CHEMICAL EFFECT OF ELECTRIC CURRENT
When current flow through materials, it has or produce chemical effects on the materials, either on their physical properties or chemical properties, or on both. These effects are studied under electrolysis.
When current flow through materials, it has or produce chemical effects on the materials, either on their physical properties or chemical properties, or on both. These effects are studied under electrolysis.
ELECTROLYSIS:
It is the decomposition of a substance in its molten form when electric current flow through the substance.
ELECTROLYTES:’
Electrolytes are substances or compounds which conduct electricity and are decomposed. They contain ions.
EXAMPLES OF ELECTROLYTES:
I. Acid Solution.
II. Bases.
III. Salt solution.
Iv. Acidulated water.
II. Bases.
III. Salt solution.
Iv. Acidulated water.
NON –ELECTROLYTES:
Non electrolytes are substances which either in their molten form or solution do not conduct electricity and are not decomposed. No electrolytes do not contain ions and they are not decomposed.
EXAMPLES OF NON ELECTROLYTES:
I. Organic solvent.
Ii. Benzene
iii. Paraffin
iv. Sugar solution.
Ii. Benzene
iii. Paraffin
iv. Sugar solution.
ELECTRODES:
Electrodes are the positive anode and the negative cathode dipped inside the electrolyte. They are conductors where current enters and leaves the electrolyte.
POSITIVE ELECTRODE – ANODE:
Anode is the positive electrode at which current enters the electrolyte.
NEGATIVE ELECTRODE – CATHODE:
Cathode is the negative electrode at which current leaves the electrolyte.
VOLTAMETER:
Voltammeter is the apparatus or vessel or container which contains the electrolyte and the electrodes. Iot is the container in which electrolysis take place.
IONS:
Ions are charged particles which exist in the electrolyte and take part in the electrolysis. They are the product of electrolysis formed by the decomposition of the electrolyte by loosing of electron and gaining of electrons.
TYPES OF IONS:
There are two types of ions:
i. Anions: these are negatively charged ions which migrate to the anode.
ii. Cations: these are positively charged ions which migrate to the cathode.
IONIC THEORY OF ELECTROLYSIS:
The ionic theory of electrolysis explains that when current is passed through electrolyte, the electrolyte is decomposed into positive ion (cation) and negative ion (anion). The positive ion (cation) migrates to the cathode, gain electron and become a neutral atom while the negative ion (anion) migrates to the anode, donate itself to the anode and the anode become a neutral atom.
EXPLANATION OF FARADAY’S SECOND LAW OF ELECTROLYSIS
Diagram here:
The set up consists of a copper voltameter, silver voltameter, and a water voltameter are connected in series as shown in e set up diagram.
The copper and the silver cathodes are weighed and their masses are recorded. The same quantity of electricity is passed through the three voltameters for about 30 minutes. The copper and silver are then rinsed with distilled , dried reweighed to know the masses of copper and silver deposited. The volumes of hydrogen and oxygen are read andverted to their masses using their densities.
If their masses are compared, they will be in the ratio,
Copper : Silver : Hydrogen : Oxygen = 32 : 108 : 1 : 8.
These ratios are the same is the same as the ratio of their chemical equivalents.
FORMULA FOR CALCULATION IN FARADAY’S SECOND LAW OF ELECTROLYSIS
Note that in Faraday’s second law of electrolysis, two substances are involved. Therefore, the formula is as follow:
Mass of A / mass of B = z of A / z of B.
A and B are the two substances while Z is their electrochemical equivalents.
Also, the formula can be written as w,
Mass of A /mass of B = (rtm of A / valency of A) / (rtm of B/valency of B)
Worked Examples:
• A copper and water voltameter are in series, and at the end of a period of time 3.0 g of copper is deposited, calculate the mass of oxygen deposited.
Solution:
Mass of copper = 3g, mass of Oxygen = ?,rtm of Copper = 63, valency = 2, rtm of Oxygen =16,valency = 2.
Formula:
M of C/M of O =(rtm of C/valency of C)/rtm of O/valency of O)
Let us substitute for the variables,
3/M of O = (63/2) / (16/2) ➡ 3/M of O = 31.5 / 8. Make No the subject.
M of O = 3 * 8 / 31.5 ➡ M of O = 24 / 31.5. Then, M of O = 0.7619 g.
