The Electrolysis of Aqueous NaCl

The figure below shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed.

Once again, the Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and water molecules.

Cathode (-): Na+ + e- Na Eo

red = -2.71 V 2 H2O + 2 e- H2 + 2 OH- Eo

red = -0.83 V

Because it is much easier to reduce water than Na+ ions, the only product formed at the cathode is hydrogen gas.

Cathode (-): 2 H2O(l) + 2 e- H2(g) + 2 OH-(aq)

There are also two substances that can be oxidized at the anode: Cl- ions and water molecules.

Anode (+): 2 Cl- Cl2 + 2 e- Eo

ox = -1.36 V 2 H2O O2 + 4 H+ + 4 e- Eo

ox = -1.23 V

The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl2 and O2gas collect at the anode. In practice, the only product of this reaction is Cl2.

Anode (+): 2 Cl- Cl2 + 2 e-

At first glance, it would seem easier to oxidize water (Eoox = -1.23 volts) than Cl- ions (Eo

ox = -1.36 volts). It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl- ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to O2 gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.

In summary, electrolysis of aqueous solutions of sodium chloride doesn't give the same products as electrolysis of molten sodium chloride. Electrolysis of molten NaCl decomposes this compound into its elements.

electrolysis 2 NaCl(l) 2 Na(l) + Cl2(g)

Electrolysis of aqueous NaCl solutions gives a mixture of hydrogen and chlorine gas and an aqueous sodium hydroxide solution.

electrolysis 2 NaCl(aq) + 2 H2O(l) 2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g)

Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

The dotted vertical line in the above figure represents a diaphragm that prevents the Cl2 produced at the anode in this cell from coming into contact with the NaOH that accumulates at the cathode. When this diaphragm is removed from the cell, the products of the electrolysis of aqueous sodium chloride react to form sodium hypo-chlorite, which is the first step in the preparation of hypochlorite bleaches, such as Chlorox.

Cl2(g) + 2 OH-(aq) Cl-(aq) + OCl-(aq) + H2O(l)

Electrolysis of Water

A standard apparatus for the electrolysis of water is shown in the figure below.

electrolysis 2 H2O(l) 2 H2(g) + O2(g)

A pair of inert electrodes are sealed in opposite ends of a container designed to collect the H2 and O2 gas given off in this reaction. The electrodes are then connected to a battery or another source of electric current.

By itself, water is a very poor conductor of electricity. We therefore add an electrolyte to water to provide ions that can flow through the solution, thereby completing the electric circuit. The electrolyte must be soluble in water. It should also be relatively inexpensive. Most importantly, it must contain ions that are harder to oxidize or reduce than water.

2 H2O + 2 e- H2 + 2 OH- Eored = -0.83 V

2 H2O O2 + 4 H+ + 4 e- Eoox = -1.23 V

The following cations are harder to reduce than water: Li+, Rb+, K+, Cs+, Ba2+, Sr2+, Ca2+, Na+, and Mg2+. Two of these cations are more likely candidates than the others because they form inexpensive, soluble salts: Na+ and K+.

The SO42- ion might be the best anion to use because it is the most difficult anion to oxidize. The

potential for oxidation of this ion to the peroxydisulfate ion is -2.05 volts.

2 SO42- S2O8

2- + 2 e- Eoox = -2.05 V

When an aqueous solution of either Na2SO4 or K2SO4 is electrolyzed in the apparatus shown in the above figure, H2 gas collects at one electrode and O2 gas collects at the other.

What would happen if we added an indicator such as bromothymol blue to this apparatus? Bromothymol blue turns yellow in acidic solutions (pH < 6) and blue in basic solutions (pH > 7.6). According to the equations for the two half-reactions, the indicator should turn yellow at the anode and blue at the cathode.

Cathode (-): 2 H2O + 2 e- H2 + 2 OH-

Anode (+): 2 H2O O2 + 4 H+ + 4 e-

Faraday's Law

Faraday's law of electrolysis can be stated as follows. The amount of a substance consumed or produced at one of the electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell.

In order to use Faraday's law we need to recognize the relationship between current, time, and the amount of electric charge that flows through a circuit. By definition, one coulomb of charge is transferred when a 1-amp current flows for 1 second.

1 C = 1 amp-s

Example: To illustrate how Faraday's law can be used, let's calculate the number of grams of sodium metal that will form at the cathode when a 10.0-amp current is passed through molten sodium chloride for a period of 4.00 hours.

