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Electrolysis

When an electric current is passed through an electrolyte solution, the ions of the electrolyte undergo chemical changes at the respective electrodes. The chemical reaction carried out by passing electricity is called electrolysis.

Electrolytic process is useful in reduction as well as purification of metals. Electrolytic reduction of metals is useful in the metallurgy of highly reactive metals like sodium, potassium, cesium, calcium, magnesium and aluminum. The ores are mostly fused before electrolysis is done as only fused ores conduct electricity.

The metal is deposited on the respective cathode while in the anode the gases are liberated. This method is highly useful for the reduction of most reactive metals because they cannot be reduced by any other method.

Definition of Electrolysis

The phenomenon of chemical change taking place by the passage of electrical energy from an external source is called electrolysis. The devices or cells used to carry out electrolysis are called Electrolytic Cells.

Example:

When an electric current is passed through molten sodium chloride, sodium is produced at the cathode and chlorine is produced at the anode.

electric
2NaCl 2Na(s) + Cl2(g)
current

Electrolysis is used to extract many metals such as Na, K, Ca, Sr, Mg, Al, etc., and for the manufacture of chemicals such as NaOH, Cl2, F2, etc.

Mechanism of Electrolysis

The process of electrolysis was explained by the theory of ionization. According to ionic theory, electrolytes are present as ions in solution or in the molten state and the function of electricity is to direct these ions to their respective electrodes. So, electrolysis is possible either in molten state or in the form of a solution.
Let us discuss some examples :

Electrolysis of molten sodium chloride

The molten NaCl has Na+ and Cl- ions

NaCl $\to$ Na+ + Cl-

During electrolysis Na+ ion moves towards cathode and Cl- moves towards anode. Na+ takes up one electron from cathode and becomes sodium metal and Cl- gives up one electron to anode and becomes chlorine atom. Since chlorine atom is not stable, two chlorine atoms combine to form a chlorine molecule (Cl2).

The reactions may be expressed as,

At cathode
Na+ + e- $\to$ Na(s) (Reduction)
At anode
Cl- - e- $\to$ Cl (Oxidation)
Cl + Cl ? Cl2(g)

Electrolysis of water

Pure water is a poor conductor of electricity. However, when few drops of H2SO4 are added to water, it becomes an electrolyte.

H2O(l) $\to$ H+(aq) + OH-(aq)
At cathode
4H+(aq) + 4e- $\to$ 2H2(g)
At anode
4OH-(aq) - 4e- $\to$ 2H2O(l) + O2(g)

Michael Faraday performed a large number of experiments on electrolysis and summarized the results in the form of two laws known as Faraday’s laws of electrolysis.

1. Faraday’s first law of electrolysis

"The amount of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolytic solution."

The amount of any substance obtained, gives the amount of chemical reaction which occurs at any electrode during electrolysis.

Thus, if w gram of the substance is deposited on passing Q coulombs of electricity, then

w a Q

or, w = ZQ

Where Z is proportionality constant and is called electrochemical equivalent.

If a current I is passed for t seconds, then

Q= I x t

So that w = Z x I x t

If I = 1amp and t =1sec, then w =Z

Hence electrochemical equivalent of a substance may be defined as the mass of the substance deposited when 1 amp of current is passed for 1 second, i.e., quantity of electricity equal to 1 coulomb is passed.

2. Faraday’s second law of electrolysis

"It states that when similar or equal quantity of electricity is passed through different electrolytic solutions connected in series, the weights of the substances that are produced at the electrodes are directly proportional to their chemical equivalent weights."

Example: When same quantity of electricity is passed through two electrolytic solutions, containing CuSO4 and AgNO3 connected in series, the weights of copper and silver deposited are :

$\frac{Weight\ of\ Copper\ deposited}{Weight\ of\ Silver\ deposited}$ = $\frac{Eq\ wt\ of\ Copper}{Eq\ wt\ of\ Silver}$

Units of Charge and their relationship
SI unit of charge is Coulomb. The most common unit of charge used in electrolysis is Faraday.

1 Faraday = charge of 1 mole of electrons.

= 6.022 x 1023 x 1.6 x 10-19 Coulomb

= 96458 Coulomb or approximately 96500 Coulomb.

Electrolytic Refining Process of Metal

Electrolytic refining is a process in which pure metals are obtained from impure metals. This is considered as one of the most convenient and exclusive methods of refining and gives a metal of very high purity. This method is applicable to many metals such as Cu, Ag, Pb, Au, Ni, Sn, Zn, etc. The blocks of impure metal form the anode and thin sheets of pure metal form the cathode. A solution of a salt of the metal is taken as an electrolyte.