• A copper and a silver voltameter are connected in series, and at the d of a period of time, 4.0 g of copper was , calculate the Mass of silver deposited at the me time. Chemical equivalent of copper = 31.5, that of silver = 108
Solution:
M of C = 5g, M of S = ?, Z of C = 31.5, Z of S = 108.
Formula:
M of C/M of S = Z of C / Z of S. Let us substitute for the variables. Then,
5 / M of S = 31.5 / 108. Let us make M of S the subject of the formula.
5 * 108 / 31.5 = M of S. ➡ M of S = 5 * 108 / 31.5.
M of S = 17.1428 g.
Note that chemical equivalent = relative atomic mass / 2.
FARADAY’S CONSTANT F.
Faraday’s constant is the quantity of charge which will liberate one mole of a monovalent element. The symbol is F.
Faraday’s constant F = 96500 c/mol.
CALCULATION ON FARADAY’S CONSTANT
Worked Example:
The mass of silver deposited when a ready current of 2A flow for 30 minutes through a silver voltameter.
Solution:
Current = 2A, time = 30 minutes, ass of silver = ?
First, we will calculate the quantity of charge produced by 2A that flow for 30 minutes. then,
Quantity of charge produced = I * t ➡ quantity of charge = 2 * 30 min
Quantity of charge = 2 * 30 * 60 seconds
= 3600 Coulomb.
According to Faraday’s law, 1 F will liberate one mole of a monovalent element. Silver is a monovalent element. Then
1 F will liberate mole of silver. One mole of silver is 108 g mass. Then,
If 1 F = 96500 C liberate 108 g of .
Then, 3600 C will liberate x g of silver.
96500 / 3600 = 108 / x. Let is make x the subject of the formula.
26.81= 108 / x ➡ x = 108 / 26.81 ➡ x = 4.0283g
Therefore, 2A of current for 30 minutes will liberate 4.0283 g of silver.
Questions on Electrolysis
*What is the part played by electrolytein electrolysis?
It is required to cover the surface of an oobject of total area 600 cm² with nickle of thickness 0.10 mm using a constant current of 5.0 A. Determine the time for which the rrent must flow if the mass of the nickel deposited per coulomb is 0.00030 g/c and its density is 9.0 g/ cm cube. If PD of 40 volt is employed, determine the cost at 2p per kilowatt-hour.
*State Faraday laws of electrolysis.
Describe an experiment to determine the mass of copper deposited per Coulomb.
State what precaution you would take to ensure that a reliable value. is obtained.
A potential difference of 6 V is maintained between the plate of a copper voltmeter and 0.66g of copper is deposited. Calculate the electrical energy consumed by the àq11q11ameter.( mass of copper deposited per = 0.00033g/c).
*Describe an experiment to check the accuracy of the 1A reading of ammeter, assuming that the value for the mass deposited coulomb of copper is available.
What is the cost of depositing a layer of silver 2mm thick on an object of total surface area 150 cm² if a current of 1.0A used for an hour s 2p? ( density of silver is 10.5g per cmsqr ; mass of silver deposited per coulomb is 0.00112g/c).
*Describe an experiment to determine how the mass of copper deposited in a copper voltameter varies with the quantity of electricity passing through it. Give à circuit diagram and full experimental details.
An electric current passes through two voltmeter in series, containing sulphate and silver nitrate respectively. What is the mass f silver deposited in a given time if the mass of copper deposited is in that time is 1g? ( R.A.M of cooper is 63, f silver 108, valency of copper is 2, valency of silver is 1).
*Describe an experiment to decompose water y electrolysis.
Calculate the current required to liberat 10 m cube of hydrogen per min in electrolysis. (1 liter of hydrogen weighs 0.09 g at the temperature and pressureat which the volumeis measured in this case. The mass of hydrogen deposited per Coulomb is 0.0000105g/).
*Explain the term electrolysis, ionization,electrochemical equivalent.
*State Faraday’s laws of electrolysis and describe how you would attenp to test the second law experimentally.
*Calculate the time for which a steady current of 0.2A Must pass through a water voltameter in other to liberate a quantity of hydrogen which would occupy a volume of 50 cm cube at STP. Take the mass of hydrogen deposited per Coulomb to be 0.00001045g/c and its density at STP to be 0.00009g/cm cube).
*Explain how the ionic theory account for Faraday’s law of electrolysis.
* Give an account of the processes when dilute sulphuric acid is electrolysed between platinum electrodes.