We start by calculating the amount of electric charge that flows through the cell.

Before we can use this information, we need a bridge between this macroscopic quantity and the phenomenon that occurs on the atomic scale. This bridge is represented by Faraday's constant, which describes the number of coulombs of charge carried by a mole of electrons.

Thus, the number of moles of electrons transferred when 144,000 coulombs of electric charge flow through the cell can be calculated as follows.

According to the balanced equation for the reaction that occurs at the cathode of this cell, we get one mole of sodium for every mole of electrons.

Cathode (-): Na+ + e- Na

Thus, we get 1.49 moles, or 34.3 grams, of sodium in 4.00 hours.

The consequences of this calculation are interesting. We would have to run this electrolysis for more than two days to prepare a pound of sodium.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php

The quantitative laws of electrolysis, relating the masses of the ions deposited to the charge transported through the electrolyte, were discovered by Faraday (Experimental Researches in Electricity; reprinted in Everyman's Library.). In his first memoir (1832) he made out the relation

between the mass of any particular ion deposited and the amount of electricity required for its deposition. The two were proportional, or in Faraday's words: "the chemical action of a current of electricity is in direct proportion to the absolute quantity of electricity which passes." This is the first law of electrolysis.

In his second memoir (1833) he made out the relation between the masses of various ions deposited by the same quantity of electricity. These were in the ratios of the chemical equivalents.

Since galvanometers and units of current were then not related, Faraday used the amount of gas produced by the electrolysis of acidulated water as a measure of the quantity of electricity which passes, and he determined the quantities of other elements liberated in electrolysis by the same current as liberates 1 gram of hydrogen.

These quantities were the ordinary chemical equivalents, and hence he discovered the second law of electrolysis:the weights of the ions deposited by the passage of the same quantity of electricity are in the proportion of their chemical equivalents.

To illustrate Faraday's laws we may connect in series a number of electrolytic cells, containing different electrolytes, with a battery as shown in

Fig: Diagram of electrolytic circuit

Suppose that the first cell contains water acidulated with sulphuric acid, the second a solution of copper sulphate, and the third fused stannous chloride. Fused salts are electrolytes, as well as their solutions.

After the current has passed for a certain time, the volumes of hydrogen and oxygen liberated from the. acidulated water, and the weights of copper and tin deposited from the solution of copper sulphate and the, fused stannous chloride, respectively, are ascertained. If the weights of the other ions which are deposited in the cells whilst 1 gm. of hydrogen is liberated in the first are determined, they are found to be equivalent weights: 7.94 gm. of oxygen, 35.2 gm. of chlorine, 31.5 gm. of copper and 59 gm. of tin.

The quantity of electricity which has passed through the solution is measured by the current strength multiplied by the time. The current strength is measured in amperes, and one ampere passing for one second corresponds with unit quantity of electricity, or one coulomb. A current of C amperes flowing for t seconds conveys Ct coulombs. Hence the weight of an ion deposited in a given time is proportional to the strength of the current. This is Faraday's First Law of Electrolysis.

The international ampere is defined as that current which, flowing uniformly for 1 second, deposits under specified conditions 0.001118 grams of silver from a solution of silver nitrate. This is called the electrochemical equivalentof silver; the weight z gm. of any ion deposited by 1 coulomb is its electrochemical equivalent; hence the weight deposited by a uniform current of C amperes flowing for t seconds is:

W = Czt.

Since the chemical equivalent of silver (O = 16.000) is 107.880, the quantity of electricity required to deposit this amount will be (by Faraday's first law) 107.880/0.001118 = 96,500 coulombs per gm. equiv. very nearly. This fundamental charge is called a faraday, denoted by F (Not to be confused with the farad, the unit of electric capacity.). Faraday's second law shows that 1F will deposit 1 chemical equivalent of any ion. Since Faraday's law will enable us to find the equivalent of an element (e.g., oxygen, copper, silver), it: will enable us to determine the valency of the element in the state of the ion investigated: valency = at. wt. / equivalent.

The laws of electrolysis are conveniently summarised in the statement that 96,500 coulombs liberate one gram-equivalent of any ion in electrolysis.

Thus, one F liberates 1 gm. atom of a univalent element, and nF liberate 1 gm. atom of an n-valent element.

Example. - Find the weight of copper deposited from a solution of copper sulphate by a uniform current of 0.25 amp. flowing for one hour.

Quantity of electricity passed through electrolyte = 0.25 x 60 x 60 =900 cmb.