By passing an electric current through the solution, pure metal dissolves from the anode and deposits on the cathode. By this process, more metal ions undergo reduction and pure metal is deposited at the cathode. The insoluble impurities either dissolve in the electrolyte or fall at the bottom and get collected as anode mud.

For example, in the refining of copper, impurities like Fe and Zn dissolve in the electrolyte, while Au, Ag and Pt are left behind as anode mud. During the electrolytic refining of copper, a thick block of impure copper is made the anode, and a thin plate of pure copper is made the cathode. Copper sulphate solution is used as an electrolyte. Normally, the pure metal to be refined is made as cathode and the impure metal is made as anode.

The Process of Electrolysis

According to the theory of electrolytic dissociation, electrolytes can be electrolyzed only in the dissolved or molten state. These charged ions are directed towards the respective electrodes by the electricity supplied.

Let us examine the process of electrolysis in an electrolytic cell when the electrolyte used is molten sodium chloride.

Electrolysis of Molten Sodium Chloride

When sodium chloride is formed, the sodium atom donates an electron to the chlorine atom to form an electrovalent bond. Hence in the process,
• Sodium atom loses an electron and becomes positively charged (Na+).
• Chlorine atom gains an electron and becomes negatively charged (Cl-).
• The electrolyte consists of sodium (Na+) and chloride ions (Cl-) as shown below :

Before the electric current is switched on, the ions in molten state, though mobile, have no direction of movement. They move only randomly. But when molten sodium chloride is electrolysed (the current is switched on), the ions follow a particular direction of movement.

The cell set up is shown below. The electrodes in this case are made of carbon.

The electrolysis of molten sodium chloride is as follows :
Nacl (s) $\rightarrow$ Na+ Cl- (l) $\rightarrow$ Na+ + Cl-

In solution or molten state when dissociation takes place, the sodium ion has a positive charge and the chloride ion has a negative charge.

These ions are capable of carrying the current. The positively charged ions Na+ migrate to the cathode and the negatively charged ions Cl- migrate to the anode. On reaching the respective electrodes these ions get involved in electrochemical reactions at the electrode and form their respective products.

Electrolysis At Cathode

The cathode is a region of surplus electrons. The ion that migrates to the cathode is the positively charged ion or cation, i.e., sodium ion.

Nao - 1e- $\rightarrow$ Na+

At the cathode, the ions accept electrons. The sodium ion takes in one electron from the cathode, gets converted into a sodium atom, and gets deposited at the cathode. Hence reduction takes place at the negative electrode or cathode.
Na+ + 1e- $\rightarrow$ Nao
The product obtained at the cathode is sodium metal.

In general, the number of electrons gained by the cation depends upon the number of positive charges it carries. For example, an aluminium ion has three positive charges; hence it will gain three electrons to get converted into an aluminium atom.
Al3+ + 3e- $\rightarrow$ Al0

Electrolysis At Anode

The anode is a region of deficiency of electrons and hence anions give up electrons here. They migrate to the anode to give up as many electrons as they have negative charges. Hence the electrode at which oxidation takes place is anode (the positively charged electrode). In this case, the chloride ion has one negative charge that it has gained from sodium. This electron is given up to the anode and chlorine is deposited in the form of bubbles of greenish yellow chlorine gas.
Cl- - 1e- $\rightarrow$ Cl

Clo + Clo $\rightarrow$ Cl2

In other words, the product obtained at anode is chlorine gas.

It should be very clear that in the above electrolysis, the electrons flowing through the bulb are not the electrons that are produced by the battery. These electrons are accepted by the cations from the electrolyte. The electrons passing through the bulb are the ones given up by the anions.

Application of Electrolysis

The process of electrolysis has many applications. Some of them are listed below :
1. Electroplating on metals
2. Electrorefining of metals
3. Electrometallurgy
4. Hair removing therapy

Electrolysis Problems

Below you could see problems

Solved Examples

Question 1: How many coulombs are required to deposit 50g of aluminum when electrode reaction is
Al3+ + 3e- → Al
Solution:
1 mole Al = 27g Al

27g Al requires = 3F charge

1g Al requires = $\frac {3} {27}$ F charge

50g Al requires = $\frac {3 * 50} {27}$ F charge

= $\frac {3 * 50 * 96500} {27}$ Coulomb

= 536111 Coulomb

Question 2: How many hours does it take to reduce 3 mole of Fe3+ to Fe2+ with 2.0 A current?
Solution:
The reaction is Fe3+ + e- → Fe2+

Reduction of 1 mol of Fe3+ ion require = 1F

Reduction of 3 mol of Fe3+ ion require = 3F

= 3 * 96500 C
= 2.895 * 105 C

Time = Charge/ current

= $\frac {2.895 * 10^{5}} {2}$ s

= $\frac {(2.895 * 10^{5})} {2 * 3600}$ h

= 40.21 h

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