Electrochemistry : 10.3 : Electrolysis cell

VOLTAN CELL VS ELECTRIOLYSIS CELL .

 Voltaic cell :

• use a spontaneous reaction to generate electric energy. 

Electrolysis :

• use electric energy to drive non- spontaneous energy.

VOLTAIC CELL.	ELECTROLYTIC
electrons generate at anode (-)	Electrons remove from anode (+)
electrons consumed at cathode (+)	Electrons supplied to cathode (-)
electrons flow from anode(-) to cathode (+)	Electron flow anode (+) to cathode (-)

ELECTROLYSIS 

• Splitting ('lysing') of a substance by input of electrical energy.
• To decompose a compound into it element.

 

ELECTROLITE IN AN ELECTROLYTIC CELL.

1. Can be :

• pure compound. 
  

    ex : H₂O , molten salt

 

• Aqueous solution of salt.

    ex : NaCI(aq) Na₂SO₄ (aq)

 

ELETROLYSIS OF PURE MOLTEN SALT

 

Example : molten NACI 

   

Anode     (oxidation)    :                              2Cl (l) ------ Cl₂(g) + 2e^-

Cathode (reduction)    :                 2Na⁺ (l) + 2e^- ------ 2Na (s)

Overall                           :            2Na⁺ (l) + 2Cl (l) ------ 2Na + Cl₂ (g) 

 

    *anion oxidised at anode!

    *cation reduced at cathode!

 

ELECRTROLYSIS OF WATER

 

    Anode (oxidation)                     2H₂O(l) ------ O₂(g) + 4H⁺(aq) + 4e^-

    Cathode (reduction)         4H₂O(l)+ 4e^- ------ 2H₂(g) + 4OH^-(aq)

    Overall                                        2H₂O(l) ------ 2H₂(g) + O₂(g)

                      [Note : 4H⁺(aq) + 4OH^-(aq) ------ 4H₂O(l)]

• volume of H₂ : O

                          = 2 : 1

• electrode : inert (Pt, etc )
• H₂O oxidized at cathode produce O₂
• H₂O reduce at cathode to produce H₂
• water can be oxidized and reduced.

ELECTROLYSIS OF AQUEOUS IONIC SOLUTION 

• Aqueous solution of salt are mixture of many species (ions and HO2)
• So we have to compare various electrode potentials (E^◦) to predict.
   

PREDICTING ELECTROLYSIS PRODUCT

• When two half- reaction are possible at an electrode.

     Cathode    : reduction with more positive E° occurs.

     Anode    : oxidation with more negative E° occurs.

 

CATION OF ACTIVE METALS (that cannot be reduced) 

• Cations of metal in group (1) and (2) and A 1
• They are not reduced (E° more negative)
• H₂O reduced to H₂ and OH 

Example: Na₂SO₄(aq)

Species present in solution: Na⁺ , SO₄^2-, H₂O 

At cathode(-):

 Na⁺(aq) + e^- ------ Na(s)                             E= -2.71 V

2H₂O(l) + 2e^- ------ H₂(g) + 2OH^- (aq)        E= -0.83 V

ANION (OXOANIONS AND F^- ) (cannot be oxidised)

• Common oxoanion such as SO4^2-, CO3^2-, HO^3-, AND PO4^3-, (AND F) are not oxidized.
• Because the central atom already at it highest oxidation state.
• H₂O oxidixed to O₂ and H⁺

EXAMPLE :Na₂SO₄(aq) 

Special present in the solution: Na⁺, SO₄^2-, H₂O

At anode(+): 

2H₂O ------ O(g) + 4H⁺(aq) + 4e^-

CATION OF LESS ACTIVE METALS (can be reduce)

• Cation of Au, Ag, Cu, Cr, Pb and Cd
• They are reduce at cathode (E° more +ve)
  

Example: AgNO(aq) 

Species present in the solution: Ag⁺ , NO^-, H₂O

At cathode(-):

 Ag⁺(aq) + e^- ------ Ag(s)

2H₂O(l) + 2e^- ------ H(g) + 2OH^-(aq)

HALIDES THAT CAN BE OXIDIZER

• I^-, Br^-, Cl^-, except F
• The concentration must be high

Example : NaCl(aq)

Species present in the solution : Na⁺, Cl^-, H₂O

At cathode(-):

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

At anode(+):

2Cl^-(aq) ------ Cl₂(g) + 2e^-

SUMMARY ON PREDICTING ELECTROLYSIS PRODUCT

• Which species will be reduced : Au³⁺(aq) or H₂O

ans : Au³⁺ 

Au³⁺(aq) + e^- ------ Au(s)

Au³⁺ is ion of less active metal (E more positive).

• Which species will be reduces : NA⁺(aq) or H₂O

ans : H₂O    

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

Na+ is ion of active metal (E more negative).

• Which species will be oxidised:

ans : H₂O    

2H₂O(l) + O(g) ------ 4H⁺(aq) + 4e^-

SO­­­­₄^2- is an oxoanion (cannot be oxidised because the central atom alreary at the highest oxidation states).

• Which species will be oxidized: CL (aq) or H₂o

ans : Cl        

2Cl^-(aq) ------ Cl₂(g) + 2e^-

hallides (except F^-) can be oxidized (due to overvoltage).

EFFECT OF CONCENTRATION

Ex: Concentration NaCl(aq)

At cathode (-)    : 2H₂O(l) + 2e^- ------ H₂(g) + 2OH^-(aq)

At anode (+)      :          2Cl^-(aq) ------ Cl₂(g) + 2e^-

 

EXAMPLE : dilute NaCl(aq)

At cathode (-)      4H₂O(l) + 4e^- ------ 2H₂(g) + 4OH^-(aq)

At anode (+)                  2H₂O(l) ------ O₂(g) + 4H⁺(aq) + 4e^-

Overall                            2H₂O(l) ------ 2H₂(g) + O₂(g)

TYPE OF ELECTRODE 

• Inactive electrode such as graphite and Pt(s) are normally use in electrolysis.
  -They do not involve in the reaction
• Active electrodes such as metal (anode) dissolve to form metallic ions
  
  EXAMPLE : Electrolysis of using inert electrode
  
  
• Cathode :               Cu²⁺(aq) + 4e^- ------ 2Cu (s)
• Anode     :                           2H₂O(l) ------ O₂(g) + 4 H⁺ (aq) + 4e^-        
• Overall    :      2H₂O(l) + 2Cu²⁺(aq) ------ 2Cu(S) + O₂(g) + 4 H⁺ (aq)
   

FARADAY LAW

• Amount of substance produce of each electrode is directly proportional to quantity of charge flowing through the cell
  
• Also called Faraday's First Law of electrolysis.
  

CALCULATING USING FARADAY'S LAW

• Main steps
  • Balance half reaction to find number of moles of electrons.
  • Needed per mole product.
  • Use Faraday's constant (96500C / mol ) to find corresponding charge.
  • Use molar mass / mole to find charge needed for a given mass / mole of product

ELECTRIC CHARGE (Q)

Charge ( Q )         =    Current ( I )    ×    time ( t )

Unit Coulomb , C        Ampere , A        Second , s