Copper is bivalent, hence equivalent weight =at. wt. / 2 = 63.5/2 = 31.75.

96,500 cmb. liberate 31.7 gm. of Cu, hence wt. of copper liberated by 900 cmb. = 31.75 x 900/96,500 = 2.95 gm.

http://chemistry.proteincrystallography.org/article92.html

Electrolysis is a process by which electrical currents can be passed through solutions. Pure water is a non-conductor of electricity. But if a salt such as NaCl is dissolved in it, the solution will start conducting if two electrodes are placed in it along with a battery that provides the

initial current. In the present chapter we will see what is electrolysis and how it is useful in our day to day life.

What we will study in this chapter 1. Process of electrolysis2. Conductor versus electrolyte3. Electrolysis of water4. Definition of standard solutions5. Faraday’s Law of Electrolysis6. Application of electrolysis

1. Process of electrolysisWhen an electric current is passed through a chemical compound, some compounds are able to conduct electricity. The compound dissociates into ions under the influence of the electric current. The electrical current initiates a chemical reaction or a break up.

To see how electricity is conducted through a solid compound, a molten compound and an aqueous solution of the compound, do the following. Take two metal rods, a 6V battery, a bulb or an ammeter, wires. Take sodium chloride in a petri dish. Keep a burner and water in a beaker also. Connect the circuit as shown in the figure below. Bulb or an ammeter will visually indicate the passage of current.

You notice the following :

No current flows through the circuit when the sodium chloride is in a solid form. When the salt is molten, current flows. When the salt is in an aqueous solution, the current flows and the bulb glows brightly.

Also bubbles are seen at the electrodes, indicating that some chemical reaction (or dissociation of the compound) is taking place.

In the above experiment, instead of salt solution, try other solutions like sugar solution, glycerin, alcohol, dilute sulphuric acid, copper sulphate solution, acetic acid.

You will notice the following :

Current flows through dilute sulphuric acid and copper sulphate solution. A weak current flows through acetic acid. No current flows through sugar solution, glycerin, alcohol.

The above experiments show that there is a relationship between the passage of electricity and the chemical reactions taking place due to it. All compounds do not conduct electricity; and the state of the compound, whether molten or aqueous, is an important factor.

Some definitions regarding electrolysis1.Electrolysis : It is a chemical process where a substance in its molten state or in an aqueous solution is decomposed by the passage of electric current.

2.Electrolyte : A compound that allows electric current to pass through itself, when either in a molten state or in an aqueous solution, is called an electrolyte. In the above experiment, solutions of sodium chloride, copper sulphate, dilute sulphuric acids, acetic acid are electrolytes. Strong electrolytes are those that allow large electric currents to be passed through them. Solutions of sodium chloride, copper sulphate, dilute sulphuric acid are examples of strong electrolytes. Weak electrolytes are those compounds which are poor

conductors of electricity when they are in a molten state or in an aqueous solution. The solution of acetic acid in the experiment above, showed that it is a weak electrolyte.

3. Non-electrolyte : A compound which does not allow electric current to pass through itself in any state, molten or aqueous, is called a non-electrolyte. In the above experiments we have seen that sugar solution, glycerin, alcohol, are non-electrolytes.

4.Electrodes : The strips of metals inserted in the electrolytes for conduction of electricity are called electrodes. The metal electrode connected to the positive terminal of the battery is called the anode (+). The metal electrode connected to the negative terminal of the battery is called the cathode (-).

5.Electrolytic cell : The complete set-up for electrolysis is called the electrolytic cell. This consists of the vessel containing the electrolyte, anode, cathode, battery and wires. Electrolytic cell is also known as a voltameter, since it generates voltage (or current) at its two terminals.

Mechanism of electrolysisIn the example of NaCl above, we saw that NaCl does not conduct electricity as it is. NaCl is an ionic compound and both the Na+and the Cl- ions are strongly attracted to each other by electrostatic attraction. The strength of the electrical current is unable to break the ionic bond. Not only breaking of the ionic bond is needed, the flow of charges also has to take place. This does not happen in solid NaCl. Thus solid NaCl is not an electrolyte.

On the other hand, in case of molten NaCl, the bond length between the Na+ and the Cl- ions has loosened. The bond is weakened. Hence the ions can become mobile and conduct electricity.

In an aqueous solution of NaCl, water molecules separate the Na+ and the Cl- ions. This makes them very mobile. The mobility is enhanced when two electrodes in the form of anode (+) and

cathode (-) are inserted in the salt solution. The Na+ ions get attracted toward the cathode and the Cl- ions get attracted toward the anode. The aqueous solution of NaCl is therefore a good electrolyte.

