Skip to main content

Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
Post a Comment

Faraday's Laws of electrolysis

 


Faraday’s Laws of Electrolysis


Faraday put forward two laws of electrolysis and he also discovered from his experiment that the quantity of products of electrolysis depends on the following

  • Time of the flow of electric current
  • The ionic charge of the element deposited
  • The amount of current


Faraday’s First Law of Electrolysis

The first law states that the amount of the element deposited during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte (electrolytic solution).

Mathematically

      M ∞ Q

Where 

Q = quantity of electricity measured in Coulomb

M = mass of the substance deposited

Since the quantity of electricity is given by

Q = current x time in seconds

Faraday’s first law can also be stated as “the mass of element deposited during an electrolysis is proportional to the size of the current and the time taken for the flow of the current.”  

 M ∞ I x t 

Or 

M = є I t

Where є is a constant called electrochemical equivalent.

I stands for the amount of current and t stands for the time taken for the flow.


Faraday’s Second Law of Electrolysis

The second law states that when the same quantity of electric current is passed through electrolytes at the same time, the masses of the different substances deposited at the respective electrodes are inversely proportional to the charges on the ions of the elements.


Faraday’s Law of Electrolysis Calculations

  1. Calculate the amount of gold deposited when a current of 10A is passed through a solution of gold salt for 4 hours. If the same current is used, find the time taken for 12g of gold to be deposited.

(Au = 197, 1faraday = 96500C)

 

Solution

 

             Au+ + e- → Au

From the ionic equation above, gold requires one mole of an electron to deposit one mole of a gold element. It also means that one faraday is required to deposit one mole of gold.

 

1F = 1mole of gold or 197g of gold

96500C = 197g of gold

 

Now, To find the quantity of electricity that will deposit the required mass of gold.

Quantity of electric current = current x time in seconds

                                                 

= 10 x 4 x 60 x60 = 144000C

96500C = 197g of gold

144000 = 197/96500 x 144000 = 293.97g of gold.

 

To solve the second question, use the same quantity of electricity. 

If 293.97g of gold is required = 144000C

12g of gold will need = 144000/ 293.97 x 12 = 5878.2C

To calculate the time, we are going to use the relation below;

Q = It

   Where Q is 5878.2C , I is 10A

5878.2 = 10 x t

time = 5878.2/10  = 587.8 seconds or 9.8 minutes.

 

  1. 0.221g of a divalent metal is deposited when a current of 0.45A is passed through a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. ( 1F = 96500C)

Solution

           Let the symbol of the divalent metal be = X

                              X2+ + 2e- → X

           Since X is divalent metal, it will need two electrons to deposit one mole of     

The metal. The first approach is to calculate the quantity of electricity required to deposit 0.221g of the metal

Q = 0.45 x 25 x 60 = 675C

675C = 0.221g of X metal

2x 96500C =  

0.221/675 x 139000 =63.2

 

 

 

 

 


Comments

Post a Comment

Popular posts from this blog

Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
Post a Comment
Read more

Qualitative Analysis of inorganic Compound

  This is a type analysis which involves the identification of the ions ( cation and anion) in a given inorganic substance. Thus, Qualitative analysis deals with the identification of the compound. To effectively identify the ions, it is necessary to be able to observe the presence of any chemical reaction which is normally recognized by ·         Colour change ·         Evolution of gas ·         Precipitation      Colour Change : colour change is associated with transition metal ions. The major cause of the colour in transition metal ions is electronic transition within the d-block level. The colour of light which show, is the colour of light which is reflected by the ion. This change of transition metal ions is common when they form a bond with water or ammonia. It is important to note that zinc does not form coloured ion, this is because zinc has completely filled the d orbital, but zinc is yellow when hot and white when cold. Evolution of gas : This is identified by
Post a Comment
Read more

Qualitative Analysis of Ammonium Trioxocarbonate (IV)

              Ammonium trioxocarbonate(IV) is an electrovalent compound just like any other ammonium salts. As an electrovalent compound, it has NH + (ammonium ion) as the cation and CO 3 2- (trioxocarbonate IV ion or radical) as the anion.                          (NH 4 ) 2 CO 3   -------    2NH 4 + + CO 3 2- Ammonium trioxocarbonate IV is a white crystal salt and it is very soluble in water like all other ammonium salts. It decomposes on heating to produce ammonium, water and carbon (IV)oxide.                      (NH 4 + ) 2 CO 3   ----------   2NH 3(g) + H 2 O (I) + CO 2(g)          Test for the Cation in Ammonium Trioxocarbonate IV To test the unknown sample, put the sample into a boiling tube, add a base or alkali into the boiling tube and heat gently.   Note: All ammonium salts liberate ammonia when heated with base or alkali.         (NH 4 ) 2 CO 3(s) + 2NaOH (aq) ----------   Na 2 CO 3 (aq) + H 2 O (l) + 2NH 3(g) Test the gas liberated with damp li
4 comments
Read more