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

Electrochemistry : 10.2 Nernst Equation

NERNST EQUATION 

• Cell potential (E°_cell) under any condition

                  E_cell = E°_cell – (RT/nF) ln Q

R: universal gas constant.            Q : reaction quontient.

N: no. of e^- transferred in             T : absolute temperature (X)

balanced redox reaction.

F : charge of 1 mol of e"96.500 C/ mol e-


Example:


Cd (s) + 2Ag+ (aq) ------ Cd²⁺(aq) + 2Ag (s)



    Q=[Cd²⁺]/[Ag⁺]²

 
Ecell At 25°C (298K)


 
    E_cell = E°_cell - (RT/nF) ln Q         (convert to logarithm)


    E_cell = E°_cell - (0.0592/N) log Q  

 
            

EFFECT OF CONC . ON CELL POTENTIAL.

  

 E_cell = E°_cell - (RT/nF) ln Q  
          

EXAMPLE 1:


 
Zn (S) +Cu2+ (aq) ------ Zn2+ (aq) + Cu (s)


    Q=[Cd²⁺]/[Ag⁺]²


WHEN Q < 1 = [reactant] > [product]

                       = in Q < 0 , so E_cell > E°_cell



WHEN Q > 1 = [reactant] = [product]

                       = in Q = 0 , so E_cell = E°_cell



WHEN Q > 1 = [reactant] < [product]

                       = in Q > 0 , so E_cell < E°_cell


 
WHEN Q = K_c


    E_cell = E°_cell - (0.0592/N) log Q
        

• E_cell = O                           
  
• The system reach equilibrium.
  
• No more energy release.
  
• Cell can do no more wore.
  

Electrochemistry : 10.1 Galvanic Cell (continued)

 SPONTANEOUS REACTION

• occurs as the result of different ability of metal to give up their electron to flow through the circuit.
  

 
CELL POTENTIAL (E_Cell)

• different in electrical potential of electrodes
  
• also called voltage or electromotive force (e.m.f)
  

 
E_cell > 0

• sponteneous reaction
  
  • The more positive E_cell
    
  • The more work the can do
    
• The further the reaction proceed to right
  

 
E_cell < 0

• Non spontaneous cell reaction
  

 
E_Cell = 0

• The reaction has reach equilibirum
  
• The cell can do no more work.
  

 
SI UNIT CELL POTENTIAL

• unit = volt (V)
  
• 1V = 1J/C
  

C = coulumb (SI unit of electrical charge)


 
STANDARD CELL POTENTIAL (E⁰_cell)

• Different in electrical potential of electrodes measured at a specified temperature (usually 298k) with all components in their standard states.
  
• standard state 
• 1 atm for gaseous
• 1 M for solution
• Pure solid for electrodes

STANDARD ELECTRODE (HALF CELL) POTENTIAL (E⁰_half-cell)

• potential associate with a given half-reaction (electrode compartment) when all component are in their standard states.
  

E⁰_{half cell =} E⁰_anode or E⁰_cathode

• also call standard reduction potential.
  
• example : 

                         Zn²⁺(aq) + 2e^- ------ Zn(s) E⁰_zinc (E⁰_anode)

                        Cu²⁺(aq) + 2e^- ------ Cu(s) E⁰_copper(E⁰_cathode)

                    Zn(s) + Cu²⁺ (aq) ------ Zn²⁺ (aq) + Cu(s)


*changing the balancing coefficients of a half-reaction does not change E⁰ value because electrode potential are intensive properties –does not depend on amount



E⁰_cell AND E⁰_{half cell}

• Example : 
  

Half cell reation

                            Cu²⁺(aq) + 2e^- ------ Cu(s)
                         Zn(s) + Cu²⁺(aq) ------ Zn²⁺(aq) + Cu(s)

                         Zn(s) + Cu²⁺(aq) ------ Zn²⁺(aq) + 2e^-
 

• E⁰_cell = E⁰_cathode – E⁰_anode
  

 
STANDARD HYDROGEN ELECTRODE

• specially prepared platinum electrode immersed in a 1M aqueous solution of a strong acid,H⁺(aq) or H₃O⁺(aq), through which H₂ gas at 1 atm is bubled
  

DETERMINING E⁰_{half cell}

• Use standard hydrogen electrode (SHE)
• standard reference half-cell
  
  2H⁺ (aq, 1m) + 2e^- ------ H₂ (g,1atm) E⁰ ref= OV
  

         Zn²⁺ (aq, 1m) + 2e^- ------ Zn (s, 1atm) E zinc= ?