We can conclude from the above discussions that the movement of ions is responsible for the flow of current in an electrolytic cell.

Steps occurring during the passage of electricity in an electrolytic cell with NaCl aqueous solution :

1. Dissociation of NaCl : NaCl Na + + Cl- Cation Anion

2. Reaction at the cathode : Na + + 1e- Na (neutralization)

Reaction at the anode : Cl- 1e- + Cl (neutralization)

Cl + Cl Cl2

To summarize the process of electrolysis, we can say the following

Electrolytes dissociate to form negatively charged anions and positively charged cations. The ions conduct electricity through the electrolyte. Cations are attracted towards the negative electrode. They take the excess electrons

from the electrode and neutralize themselves. Anions are attracted towards the positive electrode. They give up the excess electrons

from the electrode and neutralize themselves. The electrolyte dissociates and the constituent elements of the salt are liberated at the electrodes.

http://educationalelectronicsusa.com/c/electrolysis-I.htm

2. Conductor versus electrolyteWe have seen what are conductors and insulators, in the chapter on Electricity. Electrolyte, as studied in the last section, is a solution that can conduct electricity due to ions present in it. Now the question you might ask is what is the difference between a conductor and an electrolyte?

The figure below shows two circuits : one for a conductor and one for an electrolyte. The differences between the two are as follows:

In a conductor, electrical current is flowing through the material in its solid form. Conductors are generally metals. On the other hand, an electrolyte can carry current in its molten state or in a solution form. Electrolytes are generally made from ionic compounds and not covalent compounds (HCl and NH3 are a few of the exceptions). When a current passes through a conductor, it may get heated up because of its inherent resistance. Other than this physical change, there is no other change in the conductor. Once, the current stops flowing, the conductor returns to its original state. On the other hand, in an electrolyte, the cations and anions go in opposite directions and get neutralized at the positive and negative electrodes respectively. The chemical change thus produced is irreversible. After the current stops, the electrolyte may change in strength than what it had started with.

The current through a conductor is due to free electrons that flow and complete the circuit. The current through an electrolytic cell is because of the flow of ions.

For a given conductor and a cell, the current flowing through the circuit is constant. In an electrolytic cell, the current depends on the strength of the electrolyte, which may change over time.

3. Electrolysis of waterThe apparatus or the electrolytic cell, required for performing electrolysis of water is shown in the figure below. The cell is called Hoffman’s voltameter.

Since water is a covalent compound, pure or distilled water is a non-electrolyte. A few drops of ionic compound like dilute sulphuric acid are enough to make the water become an electrolyte. The Hoffman voltameter consists of platinum electrodes. The anode and cathode are connected to a battery. The cell produces a small current of the order of a few milliamps and you will see bubbles appearing in the two arms of the voltameter. The anode collects oxygen and the cathode arm collects hydrogen gas.

The overall reaction that takes place is :

H2SO4 2H+ + SO4-- (dissociation, reversible reaction)

In the presence of these ions, water also becomes capable of dissociation.

H2O H+ + OH- (dissociation, reversible reaction)

Reaction at the cathode

H+ + e- H , H + H H2

Reaction at the anode

OH- OH + e-

4OH H2O + O2

At the anode OH- ions are released in preference to SO4-- ions. This is because it is easier for

an OH- ion to give up an electron quickly than for the SO4-- ion to do so. Since the sulphuric

acid itself does not participate in the chemical reaction that is taking place, it can be called as a catalyst of the reaction.

Most of the ionic compound will also be able to initiate electrolysis of water, but acidulated water is the best way to achieve electrolysis of water.

4. Definition of standard solutionsIn the process of electrolysis, we have seen that the electrolyte has to be in a molten state or in the form of a solution. By solution, we mean that the solute is completely dissolved in the solvent. A solution thus is a homogeneous mixture. Example of solution is brine, sugar dissolved in water, water in alcohol, etc. Oil in water is a heterogeneous mixture and hence is not called as a solution.

In a solution, one would be interested in knowing the amount of solute dissolved. Therein comes the concept of strength of a solution. Normally you can say that x grams of solute is dissolved in y grams of the solvent. The x and y are variables and can take any values. But this may cause confusion for electrolysis, as the entire process parameter for electrolysis namely, the current, the type of electrodes used, etc. all will depend on the strength of the solution. To simplify matters, standard solutions are used. Standard solutions are a fixed weight of solutes, mixed or dissolved in a fixed weight of the solvent.

