Passnownow

Rated 4.8/5 by parents & students

Classwork Series and Exercises {Physics- SS3}: Electrolysis

Definition Of Electrolysis

Electrolysis is the process by which the movement of an electric current through a solution librates electrons. Chemicals in a container are decomposed in order to generate current.

Michael Faraday studied this process extensively and laid the foundation for the theory of electrolysis. Some liquids are good conductors while others are poor conductors of electricity. Good conductors are known as electrolytes, poor conductors are non-electrolytes.

Liquids such as solutions of acids, bases and salts are generally good conductors. Liquids such as benzene and paraffin or kerosene are poor conductors. Organic compounds are generally poor conductors. While pure water is also a poor conductor, water containing some dissolved salts conducts moderately.

A Voltameter is device for measuring the quantity of electricity passing through a conductor by the amount of electrolytic decomposition it produces, or for measuring the strength of a current by the amount of such decomposition in a given time.

  1. The Electrolytes is the liquid or molten substance which conducts a current and is decomposed by it. i.e. it contains mobile ions and undergoes decomposition and is called electrolyte, e.g. acids, bases, common salts, etc.
  2. Non-electrolyte this is a substance which, either in molten state or in solution, does not allow the flow of an electric current. Thus, it is a substance that does not conduct current or undergo decomposition, e.g. organic solvents such as benzene, paraffin, sugar, salt, etc.
  1. Electrodes are materials in the form of a rod or plate through which current enters or leaves the electrolyte. There are two electrodes: (a) the positive electrode through which current enters the electrolyte is called the anode. (b) The negative electrode through which liquid leaves the electrolyte is called the Cathode.
  2. Anode is a positive (+ve) electrode at which the electron enters and current leaves the electrolyte.
  3. Cathode is a negative (-ve) electrode at which the electron enters and current leaves the electrolyte
  4. Voltameter: The whole apparatus consisting of the vessel, electrolyte and electrodes is called the voltameter.
  5. Ions: They are charged particles which exists in electrolytes and take part in electrolysis These are the immediate products of decomposition of an electrolyte. The ions which go to the anode are called anions, those ions which go to the cathode are called cations

Current Electricity

Current electricity consists of fast moving negatively charged electrons. Currents travels in material which allow the flow of electrons called conductors. Current is produced in a simple circuit consisting of a battery (which is source), a bulb (which is the lighting) and a tap key (which is used in controlling the simple system).

The current in the battery due to the force applied on it. The force is not visible but it is performed by some chemicals in the battery. The electrical pressure is called the voltage.

Dynamics of charged particles (ions) in electrolytes

The Ionic theory

In an electrolyte there are positively and negatively charged particles called ions. The molecules that constitute the electrolyte are split in solution into these ions through the process known as electrolytic dissolution. The dissolution of an electrolyte occurs irrespective of whether or not an electric field is applied to the electrolyte. Ions in an electrolyte execute random movements until a battery is connected to the electrodes of a voltmeter. As soon as a p.d. is set up across the electrodes the positive ions drift to the cathode, which is at a negative potential while the negative ions drift to the anode which is at a positive potential.

This directional movement of ions is the electric current flowing through the electrolyte. Such movement ceases as soon as the battery is disconnected, and the ions move randomly once again.

Thus electrolytic solutions are able to conduct electricity because the electrolytes in solution can dissociate into ions Non electrolytes in solution do not easily dissociate into ions.

Examples of Electrolysis

Pure water is a poor electrolyte. Because of that, a few drops of sulfuric acid will help to increase its conductivity by adding more mobile electrons

As such the ions present in the electrolyte now are

From Sulfuric Acid: H+ and SO42-
From Water: H+ and OH

Inert electrodes like Carbon or Platinum electrodes are used so that the acid solutions do not corrode the electrodes easily.

At the Cathode:

H+ migrate to the cathode and accepts an electron to become a Hydrogen atom. After that, the Hydrogen atoms form a covalent bond with other hydrogen atoms to become Hydrogen gas molecules

2H+ + e–> H2

Observation:
Effervescence of colourless, odourless gas at the cathode

At the Anode:

SO42- and OH migrate to the anode, where OH is preferentially discharged due to it having a lower position in the Electrochemical series as compared to SO42-

4OH –> 2H2O + O2 + 4e

Therefore Overall Equation:

2H2O(l) –> 2H2 (g)+ O2(g)

Michael Faraday’s Laws of Electrolysis

First Law: It shows that the mass (m) of substance deposited at the cathode during electrolysis is directly proportional to the quantity of electricity (total charge q) passed by the electrolyte.

 i.e.  M = ZQ, where Q = It, M = ZIt  , Z = M/It, where Z is the constant of proportionality and is known as electrochemical equivalent (E.C.E.) of the substance.

Thus, electrochemical equivalent (ECE) may be defined as “the mass of the ion deposited by passing a current of one Ampere for one second (i.e., by passing Coulomb of electricity)”. It’s unit is gram per coulomb.

Coulomb is the unit of electrical charge.

96500Coulombs electrons = 1 mole electrons.