*A cell with copper electrodes and containing copper sulphate solution is in series with one containing dilute sulphuric acid and platinum electrodes. When 3.175g of copper have been liberated, what volume of hydrogen measured at STP is released? ( rtm of copper = 63.5, hydrogen = 1.008, valency of copper = 2, density of hydrogen at STP , 0.0899g per ).
*State the laws of electrolysis and describe two experiments to illustrate these laws.
*what are the chief differences between the passage o electricity through metals and through ionized solutions.
*A steady current pass through a silver voltameter and a 10 ohms resistance arranged in series. A high-resistance voltmeter placed across the terminals f the 10 ohms coil reads 6V. Given that 0.8048 g of silver is eposited on the cathode of the voltameter in 20 minutes, calculate the mass of silver deposited per Coulomb.
*State the relationship between the mass of a substance liberated during electrolysis and the charge that passed . Describe an experiment by which this relationship can be established for a metal such as copper.
*An ammeter and a copper voltameter are placed in eries in circuit nd after 20 minutes the mass of copper deposited is 0.36 g. That ammeter reads 0.8 A. What can you deduce from the ammeter reading?
( Assuming mass of copper deposited per Coulomb = 0.0003g/c )
... To be completed...
EXPLANATION OF FARADAY’S SECOND LAW OF ELECTROLYSIS
Diagram here:
The set up consists of a copper voltameter, silver voltameter, and a water voltameter are connected in series as shown in e set up diagram.
The copper and the silver cathodes are weighed and their masses are recorded. The same quantity of electricity is passed through the three voltameters for about 30 minutes. The copper and silver are then rinsed with distilled , dried reweighed to know the masses of copper and silver deposited. The volumes of hydrogen and oxygen are read andverted to their masses using their densities.
If their masses are compared, they will be in the ratio,
Copper : Silver : Hydrogen : Oxygen = 32 : 108 : 1 : 8.
These ratios are the same is the same as the ratio of their chemical equivalents.
FORMULA FOR CALCULATION IN FARADAY’S SECOND LAW OF ELECTROLYSIS
Note that in Faraday’s second law of electrolysis, two substances are involved. Therefore, the formula is as follow:
Mass of A / mass of B = z of A / z of B.
A and B are the two substances while Z is their electrochemical equivalents.
Also, the formula can be written as w,
Mass of A /mass of B = (rtm of A / valency of A) / (rtm of B/valency of B)
Worked Examples:
• A copper and water voltameter are in series, and at the end of a period of time 3.0 g of copper is deposited, calculate the mass of oxygen deposited.
Solution:
Mass of copper = 3g, mass of Oxygen = ?,rtm of Copper = 63, valency = 2, rtm of Oxygen =16,valency = 2.
Formula:
M of C/M of O =(rtm of C/valency of C)/rtm of O/valency of O)
Let us substitute for the variables,
3/M of O = (63/2) / (16/2) ➡ 3/M of O = 31.5 / 8. Make No the subject.
M of O = 3 * 8 / 31.5 ➡ M of O = 24 / 31.5. Then, M of O = 0.7619 g.
• A copper and a silver voltameter are connected in series, and at the d of a period of time, 4.0 g of copper was , calculate the Mass of silver deposited at the me time. Chemical equivalent of copper = 31.5, that of silver = 108
Solution:
M of C = 5g, M of S = ?, Z of C = 31.5, Z of S = 108.
Formula:
M of C/M of S = Z of C / Z of S. Let us substitute for the variables. Then,
5 / M of S = 31.5 / 108. Let us make M of S the subject of the formula.
5 * 108 / 31.5 = M of S. ➡ M of S = 5 * 108 / 31.5.
M of S = 17.1428 g.
Note that chemical equivalent = relative atomic mass / 2.
FARADAY’S CONSTANT F.
Faraday’s constant is the quantity of charge which will liberate one mole of a monovalent element. The symbol is F.
Faraday’s constant F = 96500 c/mol.
CALCULATION ON FARADAY’S CONSTANT
Worked Example:
The mass of silver deposited when a ready current of 2A flow for 30 minutes through a silver voltameter.
Solution:
Current = 2A, time = 30 minutes, ass of silver = ?
First, we will calculate the quantity of charge produced by 2A that flow for 30 minutes. then,
Quantity of charge produced = I * t ➡ quantity of charge = 2 * 30 min
Quantity of charge = 2 * 30 * 60 seconds
= 3600 Coulomb.