There are two types of standard solutions

Molar solution Normal solutions

Molar Solution (M)A Molar solution (M) is a solution that contains 1 mole of solute in each litre of solution. A mole is the molecular weight (MW) expressed as grams (sometimes referred to as the ‘gram molecular weight’ (gMW)).

Thus 1 M = 1 gMW of solute per litre of solution.

For example, how much sodium chloride is needed to make 1 litre of a 1 M solution?

First, we find out the molecular weight (MW) of sodium chloride. This is calculated as follows :

The atomic weight of sodium (Na) is 23.The atomic weight of chlorine (Cl) is 35.5

So the molecular weight of sodium chloride (NaCl) is:Na (23) + Cl (35.5) = NaCl (58.5)

Therefore, a 1 M solution of sodium chloride contains 58.5 grams of sodium chloride in 1 litre of solution.

Similarly, a 2M solution contains 117 grams of sodium chloride per litre.

And a 0.1M solution contains 5.85 grams/litre of sodium.

Normal Solution (N)A Normal solution (N) is a solution that contains 1 ‘gram equivalent weight’ (gEW) of solute per litre of solution. The gram equivalent weight is equal to the molecular weight expressed as grams divided by the ‘valency’ of the solute.

To understand valency, consider the following acids:

Hydrochloric acid (HCl) has one replaceable hydrogen ion (H), sulphuric acid (H2SO4) has two replaceable hydrogen ions (H2) and phosphoric acid (H3PO4) has three replaceable hydrogen ions (H3). The valencies of these acids are determined by their respective replaceable hydrogen ions:

HCl, Valency = 1H2SO4, Valency = 2H3PO4, Valency = 3

So, for 1N HCl the MW is 36.5, the EW is 36.5 and therefore 1N would correspond to 36.5 grams/litre. In case of HCL 1M solution is same as 1N solution.

For 1N H2SO4 the MW is 98, the EW is 98/2 = 49 (that is, valency = 2) and so a 1 N solution would be 49 grams/litre.

Similarly, for 1N H3PO4 the MW is 98, the EW is 98/3 = 32.7 and 1N would be 32.7 grams/litre.

In case of alkalis, to understand valency, consider the following alkalis:

Sodium Hydroxide (NaOH) has one replaceable hydrogen ion (OH), calcium hydroxide (Ca(OH)2) has two replaceable hydrogen ions ((OH) 2) The valencies of these acids are determined by their respective replaceable hydrogen ions:

NaOH, Valency = 1Ca(OH)2, Valency = 2

So, for 1N NaOH the MW is 40, the EW is 40 and therefore 1N would correspond to 40 grams/litre. In case of NaOH 1M solution is same as 1N solution.

For 1N Ca(OH)2 the MW is 74, the EW is 74/2 = 37 (that is, valency = 2) and so a 1 N solution would be 37 grams/litre.

Since normal solutions are a bit confusing, these days the practice is to quote strength of solutions in molar terms.

5. Faraday’s Law of ElectrolysisMichael Faraday is a renowned English scientist who had worked in the field of electricity and has made some pioneering contributions. He established the relationship between the electric current that is passed through an electrolyte and the mass of the substance released at the anode or the cathode. The relationship is known as the Faraday’s law of electrolysis. The law states that the mass of substance released at any electrode is directly proportional to the electric charge that is passed through the electrolyte.

Thus if m = mass of the substance released at the electrode

And Q = amount of electric charge that is passed through the electrode

Then according to Faraday’s Law of Electrolysis

m Q

If I is the current in amperes then Q = I x t (t = time duration in seconds)

Thus m It

M = Z x It

Z is the constant of proportionality and is known as the electromechanical equivalent.

Thus the electromechanical equivalent Z of a substance is defined as the amount of substance in grams liberated at any electrode when one coulomb charge is passed through an electrolyte.

6. Application of electrolysisThe process of electrolysis has many applications. But before we consider a few applications, let us first look at the factors that influence discharge of ions at the electrodes.

Suppose your electrolyte is a solution of NaCl in water. At the cathode, there will be a competition between release on Na of H ions as both of them are positively charged. The question that might arise in your minds, is which of the ion will be released at the cathode?