Zn(s) │ Zn²⁺ (1M) ││ H⁺ (1M) │ H₂ (1atm) │ Pt (s)

E⁰ cell  = E⁰ cathode – E⁰ anode

              = E⁰ ref – E⁰ zinc

0.76 V  = 0.00 V – E⁰ zinc

E⁰ zinc  = -0.76 V

Pt(s) │ H₂ (1atm) │ H⁺ (1M) │ Cu²⁺ (1M) │ Cu (s)

  

Anode : H₂ (1atm) 2H⁺ (1M) +2e^-

Cathode : 2e^- + Cu²⁺ (1M)     Cu (s)

H₂ (1atm) + Cu²⁺ (1M)     Cu (s) + 2H⁺ (1M)

 

RELATIVE STRENGH OF OXIDIZING AND REDUCTING AGENT

• Example:
  

Cu²⁺(aq) + 2e^- ------ Cu (s) E⁰ = 0.34 V

2H⁺ (aq) + 2e^- ------ H₂ (g) E⁰ = 0.00 V

Zn²⁺ (aq) + 2e^- ------ Zn (s) E⁰ = -0.76 V

• The more positive E⁰ value, the more tendency to be reduced
  
• Strength of oxidizing agent (reactant)
  
  Cu²⁺ > H⁺ > Zn²⁺
• Strength of reducing agent (product)
  
  Zn > H₂ > Cu
  
  

STANDARD REDUCTION POTENTIAL

• All value are relative to hydrogen electrode
  
• Strength of oxidizing agent – increase up
  
• Strength of reducing agent – increase down
  
• Half-cell component usually appear in the same order as in the half-reaction
  
                                               Cu²⁺ (aq) + 2e^- ------ Cu (s)                        (reducing)
  
                                                                Zn (s) ------ Zn²⁺ (aq) + 2e^-        (oxidizing)
  
                                           Cu²⁺ (aq) + Zn (s) ------ Zn²⁺ (aq) + Cu(s)    (overall)

Cell notation : the coefficient is not involve


 
WRITING SPONTANEOUS REDOX REACTION



Zn(s) + Cu²⁺ (aq) ------ Zn²⁺ (aq) + Cu(s)


 
Zn     - stronger reducing agent

Cu²⁺ - stronger oxidizing agent

Zn²⁺  - weaker (0.9)

Cu     - weaker (0.9) 

• stronger oxidizing agent : E⁰ larger (more positive)
  
• stronger reducing agent : E⁰ smaller (more negative)
  

Cu²⁺ (aq) + 2e^- Cu(s) E⁰: 0.34 V

Zn²⁺ (aq) + 2e^- Zn(s) E⁰: - 0.76 V

• How to determine anode (oxidation) and cathode (reduction) for spontaneous reaction, E⁰_cell > 0 ?
  
• Strong reducing agent = E⁰ larger (positive)
  

                                                 = E⁰ cathode (reduction)


Reduction: Cu²⁺ (aq) + 2e^- ------ Cu (s) E⁰ =0.34V

Oxidation :                  Zn (s) ------ Zn²⁺ (aq) + e^- E = - 0.76V

• Stronger reducing agent     = E⁰ smaller
  

                                                         = E⁰ anode (oxidation)



PREDICTING SPONTANEOUS REDOX REACTION USING DIAGONAL RULE

• Under standard – state condition, any species on the left of a given half-cell reaction will react spontaneously with a species that appear on the-right of any half-cell reaction located below it,
  
• diagonal rule !!!
  

Cu²⁺ (aq) + 2e^- ------ Cu (s)     E⁰ = 0.34V

 Zn²⁺ (aq) + 2e^- ------ Zn (s)      E⁰ = - 0.76V

Electrochemistry : 10.1 Galvanic Cell

Electrochemistry

• Study of relationship between chemical change & electric work
  
  
  
  
  
  Oxidation
  
• Loss of electron by species accompanied byn an increase in oxidation number
  
  Ex:
  
  
  
  
  
  
  
  
  
  Reduction
  
• Gain electron by a species accompanied by a decrease number of oxidation
  
  Ex:
  
  
  
  
        Redox reaction
  
• Process which there are net movement of electrons from one reactant to another
  
• Also called oxidation – reduction process.
  
• Oxidation and reduction occur at the same time.
  