According to Faraday’s law, 1 F will liberate one mole of a monovalent element. Silver is a monovalent element. Then
1 F will liberate mole of silver. One mole of silver is 108 g mass. Then,
If 1 F = 96500 C liberate 108 g of .
Then, 3600 C will liberate x g of silver.
96500 / 3600 = 108 / x. Let is make x the subject of the formula.
26.81= 108 / x ➡ x = 108 / 26.81 ➡ x = 4.0283g
Therefore, 2A of current for 30 minutes will liberate 4.0283 g of silver.
Questions on Electrolysis
*What is the part played by electrolytein electrolysis?
It is required to cover the surface of an oobject of total area 600 cm² with nickle of thickness 0.10 mm using a constant current of 5.0 A. Determine the time for which the rrent must flow if the mass of the nickel deposited per coulomb is 0.00030 g/c and its density is 9.0 g/ cm cube. If PD of 40 volt is employed, determine the cost at 2p per kilowatt-hour.
*State Faraday laws of electrolysis.
Describe an experiment to determine the mass of copper deposited per Coulomb.
State what precaution you would take to ensure that a reliable value. is obtained.
A potential difference of 6 V is maintained between the plate of a copper voltmeter and 0.66g of copper is deposited. Calculate the electrical energy consumed by the àq11q11ameter.( mass of copper deposited per = 0.00033g/c).
*Describe an experiment to check the accuracy of the 1A reading of ammeter, assuming that the value for the mass deposited coulomb of copper is available.
What is the cost of depositing a layer of silver 2mm thick on an object of total surface area 150 cm² if a current of 1.0A used for an hour s 2p? ( density of silver is 10.5g per cmsqr ; mass of silver deposited per coulomb is 0.00112g/c).
*Describe an experiment to determine how the mass of copper deposited in a copper voltameter varies with the quantity of electricity passing through it. Give à circuit diagram and full experimental details.
An electric current passes through two voltmeter in series, containing sulphate and silver nitrate respectively. What is the mass f silver deposited in a given time if the mass of copper deposited is in that time is 1g? ( R.A.M of cooper is 63, f silver 108, valency of copper is 2, valency of silver is 1).
*Describe an experiment to decompose water y electrolysis.
Calculate the current required to liberat 10 m cube of hydrogen per min in electrolysis. (1 liter of hydrogen weighs 0.09 g at the temperature and pressureat which the volumeis measured in this case. The mass of hydrogen deposited per Coulomb is 0.0000105g/).
*Explain the term electrolysis, ionization,electrochemical equivalent.
*State Faraday’s laws of electrolysis and describe how you would attenp to test the second law experimentally.
*Calculate the time for which a steady current of 0.2A Must pass through a water voltameter in other to liberate a quantity of hydrogen which would occupy a volume of 50 cm cube at STP. Take the mass of hydrogen deposited per Coulomb to be 0.00001045g/c and its density at STP to be 0.00009g/cm cube).
*Explain how the ionic theory account for Faraday’s law of electrolysis.
* Give an account of the processes when dilute sulphuric acid is electrolysed between platinum electrodes.
*A cell with copper electrodes and containing copper sulphate solution is in series with one containing dilute sulphuric acid and platinum electrodes. When 3.175g of copper have been liberated, what volume of hydrogen measured at STP is released? ( rtm of copper = 63.5, hydrogen = 1.008, valency of copper = 2, density of hydrogen at STP , 0.0899g per ).
*State the laws of electrolysis and describe two experiments to illustrate these laws.
*what are the chief differences between the passage o electricity through metals and through ionized solutions.
*A steady current pass through a silver voltameter and a 10 ohms resistance arranged in series. A high-resistance voltmeter placed across the terminals f the 10 ohms coil reads 6V. Given that 0.8048 g of silver is eposited on the cathode of the voltameter in 20 minutes, calculate the mass of silver deposited per Coulomb.
*State the relationship between the mass of a substance liberated during electrolysis and the charge that passed . Describe an experiment by which this relationship can be established for a metal such as copper.
*An ammeter and a copper voltameter are placed in eries in circuit nd after 20 minutes the mass of copper deposited is 0.36 g. That ammeter reads 0.8 A. What can you deduce from the ammeter reading?
( Assuming mass of copper deposited per Coulomb = 0.0003g/c )
... To be completed...
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