The ion that will be released at the cathode or the anode will depend on the following factors :

Relative position of the ion in the electromechanical series. The electromechanical series is a representation of how reactive the ion is. For example in a solution that contains Na+ and Hg++ ions, Na+ ions will be preferentially released as Na accepts electrons easily (see its electronic configuration). But this happens if the concentration of Hg++ ions is comparatively small. (If the electrolyte is made up of NaCL in H2O, then H ions will be preferably released, as H is more reactive than Na). Table below gives some examples of the electrochemical series.

Concentration of ions in the electrolyte. In the above example if the concentration of Hg++ ions is very large, it will be preferentially released.

The nature of the electrodes. Some electrode materials such as graphite or platinum are non-corrosive and are not affected by ions surrounding it. Some electrodes such as copper, may enhance the release of ions especially ions like Cu, Ag, Ni, etc.

The Electrochemical Series in order of increasing preference:

Cations Anions

K+ SO4- -

Ca++ NO3 -

Na+ Cl -

H+ Br -

Cu++ I -

Ag++ OH -

In the light of the above discussions, we will see how the process of electrolysis is applied :

1. Electroplating of metals 2. Electro-refining of metals 3. Extraction of metals or Electro-metallurgy 4. Battery

Electroplating of metalsElectroplating is a process whereby a thin coating of desired material is applied on a required material. This is mostly done on stainless steel to prevent rusting, or on some decorative items, so that they look attractive. On stainless steel, generally nickel-chromium plating is done. On decorative items, such as spoons, plates, jewelry items, silver, gold or other plating is done. Electroplating is cheap and cost effective. It enhances the life of the object and makes it look better in appearance.

The following method is adapted :

First the item to be electroplated is smoothened and cleaned thoroughly. It should not have any oily or dirt marks on it.

An electrolyte is selected whose ions are required to be deposited on the item. Direct current is preferred to alternating current, as alternating current may result in

non-smooth deposit. The item to be electroplated forms the anode or cathode of the electrolytic cell. This is

the drawback of the electroplating process. The item has to be electrically conducting, or has to be made electrically conducting.

For a smooth coating, the electrolytic process has to be optimized for time, temperature and current in the cell.

Figure below is a conventionally used electrolytic cell for depositing silver. Let the item to be silver-plated be a spoon.

Electrolyte used is sodium silver cyanide solution [Na{Ag(CN)2}]. Silver nitrate is not used as it is observed that the electrolyte does not give a uniform deposition. The cathode is made out of the item on which the electro deposition is to be done, in this case it is a spoon. Anode is made of a block of silver.

The [Na{Ag(CN)2}] dissociates as

Na{Ag(CN)2} Na+ + Ag + +2 CN

Reaction at the cathode

Ag + + 1 e- Ag

The positively charged Ag + ions are attracted to the cathode (-) and accept one electron and get deposited as a thin film over the cathode material, in this case the spoon.

Reaction at the anode

Ag - 1 e- Ag +

The silver atoms at anode loose electrons and enter into the electrolyte as an ion. This ensures that the concentration of the Ag in the solution remains constant.

The electroplating is complete when a desired thickness of the silver film is deposited.

The electroplating process parameters, such as the type of electrolyte, the type of anode or cathode, the temperature of the electrolyte, the strength of the current, the time for electroplating, etc. are very well documented in standard literature and these are usually followed.

Electro-refining of metalsSimilar to the process of electro-deposition, electrolysis can be used to purifying metals that are obtained from the ores. The process is known as electro-refining of metals. The metals that are generally refined by this process are Zn, Ag, Ni, Cu, Pb, Al, etc.

Figure below is a conventionally used electrolytic cell for purifying metal ores, in this case the metal is impure copper.

Let the cathode be made of pure copper sheets and the anode be made of impure copper block. The electrolytic solution is copper sulphate solution.

The electrolyte dissociated into

CuSO4 Cu ++ + SO4- -

Reaction at the cathode

Cu + + + 2 e- Cu

The positively charged Cu ++ ions are attracted to the cathode (-) and accept two electrons and get deposited as a thin film over the cathode material, in this case the pure copper sheets.

Reaction at the anode

Cu - 2 e- Cu ++

The copper atoms at the anode loose electrons and enter into the electrolyte as an ion. This ensures that the concentration of the Cu in the solution remains constant. Besides the copper atoms, other atomic impurities such as Ag and Au also get into the electrolytic solution. Thus the impure block at the anode gets used up and pure copper is deposited at the cathode.

