Electrolysis of acidified water (dilute sulphuric acid)

          The electrolysis of dilute sulphuric acid is described and explained. The electrolysis of an aqueous solution of dilute sulphuric acid is often carried out in a  Hofmann Voltammeter. Gases are produced at both the anode and the cathode. For every hydrogen ion discharged at the cathode, another hydrogen ion is formed at the anode. The net result is that the concentration of the sulphuric acid remains constant and this electrolysis consists of the decomposition of water :

                                  

                                 Electrolysis of Brine

                      Electrolysis of concentrated solution of NaCl

          The electrolysis of brine results in hydrogen (H₂) and chlorine gas. It  is a large-scale process used to manufacture chlorine (Cl₂). The other useful co-product is sodium hydroxide (NaOH).

      At the cathode, water is reduced to hydroxide ion and hydrogen gas. (The more reactive a metal, the less readily its ion is reduced on the electrode surface.) 
At the anode, chloride (Cl^-) ions are preferentially oxidized to chlorine.

Reduction at Cathode :

                  2H₂O(l) + 2e^–   =   H₂(g)  + 2OH^-(aq) 

Oxidation at Anode : 
                   2Cl^–(aq)  =  Cl₂(g) + 2e^– 

Overall reaction :

     2H₂O(l)  + 2Cl^–(aq)  =  2OH^–(aq)  + H₂(g) + Cl₂(g) 
             
              (The sodium ions are spectator ions)

               Electrolysis of Water

        Pure water is an electrical insulator. Addition of a little bit of acid, base or salt (such as Na₂SO₄) in water, the soln. behaves as a good conductor. Electrolysis of such aq. solution results in the decomposition of water at both electrodes.

At  pH=7 (neutral condition), the redox reactions are :

Reduction at Cathode  :

   4 H₂O(l) + 4e  = 2 H₂(g) + 4OH^-(aq)    E⁰ = - 0.41v   

Oxidation at Anode  :
  
   2 H₂O(l)  =  4 H⁺(aq) +  O₂(g) + 4e       E⁰ = +0.82v

            E⁰_cell  = E⁰_cathode – E⁰_anode =  (-0.41) – (+0.82) 
                                                    = -1.23v

The calculation shows that the electrolytic cell potential of water is -ve. Therefore, the redox reactions are not possible without any external energy. To derive a spontaneous redox reaction, the cell potential should be +ve. So unlike voltaic cell, the electrolytic cell needs potential input.The minimum voltage necessary to bring about electrolysis should be greater than +1.23v, as the electrolytic cell potential of water at pH 7 is -1.23v. 

                                        ELECTROLYSIS

       
          Electrolysis is a process by which electrical energy is used to make a chemical change.

Discovery of Sodium & Potassium by Electrolysis :

       The electrolysis of molten sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements.

    

Mechanism of Electrolysis :

During electrolysis, ionic substances are broken down (decomposed) into simpler substances by the influence of electricity.Generally, metals & gases may form at the electrodes.

               For electrolysis to work, the ions must be free to move. Ions are free to move when an ionic substance (i.e., an electrolyte) is dissolved in water or molten (melted). 
A common example of an electrolyte is table salt, sodium chloride. Solid sodium chloride (NaCl) and pure water both are non-conductor of electricity, but a soln. of the salt in water is a good conductor of electricity. A soln. of sugar in water, by contrast, is non-conductor and, therefore, sugar is a non-electrolyte. 

         An ionic solid such as NaCl is composed of charged particles, but these are held so tightly in the crystal lattice that they are unable to move about, so the second requirement mentioned above is not met and solid salt is not a conductor. If the salt is melted or dissolved in water, the ions can move freely and the molten liquid or the solution becomes a conductor. 

        Since positively-charged ions are attracted to a negative electrode that is traditionally known as the cathode, these are often referred to as cations. Similarly, negatively-charged ions, being attracted to the positive electrode, or anode, are called anions.

 The equipment used for electrolysis :

•   A source of   DC i.e., battery;
•  
• Two electrodes (Cathode & anode);
• 
  
•   An electrolyte. 

 










During electrolysis:
   Positively charged ions move to the negative electrode (cathode) during electrolysis. They receive electrons and are reduced.
Negatively charged ions move to the positive electrode (anode) during electrolysis. They lose electrons and are oxidised.


Electrolysis of Dilute aqueous sulphuric Acid 


Pure water is an electrical insulator. Addition of a little bit of H₂SO₄, water behaves as a good conductor. Electrolysis of such dilute acid soln. results in the decomposition of water at both electrodes. 

Reduction at Cathode :

   H⁺(aq) + e  =  ½ H₂(g)    E⁰ = 0.00v                  ........ (1)

H₂O(l) + e  =  ½ H₂(g) + OH^-(aq)    E⁰ = -0.83v     ........ (2)

As the standard reduction potential of reaction-(1) is greater than that of reaction-(2), H⁺(aq) is reduced instead of H₂O(l) & hydrogen gas is evolved at the cathode.

Oxidation at Anode :

H₂O(l)  =  2H⁺(aq) + ½ O₂(g) + 2e    E⁰ = +1.23v     ........ (3)

2SO₄^2-(aq)  =  S₂O₈^2- (aq)  + 2e   E⁰ = +2.00v       ........ (4)

As the standard reduction potential of reaction-(3) is less than that of reaction-(4) i.e., as the standard oxidation potential of reaction-(3) is greater than that of reaction-(4),  H₂O(l) is oxidised instead of SO₄^2-(aq) and oxygen gas is evolved at the anode.

 

When metal ions and hydrogen ions both are present in the soln., during electrolysis the ions which will be reduced at the cathode depends on the position of the metal in the reactivity series.

The reactivity series:

                                          

•    The metal will be produced if it is less reactive than hydrogen.

•    Hydrogen will be produced if the metal is more reactive than H.      

Electronic Configuration of Cr & Cu : 

        Using the Aufbau or build-up principle (schematically represented as below) of electrons in atomic orbitals, one can easily write theelectronic configuration of atoms. But there are few exceptions (e.g., Cr, Cu). This is due to the fact that the energies of 3d & 4s levels in Sc - Zn are so close that frequent exchange of electrons occur between them and then the electronic arrangement is controlled by the phrase "half-filled & full-filled orbits are highly stable".

Exchange Energy : Cause of Extra Stability of Half-filled & Completely Filled Orbitals

Exchange Energy : Cause of Extra Stability of Half-filled & Completely Filled Orbitals :

In case of  3d⁴ 4s²  configuration in Cr : 
The electron no.1 can exchange its position with electrons nos. 2,3 & 4  i.e., in 3 ways. 
The electron no.2 can exchange 2 ways with electrons 3 & 4 (with electron 1 has already been considered). 
The electron no.3 can exchange only in 1 way (as exchanges with electrons 1 & 2 has already been calculated).
 Hence there are 3+2+1=6 ways of exchange are possible in  3d⁴ arrangement.
Similarly, in 3d⁵ configuration, the total possible ways of exchange is 4+3+2+1=10. 
Therefore, 3d⁵ configuration is more stable than  3d⁴ configuration.