent weight. Mass of any deposited substance can be, mathematically related as;
Equivalent mass of an unknown metal or substance can be calculated by passing a known current through the solutions and determining the mass of substances (M1 and M2) deposited in their respective cells. If the equivalent of one substance is known, the equivalent mass of the unknown substance can be calculated from the above equation.
Electrolysis of Molten Salts
Metallurgy of alkali and alkaline earth and third group metals ores of metal are concentrated and converted mostly to oxides. Oxides are reduced with reducing agents such as carbon, aluminium etc. Since, alkali and alkali earth metals have the largest reduction potentials they cannot be, reduced by any other metals or their compounds.
The only way of isolation of alkali and alkali earth metals is to directly electrolyze their molten chlorides. Mixing with other halides like calcium chlorides reduces the melting point of pure halides.
Electrorefining – Purification of Metals
Metals obtained after concentration and reduction of ores have a purity of about 90 -99%. An aqueous solution of the metal salt with the impure metal as anode and a pure metal as the cathode is electrolyzed. Pure metal, of more than 99% purity, deposits on the cathode and the impurities are collected at the bottom as mud. Copper and nickel are some examples of the metal purified by electrorefining.
An object can be, coated to the required thickness with a select metal by electrolysis. The object to be, coated is, made cathode. An aqueous solution of the metal salt to be coated is the electrolyte. The same metal or any inert metal can be the anode. On electrolysis, metal ion from the electrolyte deposits on the object. The loss of metal ions in the solution will be, compensated if the same metal is made the anode.
The deposition can be, used to protect the metal from corrosion are for making ornaments etc. Coating of iron with metals like zinc, lead, chromium, nickel improves the corrosion resistance of iron. Gold and silver coating on cheaper metals is used for making ornaments.
Electroforming is a process of making a replica of objects using electrolysis. The object to be replicated is pressed in wax to make a mould. Graphite powder is, coated uniformly to make it conducting. This is, used as a cathode and the salt of the metal to be deposited is taken as the electrolyte. After getting the required coating by electrolysis, wax and the graphite are, melted away.
Manufacture of Pure Gases
Aqueous salts on hydrolysis yield different products depending on the relative concentrations of salt and water. Electrolysis of concentrated brine (sodium chloride) forms pure hydrogen and chlorine gases. Pure chlorine gas is, collected in the Chlor-alkali industries by the electrolysis of brine aqueous solution.
Pure hydrogen and oxygen are, obtained by hydrolysis of water in the presence of acid or base or inert salt of alkali and alkaline earth metals. Percentage of hydrogen for commercial use is manufactured by the electrolysis of water, worldwide.
Continuous electrolysis of water, removes all the normal hydrogen isotopes leaving the deuterium ions. The deuterium oxide leftover after electrolysis of normal water is ‘Heavy water’. Heavy water is used as a moderator in nuclear reactors producing electrical energy from nuclear reactions.
Manufacture of Compounds
Compounds like sodium hydroxide, sodium hydrosulphite, potassium permanganate, potassium chlorate, ammonium per-sulphate, heavy water etc. are manufactured by electrolysis. Sodium hydroxide is a side product in the chloralkali industries preparing chlorine gas by the electrolysis of brine.
Electrolysis Problems With Solutions
1. An iron pipe with 14cm diameter and length of 1 meter is to be galvanized to a thickness of 0.01cm using zinc nitrate solution and a current of 25amp. What will be the loss of mass from zinc anode and what will be the time required for the electrolysis? The density of zinc is 7.14g/cm3. The equivalent weight of zinc is 32.8
Volume of the zinc coating = 2πrxl × 0.01 cm3
Mass of the zinc to be coated= V × d
Mass of the zinc lost from the anode = 440g.
By Faraday’s first Law of electrolysis,
2. A current of 3 amperes is, passed through neutral water containing a small amount of sodium Sulphate for one hour. Calculate the amount of hydrogen liberated at one atmosphere.
Number of coulombs passed = 3 × 60 × 60
Half reactions of the hydrolysis of water are-
At cathode: 2H2O + 2e– → H2(g) + 2OH–
At anode: 2H2O → O2(g) + 4H+ + 4e–
The net reaction of electrolysis of water is, 2H2O → 2H2(g) + O2(g)
Four faradays or 4 x 96485 coulomb of electric current liberates 2 moles or 2× 22.4L of hydrogen gas.
10800 coulombs of electric current liberates Hydrogen =
Total pressure above water = pressure of water vapour + pressure of hydrogen
The pressure of hydrogen = Total pressure above water – pressure of water vapour
The pressure of hydrogen = 1atm – 0.0316atm =0.9684atm
So, volume of hydrogen liberated =