Extraction of metals or Electro-metallurgyExtraction of metals by the process of electrolysis is known as electro-metallurgy. This process is used in case highly reactive metals such as sodium. An ore containing sodium is used in a molten form. This forms the electrolyte. Anode and cathodes are generally carbon rods or steel. The Na atoms get attracted to the cathode of the cell and then the entire cathode with its coating is stored for further use.

BatteryAll batteries that we come across in our day to day use, including car batteries, dry cells used in torches, calculators, hand-sets, etc. are all examples of an electrolytic cell. But in this case the reverse of an actual electrolytic process is being used. The chemicals inside the cells produce current (and voltage) which is utilized.

In a car battery, for example, two grids are used as anode (Pb) and cathode (PbO2). The solution is H2SO4 of generally about 6 M in concentration.

The overall reaction is:

PbO2 + Pb + 2 SO4-2 + 4 H+1 2 PbSO4 + 2 H2O

During the discharge process, the metallic lead atom (Pb) at the anode loses two electrons and becomes positively charged. This process is called "oxidation." The electrons flow from the anode through the bath to the PbO2 cathode. At the cathode, the positively charged lead ion in

the lead dioxide (Pb+4 O-22) accepts two electrons through the external circuit. This electron

acceptance process is called "reduction."

The discharge product on both electrodes is lead sulfate (PbSO4). The overall voltage of these cells is about 2.0 V, so a 6 V battery has three of these sets of cells in series, while a 12 V battery has six of them.

Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds:

"By means of a flame which was thrown on a spoon containing potash, this alkali was kept for some

minutes at a strong red heat, and in a state of perfect fluidity." One pole of a battery of copper-zinc

cells was connected to the spoon, and the other was connected to platinum wire which dipped into the

melt. "By this arrangement some brilliant phenomena were produced. The potash appeared to be a

conductor in a high degree, and as long as the communication was preserved, a most intense light was

exhibited at the negative wire, and a column of flame, which seemed to be owing to the development

of combustible matter, arose from the point of contact." The flame was due to the combustion in the

air of metallic potassium. In another experiment, Davey observed "small globules having a high

metallic lustre, precisely similar in visible characters to quicksilver, some of which burnt with explosion

and bright flame, as soon as they were formed, and others remained, and were merely tarnished, and

finally covered by a white film which formed on their surfaces."

Electrolysis of molten alkali halides is the usual industrial method of preparing the alkali metals:

cathode:  Na+ + e– → Na(l) E° = –2.71 v

anode:   Cl– → ½ Cl2(g) + e– E° = –1.36 v

net:   Na+ + Cl– → Na(l) + ½ Cl2(g) E° = –4.1 v

Ions in aqueous solutions can undergo similar reactions. Thus if a solution of nickel chloride undergoes electrolysis at platinum electrodes, the reactions are

cathode:  Ni2+ + 2 e– → Ni(s) E° = –0.24 v

anode:   2 Cl– → Cl2(g) + 2 e– E° = –1.36 v

net:   Ni2+ + 2 Cl– → Ni(s) + Cl2(g) E° = –1.60 v

Both of these processes are carried out in electrochemical cells which are forced to operate in the "reverse", or non-spontaneous direction, as indicated by the negative for the above cell reaction. The free energy is supplied in the form of electrical work done on the system by the outside world (the surroundings). This is the only fundamental difference between an electrolytic cell and the galvanic cell in which the free energy supplied by the cell reaction is extracted as work done on the surroundings.

A common misconception about electrolysis is that "ions are attracted to the oppositely-charged electrode." This is true only in the very thin interfacial region near the electrode surface. Ionic motion throughout the bulk of the solution occurs mostly by diffusion, which is the transport of molecules in response to a concentration gradient. Migration— the motion of a charged particle due to an applied electric field, is only a minor player, producing only about one non-random jump out of around 100,000 random ones for a 1 volt cm–1 electric field. Only those ions that are near the interfacial region are likely to undergo migration.

Electrolysis in aqueous solutions

Water is capable of undergoing both oxidation

H2O → O2(g) + 4 H+ + 2 e–      E° = -1.23 v

and reduction

2 H2O + 2 e– → H2(g) + 2 OH–      E° = -0.83 v

Thus if an aqueous solution is subjected to electrolysis, one or both of the above reactions may be able to compete with the electrolysis of the solute.