  Ex:
  
  
  
  
  
  
  
  Oxidizing agent
  • Substance that accepts electron in redox reaction and undergoes decrease number of oxidation.
    
    
    O.A is Fe₂O₃
    
    
    
    
    Reducing agent
    
  • Substance that donate electron in redox reaction and undergoes an increase oxidation number.
    
    Ex:
    
    
    
    CO is an reducing agent
    
    
    
    Three key point
    
  • Oxidation always accompany by reduction
    
  • Oxidizing agent reduced reducing agent oxidised

ELECTROCHEMICAL CELL

There are two type :

• Voltaic cell
  
• Electrolytic cell
  

VOLTAIC CELL

• Use spontaneous reaction to generate electric energy
  
• System does work on surrounding
  
• All batteries contains voltaic cell
  

ELECTROLYTIC CELL

• Use electrical energy to drive non-spontaneous reaction
  
• Surrounding do work on system
  
• Ex: electroplating & recovering metal from ores
  

ELECTRODES

• Object that conduct electricity between cell and surrounding
  
• 2 electrode (anode & cathode) are dipped into electrolyte
  
• ACTIVE ELECTRODES
  
• Involve in half reaction
  
• Ex: zinc (Zn), copper (Cu), Iron (Fe)

INACTIVE ELECTRODE

• If no reactant or product can be uses as electrode
  
• Ex: graphite (C), Platinum (Pt)
  
• The electrolyte solution contain all species involved in the half – reaction 

ANODE & CATHODE

• Oxidation half reaction takes place at cathode
  
• Electron given up by substance being oxidised (reducing agent) and leave the cell at anode
  
• Reducing half reaction takes place at cathode
  
• Electron are taken up by substance being reduced (oxidizing agent) & enters the cell at cathode
  

ELECTROLYTE

• Mixture of ions (usually in aqueos solution) that are involved in reaction or that carry charge.
  

ESSENTIAL IDEA OF VOLTAIC CELL

• Component of each half reaction (half – cell) are placed in a seperate container
  
• The two half cell are joined by circuit which consist a wire and a salt bridge
  

OXIDATION HALF CELL

• Metal bar (anode) is immersd in Zn²⁺ electrolyte
  
  Ex: ZnSO₄
  
• Zn is reactant in oxidation half reaction
  
• Zn conduct electricity and release electron out of its half cell
  
   
• Mass of Zn electrode decrease
  

REDUCTION HALF CELL

• A Cu metal bar (cathode) is immersed in a Zn²⁺ electrolyte
  
• Ex: CuSO₄ solution
  
• Cu is a product in reduction half reaction
  
• Cu conducts electricity and released electron into itself.
  
  
• Mass of copper electrode increase
  

RELATIVE CHARGE ON ELECTRODES

• Anode:
  

  
• Electron flow from anode to cathode through wire
  
• Cathode
  
  
• Electron are continously generated at anode * consume at cathode
  

SALT BRIDGE

• Contains a solution of non-reacting ion such as KCl,KNO_3, Na₂SO₄ acts as liquid wire (allowing ions to flow & complete the circuit.
  
  
  
  
  
  
  HOW DOES THE CELL MAINTAIN ITS ELECTRICAL NEUTRALITY 

Anode Half Cell	Cathode Half Cell
Zn 2+ ions enters the solution causing the excess of positive charge.	Cu2+ ions leave the solution causing the excess of negative charge
Cl- ions from the salt bridge move into anode (Zn) half cell	K+ ions from salt bridge move into cathode Cu half cell

  

CELL NOTATION

• Ex : Zn(s) l Zn²⁺ (aq) l l Cu²⁺ (aq) l Cu(s)
  
• Anode left, cathode right
  
• 'l' represent phase boundary
  
• Use ',' for component that are in the same phase
  
  Graphite l I^- (aq) l I(s) l l H⁺ (aq), MnO₄^- (aq), Mn²⁺ (aq) l graphite
  
• Sometimes we specify the concentration of dissolved component
  
  Zn(s) l Zn²⁺ (1 M) l Cu²⁺ (1 M) l Cu(s)
  
• Electrode appear ait far right and left notation.
  
• Half-cell component usually appear in the same order as in the half-reaction
  
  
  
      Cu ²⁺ (aq) + 2^e- ------ Cu (s)        (reducing)
                      Zn(s) ------ Zn²⁺ (aq) + 2e^- (oxidizing)
  
  Cu ²⁺ (aq) + Zn(s) ------ Cu (s) + Zn²⁺ (aq) (overall)

Cell notation: the coefficient is not involved.