1 Coulomb = 6.023×1023/96500 = 6.85 × 1018 electrons, or 1 electronic charge 1.6 × 10–19 Coulomb.

Second law: It states that, “When the same quantity of electricity is passed through different electrolytes, the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalents (Equivalent weights).” i.e.,

M1/W2 = E1/E2 or Z1It/Z2It or Z1/Z2 = E1/E2 (∴ W = ZIt)

Thus the electrochemical equivalent (Z) of an element is directly proportional to its equivalent weight (E), i.e., E ∝ Z or E = FZ or E = 96500 × Z

where, F = Faraday constant = 96500 C mol–1

So, 1 Faraday = 1F =Electrical charge carried out by one mole of electrons.

1F = Charge on an electron × Avogadro’s number.

1F = e × N = (1.602 × 10–19c) × (6.023 × 1023 mol–1)

Number of Faraday = number of electrons passed/6.023 × 1023

Faraday’s law for gaseous electrolytic product For the gases, we use V = It Ve/96500

where, V = Volume of gas evolved at S.T.P. at an electrode

Ve = Equivalent volume = Volume of gas evolved at an electrode at S.T.P. by 1 Faraday charge

Quantitative aspects of electrolysis: We know that, one Faraday (1F) of electricity is equal to the charge carried by one mole (6.023 × 1023) of electrons. So, in any reaction, if one mole of electrons is involved, then that reaction would consume or produce 1F of electricity. Since 1F is equal to 96,500 Coulombs, hence 96,500 Coulombs of electricity would cause a reaction involving one mole of electrons.

If in any reaction, n moles of electrons are involved, then the total electricity (Q) involved in the reaction is given by, Q = nF = n × 96500 C

Thus, the amount of electricity involved in any reaction is related to,

(i) The number of moles of electrons involved in the reaction,

(ii) The amount of any substance involved in the reaction.

Therefore, 1 Faraday or 96,500 C or 1 mole of electrons will reduce,

(a) 1 mole of monovalent cation,(b) 1/2mole of divalent cation,

(c) 1/3 mole of trivalent cation, (d) 1/n mole of n valent cations.

Uses of Electrolysis

  1. Extraction or isolation of metals: Electrolysis is a process used in extracting metals. Such elements are usually very reactive, e.g. sodium, potassium, chlorine, oxygen, aluminium, etc. Such elements are found at the top reactivity series. They react too readily and cannot be prepared by the electrolysis in an aqueous solution of one of the salts. They are prepared by the electrolysis of their fused salts.
  2. Purification of Metals: many do not in free or uncombined state. They exist in a state of combination with other elements. To extract such metals from their ores, they have to be purified. The overall result of the experiment is that the metal is transformed from its impure to its pure state.
  3. Electroplating of Metals: Electroplating is the process of coating a substance with the layer of another substance. It is done through electrolysis. Object to be plated is fairly cleaned to make sure that the anode deposits sticks firmly. If it is not a conductor, it is first coated with graphite, so that what is to be coated on it sticks firmly. The anode is the pure metal to be deposited, while the cathode is made the electrolytic cell. The electrolyte is a solution of soluble salt and the pure metal used as the anode.

Examples

  1. Calculate the time in minute, required to plate a substance a 300cm2, a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm3 and one coulomb liberates 0.00033g copper.

Solution

Given that Area = 300cm2, thickness = 0.6mm = 0.06cm

Mass = 0.00033g, density = 8.8g/cm3

Density = mass/volume, mass = density x volume

Mass = 8.8 x 300 x 0.06 = 158.4g

Using M = Zit

T = m/ZI = 158.4/2 x 0.00033 = 24000secs = 4000mins

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

  1. Find the mass of copper deposited on the cathode of a copper voltameter if a current of 0.53A is placed through it for 30 minutes (e.c.e. of copper = 3.3 x 10-4 gC-1) A. 0.465 g B. 0.425 g C. 0.325 g D. 0.315 g
  2. Calculate the time in minutes required to electroplate an article of area 300 cm2 with a layer of copper 0.06 cm thick if a constant current of 2 A is maintained. Assumed that the density of copper is 8.8 gcm-3 and that one coulomb liberates 0.00033 g of copper. A. 400 minutes B. 4000 minutes C. 4500 minutes D. 5000 minutes
  3. A copper and a silver voltameter are connected in series, and at the end of a period of time, 5.0 g of copper was deposited, calculate the mass of silver deposited at the same time. Chemical equivalent of copper = 31.5. Chemical equivalent of silver = 108. A. 47 g B. 17.14 g C. 20.17 g D. 23 g
  4. The electrochemical equivalent of a metal is 0.126 x 10-6 kgC-1. The mass of the metal that a current of 5 A will deposit from a suitable bath in 1 hour is A. 0.0378 x 10-3 kg B. 0.227 x 10-3 kg C. 0.378 x 10-3 kg D. 0.595 x 10-3 kg
  5. Ions are ……………………………. particles which exists in electrolytes and take part in electrolysis A. Non charged B. Charged C. Neutral D. Positive

Read More

Photo Credit

3 thoughts on “Classwork Series and Exercises {Physics- SS3}: Electrolysis”

  1. Renaissant Da Vinci

    Thanks a million….. This article was quite helpful…. Students shouldn’t miss this lesson…. It met to all my queries

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top