For example, if we try to electrolyze a solution of sodium chloride, hydrogen is produced at the cathode instead of sodium:

cathode:  H2O + 2 e– → H2(g) + 2 OH– E =+0.41 v ([OH–] = 10-7 M)

anode:   Cl– → ½ Cl2(g) + e– E° = –1.36 v

net:   Cl– + H2O → 2 H2(g) + ½ Cl2(g) + 2 OH– E = –0.95 v

 

[This illustration is taken from the excellent

Purdue University Chemistry site]

Reduction of Na+ (E° = –2.7 v) is energetically more difficult than the reduction of water (–1.23 v), so in aqueous solution the latter will prevail.

Electrolysis of salt ("brine") is carried out on a huge scale and is the basis of the chloralkali   industry .

Electrolysis of water

Pure water is an insulator and cannot undergo signifigant electrolysis without adding an electrolyte. If the object is to produce hydrogen and oxygen, the electrolyte must be energetically more difficult to oxidize or reduce than water itself. Electrolysis of a solution of sulfuric acid or of a salt such as NaNO3 results in the decomposition of water at both electrodes:

cathode:  H2O + 2 e– → H2(g) + 2 OH– E =+0.41 v ([OH–] = 10-7 M)

anode:   2 H2O → O2(g) + 4 H+ + 2 e– E° = -0.82 v

net:   2 H2O(l) → 2 H2(g) + O2(g) E = -1.23 v

Electrolytic production of hydrogen is usually carried out with a dilute solution of sulfuric acid; see here for a concise summary of the chemistry and energetics. This process is generally too expensive for industrial production unless highly pure hydrogen is required. However, it becomes more efficient at higher temperatures, where thermal energy reduces the amount of electrical energy required, so there is now some interest in developing high-temperature electrolytic processes. Most hydrogen gas is manufactured by the steam reforming of natural gas.

Faraday's laws of electrolysis

One mole of electric charge (96,500 coulombs), when passed through a cell, will discharge half a mole of a divalent metal ion such as Cu2+. This relation was first formulated by Faraday in 1832 in the form of two laws of electrolysis:

1. The weights of substances formed at an electrode during electrolysis are directly proportional to the quantity of electricity that passes through the electrolyte.

2. The weights of different substances formed by the passage of the same quantity of electricity are proportional to the equivalent weight of each substance.

The equivalent weight of a substance is defined as the molar mass, divided by the number of electrons

required to oxidize or reduce each unit of the substance. Thus one mole of V3+ corresponds to three

equivalents of this species, and will require three faradays of charge to deposit it as metallic

vanadium.

Most stoichiometric problems involving electrolysis can be solved without explicit use of Faraday's laws. The "chemistry" in these problems is usually very elementary; the major difficulties usually stem from unfamiliarity with the basic electrical units:

current (amperes) is the rate of charge transport; 1 amp = 1 c/sec.

power (watts) is the rate of energy production or consumption;

1 w = 1 J/sec = 1 volt-amp; 1 watt-sec = 1 J, 1 kw-h = 3600 J.

Problem Example 1

A metallic object to be plated with copper is placed in a solution of CuSO4. a) To which electrode of a direct current power supply should the object be connected?b) What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?

Solution:

a) Since Cu2+ ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!)

b) The amount of charge passing through the cell is

(0.22 amp) × (5400 sec) = 1200 cor

(1200 c) ÷ (96500 c F–1) = 0.012 F

Since the reduction of one mole of Cu2+ ion requires the addition of two moles of electrons, the mass of Cu deposited will be

(63.54 g mol–1) (0.5 mol Cu/F) (.012 F) = 0.39 g of copper

Problem Example 2

How much electric power is required to produce 1 metric ton (1000 kg) of chlorine from brine, assuming the cells operate at 2.0 volts and assuming 100 % efficiency?

Solution:

moles of Cl2 produced: (106 g) ÷ 70 g mol–1 = 14300 mol Cl2

faradays of charge: (2 F/mol) × (14300 mol) = 28600 F

charge in coulombs: (96500 c/F) × (28600 F) = 2.76 × 109 c

duration of electrolysis: (3600 s/h) x (24 h) = 86400 s

current (rate of charge delivery): (2.76 × 109 amp-sec) ÷ (86400 sec) = 32300 amps

power (volt-amps): (2.0 v) × (32300 a) = 64.6 kw

energy in kW-h: (64.6 kw) × (24 h) = 1550 kw-h

energy in joules: (1550 kw-h) × (3.6Mj/kw-h) = 5580 Mj (megajoules)

(In the last step, recall that 1 w = 1 j/s, so 1 kw-h = 3.6 Mj)

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