Thermochemistry : 9.3 Born-Haber Cycle

• First Ionization Energy (IE₁)
  • Energy required for 1 mol of gaseous atom to lose 1 mol of electrons.
    
    
• Affinity Electron (EA) 
  • Energy change that occurs when 1 mol of gaseous atom gains 1 mol of electrons.
    
    
• Lattice Energy
  • Energy change when 1 mol of solid ionic compound formed from its gaseous ions
    
    
• Magnitude Lattice Energy
  • Indicates the strength of ionic bonding
  • The more negative lattice energy, the stronger ionic bonding
  • It influences : 
    • Melting point
    • Hardness
    • Solubility
      
      
• Determining Lattice Energy
  • Lattice energy cannot be measured directly.
    • Can be calculated by Born-Haber cycle
      
      
• Born-Haber Cycle
  • A series of chosen steps from elements to ionic compound for which all the enthalpies are known , except lattice energy.
    ΔH_f = sum of ΔH^o for multistep path
  • Calculation Step
  
  
  Example : LiF
  
  
  Overall : Formation of LiF compound
      Li(s) + ½ F₂(g)     LiF(s)        ΔH^o
  _overall = 617 kJ
          ΔH^o_overall = ΔH^o_f

    Step 1 :- atomization of lithium

        Li(s)     Li(g)                ΔH_f = 161 kJ

            ΔH₁^o = ∆H^o_atom or ∆H^o _sublim

    Step 2 :- atomization of fluorine

        ½ F₂(g)     F(g)             ∆H₂^o = 79.5 kJ

            ∆H₂^o = ∆H^o _atom

    Step 3 :- ionization of lithium atom

        Li(g)         Li⁺(g) + e^-         ∆H₃^o = 520 kJ

            ∆H₃^o = IE₁

    Step 4 :- Electron affinity of F atom

        F(g) + e^-     F^- (g)            ∆H₄^o = -328 kJ

            ∆H₄^o = EA

    Step 5 :- Formation of LiF(s) from its gaseous ions

        Li⁺(g) + F^-(g)         LiF(s)        ∆H₅^o = ?

            ∆H₅^o = ∆H^o_lattice of LiF

• Factors Affecting Lattice Energy
  • From Coulomb's Law

                             ΔE  α  (n⁺Q⁺)(n^-Q^-)

                                                    ---------------------------                                                          

                                                d

∆H : electrostatic energy

n⁺ : number
of positive charge

n^- : number of negative charge 

Q⁺ : electrostatic charge (+)

Q^- : electrostatic charge (-)

d : distance between the ions = r⁺ + r^- (radius of ions)

• Ionic bond strength increase when
  • Q, charge (Q) increase
  • Size of atoms (d) decrease

• Hydration Process of Ionic Crystal in Water
  • 2 main process 
    • Breaking lattice energy of ionic crystal
    • Solvation (hydration)
      
      

Thermochemistry : 9.2 Hess Law

• Hess's Law of Heat Summation
  

When reactants are converted to products, the change in enthalpy is the same whether the reaction take step in one step or in a series of step.

    ΔH₁ =ΔH₂ + ΔH₃

• Application Using Hess's Law
  
  • To find ΔH of any reaction for which we can write on equation, even if it is impossible to carry out.
    
  • Two methods of calculation
    
    • Algebraic method
      
    • Energy cycle method
      
• Algebraic Method
  
  • Important steps:
    
    • Identify the target equation whose ΔH unknown.
      
    • Rearrange the equations given mole of reactants and products are on the correct side.
      
    • Add the equation to obtain the target equation all other substances must cancel.
      
• Standard Enthalpy of Formation (ΔH_f)
  
  • Heat change when 1 mol of a compound is produced from its elements in their standard states.
    
• ΔH_f
  of Elements
  
  • In its standard states
    
• ΔH_f of Compounds
  
  • Most compound have negative ΔH_f
    
    • The compound is more stable than its components elements.
      
• Determining ΔH_rxn From ΔH_f
  
  aA + bB = cC + dD
  
  H_rxn = [cΔH_f (C) + dΔH_f (D) – aΔH_f (A) + bΔH_f (B)]
  
  H_rxn = ( sum of ΔH_f of all of the products ) – ( sum of ΔH_f of all of the reactant )